SOLUTIONS MANUAL For Contemporary Engineering Economics 7th [7 ed.]

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Contemporary Engineering Economics, 7th ed. ©2023

Chapter 1 Engineering Economic Decisions 1.1 •

•

Lease o Deposit (typically one month worth of deposit) refundable when lease expires. o Monthly lease payment o Monthly maintenance fees o Monthly utility expenses Buy o Closing fees o Down payment o Monthly mortgage payments o Property taxes o Monthly utility fees o Monthly maintenance fees o Repair expenses o Homeowners’ association fee (if applicable)

1.2 •

Option 1: o Total amount at the end of two years: $1,150

•

Option 2: o Loan $500 to a friend for one year and receive $600 o Deposit $500 (left over) in a back at 3% for two years: $500(1.03)(1.03) = $530.45 o Deposit $600 received from your friend at 3% per year for a year: $600(1.03) = $618 Total amount at the end of two years: $530.45 + $618 = $1,148.45 These two options are about the same. But considering the trustworthiness, you could go with Option 2.

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Contemporary Engineering Economics, 7th ed. ©2023

Chapter 2 Accounting Information for Engineering Economic Decisions 2.1 (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 2.2 (7), (8), (1), (11), (3), (9) 2.3 (a) • • • •

Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000

(b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15 Market price = $15 + $15 = $30 per share

2.4 (a) Shareholder’s equity in 2021 = $700 - $510 = $190(M) Shareholder’s equity in 2022 = $900 - $640 = $260(M) (b) Net working capital in 2021 = $100 - $60 = $40(M) Net working capital in 2022 = $200 - $90 = $110(M) (c) The income taxes in year 2022: ($2,350 - $1,130-$420-$210) *0.35 = $206.5(M) (d) $383.50 + $420=$803.50 (M) (Cash from Operating activities = Net income + Depreciation)

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Contemporary Engineering Economics, 7th ed. ©2023 2.5 (a) ROE (= Net income/Equity) ROA (= Net income + interest expense (1-tax rate)/Average total assets)

Company A 26.03%

Company B 22.29%

17.34%

12.59%

(b) Company A has performed better in terms of profitability. (c) If two companies were merged, the impact on the results of ROE could be positive under the situation where the Company A leads the acquisition using a stock swap instead of issuing new stocks for M&A cost. If Company A uses a stock swap, the stock value wouldn’t be decreased in terms of scarcity. 2.6 Inventory turnover ratio (2021) = Sales/Average inventory balance = $3,776,395 / ($202,794 + $231,313)×0.5 = 17.4 times Inventory turnover ratio (2022) = 15.6 times This ratio shows how many times the inventory of a firm is sold and replaced over a specific period. From the data, Metronix was holding more stocks of inventory than last year; having more inventories on stock is unproductive. 2.7 (b) 2.8 (b) 2.9 (d) 2.10 Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets •

•

Net Income = $8 X Where X = the number of outstanding shares EPS =

Book value =

Total shareholders' equity = $80 X

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2.11

•

Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = 6 million shares

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From the book value per share, we know that the total shareholders’ equity = 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million

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Debt ratio = $240 million/$720 million = 0.33

(a) Debt ratio (= Total debt/Total assets) = $19,483,000/$38,599,000 = 50.48% (b) Times-interest-earned ratio (= EBIT/Interest expense) = Not defined (c) Current ratio (= Current assets/Current liabilities) = 29,021,000/19,483,000 = 1.49 (d) Quick (acid test) ratio (= (Current assets - Inventories)/Current liabilities)) = (29,021,000-1,301,000)/19,483,000 = 1.42 (d) Inventory-turnover ratio (= Sales/Avg. inventory balance) = 61,494,000/ ((1,301,000+1,051,000)×0.5) = 52.29 (f) Days-sales-outstanding ratio (= Receivables/ (Annual sales/365)) =10,136,000/ (61,494,000/365) = 60.16 (g) Total-assets-turnover ratio (= Sales/Total assets) = 61,494,000/38,599,000 = 1.59 (h) Profit margin on sales (= Net income available to common stockholders/Sales) = 2,635,000/61,494,000 = 4.28% (i) Return on total assets (= (Net income + interest expense (1-tax rate))/Avg. total assets)

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Contemporary Engineering Economics, 7th ed. ©2023 = 2,635,000/ ((38,599,000 + 33,652,000)×0.5) = 7.29% (j) Return on common equity (= (Net income available to common stockholders)/Avg. common equity) = 2,635,000/ ((7,766,000 + 5,641,000)×0.5) = 39.31% (k) Price/earnings ratio (= Price per share/Earnings per share) =13.47/ (3,350,000/1,944,000) = 7.82 (l) Book value per share (= (Total stockholders’ equity-Preferred stock)/Shares outstanding) = 7,766,000/1,944,000 = $3.99 To make an informed analysis of the firm’s financial health, we need to calculate the various financial ratios of the firm’s competitors along with the S&P 500. 2.12 Income Statement: A

B

C

D

E

F

$900,000

$585,000

$315,000

$270,000

$108,000

$162,000

Balance Sheet:  $160,000











$120,000

$320,000

$600,000

$900,000

$1,500,000











$450,000

$700,000

$100,000

$700,000

$800,000

•

•

From Current ratio Total current assets = 2.4 × $250,000 = $600,000 ----------------------------------- ③ Plant and equipment, net = $1,500,000-$600,000=$900,000-----------------------  From Quick ratio

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Contemporary Engineering Economics, 7th ed. ©2023 Inventory = $600,000 - (1.12 × $250,000) = $320,000 -----------------------------② •

From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) × 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 ------- A

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From DSO Accounts receivable = 24.3333 × ($1,800,000 ÷365) = $120,000 ------------------ ① Cash = ③-(②+①) = $160,000 -----------------------------------------------------------ⓞ

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From interest expense of income statement Bond = $450,000 ----------------------------------- ⑥ 250,000 + ⑥ = $700,000 --------------------------⑦

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From Debt-to-Equity ratio Total Equity ⑩ = $700,000 ÷ 0.875 =$ 800,000 ------------- ⑩ Total assets or Total liabilities and equity = ⑦ + ⑩ = $1,500,000 ------⑤

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From Return on total assets Net income F = 14%× ($1,350,000) - ($45,000) (0.6) = $162,000

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From F, D =F ÷ 0.6 = $270,000, E = D× (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000

•

From EPS Stock Outstanding = F ÷ 4.05 = 40,000 shares Common stock = $2.50 × 40,000 = $100,000 -------------------------------⑧ Retained Earnings = ⑩ - ⑧ = $700,000 ------------------------------------

2.13 • • •

•

Accounts receivable = DSO × Sales/365 = 45 days × ($1,200)/365 days) = $147.945 Current assets = (Cash and marketable securities) + (Accounts receivable) + Inventory = $427.945 Long-term debt = (Total assets) – (Current liabilities) – (Common equities) = $427.945 + $280 – (current assets/current ratio) - $500 = ($207.945) – (427.945/3.2) = $74.212 Total assets turnover = Sales/Total assets = $1,200/ ($427.945 + $280) = 1.695 times 5 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

2.14

(a) Find Tiger’s accounts receivable. DSO = 91.25 =

AR  AR = $50, 000 200, 000 / 365

(b) Determine the amount of current liabilities. CA = Cash + Inventory + AR = $10, 000 + $150, 000 + $50, 000 = $210, 000 $210, 000 Current Ratio = 4.2 =  Current Liabilities = $50, 000 Current Liabilities

(c) Calculate the amount of the long-term debt. Total Asset = Current Asset + Fixed Asset = $210, 000 + $90, 000 = $300, 000 $300, 000 = ($50, 000 + Long term debt ) + $200, 000

 Long term debt = $50, 000

(d) Calculate the Return on Common Equity. ROE =

net income $15, 000 = = 0.075  7.5% equity $200, 000

2.15 (a) Find Fisher’s accounts receivable. 𝐴𝑅 𝐷𝑆𝑂 = → 𝐴𝑅 = 𝟏𝟒𝟕. 𝟗𝟓𝑴 1,200 365 (b) Calculate the amount of current assets. 𝐶𝐴 = 𝑐𝑎𝑠ℎ + 𝐼𝑛𝑣. +𝐴𝑅 = 100 + 180 + 147.95 = 𝟒𝟐𝟕. 𝟗𝟓𝑴 (c) Determine the amount of current liabilities. 𝐶𝐴 427.95 𝐶𝑅 = 3.2 = = → 𝐶𝐿 = 𝟏𝟑𝟑. 𝟕𝟑𝑴 𝐶𝐿 𝐶𝐿 (d) Determine the amount of total assets. 𝑇𝐴 = 𝐶𝐴 + 𝐹𝐴 = 427.95 + 280 = 𝟕𝟎𝟕. 𝟗𝟓𝑴

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(e) Calculate the amount of the long-term debt. 707.95 = 133.73 + 𝐿𝐵 + 500 → 𝐿𝐵 = 𝟕𝟒. 𝟐𝟐𝑴 (f) Calculate the profit margin. 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑝𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑟𝑔𝑖𝑛 = = = 𝟐𝟗. 𝟖𝟑% 𝑠𝑎𝑙𝑒𝑠 1,200 (g) Calculate the Return on Common Equity 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑅𝑂𝐸 = = = 𝟕𝟏. 𝟔% 𝑒𝑞𝑢𝑖𝑡𝑦 500

Short Case Studies with Excel ST2.1 Not provided ST2.2 (a) Working capital = Current assets – Current liabilities Working capital requirements = Changes in current assets – Changes in current liabilities: WC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets. (b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000

ST2.3

(d) Net cash flow: • Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000 • Investing activities = equipment purchase = ($400,000) • Financing activities = borrowed fund = $200,000 • Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 Not provided (Visit the websites and get the most recent financial statements available)

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Contemporary Engineering Economics, 7th ed. ©2023

Chapter 3: Interest Rate and Economic Equivalence Types of Interest 3.1

I = (iP) N = (0.06)($2, 000)(5) = $600

3.2 •

Simple interest:

$20,000 = $10,000(1 + 0.075N ) (1 + 0.075N ) = 2 N= •

1 = 13.33≈ 14 years 0.075

Compound interest: $20,000 = $10,000(1 + 0.07) N (1 + 0.07) N = 2 N = 10.24 ≈ 11years

3.3 •

Compound interest:

F = $1, 000(1 + 0.065)5 = $1,370.09 •

Simple interest: F = $1, 000(1 + 0.068(5)) = $1, 340 The compound interest option is better.

3.4 •

Simple interest (John):

I = iPN = (0.1)($1,000)(5) = $500 •

Compound interest (Susan):

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I = P (1 + i ) N − 1 = $1, 000 (1 + .095)5 − 1 = $574.24 •

Susan’s balance will be greater by $74 (or $74.24 to be exact)

•

Simple interest:

3.5 I = iPN = (0.10)($10, 000)(5) = $5, 000

•

Compound interest:

I = P[(1 + i ) N − 1] = $10, 000(1.6105 − 1) = $6,105 3.6 •

Option 1: Compound interest with 8%:

F = $4,500(1 + 0.085)5 = $5, 000(1.4693) = $6, 766.45 •

Option 2: Simple interest with 9.5%: $4, 500(1 + 0.095 × 5) = $5, 000(1.475) = $6, 637.50

∴ Option 1 is still better. 3.7 End of Year

Principal Repayment

Interest payment

0 1

$4,620.50

$1,200.00

Remaining Balance $15,000.00 $10,379.50

2

$4,990.14

$830.36

$5,389.36

3

$5,389.35

$431.15

$0

Equivalence Concept 3.8

3.9

P = $22, 000( P / F , 5%, 5) = $22, 000(0.7835) = $17, 237.58

F = $30, 000( F / P, 9%, 3) = $30, 000(1.295) = $38,850.87

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Contemporary Engineering Economics, 7th ed. ©2023 3.10 F = $100( F / P,10%,10) + $200( F / P,10%,8) = $688

3.11 $1, 000( F / P, i, 2) = $1, 200 $1, 000(1 + i ) 2 = $1, 200

i = 1.2 − 1 i = 9.54%

Single Payments (Use of F/P or P/F Factors) 3.12

i = 10.5% , two-year discount rate is (1 + 0.105) 2 = 1.221 (or 22.1%) 3.13

F = 2 P = P(1 + 0.06) N log 2 = N log 1.06 N = 11.896 years (or 12 years) 3.14 F = $1(1.08)394 = $14, 755, 694, 730, 611

3.15

P = $450, 000( P / F ,5%,5) = 450, 000(0.7835) = $352,575

3.16 F = $250, 000( F / P, 6%,10) = $447, 712

3.17 (a) F = $5, 000( F / P , 7%, 5) = $7, 013 (b) F = $7, 250( F / P , 9%,15) = $26, 408 (c) F = $9, 000( F / P , 6%, 33) = $61,565 (d) F = $12, 000( F / P, 5.5%,8) = $18, 416

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Contemporary Engineering Economics, 7th ed. ©2023 3.18 P = $300, 000( P / F ,8%,10) = $138, 958

3.19 (a) P = $25,500( P / F ,12%,8) = $10, 299 (b) P = $58, 000( P / F , 4%,12) = $36, 227 (c) P = $25, 000( P / F , 6%, 9) = $14, 797 (d) P = $35, 000( P / F , 9%, 4) = $24, 795 3.20 (a) P = $12, 000( P / F ,13%, 4) = $7, 360 (b) F = $30, 000( F / P,13%, 5) = $55, 273 3.21 F = 3P = P (1 + 0.08) N log 3 = N log(1.08) N = 14.27 → 15 years 3.22 F = 2 P = P(1 + 0.06) N

•

log 2 = N log(1.06) N = 11.90 years  12 years

•

Rule of 72: 72 / 6 = 12 years

3.23 ($16.50)(100)( F / P, i, 44) = $77.50(204,800) $15,872, 000 ( F / P, i, 44) = = 9, 619.39 $1, 650 i = 23.18%

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Contemporary Engineering Economics, 7th ed. ©2023 3.24 (a)

$18( F / P, i, 49) = $190,500 $190,500 ( F / P, i, 49) = = 10,583.33 $18 i = 20.82% (b)

F = $304.8 B( F / P , 20.82%,17) = $7, 592.33B

Uneven Payment Series 3.25

P= 3.26

$2, 000 $800 $1, 000 + + = $3, 230.65 1.11 1.12 1.13

P = $35, 000( P / F ,9%, 4) + $10, 000( P / F ,9%, 2) = $35, 000(0.7084) + $10, 000(0.8417) = $33, 211

3.27 $180, 000 = $20, 000( P / A,9%,5) − $10, 000( P / F ,9%,3) + X ( P / F ,9%, 6)180, 000 − 20, 000(3.8897) + 10, 000(0.7722) = X (0.5963) X = $184,350.16

3.28

P=

$60, 000 $77, 000 $65, 000 $57, 000 45, 000 + + + + = $212,873.89 1.14 1.142 1.143 1.144 1.145

3.29 $1,000 +

X $1,000 $1,500 $1,210 + = + 4 3 2 1.1 1.1 1.1 1.1 X = $2,981

3.30

P=

$15, 000 $23, 000 $36, 000 $48, 000 + + + = $93,564 1.07 2 1.073 1.07 4 1.075

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Contemporary Engineering Economics, 7th ed. ©2023 3.31 F = $2, 000( F / P, 6%,10) + $2,500( F / P, 6%,8) + $3, 000( F / P ,6%,6) = $11,822

3.32

P = $3, 000, 000 + $2, 400, 000( P / F ,8%,1) +  +$3, 000, 000( P / F ,8%,10) = $20, 734, 618

Or, P = $3, 000, 000 + $2, 400, 000( P / A,8%,5) +$3, 000, 000( P / A,8%,5)( P / F ,8%,5) = $20, 734, 618

3.33 P = $9, 000( P / F ,8%, 2) + $6, 000( P / F ,8%, 5) + $3, 000( P / F ,8%,7) = $13, 550

Equal Payment Series 3.34

A = $250, 000( A / F ,5%,5) = $250, 000(0.1810) = $45, 250 F = $5,000( F / A,5%, 7) = $5, 000(8.1420) = $40, 710 F = $5, 000( F / A, 5%, 7)(1.05) = $5, 000(8.1420)(1.05) = $42, 745.50

(a) (b) 3.35 •

Equal annual payment amount:

A = $20, 000( A / P,10%,3) = $20, 000(0.4021) = $8, 042 •

Loan balance calculation:

End of period 0 1 2 3

Principal Payment $0.00 $6,042.00 $6,646.20 $7,310.82

Interest Payment $0.00 $2,000.00 $1,395.80 $731.18

Interest payment for the second year = $1,395.80

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Remaining Balance $20,000.00 $13,958.00 $7,311.80 $0

Contemporary Engineering Economics, 7th ed. ©2023

3.36

F = $500( F / A, 7%,15)(1.07) = $500(25.1290)(1.07) = $13, 444.02 3.37 (a) With deposits made at the end of each year F = $3, 000( F / A, 9%,15) = $88, 083

(b) With deposits made at the beginning of each year F = $3, 000( F / A,9%,15)(1.09) = $96, 010

3.38 F = $10, 000( F / A, 6%,20) = $367,856

3.39 (a) F = $8, 000( F / A,11.75%, 5) = $50, 571 (b) F = $2, 000( F / A, 4.25%,12) = $30, 486 (c) F = $7, 000( F / A, 6.45%, 20) = $270, 309 (d) F = $4, 000( F / A, 7.75%,12) = $74, 793 3.40 (a) A = $45, 000( A / F ,8%,11) = $2, 703 (b) A = $35, 000( A / F , 6%,18) = $1,132 (c) A = $25, 000( A / F , 7%, 6) = $3, 495 (d) A = $12, 000( A / F ,11%,13) = $458 3.41 $50, 000( A / F , 6%,10) = $3, 793.40

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Contemporary Engineering Economics, 7th ed. ©2023 3.42

$55, 000 = $3, 000( F / A, 6%,N ) ( F / A, 6%, N ) = 18.33 N = 13years

3.43 $15, 000 = A( F / A,11%,5) A = $2, 408.57

3.44

$10,000 = $1,000( F / P,7%,5) + A( F / A,7%,5) A = $1,536.57 3.45 (a) A = $15, 000( A / P,3.5%, 6) = $2,815.02 (b) A = $7,500( A / P, 7.5%, 7) = $1, 416 (c) A = $2, 500( A / P,5.25%,5) = $581.43 (d) A = $12, 000( A / P , 6.25%,15) = $1, 255.81 3.46 (a) The capital recovery factor ( A / P, i, N ) for N 35 40

6% 0.0690 0.0665

7% 0.0772 0.0750

To find ( A / P, 6.25%, 38) , first, interpolate for N = 38 : N 38

6% 0.0675

7% 0.0759

Then, interpolate for i = 6.25% ; (A / P, 6.25%, 38) = 0.0696 :

As compared to the value from the interest formula: (A / P, 6.25%, 38) = 0.0694

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Contemporary Engineering Economics, 7th ed. ©2023 (b) The equal payment series present-worth factor ( P / A, i, 85) for i

9% 11.1038

10% 9.9970

Then, interpolate for i = 9.25% : ( P / A, 9.25%, 85) = 10.8271

As compared to the value from the interest formula: ( P / A, 9.25%, 85) = 10.8049

3.47 •

Equal annual payment: A = $50, 000( A / P,12%, 3) = $20,817.45

•

Interest payment for the second year: End of Year 0 1 2 3

3.48

Principal Repayment

Interest payment

$14,817.45 $16,595.54 $18,587.01

$6,000 $4,221.91 $2,230.44

A = $15, 000( A / P , 9%,10) = $2, 337.30

3.49 (a) P = $1, 000( P / A, 7.2%,8) = $5,925.29 (b) P = $4, 500( P / A, 9.5%,12) = $31, 427.28 (c) P = $1, 900( P / A,8.25%,13) = $14,812.86 (d) P = $19, 300( P / A, 7.75%,8) = $111, 970.11

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Remaining Balance $50,000 $35,182.55 $18,587.01 0

Contemporary Engineering Economics, 7th ed. ©2023 3.50 P = $35,000( P / A,12%,10) = $197,758 Since $200,000 > $197,758, You should not purchase the equipment.

3.51 (a)

P = $9, 600, 000 + $15,100, 000( P / F , 6%,1) +$17,100, 000( P / F , 6%, 2)... + $19, 250, 000( P / F , 6%,5) +$400, 000( P / A, 6%,5) = $89,970,959 (b) P = $7, 600, 000 + $5, 600, 000( P / A, 6%, 4) = $27, 004,591 Since $27, 004,591 > $25,000,000, the prorated payment option is better choice.

Linear Gradient Series 3.52

F = $20, 000( F / A, 6%,5) + $5, 000( F / G , 6%, 5) = $20, 000( F / A, 6%,5) + $5, 000( A / G , 6%,5)( F / A, 6%, 5) = $165,833

3.53

F = $10, 000( F / A,8%,5) − $2, 000( F / G ,8%,5) = $10, 000( F / A,8%,5) − $2, 000( P / G ,8%,5)( F / P, 8%, 5) = $37, 001

3.54 P = $100 + [$100( F / A,9%, 7) + $50( F / A,9%, 6) +$50( F / A,9%,4) + $50( F / A, 9%, 2)]( P / F , 9%, 7) = $991.32

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Contemporary Engineering Economics, 7th ed. ©2023 3.55 A = $15, 000 − $1, 000( A / G , 8%, 12) = $10, 404.25

3.56

P = $1, 000( P / A, 6%, 5) + $250( P / G , 6%, 5) = $6,196

Geometric-Gradient Series 3.57

$1, 000, 000 = A( F / A, 6%,30) = A(79.0582)  A = $12,649 should be set aside on the account a) $1, 000, 000 = A( P / A, 6%, 20) = A(11.4699)  A = $87,185 / year b) $1, 000, 000 = A1 ( P / A1, 3%, 6%, 20) 1 − (1.03) (1.06 ) = A1 0.06 − 0.03 = $68, 674 / year 20

−20

3.58 Using the geometric gradient series present worth factor, we can establish the equivalence between the loan amount $150,000 and the balloon payment series as $150, 000 = A1 ( P / A1 ,10%, 7%,5) = 4.9424 A1

A1 = $30,349.63 Payment series N 1 2 3 4 5

Payment $30,349.63 $33,384.59 $36,723.05 $40,395.35 $44,434.89

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Contemporary Engineering Economics, 7th ed. ©2023

F = $8, 000( P / A1 ,5%, 7%,30)( F / P, 7%,30) = $172,895.56(7.61226) = $1,316,126 3.60 (a) P = $6,000,000( P / A1 ,−10%,12%,7) = $21,372,076 (b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows: An = $60(1 + 0.05) n −1100, 000(1 − 0.1) n −1 = $6, 000, 000(0.945) n −1 = $6, 000, 000(1 − 0.055) n −1

This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%. So,

P = $6, 000, 000( P / A1 , −5.5%,12%,7) = $23,847,897 (c) Computing the present worth of the remaining series ( A4 , A5 , A6 , A7 ) at the end of period 3 gives A4 = 6, 000, 000(1 − 0.055)3 = 5, 063, 451.75 P = $5, 063, 451.75( P / A4 , −5.5%,12%, 4) = $14, 269, 627.82

3.61 20

P =  An (1 + i) − n n =1 20

=  (2, 000, 000)n(1.06) n −1 (1.06) − n n =1

20 1.06 n = (2, 000, 000 /1.06) n( ) 1.06 n =1 = $396, 226, 415

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(a) The withdrawal series would be Period 11 12 13 14 15

Withdrawal $15,000 $15,000(1.08) $15,000(1.08)(1.08) $15,000(1.08)(1.08)(1.08) $15,000(1.08)(1.08)(1.08)(1.08)

Amount $15,000 $16,200 $17,496 $18,896 $20,407

P10 = $15, 000( P / A1 ,8%,9%,5) = $67,556 Assuming that each deposit is made at the end of each year, then: $67,556 = A( F / A,9%,10) A = $4, 446.54

(b) P10 = $15, 000( P / A1 ,8%, 6%,5) = $73, 476 $73, 476 = A( F / A, 6%,10) A = $5, 574.47

Various Interest Factor Relationships 3.63 (a) (P / F, 8%, 67) = (P / F, 8%,50)(P / F,8%,17) = 0.0058

(P / F, 8%,67) = (1 + 0.08)−67 = 0.0058 i 1 − ( P / F , i, N ) ( P / F , 8%, 42) = ( P / F , 8%,40)( P / F ,8%,2) = 0.0394 0.08 ( A / P, 8%, 42) = = 0.0833 1 − 0.0394

(b) ( A / P, i, N ) =

( A / P, 8%, 42) =

(c) ( P / A, i, N ) =

0.08(1.08) 42 = 0.0833 (1.08) 42 − 1

1 − ( P / F , i, N ) 1 − ( P / F ,8%,100)( P / F ,8%,35) = = 12.4996 i 0.08

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(P / A,8%,135) =

(1.08)135 − 1 = 12.4996 0.08(1.08)135

3.64 (a) ( F / P, i, N ) = i( F / A, i, N ) + 1 (1 + i ) N − 1 +1 i = (1 + i) N − 1 + 1

(1 + i) N = i

= (1 + i ) N (b)

( P / F , i, N ) = 1 − ( P / A, i, N )i (1 + i ) − N = 1 − i =

(1 + i ) N − 1 i (1 + i ) N

(1 + i ) N (1 + i ) N − 1 − (1 + i ) N (1 + i ) N

= (1 + i ) − N (c)

( A / F , i , N ) = ( A / P, i, N ) − i i i (1 + i ) N i (1 + i ) N i[(1 + i ) N − 1] i = − = − (1 + i ) N − 1 (1 + i ) N − 1 (1 + i ) N − 1 (1 + i ) N − 1 i = (1 + i ) N − 1

(d) ( A / P, i , N ) =

i [1 − ( P / F , i, N )]

i (1 + i ) N i = N N (1 + i ) − 1 (1 + i ) 1 − N (1 + i ) (1 + i ) N =

i (1 + i ) N (1 + i ) N − 1

(e) , (f) , (g) Divide the numerator and denominator by (1 + i) N and take the limit N →∞. 14 Copyright © 2023 Pearson Education, Inc.

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Equivalence Calculations 3.65 P = [$100( F / A,12%, 9) + $50( F / A,12%, 7) + $50( F / A,12%, 5)]( P / F ,12%,10) = $740.49

3.66 P (1.08) + $200 = $200( P / F ,8%,1) + $120( P / F ,8%, 2) + $120( P / F ,8%,3) + $300( P / F ,8%, 4) P = $373.92

3.67

A( P / A,15%,5) = $100( P / A,15%,5) + $20( P / A,15%,3)( P / F ,15%, 2) 3.35216 A = $369.74 A = $110.30 3.68 P1 = $200 + $100( P / A,8%,5) + $50( P / F ,8%,1) +$50( P / F ,8%, 4) + $100( P / F ,8%,5) = $750.37 P2 = X ( P / A,8%,5) = $750.37 X = $187.93

3.69

P = $20( P / G ,10%,5) − $20( P / A,10%,12) = $0.96

$40

$60

$80

$20 0

1

2

3

4

5

6

7

$20 15 Copyright © 2023 Pearson Education, Inc.

8

9 10

11

12

Contemporary Engineering Economics, 7th ed. ©2023 3.70 Establish economic equivalent at N = 8 : C ( F / A,8%,8) − C ( F / A,8%,2)( F / P,8%,3) = $6,000( P / A,8%,2) 10.6366C − (2.08)(1.2597 )C = $6,000(1.7833) 8.0164C = $10,699.80 C = $1,334.73

3.71 The original cash flow series is

N

AN

N

AN

0

0

6

$900

1

$800

7

$920

2 3

$820 $840

8 9

$300 $300

4

$860 10 $300 − $500

5

$880

3.72 $300( F / A,10%,8) + $200( F / A,10%,3) = 2C ( F / P,10%,8) + C ( F / A,10%,7) $4,092.77 = 2C ( 2.1436) + C (9.4872) C = $297.13

3.73

Establishing equivalence at N = 5 $200( F / A,8%,5) − $50( F / P,8%,1) = X ( F / A,8%,5) − ($200 + X )[( F / P,8%, 2) + ( F / P,8%,1)] $1,119.32 = X (5.8666) − ($200 + X )(2.2464) X = $433.29

3.74

Computing equivalence at N = 5 X = $3, 000( F / A,9%,5) + $3, 000( P / A,9%,5) = $29, 623.08

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3.75 (b), (d), and (f) 3.76 (b), (d), and (e) 3.77 A1 = ($50 + $50( A / G ,10%,5) − [$50 + $50( P / F ,10%,1)]( A / P,10%,5) = $115.32 A2 = A + A( A / P,10%,5) = 1.2638 A A = $91.25

3.78 (a) 3.79 (b) 3.80 (b) $25, 000 + $30, 000( P / F ,10%, 6) = C ( P / A,10%,12) + $1, 000( P / A,10%, 6)( P / F ,10%, 6) $41,935 = 6.8137C + $2, 458.43 C = $5, 794

Solving for an Unknown Interest Rate of Unknown Interest Periods 3.81 P1 = 30, 723( P / F , i %,5) P2 = A( P / A, i %,10)  (1 + i )10 − 1  $50, 000(1 + i ) −5 = $5, 000  10   i (1 + i )  ∴ i = 13.06%

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3.82 2 P = P(1 + i )5 21/5 = 1 + i i = 14.87% 3.83

Establishing equivalence at n = 0

$2,000( P / A, i,6) = $2,500( P / A1 ,−25%, i,6) By Excel software, i = 92.36% 3.84 $40, 000 = $15, 000( F / P, i, 5) = $15, 000(1 + i )5 i = 21.67%

3.85 $1, 000, 000 = $2, 000( F / A, 6%, N ) (1 + 0.06) N − 1 0.06 31 = (1 + 0.06) N log 31 = N log1.06 500 =

N = 58.93 ≈ 59years

3.86 Option 1: $100, 000( F / A, 7%, 7)( F / P, 7%,13) = 2, 085, 484.95 Option 2: $100, 000( F / A, 7%,13) = 2, 014, 064.29

$100, 000( F / A, i, 7)( F / P, i,13) = $100,000( F / A, i,13) i = 6.6% 3.87 Assuming that annual renewal fees are paid at the beginning of each year, (a) $15.96 + $15.96( P / A,6%,3) = $58.62 It is better to take the offer because of lower cost to renew.

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(b)

$57.12 = $15.96 + $15.96( P / A, i,3) i = 7.96% 3.88 If premiums paid at the end of each year, the maximum amount to invest in the prevention program is P = $14,000 ( P / A,12%,5) = $50,467 .

If the premiums paid at the beginning of each year, the solution changes to P = $14, 000 + $14, 000( P / A,12%, 4) = $56,523.

Short Case Studies ST 3.1 (a)

(b)

P = 280, 000( P / A,8%,19) = 2, 689, 007.78

280, 000( P / A, i,19) = 5, 600, 000 − 283, 770 i = 0.00709%

ST 3.2 Establish the following equivalence equation: $140,000 = $32,639 (P / A,i, 9) . The interest rate makes two options equivalent is i = 18.10% by Excel. So, if her rate of return is over 18.10%, it is a good decision.

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ST 3.3 (a) PContract = $5, 600, 000 + $7,178, 000( P / F , 6%,1) +$11, 778, 000( P / F , 6%, 2) +  +$17, 778, 000( P / F , 6%,9) = $97,102,826.86

(b) PBonus = $5,000,000 + $5,000,000( P / A,6%,5) + $778,000( P / A,6%,9)

= $31,353,535.52 > $23,000,000

It is better stay with the original plan.

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Contemporary Engineering Economics, 7th ed. ©2023

Chapter 4 Understanding Money and Its Management Nominal and Effective Interest Rates 4.1 •

Nominal interest rate:

r = 1.3% ×12 = 15.6% •

Effective annual interest rate: ia = (1 + 0.013)12 − 1 = 16.77%

4.2

(a) Monthly interest rate: i = 17.85% ÷12 = 1.4875% Annual effective rate: ia = (1 + 0.014875)12 − 1 = 19.385% (b) $2,500(1 + 0.014875)2 = $2,574.93

4.3 Assuming weekly compounding:

r = 6.89% 0.0689 52 ia = (1 + ) − 1 = 0.07128 52 4.4 •

Nominal interest rate:

r = 1.3% ×12 = 15.6% •

Effective annual interest rate: ia = (1 + 0.013)12 − 1 = 16.77%

4.5.

$20,000 = $520( P / A, i, 48) ( P / A, i, 48) = 38.4615 Use Excel to calculate i : 1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 i = 0.9431% per month r = 0.9431*12 = 11.32%

4.6.

$16,000 = $517.78( P / A,i,36) ( P / A,i,36) = 30.901155 i = 0.85% per month r = 0.85 × 12 = 10.2%

4.7 •

Interest rate per week Given : P = $550, A = $42, N = 16 weeks

$550 = $42( P / A, i,16) i = 2.46% per week •

Nominal annual interest rate:

r = 2.46% × 52 = 127.95% •

Effective annual interest rate: ia = (1 + 0.0246)52 − 1 = 253.86%

4.8 The effective annual interest rate: ia = e0.087 − 1 = 9.09%

4.9 Interest rate per week:

$450 = $400(1 + i) i = 12.5% per week (a) Nominal interest rate:

r = 12.5% × 52 = 650% (b) Effective annual interest rate

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ia = (1+ 0.125)52 −1 = 45,602% 4.10 The effective annual interest rate : ia = e 0.06 − 1 = 6.184%

4.11 24-month lease plan with 40,000 miles over 2 years: P = ($2,399 + $189) + $189( P / A, 0.5%, 23) + ($350 + $0.18 × (40, 000 − 24, 000))( P / F , 0.5%, 24) = $9,869.70

4.12

The three options: a) ia = r = 6.12% 4

 0.06  b) ia =  1 +  − 1 = 6.136% 4   c) ia = e0.059 −1 = 6.078%

Bank B is the best option. 4.13 12

 0.06  ia = 1 +  − 1 = 6.168% 12   ia = er − 1 = 0.06168 er = 1.06168 r = 5.985% 4.14  0.05  Bank A: ia =  1 +  365  

365

− 1 = 5.127%

Bank B: ia = e0.046 − 1 = 4.707% The difference between two banks after 2 years:

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$3,000[ ( F / P,5.127%, 2) − (F / P,4.707%, 2)] = $26.44

Compounding More Frequent than Annually 4.15 0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12

a) i = (1 +

4.16

iquarter = e0.09/4 − 1 = 0.022755 (or 2.28%) 4.17 0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12

a) i = (1 +

4.18

$25, 000 = $563.44( P / A, i, 48) ( P / A, i, 48) = 44.3703 i = 0.3256% per month

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4.19 0.11 1 ) − 1 = 11% 1 0.08 2 b) i = (1 + ) − 1 = 8.16% 2 0.095 4 c) i = (1 + ) − 1 = 9.844% 4 0.075 365 d) i = (1 + ) − 1 = 7.788% 365 a) i = (1 +

4.20

F = PerN = $5,000e(0.06×10) = $9,110.59 4.21

P = Fe− rN = $5,000e−0.06×5 = $3,704.09 4.22

F = PerN = $5, 000e(0.09×5) = $7,841.56 4.23 2 P = Pe 0.06 N ln 2 = 0.06N N = 11.55 years

4.24 (a) Nominal interest rate:

r = 2.32% ×12 = 27.84% (b) Effective annual interest rate: ie = (1 + 0.0232)12 − 1 = 31.68%

(c)

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3P = P(1 + 0.0232) N log 3 = N log1.0232 N = 47.90 months ≅ 4 years (d)

4 = e0.0232 N ln(4) = 0.0232 N N = 59.75 months ≅ 5 years 4.25 F = $15, 000(1 +

0.08 8 ) = $15, 000( F / P, 2%,8) 4

= $17,575

4.26

ia = (1 +

0.06 365 ) − 1 = 6.183% 365

F = $15, 000( F / P, 6.183%,12) = $15, 000( F / P,

6 %,12 × 365) 365

= $ 30,815 4.27 (a) 0.082 24 ) = $9,545( F / P, 4.1%, 24) 2 = $25, 037.64

F = $9,545(1 +

(b) 0.06 40 ) = $6,500( F / P,1.5%, 40) 4 = $11, 791.12

F = $6,500(1 +

(c) F = $42,800(1 +

0.09 96 ) = $42, 000( F / P, 0.75%,96) 12

= $87, 693.83

4.28 (a) Quarterly interest rate = 2.25%

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3P = P(1 + 0.0225) N log 3 = N log1.0225 N = 49.37quarters = 12.34 years (b) Monthly interest rate = 0.75%

3P = P(1 + 0.0075) N log 3 = N log1.0075 N = 147.03 months = 12.25 years (c)

3 = e0.09 N ln(3) = 0.09 N N = 12.20 years 4.29. (a)

F = $10, 000( F / A, 4%, 20) = $297, 781 (b)

(c)

F = $9, 000( F / A, 2%, 24) = $273, 796.76 F = $5, 000( F / A, 0.75%,168) = $1,672,590.40

4.30 (a)

(b)

(c)

A = $11,000( A / F , 4%, 20) = $369.60 A = $3, 000( A / F ,1.5%, 60) = $31.18 A = $48, 000( A / F ,0.6125%, 60) = $484.46

4.31 (a) Quarterly effective interest rate = 1.5%

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 F = $10, 000( F / A,1.5%, 60) = $962,147

(b) Quarterly effective interest rate = 1.508% F = $10, 000( F / A,1.508%, 60) = $964, 722

(c) Quarterly effective interest rate = 1.511% F = $10, 000( F / A,1.511%, 60) = $965, 690

4.32

F = $7, 500( F / A, 0.669%, 60) = $551, 479

4.33 (a) Quarterly effective interest rate = 2.25% F = $4,000( F / A,2.25%,40) = $255,145

(b) Quarterly effective interest rate = 2.2669% F = $4,000( F / A,2.2669%,40) = $256,093

(c) Quarterly effective interest rate = 2.2755% F = $4,000( F / A,2.2755%,40) = $256,577

4.34

(d)

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Effective interest rate per payment period

i = (1 + 0.01)3 – 1 = 3.03%

0

1

2

3

4

5

6

7

$1,000

4.35 ia = e0.086/4 − 1 = 2.1733% A = $10, 000( A / P, 2.1733%, 20) = $ 621.84

4.36 (a) Monthly effective interest rate = 0.74444% F = $1,500( F / A, 0.74444%, 96) = $209,170

(b) Monthly effective interest rate = 0.75% F = $1, 500( F / A, 0.75%, 96) = $209, 784

(c) Monthly effective interest rate = 0.75282% F = $1, 500( F / A, 0.75282%, 96) = $210, 097

4.37 (b) 4.38 i = e 0.0225 − 1 = 2.2755%

9 Copyright © 2023 Pearson Education, Inc.

8

9

10

11

12

Contemporary Engineering Economics, 7th ed. ©2023 F = $5, 000( F / A, 2.2755%, 40) = $320, 721

F = $320, 721( F / P, 2.2755%, 20) = $502, 990

4.39 Effective interest rate per month = e 0.0975/12 − 1 = 0.8158% A = $48, 000( A / P, 0.8158%, 60) = $1, 014.90

4.40 Effective interest rate per quarter = e0.0688/ 4 − 1 = 1.7349% P = $2, 500( P / A,1.7349%, 20) = $41, 944

4.41 (a) F = $20, 000( F / A, 4%,10) = $240,122 (b) F = $60, 000( F / A,1.5%, 40) = $3, 256, 074 (c) F = $13, 000( F / A, 0.75%, 72) = $1, 235, 091 4.42

P = Fe− rN = $15,345.36e−0.06×10 = $ 8,421.71 4.43 (a) A = $45,000( A / F ,3.725%,20) = $1,554.80 (b) A = $25,000( A / F ,1.5875%,60) = $252.33 (c) A = $12, 000( A / F , 0.7708%, 60) = $158.06 4.44

id = (1 +

0.05 1 ) − 1 = 0.01369863% 365

F = $3.75( F / A, 0.01369863%,10950) = $95, 299 10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

4.45

$25, 000 = $489.15( P / A, i, 60) i = 0.5416% ia = (1 + 0.005416)12 − 1 = 6.69685%

4.46 $30, 000 = $500( F / A, i, 20) i = 10.4084% i = e r /4 − 1 = e0.25 r − 1 = 0.104084 r = 39.61%

4.47

$15, 000 = $409.61( P / A, i, 42) i = 0.6542% r = 0.006542 ×12 = 7.85%

4.48

A = $70,000( A / F,0.5%,36) = $1,779.54 4.49 (a) P = $5, 000( P / A, 4.5%, 24) = $72, 477 (b) P = $7, 000( P / A, 2%, 40) = $191, 488 (c) P = $3, 500( P / A, 0.5%, 60) = $181, 039 4.50 •

Equivalent future worth of the receipts:

F1 = $1, 500(F / P, 2%, 4) + $2, 500 = $4,123.65

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Contemporary Engineering Economics, 7th ed. ©2023 •

Equivalent future worth of deposits:

F2 = A( F / A, 2%,8) + A( F / P, 2%,8) = 9.7546 A ∴ Letting F1 = F2 and solving for A yields A = $422.74

4.51

C ( F / A, 6.168%, 7) + C ( F / P, 0.5%,84) = $1, 600 + $1, 400( F / P, 0.5%,12) + $1, 200( F / P, 0.5%, 24) +$1, 000( F / P, 0.5%,36) = 8.437C + 1.52C = 1, 600 + 1, 486.35 + 1,352.59 + 1,196.68 9.957C = 5, 635.62 ∴ C = $566

4.52 •

Option 1 .06 1 ) − 1 = 1.5% 4 F = $1, 000( F / A,1.5%, 40)( F / P,1.5%, 60) = $132,587 i = (1 +

•

Option 2 .06 4 ) − 1 = 6.136% 4 F = $6, 000( F / A, 6.136%,15) = $141,111 i = (1 +

•

Option 2 – Option 1 = $141,110 – 132,587 = $8,523

•

Select (b)

4.53 Given: r = 7% compounded daily, N = 25 years

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Contemporary Engineering Economics, 7th ed. ©2023 •

Since deposits are made at year end, find the effective annual interest rate: ia = (1 + 0.07 / 365)365 − 1 = 7.25%

•

Then, find the total amount accumulated at the end of 25 years: F = $3,250(F / A,7.25%,25) + $150(F / G,7.25%,25) = $3,250(F / A,7.25%,25) + $150(P / G,7.25%,25)(F / P,7.25%,25) = $297,016.95

4.54 •

The balance just before the transfer:

F9 = $22,000(F / P,0.5%,108) + $16,000(F / P,0.5%,72) +$13,500(F / P,0.5%,48) = $77,765.70 Therefore, the remaining balance after the transfer will be $38,882.85. This remaining balance will continue to grow at 6% interest compounded monthly. Then, the balance 6 years after the transfer will be:

F15 = $38,882.85( F / P,0.5%,72) = $55,681.96 •

The funds transferred to another account will earn 8% interest compounded quarterly. The resulting balance six years after the transfer will be:

F15 = $38,882.85( F / P, 2%, 24) = $62,540.63

4.55 Establish the cash flow equivalence at the end of 25 years. Let’s define A as the required quarterly deposit amount. Then we obtain the following: A( F / A,1.5%,100) = $80, 000( P / A, 6.136%,15) 228.8030 A = $770,104 A = $3,365.79

4.56 •

Monthly installment amount:

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Contemporary Engineering Economics, 7th ed. ©2023 A = $22, 000( A / P , 0.75%, 60) = $456.68

•

The lump-sum amount for the remaining balance:

P24 = $456.68( P / A,0.75%,36) = $14,361.13 4.57

$225, 000 = $5, 000( P / A, 0.75%, N ) ( P / A, 0.75%, N ) = 45 N = 55 months or 4.58 years

4.58

$20, 000 = $650.52( P / A, i,36) ( P / A, i,36) = 30.7446 i = 0.879% APR = 0.879% × 12 = 10.55%

4.59 Given r = 6% per year compounded monthly, the effective annual rate is 6.168%. Now consider the four options: 1. Renew every three-month at $45 for two years. 2. Renew every year at $160 for two years. 3. Buy a two-year subscription ($279) now. To find the best option, compute the equivalent PW for each option.

o

Poption 1 = $45 + $45( P / A,1.5075%, 7) = $341.83

o

Poption 2 = $160 + $160( P / F , 6.168%,1) = $310.70

o

Poption 3 = $279

Option 3 is the best option. 4.60

14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 To find the amount of quarterly deposit (A), we establish the following equivalence relationship: 4  0.06  ia =  1 +  − 1 = 0.06136 4   A( F / A,1.5%, 60) = $60, 000 + $60, 000( P / A, 6.136%,3) A = $219,978 / 96.2147 A = $2, 286.32

4.61 Setting the equivalence relationship at the end of 20 years gives 2

isemiannual A( F / A,

 0.06  = 1 +  − 1 = 3.0225% 4  

6% ,80) = $40, 000( P / A,3.0225%, 20) 4 152.71A = $593,862.93 A = $3,888.81

4.62 Given i =

5% = 0.417% per month 12

A = $500,000( A / P,0.417%,120) = $5,303.26 4.63 First compute the equivalent present worth of the energy cost savings during the first operating cycle:

$70 $70 0

1

2 June

3

$70 4

July Aug.

$80 $80 $80 5

6

7

8

9

10

11

12

Dec. Jan. Feb. .

P = $70( P / A, 0.5%,3)( P / F , 0.5%,1) + $80( P / A, 0.5%,3)( P / F , 0.5%, 7) = $436.35

15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Then, compute the total present worth of the energy cost savings over 5 years. P = $436.35 + $436.35( P / F , 0.5%,12) + $436.35( P / F , 0.5%, 24) +$436.35( P / F , 0.5%,36) + $436.35( P / F , 0.5%, 48) = $1,942.55

Continuous Payments with Continuous Compounding 4.64 Given i = 10%, N = 10 years, and A = $95,000 × 365 = $34,675,000 • Daily payment with daily compounding:

10% ,3650) = $219,170,331.48 365 • Continuous payment and continuous compounding: P = $95,000(P / A,

P=



10

0

Ae− rt dt

 e(0.10)(10) − 1  = $34,675,000  (0.10)(10)   0.1e  = $219,187,803.77

∴ The difference between the two compounding schemes is only $17,472.27.

4.65 Given i = 11%, N = 3, F0 = $500, 000, FN = $40, 000, and 0 ≤ t ≤ 3,

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Contemporary Engineering Economics, 7th ed. ©2023

$500,000

$40,000 0

1

2

3

t

460, 000 t 3 3 460, 000 − rt P =  (500, 000 − t )e dt 0 3 1 − e −0.11(3)  $500, 000 − $40, 000 = 500, 000  [ −0.11(3)e −0.11(3) − e −0.11(3) + 1] − 2 0.11 3(0.11)  

f (t ) = 500, 000 −

= $1, 277, 619.39 − $555, 439.67 = $722,179.72

4.66 Given r = 9% , A = $25, 000, N s = 2, N e = 7, 7

P =  25, 000e − rt dt 2

 e −0.09(2) − e −0.09(7)  = $25, 000   0.09   = $84, 077.34

4.67

yt = 5e−0.25t , yt ut = 275e

ut = $55(1 + 0.09t )

−0.25 t

+ 24.75te −0.25t

20

20

P =  275e−0.25t e −0.12t dt +  24.75te −0.25t e−0.12t dt 0

0

e − 1  24.75 24.75 = 275  + (1 − e −0.37(20) ) − (20e −0.37(20) ) 0.37(20)  2 0.37  0.37e  0.37 = $742.79 + $179.86 = $922.65 0.37(20)

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Contemporary Engineering Economics, 7th ed. ©2023

Changing Interest Rates 4.68

F = $10, 000( F / P,6%,3)( F / P,9%, 4)( F / P,7%,3) = $20,595.62 4.69 Given r1 = 6% compounded quarterly, r2 = 10% compounded quarterly, and r3 = 8% compounded quarterly, indicating that i1 = 1.5% per quarter,

i2 = 2.5% per quarter, and i3 = 2% per quarter. (a) Find P: P = $2,000( P / F ,1.5%,4) + $2,000( P / F ,1.5%,8) + $3,000( P / F ,2.5%,4)( P / F ,1.5%,8) + $2,000( P / F ,2.5%,8)( P / F ,1.5%,8) + $2,000( P / F ,2%,4)( P / F ,2.5%,8)( P / F ,1.5%,8) = $8,875.42

(b) Find F:

F = P( F / P,1.5%,8)( F / P,2.5%,8)( F / P,2%,4) = $13,186 (c) Find A, starting at 1 and ending at 5: F = A + A(F / P, 2%, 4) + A(F / P, 2.5%, 4)(F / P, 2%, 4) + A(F / P, 2.5%, 8)(F / P, 2%, 4) + A(F / P,1.5%, 4)(F / P, 2.5%, 8)(F / P, 2%, 4) = 5.9958A A=

$13,186 = $2,199.21 5.9958

4.70 (a)

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Contemporary Engineering Economics, 7th ed. ©2023 P = $300( P / F ,0.5%,12) + $300( P / F ,0.75%,12)( P / F ,0.5%,12) + $500( P / F ,0.75%,24)( P / F ,0.5%,12) + $500( P / F ,0.5%,12)( P / F ,0.75%,24)( P / F ,0.5%,12) = $1,305.26

(b)

$1,305.26 = $300( P / A, i, 2) + $500( P / A, i, 2)( P / F , i, 2) i = 7.818% per year

4.71 Since payments occur annually, you may compute the effective annual interest rate for each year.

i1 = (1 +

0.09 365 ) − 1 = 9.416% , 365

i2 = e 0.09 − 1 = 9.417%

F = $400( F / P,9.416%, 2)( F / P,9.417%, 2) + $250( F / P,9.416%,1)( F / P,9.417%, 2) +$100( F / P,9.417%, 2) + $100( F / P,9.417%,1) + $250 = $1,379.93

4.72 i1 = e0.06 − 1 = 6.18%,

i2 = e0.08 − 1 = 8.33%

F = $1, 000e0.06e0.08 = $1, 000( F / P, 6.18%,1)( F / P,8.33%,1) = $1,150.25

Amortized Loans 4.73 Loan repayment schedule for the selected 6 payments: End of month 0 1 2 13

Interest Payment $0.00 $100.00 $97.46 $68.64

Principal Payment $0.00 $508.44 $510.98 $539.80

19 Copyright © 2023 Pearson Education, Inc.

Remaining Balance $20,000.00 $19,491.56 $18,980.58 $13,188.31

Contemporary Engineering Economics, 7th ed. ©2023 24 36

$38.20 $3.03

$570.24 $605.41

$7,069.38 $0

4.74 (a) (i) $10,000 ( A / P,0.75%,24) (b) (iii) B12 = A( P / A,0.75%,12) 4.75 Given information: i = 9.45% / 365 = 0.0259% per day , N = 36 months.

•

Effective monthly interest rate, i = (1 + 0.000259)30 − 1 = 0.78% per month

•

Monthly payment, A = $13,000( A / P,0.78%,36) = $415.58 per month

•

Total interest payment, I = $415.58 × 36 − $13, 000 = $1, 960.88

4.76 (a) Using the bank loan at 9.2% compound monthly Purchase price = $22,000, Down payment = $1,800 A = $20,200( A / P,(9.2 / 12)%,48) = $504.59

(b) Using the dealer’s financing, Purchase price = $22,000, Down payment = $2,000, Monthly payment = $505.33, N = 48 end of month payments. Find the effective interest rate: $505.33 = $20, 000( A / P, i, 48) i = 0.8166% per month APR(r ) = 0.8166% × 12 = 9.80%

4.77 Given Data: P = $25,000, r = 9% compounded monthly, N = 36 month, and 20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 i = 0.75% per month. •

Required monthly payment: A = $25, 000( A / P, 0.75%, 36) = $795

•

The remaining balance immediately after the 20th payment:

B20 = $795( P / A,0.75%,16) = $11,944.33 4.78 Given Data: P = $250,000 - $50,000 = $200,000. •

Option 1: N = 15 years × 12 = 180 months APR = 4.25% ∴ A = $200, 000( A / P, 4.25% / 12,180) = $1, 504.56

•

Option 2: N = 30 years × 12 = 360 months APR = 5% ∴ A = $200,000( A / P,5% / 12,360) = $1,073.64 ∴ Difference = $1, 504.56 - $1,073.64 = $430.92

4.79 •

The monthly payment to the bank: Deferring the loan payment for 6 months is equivalent to borrowing $16, 000( F / P, 0.75%, 6) = $16, 733.64

To pay off the bank loan over 36 months, the required monthly payment is A = $16, 733.64( A / P, 0.75%,36) = $532.13 per month

•

The remaining balance after making the 16th payment: $532.13( P / A, 0.75%, 20) = $9,848.67

21 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

The loan company will pay off this remaining balance and will charge $308.29 per month for 36 months. The effective interest rate for this new arrangement is: $9,848.67 = $308.29( P / A, i,36) ( P / A, i,36) = 31.95 i = 0.66% per month

∴ r = 0.66% × 12 = 7.92% per year

4.80 (a)

A = $300, 000( A / P,

5.5% ,180) = $2, 451.25 12

(b) Remaining balance after the 59th payment:

B59 = $2, 451.25( P / A,

5.5% ,121) = $227, 276.52 12

•

Interest for the 60th payment = (5.5% /12) × B23 = $1,041.68

•

Principal payment for the 60th payment = $2, 451.25 − $1, 041.68 = $1, 409.57

•

9% ,180) = $4, 057.07 12 Total payments over the first 5 years (60 months)

4.81

A = $400, 000( A / P,

$4,057.07 × 60 = $243,424.20

•

Remaining balance at the end of 5 years: B60 = $4,057.07( P / A,0.75%,120) = $320,271.97

• •

Reduction in principal = $400,000 - $320,271.97 = $79,728.03 Total interest payments = $243, 424.20 − $79,728.03=$163,696.17

4.82 The amount to finance = $300,000 - $45,000 = $255,000 22 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

A = $255, 000( A / P, 0.5%,360) = $1,528.85

Then, the minimum acceptable monthly salary (S) should be

S=

A $1,528.85 = = $6,115.42 0.25 0.25

4.83 Given Data: purchase price = $150,000, down payment (sunk equity) = $30,000, interest rate = 0.75% per month, N = 360 months, •

Monthly payment: A = $120,000 ( A / P,0.75%,360 ) = $965 .55

•

Balance at the end of 5 years ( 60 months):

B60 = $965.55( P / A,0.75%,300) = $115,056.50 •

Realized equity = sales price – balance remaining – sunk equity: $185,000 - $115,056.60 - $30,000 = $39,943.50 Note: For tax purpose, we do not consider the time value of money on $30,000 down payment made five years ago.

4.84 Given Data: interest rate = 0.75% per month, each individual has the identical remaining balance prior to their 20th payment, that is, $80,000. With equal remaining balances, all will pay the same interest for the 20th mortgage payment. $80,000(0.0075) = $600 4.85 Given Data: loan amount = $130,000, point charged = 3%, N = 360 months, interest rate = 0.75% per month, actual amount loaned = $126,100: •

Monthly repayment: A = $130,000 ( A / P,0.75%,360 ) = $1,046

•

Effective interest rate on this loan

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Contemporary Engineering Economics, 7th ed. ©2023

$126,100 = $1,046( P / A, i,360) i = 0.7787% per month ∴ ia = (1 + 0.007787)12 − 1 = 9.755% per year

4.86 (a)

$50,000 = $7,500(P / A,i,5) + $2,500(P / G,i,5) i = 6.914%

(b) P = $50,000 Total payments = $7,500 + $10,000 + … + $17,500 = $62,500 Interest payments = $3,456.87 + … + $1,131.66 = $12,500 End of month 0 1 2 3 4 5

Interest Payment $0.00 $3,456.87 $3,177.34 $2,705.64 $2,028.48 $1,131.66

Principal Payment $0.00 $4,043.13 $6,822.66 $9,794.36 $12,971.52 $16,368.34

Remaining Balance $50,000.00 $45,956.87 $39,134.21 $29,339.85 $16,368.34 $0

4.87 Given Data: r = 7% compounded daily, N = 25 years •

The effective annual interest rate is ia = (1 + 0.07 / 365) 365 − 1 = 7.25%

•

Total amount accumulated at the end of 25 years $75, 000 × 5% = $3, 750

F = $3, 750(F / A, 7.25%, 25) + $150(F / G, 7.25%, 25) = $3, 750(F / A, 7.25%, 25) + $150(P / G, 7.25%, 25)(F / P, 7.25%, 25) = $329, 799.78

4.88 (a) The dealer’s interest rate to calculate the loan repayment schedule. (b) 24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

•

Required monthly payment under Option A: A = $26, 200( A / P, 0%, 36) = $727.78

•

Break-even savings rate $24, 048 = $727.78( P / A, i,36) i = 0.4708% per month r = 0.4708% × 12 = 5.6501% per year

The funds are earning 5% annual interest rate compounded monthly ( i = 5% / 12 = 0.4167 per month ). It means Option B is better than A. If funds can earn more than 5.6501% APR, the dealer’s financing is a leastcost alternative. (c) The dealer’s interest rate is only good to determine the required monthly payments. The interest rate to be used in comparing different options should be based the earning opportunity foregone by purchasing the vehicle. In other words, what would the decision maker do with the amount of $24,048 if he or she decides not to purchase the vehicle? If he or she would deposit the money in a saving account, then the savings rate is the interest rate to be used in the analysis.

Add-on Loans 4.89 •

The total amount you paid in interest

•

iPN = (0.18)$20,000(5) = $18,000 Monthly payment

A=

$20,000 + $18,000 = $633.33 60

4.90 (a)

$5, 000 = $189.58( P / A, i,36) i = 1.789% per month r = 1.789% × 12 = 21.4684% 25 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b)

P = $189.58( P / A,1.789%,12) = $2, 031.10

4.91

$5, 025 = $146.35( P / A, i, 48) i = 1.46% per month P = $146.35( P / A,1.46%,33) = $3,810.89

Loans with Variable Payments 4.92 0.12 1 ) − 1 = 1% per month 12 ⋅1 P = $2, 000( F / P,1%, 2) = $2, 040.20 i = (1 +

4.93 • Effective interest rate for Bank A 0.18 4 ) − 1 = 19.252% i = (1 + 4 • Effective interest rate for Bank B 0.175 365 ) − 1 = 19.119% i = (1 + 365 •

Select (c)

4.94 im = 2.9% / 12 = 0.2417%,17% /12 = 1.417% $3, 000( F / P, 0.2417%, 6)( F / P,1.417%, 6) −$100[( F / A, 0.2417%, 6)( F / P, 0.2417%, 6) + ( F / A,1.417%, 6)] = $2, 077.79 4.95 im = (1 +

0.12 1 ) − 1 = 1% 12 ⋅1

26 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) $10,000( A / P,1%, 48) = 10,000(0.0263) = $263 / month (c) Remaining balance at the beginning of 20th month is

$263( P / A,1%, 29) = $263(25.0658) = $6,592.3 1 So, the interest payment for the 20th payment is $6,592.31*1% = $65.92. (d) $263 x 48months = $12,624 So, the total interest paid over the life of the loan is $2,624. 4.96

$18, 000 = A( P / A, 0.667%,12) + A( P / A, 0.75%,12)( P / F , 0.667%,12) = A(11.4958) + A(11.4349)(0.9234) = 22.05479 A

4.97

A = $816.15

• 24-month lease plan:

P = ($2,500 + $520) + $500 + $520( P / A,0.5%,23) −$500( P / F,0.5%,24) = $13,884.13 • Up-front lease plan:

P = $12,780 + $500 − $500(P / F,0.5%,24) = $12,836.4 Select the single up-front lease plan.

4.98 (a) Amount of dealer financing = $15,458(0.90) = $13,912 A = $13, 912(A / P,11.5% / 12, 60) = $305.96

(b) Assuming that the remaining balance will be financed over 56 months, B4 = $305.96(P / A,11.5% / 12, 56) = $13, 211.54 A = $13, 211.54(A / P,10.5% / 12, 56) = $299.43 27 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(c) Interest payments to the dealer:

Idealer = $305.96 × 4 + $13,211.54 − $13,912 = $523.38 Interest payments to the credit union:

I union = $299.43 × 56 − $13, 211.54 = $3,556.54 4.99 Given: purchase price = $155,000, down payment = $25,000 •

Option 1: i = 7.5% /12 = 0.625% per month , N = 360 months

•

Option 2: For the assumed mortgage, P1 = $97, 218 , i1 = 5.5% / 12 = 0.458% per month , N1 = 300 months , A1 = $597 per month ;

For the 2nd mortgage P2 = $32,782 , i2 = 9% /12 = 0.75% per month ,

N2 = 120 months (a) For the second mortgage, the monthly payment will be A2 = P2 (A / P,i2 , N 2 ) = $32, 782(A / P, 0.75%,120) = $415.27

$130, 000 = $597( P / A, i,300) + $415.27( P / A, i,120) i = 0.5005% per month r = 0.5005% × 12 = 6.006% per year ia = 6.1741%

(b) Monthly payment •

Option 1: A = $130,000( A / P,0.625%,360) = $908.97

•

Option 2: $1,012.27 (= $597 + $415.27) for 120 months, then $597 for remaining 180 months.

(c) Total interest payment

28 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • •

Option 1: I = $908.97 × 360 − $130, 000 = $197, 229.20 Option 2: I = $228,932 .4 − $130,000 = $98,932 .4

(d) Equivalent interest rate: $908.97( P / A, i,360) = $597( P / A, i,300) + $415.27( P / A, i,120) i = 1.2016% per month r = 1.2016% × 12 = 14.419% per year ia = 15.4114%

Loans with Variable Interest Rates 4.100

$10,000 = A( P / A,0.6667%,12) + A( P / A,0.75%,12)( P / F ,0.6667%,12) = 22.05435 A ∴ A = $453.43

4.101

Given: i = 9% / 12 = 0.75% per month, deferred period = 6 months, N = 36 monthly payments, first payment due at the end of 7th month, the amount of initial loan = $15,000 (a) First, find the loan adjustment required for the 6-month grace period. $15,000 ( F / P,0.75%,6) = $15,687 .78

Then, the new monthly payments should be A = $15,687.78( A / P,0.75%,36) = $498.87

(b) Since there are 10 payments outstanding, the loan balance after the 26th payment is

B26 = $498.87( P / A,0.75%,10) = $4,788.95 (c) The effective interest rate on this new financing is

$4,788.95 = $186(P / A,i,30) i = 1.0161% per month r = 1.0161% × 12 = 12.1932% ia = (1+ 0.010161)12 − 1 = 12.90%

29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 4.102 Given: P = $120,000 , N = 360 months, i = 9% / 12 = 0.75% per month (a)

A = $120,000 ( A / P,0.75 %,360 ) = $965 .55

(b) If r = 9.75% APR after 5 years, then i = 9.75% / 12 = 0.8125% per month. •

The remaining balance after the 60th payment:

B60 = $965.55( P / A,0.75%,300) = $115,056.50 •

Then, we determine the new monthly payments as A = $115, 056.50(A / P, 0.8125%, 300) = $1, 025.31

4.103 Let Ai be the monthly payment for i th year and B j the balance of the loan at the end of j th month. Then, A1 = $475,000( A / P,8.125% / 12,360) = $3,526.85

B12 = $3,526.85( P / A,8.125% / 12,348) = $471,129.35 A2 = $471,129.35( A / P,10.125% / 12,348) = $4, 200.83 B24 = $4, 200.83( P / A,10.125% / 12,336) = $468, 291.82 A3 = $468, 291.82( A / P,12.125% / 12,336) = $4,898.83 B36 = $4,898.83( P / A,12.125% / 12,324) = $466,170.79

A4 − 30 = $466,170.79( A / P ,13.125% / 12, 324) = $5, 253.55

(a) The monthly payments over the life of the loan are A1 = $475,000( A / P,8.125% / 12,360) = $3,526.85 A2 = $471,129.35( A / P,10.125% / 12,348) = $4, 200.83 A3 = $468, 291.82( A / P,12.125% / 12,336) = $4,898.83

A4−30 = $466,170.79( A / P,13.125% /12,324) = $5, 253.55 (b)

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Contemporary Engineering Economics, 7th ed. ©2023

($3,526.85 ×12) + ($4, 200.83 ×12) + ($4,898.83 ×12) + ($5, 253.55 × 324) − $475,000 = $1,378,670 (c)

$475, 000 = $3,526.85( P / A, i,12) + $4, 200.83( P / A, i,12)( P / F , i,12) +$4,898.83( P / A, i,12)( P / F , i, 24) + $5, 253.55( P / A, i,324)( P / F , i,36) i = 1.0058% per month APR(r ) = 1.0058% ×12 = 12.0696% ia = (1 + 0.010058)12 − 1 = 12.76% per year

Investment in Bonds 4.104 Given: Par value = $1,000, coupon rate = 6%, paid $30 semiannually, N = 60 semiannual periods. Note that the maturity date is December 31, 2049. (a) Find YTM

$1, 000 = $30( P / A, i, 60) + $1, 000( P / F , i, 60) i = 3% semiannually ia = 6.09% per year Note: Since they bought and redeemed at par, we can calculate ia simply by ia = (1 + 0.03) 2 − 1 = 6.09%

(b) Find the bond price after 5 years with r = 5%: i = 2.5% semiannually, N = 2(30 – 5) = 50 semiannual periods.

P = $30( P / A, 2.5%,50) + $1,000( P / F , 2.5%,50) = $1,141.81 (c) •

Sale price after 2 ½ years later = $922.38, the YTM for the new investors:

$922.38 = $30( P / A, i,55) + $1, 000( P / F , i,55) i = 3.31% semiannually ia = 6.73%

31 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • • •

Current yield at sale = $30/$922.38 = 3.25% semiannually Nominal current yield = 3.25% × 2 = 6.5% per year Effective current yield = 6.61% per year

4.105 Given: Purchase price = $1,010, par value = $1,000, coupon rate = 9.5%, bond interest ($47.50) semiannually, required return = 10% per year compounded semiannually, N = 6 semiannual periods F + $47.5(F / A, 5%, 6) − $1, 010(F / P, 5%, 6) = 0

∴ F = $1,030 .41

4.106 Given: Par value = $1,000, coupon rate = 8%, $40 bond interest paid semiannually, purchase price = $920, required return = 9% per year compounded semiannually, N = 8 semiannual periods $920 = $40( P / A,4.5%,8) + F ( P / F ,4.5%,8)

∴ F = $933.13

4.107 •

Option 1: Given purchase price = $513.60, N = 10 semiannual periods, par value at maturity = $1,000

$513.60 = $1, 000( P / F , i,10) i = 6.89% semiannually ia = 14.255% per year •

Option 2: Given purchase price = $1,000, N = 10 semiannual periods, $113 interest paid every 6 months

$1, 000 = $113( P / A, i,10) + $1, 000( P / F , i,10) i = 11.3% semiannually ia = 23.877% per year ∴ Option 2 has a better yield.

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 4.108 Given: Par value = $1,000, coupon rate = 15%, or $75 interest paid semiannually, purchase price = $1,298.68, N = 24 semiannual periods

$1,298.68 = $75(P / A,i,24) + $1,000(P / F,i,24) i = 5.277% semiannually ia = 10.83% per year 4.109 Given: Par value = $1,000, i = 4.5% semiannually, N A = 30, N B = 2 semiannual periods

PA = $100( P / A,4.5%,30) + $1,000( P / F ,4.5%,30) = $1,895.89 PB = $100( P / A,4.5%,2) + $1,000( P / F ,4.5%,2) = $1,103 4.110 Given: Par value = $1,000, coupon rate = 8.75%, or $87.5 interest paid annually, N = 4 years (a) Find YTM if the market price is $1,108:

$1,108 = $87.5( P / A, i,4) + $1,000( P / F , i,4) i = 5.66% (b) Find the present value of this bond if i = 9.5%:

P = $87.5( P / A,9.5%,4) + $1,000( P / F ,9.5%,4) = $975.97 ∴ It is good to buy the bond at $935.

4.111 Given: Par value = $1,000, coupon rate = 12%, or $60 interest paid every 6 months, N = 30 semiannual periods (a)

P = $60( P / A,4.5%, 26) + $1,000 ( P / F ,4.5%, 26) = $1,227 .20 33 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b)

P = $60( P / A, 6.5%, 26) + $1, 000( P / F , 6.5%, 26) = $938.04

(c) Current yield = $60 / $783.58 = 7.657% semiannually. The effective annual current yield = 15.9%. 4.112 Given: Par value = $1,000, coupon rate = 10%, paid as $50 every 6 months, N = 20 semiannual periods,

P = $50( P / A,3%,14) + $1,000( P / F ,3%,14) = $1,225.91

Short Case Studies with Excel ST 4.1 (a) • Bank A: ia = (1 + 0.0155)12 − 1 = 20.27% • Bank B: i a = (1 + 0.165 / 12)12 − 1 = 17.81% (b) Given i = 6% / 365 = 0.01644% per day, the effective interest rate per payment period is i = (1 + 0.0001644)30 −1 = 0.494% per month. We also assume that the $300 remaining balance will be paid off at the end of 24 months. So, the present worth of the total cost for the credit cards is, • Bank A:

P = $20 + $4.65( P / A,0.494%, 24) + $20( P / F ,0.494%,12) = $143.85 • Bank B:

P = $30 + $4.13( P / A,0.494%, 24) + $30( P / F ,0.494%,12) = $151.55 ∴ Select bank A

(c) Assume that Anita makes either the minimum 5% payment or $20, whichever is larger, every month. It will take 59 months to pay off the loan. The total interest payments are $480.37.

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Contemporary Engineering Economics, 7th ed. ©2023

Month (n ) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Beginning Unpaid Balance

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

1,500.00 1,444.59 1,391.23 1,339.85 1,290.35 1,242.69 1,196.79 1,152.58 1,110.01 1,069.01 1,029.52 991.49 954.87 919.60 885.63 852.92 821.42 791.07 761.85 733.71 706.61 680.51 655.37 631.17 607.85

Interest Charged

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

20.63 19.86 19.13 18.42 17.74 17.09 16.46 15.85 15.26 14.70 14.16 13.63 13.13 12.64 12.18 11.73 11.29 10.88 10.48 10.09 9.72 9.36 9.01 8.68 8.36

Total Outstanding

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

35 Copyright © 2023 Pearson Education, Inc.

1,520.63 1,464.46 1,410.36 1,358.27 1,308.10 1,259.78 1,213.25 1,168.43 1,125.27 1,083.71 1,043.68 1,005.13 968.00 932.25 897.81 864.65 832.71 801.95 772.33 743.80 716.33 689.87 664.39 639.85 616.21

Minimum Payment

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

76.03 73.22 70.52 67.91 65.40 62.99 60.66 58.42 56.26 54.19 52.18 50.26 48.40 46.61 44.89 43.23 41.64 40.10 38.62 37.19 35.82 34.49 33.22 31.99 30.81

Remaining Balance $ 1,500.00 $ 1,444.59 $ 1,391.23 $ 1,339.85 $ 1,290.35 $ 1,242.69 $ 1,196.79 $ 1,152.58 $ 1,110.01 $ 1,069.01 $ 1,029.52 $ 991.49 $ 954.87 $ 919.60 $ 885.63 $ 852.92 $ 821.42 $ 791.07 $ 761.85 $ 733.71 $ 706.61 $ 680.51 $ 655.37 $ 631.17 $ 607.85 $ 585.40

Contemporary Engineering Economics, 7th ed. ©2023 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

585.40 563.78 542.95 522.90 503.58 484.98 467.07 449.82 433.20 417.20 401.79 386.95 372.27 357.39 342.30 327.01 311.50 295.79 279.85 263.70 247.33 230.73 213.90 196.84 179.55 162.02 144.25 126.23 107.96 89.45 70.68 51.65 32.36 12.81

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

8.05 7.75 7.47 7.19 6.92 6.67 6.42 6.18 5.96 5.74 5.52 5.32 5.12 4.91 4.71 4.50 4.28 4.07 3.85 3.63 3.40 3.17 2.94 2.71 2.47 2.23 1.98 1.74 1.48 1.23 0.97 0.71 0.44 0.18

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

593.45 571.53 550.42 530.09 510.51 491.65 473.49 456.00 439.16 422.94 407.31 392.27 377.39 362.30 347.01 331.50 315.79 299.85 283.70 267.33 250.73 233.90 216.84 199.55 182.02 164.25 146.23 127.96 109.45 90.68 71.65 52.36 32.81 12.98

36 Copyright © 2023 Pearson Education, Inc.

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

29.67 28.58 27.52 26.50 25.53 24.58 23.67 22.80 21.96 21.15 20.37 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 12.98

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

563.78 542.95 522.90 503.58 484.98 467.07 449.82 433.20 417.20 401.79 386.95 372.27 357.39 342.30 327.01 311.50 295.79 279.85 263.70 247.33 230.73 213.90 196.84 179.55 162.02 144.25 126.23 107.96 89.45 70.68 51.65 32.36 12.81 -

Contemporary Engineering Economics, 7th ed. ©2023

ST 4.2 (a) Annual percentage rate: o Loan solicitation offer:

$2, 009.06 = $150( P / A, i,18) i = 3.3155% per month r = APR = 3.3155% ×12 = 39.79% ia = (1 + 0.033155 ) − 1 = 47.91% o Cash advance on your credit card: 12

APR = 24.99% The cash advance option is cheaper. (b) Total finance changes: o Loan solicitation option: I = $150 × 18 − $2, 009.06 = $690.94

o Cash advance option:

$2,009.06 = $50( P / A, 24.99% /12, N ) N  88 months I = $50 × 88 − $2, 009.06 = $2, 390.94

ST 4.3 (a) It is a quite deceiving claim that you would save $146,985.60 over 10 years, as the various loans have different maturity dates assumed. For example, credit card debt will be paid over 37 months whereas the current home mortgage loan assumes a loan life of 360 months.

Interest

Outstanding Monthly

Loan Life

Rate

Balance

Assumed

Payment

Credit Card

21%

$12,000

$448.62

37 months

Auto Loan

8%

$15,000

$359.20

50 months

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Contemporary Engineering Economics, 7th ed. ©2023 Home

10%

$23,000

$488.68

60 months

8%

$170,000

$1,247.40

360 months

$220,000

$2,543.90

$220,000

$1,319.02

Improvement Loan Current Home Mortgage

Total Current Debt New Homeowner’s

6%

360 months

Bill Consolidation Loan $1,224.88

Monthly savings

(b) It is true that whenever a borrower can refinance the current debt at a lower rate of interest, it would be advantageous to the borrower. However, in this example, the loan maturity for each of debt is different, so we need to compare the current debt repayment plan with the new consolidated loan plan considering your time value of money that is your personal interest rate. Suppose your personal interest rate is 4% compounded monthly. Then, the total cost of each plan is as follows:

4% 4% ,37) + $359.20( P / A, ,50) 12 12 4% 4% , 60) + $1, 247.40( P / A, ,360) +$488.68( P / A, 12 12 = $319,929

Pold = $448.68( P / A,

PNew = $1,319.02( P / A,

4% ,360) 12

= $276, 284 Δ = $319,929 − $276, 284 = $43, 645

38 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 This indicates that the savings over 30 years is about $43,645, rather than $146,985.60 over 10 years. ST 4.4 The primary risk for consumers purchasing with ARM loans is that interest rates will rise and then their payments will increase to an amount that they are no longer able to afford. A hybrid loan can be a ticking time bomb for borrowers who plan on holding the loan for the long term. Let's say a borrower takes out a 30-year $200,000 hybrid loan that will remain at a fixed rate of 5.19% for five years, and then will switch to an adjustable-rate mortgage for the remaining period. The lender agrees to set a cap on the adjustable-rate portion of the loan, so that the rate will climb no more than five percentage points. So, conceivably, if rates are sharply higher after that initial seven-year period, a borrower could be looking at a mortgage rate of more than 10%. Under this scenario, the homeowner's monthly payment on a $200,000 mortgage would jump to $1,673 from $1,097 after the five-year term expires - a 53% increase! And despite the current quiescent rate environment, such a scenario is possible. But if there's a very real chance, you'll be looking to sell your home over the next 10 years, hybrids can make a lot of sense, since shorter-term loans usually carry the lowest rates. ST4.5 (a) (b)

A = $400, 000( A / P, 2.9% / 12, 360) = $1, 664.92

$400, 000 = $1,347.24( P / A, i,12) +$1, 414.60( P / A, i,12)( P / F , i,12) +$1, 485.33( P / A, i,12)( P / F , i, 24) +$1,559.60( P / A, i,12)( P / F , i,36) +$1, 637.58( P / A, i,12)( P / F , i, 48) +$1, 719.46( P / A, i,300)( P / F , i, 60)

Solving for i by the trial-and-error yields

i = 0.2417% ia = 2.94%

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Contemporary Engineering Economics, 7th ed. ©2023

Comments: With Excel, you may enter the loan payment series and use the IRR (range, guess) function to find the effective interest rate. If the loan amount (-$400,000) is entered in cell A1 and the following loan repayment series in cells A2 through A361, the effective interest rate is found with a guessed value of 0.2%: =IRR(A1:A361,0.2%)=0.002417

(c)

Compute the mortgage balance at the end of 5 years:



Conventional mortgage:

B60 = $1,664.92( P / A, 2.9% /12,300) = $354,973.56 

Graduated mortgage (not including the mortgage insurance):

B60 = $1,719.46( P / A, 2.9% /12,300) = $366,601.90 (d)

Compute the total interest payment for each option:



Conventional mortgage:

I = $1,664.92(360) − $400,000 = $199,371.20 

The graduated mortgage using Excel I = $205,159.83

(e)

Compute the equivalent present worth cost for each option at i = 6% /12 = 0.5% per month:



Conventional mortgage: P = $1, 664.92( P / A, 0.5%, 360) = $277, 694.69



The graduated mortgage including mortgage insurance: P = $1, 410.90( P / A, 0.5%,12) +$1, 478.26( P / A, 0.5%,12)( P / F , 0.5%,12) +$1,548.99( P / A, 0.5%,12)( P / F , 0.5%, 24) +$1, 623.26( P / A, 0.5%,12)( P / F , 0.5%,36) +$1, 701.24( P / A, 0.5%,12)( P / F , 0.5%, 48) +$1, 783.12( P / A, 0.5%,300)( P / F , 0.5%, 60) = $285, 033.90

∴ The conventional mortgage option is more desirable (least cost). 40 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 5 Present-Worth Analysis Identifying Cash Inflows and Outflows 5.1

Revenue = $45 × (0.40)(5,000) × 5 = $450,000 (No. of engineer:5)

n 0 1 2 … 8

Inflow $450,000 $450,000 … $450,000

Outflow $500,000 175,000 175,000 … 175,000

Net flow -$500,000 275,000 275,000 … 275,000

5.2 Project cash flows over the project life Cmax

Demand

Revenue

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

6,000,000 6,000,000 6,000,000 6,000,000 6,000,000 6,000,000 12,000,000 12,000,000 12,000,000 12,000,000 12,000,000 12,000,000 12,000,000 12,000,000

3,000,000 3,300,000 3,630,000 3,993,000 4,392,300 4,831,530 5,314,683 5,846,151 6,430,766 7,073,843 7,781,227 8,559,350 9,415,285 10,356,814

$16,256,976 17,882,673 19,670,941 21,638,035 23,801,838 26,182,022 28,800,224 31,680,247 34,848,271 38,333,099 42,166,408 46,383,049 51,021,354 56,123,490

15

24,000,000

11,392,495

61,735,839

$6,462,108 7,096,319 7,793,951 8,561,346 9,405,481 10,334,029 11,355,432 12,478,975 13,714,872 15,074,359 16,569,795 18,214,775 20,024,252 22,014,678

Cost of Bldg. $1,527,776 ♠ 1,545,123 1,573,302♠

NCF -$1,527,776 9,794,868 10,786,354 11,876,990 13,076,689 14,396,358 14,302,870 17,444,793 19,201,272 21,133,399 23,258,739 25,596,613 28,168,274 30,997,102 32,535,510

24,204,145

-

$37,531,693

Expense

♠: The cost of building is given as if Cmax is being built from scratch. No “credit” is given for the capacity already in place. This assumption could be rather unrealistic. In that case, what we need to do is to identify the incremental cost of adding the additional capacity above the existing capacity.

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Contemporary Engineering Economics, 7th ed. ©2023

Payback Period 5.3. (a) It will take 3 years to recover the total investment. Inflow

n 0 1 2 3 4 5

Outflow $35,500 $0 $0 $0 $0 $0

$0 $12,000 $12,000 $12,000 $12,000 $17,000

Net Cash Flow -$35,500 $12,000 $12,000 $12,000 $12,000 $17,000

Cumulative CF -$35,500 -$23,500 -$11,500 $500 $12,500 $29,500

(b) It will take 5 years to recover the total investment. n 0 1 2 3 4 5

Cash Flow -$35,500 $12,000 $12,000 $12,000 $12,000 $17,000

Cost of funds 17% $0 -$6,035 -$5,021 -$3,835 -$2,446 -$822

Cumulative CF -$35,500 -$29,535 -$22,556 -$14,391 -$4,837 $11,341

5.4 (a) Conventional payback period: $100,000/$25,000 = 4 years (b) Discounted payback period at i = 15%: n 0 1

Net Cash Flow

Cost of Funds (15%)

-$100,000 $25,000 -$100,000(0.15) =-$15,000

2 3 4 5

$25,000 $25,000 $25,000 $25,000

6 7

$25,000 $25,000

-$90,000(0.15) =-$13,500 -$11,775 -$9,791 -$7,510 -$4,886 -$1,869

Payback period = 6.54 years

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Cumulative Cash Flow -$100,000 -$90,000 -$78,500 -$65,275 -$50,066 -$32,576 -$12,463 $10,668

Contemporary Engineering Economics, 7th ed. ©2023 5.5 (a) Conventional payback period: $1,527,776/$9,794,868 = 0.1560 years (b) Discounted payback period = $1,527,776(1.15)/$9,794,868 = 0.1794 years, assuming continuous payments:

n

Net Cash Flow

0 1

-$30,000 +$40,000

Cost of Funds (15%)

Cumulative Cash Flow -$30,000 -$30,000(0.15) = -$4,500 +$5,500

5.6 •

Compute the required annual net cash profit to pay off the investment and interest.

$70,000, 000 = A(P / A,10%, 5) = 3.7908A A = $18, 465,824 •

Decide the number of shoes, X

$18, 465, 824 = X($100) X = 184, 658.24 5.7 Project A B C D

(a) Payback Period 3.40 1.67 3.18 2.60

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(b) Discounted Payback Period 3.70 1.99 3.66 3.25

Contemporary Engineering Economics, 7th ed. ©2023

NPW Criterion 5.8 (a)

PW(10%) A = −$800 + $3, 000( P / F ,10%,3) = $1, 453.9 PW(10%) B = −$1,800 + $600( P / F ,10%,1) +$900( P / F ,10%, 2) + $1, 700( P / F ,10%,3) = $766.49 PW(10%)C = −$1, 000 − $1, 200( P / F ,10%,1) +$900( P / F ,10%, 2) + $3,500( P / F ,10%,3) = $1282.3 PW(10%) D = −$6, 000 + $1,900( P / A,10%, 2) +$2,800( P / F ,10%,3) = −$598.91 (b)

Not provided.

5.9 PW(9%) = −$4, 000 + $3, 400( P / F ,9%,1) +$3, 400( P / F ,12%,1)( P / F ,9%,1) +$1,500( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) +$3,500( P / F ,13%,1)( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) +$4,300( P / F ,12%,1)( P / F ,13%,1)( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) = $7,858.34

5.10 PW(12%) = −$100, 000 + $30, 000( P / A,12%,5) = $8,143.3

5.11

PW(12%) = −$250, 000 − $20, 000 + $90, 000( P / A,12%, 5) + $75, 000( P / F ,12%, 5) = $96,986.87

Since PW(12%) > 0, this purchase should be justified.

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Contemporary Engineering Economics, 7th ed. ©2023

5.12 (a) PW(10%) = −$28,500 + $25,500( P / F ,10%,1) + $27,980( P / F ,10%, 2) + $32, 660( P / F ,10%,3) + $40, 230( P / F ,10%, 4) = $69,821.36

(b) Draw the graphs using Excel or other software (not provided).

5.13 (a)

(b)

(c)

PW(15%) = −$250, 000 + $40, 000( P / A,15%,35) + $50, 000( P / F ,15%,35) = $15, 040 > 0

PW(20%) = −$250, 000 + $40, 000( P / A, 20%,35) + $50, 000( P / F , 20%,35) = −$50, 254

PW(i ) = −$250, 000 + $40, 000( P / A, i,35) + $50, 000( P / F , i,35) = 0 i = 15.93%

∴ The correct answer is (d) as all of the statements are true.

5.14 •

Total cost in PW: $92,000,000 + $150,000( P / A1 ,5%,6%, 40) + $500,000( P / A,6%, 40) = 92,000,000 + $4,733,353 + $7,523,148 = $104, 256,501

•

Total tick revenue in PW: 100,000 X ( P / A,6%, 40) = 1,504,630 X

•

Ticket price: 1,504,630 X = $104, 256,501 X = $69.29 per ticket

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Contemporary Engineering Economics, 7th ed. ©2023

5.15 Given: Estimated remaining service life = 25 years, current rental income = $250,000 per year, O&M costs = $85,000 for the first year increasing by $5,000 thereafter, salvage value = $50,000, and MARR = 12% . Let A0 be the maximum investment required to break even. PW(12%) = − A0 + [$250, 000( F / A,12%, 25) + $25, 000( F / A,12%, 20) +$27,500( F / A,12%,15) + $30, 250( F / A,12%,10) +$33, 275( F / A,12%,5) + $50, 000]( P / F ,12%, 25) −$85, 000( P / A,12%, 25) − $5, 000( P / G ,12%, 25) =0

∴ Solving for A0 = yields

A0 = $1, 241, 461 5.16

PW(15%) = − X + $120, 000( P / A,15%, 4) + 0.1X ( P / F ,15%, 4) = 0 $342,597 = X − 0.0572 X X * = $363,382

5.17

PW(15%) = −$500, 000 − 2, 200, 00( P / F ,15%,1) + $2, 700, 000( P / A,15%,8) + $1,500, 000( P / F ,15%,8) = $10,193,077

5.18 PW(18%) = −$3, 500, 000 + [$1, 550, 000 − $350, 000 − $150, 000]( P / A,18%,10) + $200, 000( P / F ,18%,10) = $1, 257, 004 Since PW is positive, the project is justified.

5.19

PW(15%) = − A0 + $60, 000( P / A,15%,10) =0 A0 = $301,126

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Contemporary Engineering Economics, 7th ed. ©2023

Future Worth and Project Balance 5.20 FW(12%) = −$42, 000( F / P,12%,5) + $32, 400( F / P,12%, 4) +  + $32, 500( F / P,12%,1) + $33, 000 = $134, 056 > 0 → accept

5.21 (a) FW(15%) A = −$12,500( F / P,15%,3) + $6, 400( F / P,15%, 2) +$14, 400( F / P,15%,1) + $7, 200 = $13, 213 FW(15%) B = −$12,500( F / P,15%,3) − $3,000( F / P,15%, 2) +$23,000( F / P,15%,1) + $13,000 = $16, 472 FW(15%)C = $22,500( F / P,15%,3) − $7,000( F / P,15%, 2) −$2,500( F / P,15%,1) + $4,000 = $26, 087 FW(15%) D = −$14,000( F / P,15%,3) + $7,500( F / P,15%, 2) +$4,500( F / P,15%,1) + $8,500 = $2,302

∴ All projects are acceptable.

(b) Sample calculation for Project A: PB(15%)0 = −$12,500 PB(15%)1 = −$12,500(1 + 0.15) + $6, 400 = −$7,975 PB(15%) 2 = −$7,975(1 + 0.15) + $14, 400 = $5, 228.75 PB(15%)3 = $5, 228.75(1 + 0.15) + $7, 200 = $13, 213.06 The terminal project balance is the same as the net future worth of the project. 5.22

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (a) (b) (c) (d)

ia = 12.68% : $5,000( F / P,12.68%,20) = $54,438.58 i = 1% : $80( P / A,1%,120)( F / P,1%,240) = $60,737.33 i = 1% : $50( F / A,1%,240) = $49,462.77 ia = 12.68% : $15,000( F / P,12.68%,10) = $49,494.81

∴ Option B is the best choice.

5.23 (a)

PB(i )3 = −$7, 000(1 + i ) + $1,840 = −$6, 000 i = 12% (b) The original cash flows of the project are as follows.

n

An -$10,000 2,200 3,080 1,840 6,820

0 1 2 3 4

Project Balance -$10,000 -9,000 -7,000 -6,000 100

(c) No, the project is not acceptable. 5.24 (a) Using the project balance equation, 𝑃𝐵(𝑖) = 𝐴 = −$1,000 𝑃𝐵(𝑖) = 𝑃𝐵(𝑖) (1 + 𝑖) + 𝐴 = −$1,000(1 + 𝑖) + 𝐴 = −$1,100 𝑃𝐵(𝑖) = 𝑃𝐵(𝑖) (1 + 𝑖) + 𝐴 = −$1,100(1 + 𝑖) + 𝐴 = −$800 𝑃𝐵(𝑖) = 𝑃𝐵(𝑖) (1 + 𝑖) + 𝐴 = −$800(1 + 𝑖) + $460 = −$500 𝑃𝐵(𝑖) = 𝑃𝐵(𝑖) (1 + 𝑖) + 𝐴 = −$500(1 + 𝑖) + 𝐴 = $0

From PB(i)3, we find −$800(1 + 𝑖) + $460 = −$500 → 𝑖 = 20%

Therefore, A0 = -$1,000, A1 = $100, A2 = $520, A3 = $460, A4 = $600 (b) 𝑖 = 20% ← from 𝑃𝐵(𝑖)

(c)

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Since IRR = 20% > MARR (12%), Yes.

5.25 (a) No graphs are given. Period (n) 0 1 2 3 4

A -$2,500.00 -2,750.00 -3,025.00 -3,327.50 +2,539.75

Project Balance, PB(i)n B C -$3,500.00 -$2,800.00 -2,250.00 -4,880.00 325.00 -6,268.00 3,857.50 -2,394.80 6,443.25 1,865.72

D -$2,300.00 -3,530.00 -1,983.00 118.70 1,630.57

(b) Project PB2

A -$3,025

B $325

C -$6,268

D -$1,983

Project B would be preferred because it has recovered all of its investment and some surplus at the end of year 2. 5.26 (a) PW(10%) A = −$200 + $50( P / A,10%,3) −$100( P / F ,10%, 4) +$400( P / A,10%, 2)( P / F ,10%, 4) = $330.20 PW(10%) B = −$100 + $40( P / A,10%,3) +$10( P / A,10%, 2)( P / F ,10%,3) = $12.51 PW(10%)C = $120 − $40( P / A,10%,3) = $20.53 All projects are acceptable. (b) FW(10%) A = $330.20( F / P,10%, 6) = $584.97 FW(10%) B = $12.51( F / P,10%,5) = $20.15 FW(10%)C = $20.53( F / P,10%,3) = $27.32 9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

All projects are acceptable. (c) FW(i ) A = [−$200( F / P,10%,3) + $50( F / A,10%,3)]( F / P,15%,3) +[−$100( F / P,15%, 2) + $400( F / A,15%, 2)] = 574.60 FW(i ) B = [−$100( F / P,10%,3) + $40( F / A,10%,3)]( F / P,15%,3) +$10( F / P,15%, 2) + $10( F / P,15%,1) = $23.66 FW(i )C = $120( F / P,10%,3)( F / P,15%,3) −$40( F / A,10%,3)( F / P,15%,3) = $41.55 5.27 (a) •

Without software updates,

PW(15%) = $1, 200, 000( P / A1 , −25%,15%, 4) = $2, 457, 281 •

With software updates, PW(15%) = ($1, 200, 000 − $200, 000)( P / A,15%, 4) = $2,854,979

It is worth investing in software update annually. (b) Note that without software updates, you would expect $2,457,281. So, this is the minimum price that you may be willing to accept. 5.28 (a) First find the interest rate that is used in calculating the project balances. We can set up the following project balance equations: PB(i)1 = −$10, 000(1 + i) + A1 = −$11, 000 PB(i ) 2 = −$11, 000(1 + i ) + A2 = −$8, 200 PB(i)3 = −$8, 200(1 + i ) + $8, 000 = −$1,840 PB(i ) 4 = −$1,840(1 + i ) + A4 = $3, 792 PB(i)5 = $3, 792(1 + i) + A5 = $7,550

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Contemporary Engineering Economics, 7th ed. ©2023

i = 20% Substituting i into other PB(i) n yields From PB(i)3 , we can solve for i :

n

An -$10,000 1,000 5,000 8,000 6,000 3,000

0 1 2 3 4 5

PB(i)n -$10,000 -11,000 -8,200 -1,840 3,792 7,550

(b) Conventional payback period = 3 years (or 2.5 years with continuous cash flows) 5.29 (a) From the project balance diagram, note that PW(24%)1 = 0 for project 1 and PW(23%) 2 = 0 for project 2.

PW(24%)1 = −$1, 000 + $400( P / F , 24%,1) + $800( P / F , 24%, 2) + X ( P / F , 24%,3) = 0 PW(23%) 2 = −$1, 000 + $300( P / F , 23%,1) + Y ( P / F , 23%, 2) + $800( P / F , 23%,3) = 0 X = $299.58, Y = $493.49 (b) Since PW(24%)1 = 0 , FW(24%)1 = 0. (c)

ⓐ = $593.49,

ⓑ = $499.58,

ⓒ = 17.91%

5.30 (a) • PB(i )2 = −$2, 700(1 + i ) + $1, 470 = −$1,500 ∴ i = 10% • PW(10%) = $270( P / F ,10%, 5) = $167.65

(b) The original cash flows of the project are as follows

n 0 1 2

An -$3,000 ($600) $1,470 11 Copyright © 2023 Pearson Education, Inc.

Project Balance -$3,000 -$2,700 -$1,500

Contemporary Engineering Economics, 7th ed. ©2023 3 4 5

0 -$300 ($270)

($1,650) (-$300) $600

5.31 (a) No graphs are given. Period (n) 0 1 2 3 4

A -$2,500.00 -2,750.00 -3,025.00 -3,327.50 +2,539.75

Project Balance, PB(i)n B C -$3,500.00 -$2,800.00 -2,250.00 -4,880.00 325.00 -6,268.00 3,857.50 -2,394.80 6,443.25 1,865.72

D -$2,300.00 -3,530.00 -1,983.00 118.70 1,630.57

(b) Project PB 2

A -$3,025

B $325

C -$6,268

D -$1,983

Project B would be preferred because it has recovered all of its investment and some surplus at the end of year 2. 5.32 (a) Period (n) 0 1 2 3

A -$1,000.00 -1,100.00 -1,210.00 634.00

Project Balance, PB(i)n B C -$1,000.00 -$1,000.00 -506.00 -413.00 37.40 133.70 635.14 634.07

D -$1,000.00 -600.00 -60.00 634.00

FW(10%) A = FW(10%) B = FW(10%)C = FW(10%) D = $634 Note that the terminal project balances are the same (rounding errors), which are the net future worth of the projects. (b) Discounted payback period: A: 2.67 years, B: 1.93 years, C: 1.77 years, D: 2.09 years (c) & (d) Period (n) 0 1 2 Total Area

Area of Negative Project Balance, PB(i)n A B C D -$1,000.00 -$1,000.00 -$1,000.00 -$1,000.00 -1,100.00 -506.00 -413.00 -600.00 -1,210.00 -60.00 3,310.00 1,506.00 1,413.00 1,660.00 12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

It appears that Project C recovers its investment earliest among the four projects, so we can say that it has the minimum exposure to risk if other things are equal. 5.33 n 0 1 2 3 4 5

Cash flow -$152,000 $42,400 $53,400 $52,500 $62,500 $33,000

Project Balance -$152,000 -$132,400 -$98,860 -$61,189 -$7,867 $23,953

At the end of year 3, you still need to recover $61,189 from your previous investment. And you expect two more cash inflows from the project if you kept the project. The equivalent worth of these two payments at year 3 is P = $62,500(P / F ,15%,1) + $33,000(P / F ,15%,2) = $79,300 Selling price = $79,300 Net profit=$79,300 - $61,189 = $18,111

Of course, we don’t know whether or not the buyer would be willing to accept this amount but at least we know what to do when an offer is on the table. 5.34 From the annual cash flow and balance table, shown below, it will take 6.54 years approximately. n 0 1 2 3 4 5 6 7 8 9 10

Cash flow -$20,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00 $5,000.00

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Balance -$20,000.00 -$18,000.00 -$15,700.00 -$13,055.00 -$10,013.25 -$6,515.24 -$2,492.52 $2,133.60 $7,453.64 $13,571.68 $20,607.44

Contemporary Engineering Economics, 7th ed. ©2023

5.35  The present worth of the expenses already spent on the project: PW(20%) expense = −$10M( F / A, 20%,5) = −$74.416M

 The present worth of the expected revenue:

PW(20%) revenue = $100M( P / A, 20%,10) = $419.25M Net profit expected by keeping the project = -$74.416M + $419.25M - $30M = $314.83M. Here $74.416M is a kind of sunk cost but because of these expenditures, your R&D project is worth generating $419.25M in the future. Therefore, you may set your minimum asking price at $419.25M. Of course, if you don’t have to pay $30M, you could ask less than $419.25M. 5.36  −$800( F / P,10%,3) + $150( F / P,10%, 2)  FW(i ) A =   ( F / P,15%,3)  +$150( F / P,10%,1) + $350  + ( −$200( F / P,15%, 2) + $500( F / P,15%,1) + $400 ) = $150.35 FW(i ) B = ( −$500( F / P,10%,3) + $150( F / A,10%,3) ) ( F / P,15%,3) + ( $100( F / P,15%, 2) + $100( F / P,15%,1) )

= −$9.78  $200( F / P,10%,3) − $40( F / P,10%, 2)  FW(i )C =   ( F / P,15%,3)  −$60( F / P,10%,1) − $140  = $17.95

5.37 (a) Since a project’s terminal project balance is equivalent to its future worth, we can easily find the equivalent present worth for each project by PW(10%) A = $105( P / F ,10%,5) = $65.20 PW(20%)C = −$1, 000( P / F , 20%,5)

(b)

= −$401.88

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Contemporary Engineering Economics, 7th ed. ©2023 PB(10%)0 = A0 = -$1,000 PB(10%)1 = PB(10%)0 (1+0.10)+A1 = -$1,000 PB(10%) 2 = PB(10%)1 (1+0.10)+A2 = -$900 PB(10%)3 = PB(10%) 2 (1+0.10)+A3 = -$690 PB(10%) 4 = PB(10%)3 (1+0.10)+A4 = -$359 PB(10%)5 = PB(10%) 4 (1+0.10)+A5 = $105 From the project balance equations above, we derive A0 = $1, 000, A1 = $100, A2 = $200, A3 = $300, A4 = $400, and A5 = $500 .

(c)

FW(20%) = PB(20%)5 = −$1, 000 (Note that the terminal project balance is the future worth of the project.) (d) PW(i ) B = FW(i ) B ( P / F , i, N ) = $198 / (1 + i )5 = $79.57

∴Solving for i yields i = 20%

5.38 (a)

PW(0%)A = 0 PW(18%) B = $575(P / F ,18%,5)=$251.34 PW(12%)C = 0 (b)

Assume that A2 = $500. PB(12%)0 = −$1, 000

PB(12%)1 = −$1, 000(1.12) + A1 = −$530 PB(12%)2 = −$530(1.12) + $500 = X Solving for X yields X = -$93.60. (c) The net cash flows for each project are as follows:

n 0 1 2 3 4 5

Net Cash Flow A B -$1,000 -$1,000 $200 $500 $200 $500 $200 $300 $200 $300 $200 $300

(d) 15 Copyright © 2023 Pearson Education, Inc.

C -$1,000 $590 $500 -$106 $147 $100

Contemporary Engineering Economics, 7th ed. ©2023

FW(0%)A = 0 FW(18%)B = $575 FW(12%)C = 0

Capitalized Equivalent Worth 5.39 (a) PW(25%) A = −$1, 000 + $912( P / F , 25%,1) + $684( P / F , 25%, 2) + $456( P / F , 25%,3) + $228( P / F , 25%, 4) = $494.22 PW(25%) B = −$1, 000 + $284( P / F , 25%,1) + $568( P / F , 25%, 2) + $852( P / F , 25%,3) + $1,136( P / F , 25%, 4) = $492.25 Select project A. (b) Project A

n 0 1 2 3 4

Cash Flow -$1000 $912 $684 $456 $228

Cost of funds $0 -$250 -$85 $65 196

Project Balance -$1,000 -$338 $262 $783 $1,207

Project B

n 0 1 2 3 4

Cash Flow

-$1,000 $284 $568 $852 $1,136

Cost of funds $0 -$250 -$242 -$160 $13

Project Balance -$1,000 -$966 -$640 $53 $1,202

Project B is exposed to higher risk of loss if either project terminates at the end of the year 2, according to the results below.

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Contemporary Engineering Economics, 7th ed. ©2023 5.40 PW(5%) = $1, 000, 000( P / A,5%,5) + $1,300, 000( P / A,5%,5)( P / F ,5%,5) + ($1,500, 000 / 0.05)( P / F ,5%,10) = $8, 739, 412 + $18, 417,398 = $27,156,810

5.41 (a) PW(0.5%) = $2, 460( P / A, 0.5%, 240) = $343,368.70

(b) PW(0.5%) = $2, 460( P / A, 0.5%, 480) = $447, 099.06

(c)

A i $2, 460 A= 0.005 = $492, 000 Comments: Longer life means greater total benefit, but most of the benefit is collected in the first 20 years. PW(0.5%) =

5.42  Capitalized equivalent amount:

$10, 000 $20, 000(A / F,10%, 4) + 0.1 0.1 = $243, 094.16

CE(10%) = $100, 000 +

5.43  Find the equivalent annual series for the first cycle:

A = [$100(P / A,15%,2) + $40(P / A,15%,2)(P / F,15%,2) + $20(P / F,15%,5)](A / P,15%,5) = $66.13  Capitalized equivalent amount:

CE(15%) =

$66.13 = $440.88 0.15

5.44

17 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Given: r =5% compounded monthly, maintenance cost = $80,000 per year

ia = (1 +

0.05 12 ) − 1 = 5.116% 12

CE(5.116%) =

$80, 000 = $1,563, 664 0.05116

5.45 Given: Construction cost = $15,000,000, renovation cost = $3,000,000 every 15 years, annual O & M costs = $1,000,000 and i = 5% per year (a) P1 = $15, 000, 000 $3, 000, 000( A / F ,5%,15) 0.05 = $2, 780,537

P2 =

P3 = $1, 000, 000 / 0.05 = $20, 000, 000 CE(5%) = P1 + P2 + P3 = $37, 780,537

(b) P1 = $15, 000, 000 $3, 000, 000( A / F ,5%, 20) 0.05 = $1,814,555

P2 =

P3 = $1, 000, 000 / 0.05 = $20, 000, 000 CE(5%) = P1 + P2 + P3 = $36,814,555

(c) • 15-year cycle with 10% of interest:

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Contemporary Engineering Economics, 7th ed. ©2023 P1 = $15, 000, 000 $3, 000, 000( A / F ,10%,15) 0.1 = $944, 213

P2 =

P3 = $1, 000, 000 / 0.1 = $10, 000, 000 CE(10%) = P1 + P2 + P3 = $25,944, 213

• 20-year cycle with 10% of interest: P1 = $15, 000, 000 $3, 000, 000( A / F ,10%, 20) 0.1 = $523, 789

P2 =

P3 = $1, 000, 000 / 0.1 = $10, 000, 000 CE(10%) = P1 + P2 + P3 = $25,523, 789 As interest rate increases, CE value decreases.

5.46 Given: Cost to design and build = $830,000 , rework cost = $120,000 every 10 years, new type of gear = $80,000 at the end of 5th year, annual operating costs = $70,000 for the first 15 years and $100,000 thereafter $120,000( A / F,7%,10) 0.07 +$80,000(P / F,7%,5)

CE(7%) = $830,000 +

+$70,000(P / A,7%,15) $100,000 (P / F,7%,15) 0.07 = $2,166,448.62 +

Mutually Exclusive Alternatives 5.47 (a) Option A n 0 1

Net Cash Flow -$5,000 $2,000

Cumulative CF -$5,000 -$3,000

19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 2 3 4

$2,000 $2,000 - $1,000 $2,000

-$1,000 $0 $2,000

Option B Net Cash Flow -$3,500 $2,000 - $1,500 $2,000 $2,000 - $1,000 $2,000

n 0 1 2 3 4

Cumulative CF -$3,500 -$3,000 -$1,000 $0 $2,000

∴ Both options are equally likely.

(b) PW(10%) A = $2, 000( P / A,10%, 4) − $1, 000( P / F ,10%,3) − $5, 000 = $588.42 PW(10%) B = $2, 000( P / A,10%, 4) − $1, 000( P / F ,10%,3) −$1,500( P / F ,10%,1) − $3,500 = $724.78

∴ Select project B.

5.48 PW(12%) A = −$1, 200 − $1,300( P / F ,12%,1) +  + $760( P / F ,12%,10) = $1,300.23 PW(12%) B = −$3,100 − $660( P / F ,12%,1) +  + $950( P / F ,12%,10) = $1, 466.08

∴ Select project B.

5.49 (a)

20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(15%) A = −$1,800 + $1,150( P / F ,15%,1) +  +$200( P / F ,15%, 4) = −$24.85 PW(15%) B = −$1,500 + $1, 200( P / F ,15%,1) +  +$100( P / F ,15%, 4) = $590.22

∴Select Project B.

(b) PW(15%)C = −$4, 000 + $1,500( P / F ,15%,1) + X ( P / F ,15%, 2) +$1,800( P / F ,15%,3) + X ( P / F ,15%, 4) = 1.3279 X − $1,512.12

To be acceptable, it must satisfy the following condition: PW(15%)C > 0 1.3279 X − $1,512.12 > 0 X > $1,138.74

(c) PW(18%) D = $1, 400 − $450( P / A,18%, 4) = $189.47 > 0 Yes, project D is acceptable.

(d) If the interest rate (MARR) is higher than 11.76%, project D is better. Otherwise, project E is better. 5.50 PW(12%)A = −$22,500 + $18, 610( P / F ,12%,1) + $15,930( P / F ,12%, 2) + $16,300( P / F ,12%,3) = $18, 417 PW(12%)B = −$16,900 + $15, 210( P / F ,12%,1) + $16, 720( P / F ,12%, 2) + $12,500( P / F ,12%,3) = $18,907 ∴ Select project B.

5.51

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Contemporary Engineering Economics, 7th ed. ©2023 PW(i ) B-A = −$2, 000 + $1,300( P / F , i,1) + $1,500( P / F , i, 2) > 0 Δ = −$2, 000 +

$1,300 $1,500 + >0 1+ i (1 + i ) 2

= −$2, 000 + $1,300 X − $1,500 X 2 , where X = X = 0.8 → i = 25%

1 . 1+ i

∴ Select project B, if MARR < 25%.

5.52 • Model A:

CE(12%) A = $60, 000 +

$25, 000( A / F ,12%,5) = $92, 794 0.12

• Model B:

CE(12%) B = $150, 000 +

$180, 000( A / F ,12%,50) = $150, 625 0.12

∴ Project A is more economical.

5.53 • Standard Lease Option:

PW(0.5%)SL = −$5,500 − $1,150 − $1,150( P / A, 0.5%, 23) +$1, 000( P / F , 0.5%, 24) = −$30, 690 • Single Up-Front Option: PW(0.5%)SU = −$29,500 + $1, 000( P / F , 0.5%, 24) = −$28, 613

∴ Select the single up-front lease option.

5.54 • Machine A: PW(13%) A = −$75, 200 − ($6,800 + $2, 400)( P / A,13%, 6) +$21, 000( P / F ,13%, 6) = −$101,891

• Machine B:

22 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(13%) B = −$44, 000 − $11,500( P / A,13%, 6) = −$89,972

∴ Machine B is a better choice.

5.55 (a) • Required HP to produce 10 HP: -Motor A: X 1 = 10 / 0.85 = 11.765 HP -Motor B: X 2 = 10 / 0.90 = 11.111 HP • Annual energy cost: -Motor A: 11.765(0.7457)(2,000)(0.09) = $1,579.17 -Motor B: 11.111(0.7457)(2,000)(0.09) = $1,491.39

• Equivalent cost: PW(8%) A = −$1, 200 − $1,579.17( P / A,8%,15) +$50( P / F ,8%,15) = −$14, 710.11 PW(8%) B = −$1, 600 − $1, 491.39( P / A,8%,15) +$100( P / F ,8%,15) = −$14,334

∴ Motor B is preferred.

(b) With 1,000 operating hours: •

Annual energy cost: -Motor A: 11.765(0.7457)(1, 000)(0.09) = $789.58 -Motor B: 11.111(0.7457)(1, 000)(0.09) = $745.69

•

Equivalent cost:

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Contemporary Engineering Economics, 7th ed. ©2023 PW(8%) A = −$1, 200 − $789.58( P / A,8%,15) +$50( P / F ,8%,15) = −$7,942.62 PW(8%) B = −$1, 600 − $745.69( P / A,8%,15) +$100( P / F ,8%,15) = −$7,951.19

∴ Motor A is now preferred.

5.56 Given: Required service period = infinite, analysis period = least common multiple service periods (6 years) • Model A:

PW(16%)cycle = −$22, 000 + $17,500( P / F ,16%,1) + $17, 000( P / F ,16%, 2) PW(16%)total

+ $15, 000( P / F ,16%,3) = $15,330 = $15,330[1 + ( P / F ,16%,3)] = $25,151

• Model B:

PW(12%)cycle = −$27, 000 + $20,500( P / F ,16%,1) + $28, 000( P / F ,16%, 2) = $11, 481 PW(16%)total = $11, 481[1 + ( P / F ,16%, 2) + ( P / F ,16%, 4)] = $26,354 ∴ Model B is preferred.

5.57 (a) Without knowing the future replacement opportunities, we may assume that both alternatives will be available in the future with the same investment and expenses. We further assume that the required service period is 6 years. (b) With the common service period of 6 years, • Project A1:

24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(10%)cycle = −$900 − $400( P / A,10%,3)

PW(10%)total

+$200( P / F ,10%,3) = −$1, 744.48 = −$1, 744.48[1 + ( P / F ,10%,3)] = −$3, 055.13

• Project A2:

PW(10%)cycle = −$1,800 − $300( P / A,10%, 6) +$500( P / F ,10%, 6) = −$2,824.34 ∴ Project A2 is preferred.

(c)

PW(10%) A1 = −$1, 744.48 PW(10%) A2 = −$1,800 − $300( P / A,10%,3) + S ( P / F ,10%,3) = −$2,546.06 + 0.7513S Let PW(10%) A1 = PW(10%) A2 , then S = $1, 067 5.58 (a) Assuming a common service period of 15 years • Project B1:

PW(12%) cycle = −$20, 000 − $2, 000( P / A,12%,5) + $2, 000( P / F ,12%,5) PW(12%)total

= −$26, 074.7 = −$26, 074.7[1 + ( P / A, 76.23%, 2)] = −$49, 266.29

Note:

(1.12) 5 − 1 = 76.23%

• Project B2:

PW(12%)cycle = −$17, 000 − $2,500( P / A,12%,3) + $3, 000( P / F ,12%,3) PW(12%)total

= −$20,869.24 = −$20,869.24[1 + ( P / A, 40.49%, 4)] = −$59,180.43

Note:

(1.12) 3 − 1 = 40.49%

∴ Select project B1.

(b) • Project B1 with 2 replacement cycles:

25 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(12%) = −$26, 074.7 − $26, 074.7( P / F ,12%,5) = −$40,870.19

• Project B2 with 4 replacement cycles where the 4th replacement ends at the end of first operating year: PW(12%) = −$20,869.24[1 + ( P / F ,12%, 3) + ( P / F ,12%, 6)] − [$17, 000 − ($2, 500 − $9, 000)( P / F ,12%,1)]( P / F ,12%, 9) = −$50, 334.10

∴ Project B1 is still a better choice.

5.59 PW(12%) K = −$25, 000 − 11, 000( P / A,12%,10) + 3, 000( P / F ,12%,10) = −$86,186.53 PW(12%)T = −$32, 000 − 9, 700( P / A,12%,10) + S ( P / F ,12%,10) = −$86,807.16 + S (0.3220)

PW(12%) K = PW(12%)T −$86,186.53 = −$86,807.16 + S (0.3220) S = $1,927.42 5.60 Since only Model B is repeated in the future, we may have the following sequence of replacement cycles: • Option 1: Purchase Model A now and repeat Model A forever. • Option 2: Purchase Model B now and replace it at the end of year 2 by Model A. Then repeat Model A forever.

PW(12%) A = −$8, 000 + $3,500( P / A,12%,3) = $406.41 PW(12%) B = −$15, 000 + $10, 000( P / A,12%, 2) = $1,900.51 (a) • Option 1:

$406.41( A / P,12%,3) 0.12 = $1, 410.08

PW(12%) AAA = • Option 2:

26 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(12%) BAA = $1,900.51 +

$406.41( A / P,12%,3) ( P / F ,12%, 2) 0.12

= $3, 024.62 ∴ Option 2 is a better choice.

(b) Let S be the salvage value of Model A at the end of year 2. −$8, 000 + $3,500( P / A,12%, 2) + S ( P / F ,12%, 2) = $1,900.51 −$2, 084.82 + S (0.7972) = $1,900.51

Solving for S yields S = $4,999.16

5.61 • Since either tower will have zero salvage value after 20 years, we may select the analysis period of 35 years:

PW(11%) Bid A = −$137, 000 − $2, 000( P / A,11%,35) PW(11%) Bid B

= − $154, 710 = −$128, 000 − $3,300( P / A,11%,35) = − $157, 222

∴ Bid A is a better choice.

• If you assume an infinite analysis period, the present worth of each bid will be: [−$137, 000 − $2, 000( P / A,11%, 40)]( A / P,11%, 40) 0.11 = − $157,322 [−$128, 000 − $3,300( P / A,11%,35)]( A / P,11%,35) = 0.11 = − $161, 407

PW(11%) Bid A =

PW(11%) Bid B

∴ Bid A is still preferred.

5.62 (a)

PW(15%) A1 = −$15, 000 + $9,500( P / F ,15%,1) +$12,500( P / F ,15%, 2) + $7,500( P / F ,15%,3) = $7, 644.04 27 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) PW(15%) A 2 = −$25, 000 + X ( P / A,15%, 2)( P / F ,15%,1) = $9,300 X = $24, 263

(c) Note that the net future worth of the project is equivalent to its terminal project balance. PB(15%)3 = $7, 644.04( F / P,15%,3) = $11, 625.63

(d) Select A2. It has the greater present worth. 5.63 (a) Project balances as a function of time are as follows: Modified Cash Flows B C n 0 -$7,000 -$5,000 1 -2,500 -2,000 2 -2,000 -2,000 3 -1,500 2,000 4 -1,500 -2,000 5 -1,500 -2,000 6 -1,500 -2,000 7 -3,000 -2,000 8 -3,000 -3,000 PW(15%) -$16,033 -$14,302 Option B is more economical.

5.64 • Option 1: Non-deferred plan PW(12%) = −$300, 000 − $49, 000( P / A,12%,8) = − $543, 414.35 • Option 2: Deferred plan PW(12%) = −$140, 000( P / F ,12%, 2) − $14, 000( P / A,12%, 6)( P / F ,12%, 2) − $160, 000( P / F ,12%,5) − $21, 000( P / A,12%,3)( P / F ,12%,5) − $180, 000( P / F ,12%,8) = − $349, 600.72

28 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 ∴ Option 2 is a better choice.

5.65 • Alternative A: Once-for-all expansion PW(15%)A = −$25M − $0.40M( P / A,15%, 25) +$0.85M( P / F ,15%, 25) = −$27,559,839 • Alternative B: Incremental expansion PW(15%) B = −$15M − $18M( P / F ,15%,10) −$12M( P / F ,15%,15) + $1.5M( P / F ,15%, 25) −$0.25M( P / A,15%, 25) −$0.10M( P / A,15%,15)( P / F ,15%,10) −$0.10M( P / A,15%,10)( P / F ,15%,15) = −$22, 639, 067

∴ Select alternative B.

5.66 • Option 1: Tank/tower installation

PW(12%)1 = −$164, 000 • Option 2: Tank/hill installation with the pumping equipment replaced at the end of 20 years at the same cost PW(12%) = −($120,000 + $12,000) −($12,000 − $1,000)( P / F ,12%, 20) +$1,000( P / F ,12%, 40) − $1,000( P / A,12%, 40) = −$141,373

∴ Option 2 is a better choice.

5.67 (a) •

Deferred Plan:

29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(6%) deferred = $2, 000, 000 + $566, 000( P / F , 6%,1) +  +$1,140, 000( P / F , 6%,10) + $1, 260, 000( P / F , 6%,11) = $8,574, 491 •

Non-deferred Plan: PW(6%)non-deferred = $2, 000, 000 + $900, 000( P / F , 6%,1) + 

+$1,500, 000( P / F , 6%, 4) + $1,975, 000( P / F , 6%,5) = $7, 431,562 (b)

Season 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 2026

Deferred Plan (A) $ $ $ $ $ $ $ $ $ $ $ $

2,000,000.00 566,000.00 920,000.00 930,000.00 740,000.00 740,000.00 740,000.00 790,000.00 540,000.00 1,040,000.00 1,140,000.00 1,260,000.00

Cash Flows Non-deferred Plan (B) $ $ $ $ $ $

2,000,000.00 900,000.00 1,000,000.00 1,225,000.00 1,500,000.00 1,975,000.00

A-B $ $ $ $ $ $ $ $ $ $ $ $

(334,000.00) (80,000.00) (295,000.00) (760,000.00) (1,235,000.00) 740,000.00 790,000.00 540,000.00 1,040,000.00 1,140,000.00 1,260,000.00

PW(i) A-B = −$334, 000 − $80, 000( P / F , i,1) −  +$1,140, 000( P / F , i,9) + $1, 260, 000( P / F , i,10) =0 i* = 15.72%

If his earning interest rate is greater than15.72%, the non-deferred plan is better. 5.68 •

US: F = $10,000(F/P, 4%, 5) = $12,166.53

•

Euro: F = [$10,000(0.735)(F/P, 5%, 5)]/0.88 = $10,732.37

•

Japan: F = [$10,000(101.67)(F/P, 6%, 5)]/110 = $12,368.85

30 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

∴ Therefore, the best strategy is to put the money in Japanese market (c).

5.69 (a) Project balances as a function of time are as follows: n 0 1 2 3 4 5 6 7 8

Project Balances A -$2,500 -$2,100 -$1,660 -$1,176 -$694 -$163 $421 $763 $1,139

D -$5,000 -$6,000 -$7,100 -$3,810 -$1,191 $1,690 $3,859 $7,245

All figures above are rounded to nearest dollars. (b) Knowing the relationship FW(i ) = PB(i ) N ,

FW(10%) A = $1,139 FW(10%) D = $7, 245 (c) Assuming a required service period of 8 years PW(10%) B = −$7, 000 − $1,500( P / A,10%,8) −$1, 000( P / F ,10%,1) − $500( P / F ,10%, 2) −$1,500( P / F ,10%, 7) − $1,500( P / F ,10%,8) = −$17, 794 PW(10%)C = −$5, 000 − $2, 000( P / A,10%, 7) −$3, 000( P / F ,10%,8) = −$16,136 Select Project C. 5.70 •

Project A with repeating 2 times, PW(12%)A = −1, 000 − 400( P / F ,12%,1) − 400( P / F ,12%, 2) −1, 200( P / F ,12%,3) − 400( P / F ,12%, 4) −400( P / F ,12%,5) − 200( P / F ,12%, 6) = −$3,112.66

•

Project B with repeating 3 times,

31 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(12%) B = −800 − 200( P / F ,12%,1) − 1, 000( P / F ,12%, 2) −200( P / F ,12%,3) − 1, 000( P / F ,12%, 4) −200( P / F ,12%,5) − 200( P / F ,12%, 6). = −$2, 768.45 Select B to save $344 in PW. The correct answer is (a).

Short Case Studies with Excel ST 5.1 • Option 1: Process device A lasts only 4 years. You have a required service period of 6 years. If you take this option, you must consider how you will satisfy the rest of the required service period at the end of the project life. One option would subcontract the remaining work out for the duration of the required service period. If you select this subcontracting option along with the device A, the equivalent net present worth would be PW(12%)1 = −$100, 000 − $60, 000( P / A,12%, 4) +$10, 000( P / F ,12%, 4) −$100, 000( P / A,12%, 2)( P / F ,12%, 4) = −$383, 292 • Option 2: This option creates no problem because its service life coincides with the required service period.

PW(12%)2 = −$150, 000 − $50, 000( P / A,12%, 6) +$30, 000( P / F ,12%, 6) = −$340,371 • Option 3: With the assumption that the subcontracting option would be available over the next 6 years at the same cost, the equivalent present worth would be PW(12%)3 = −$100, 000( P / A,12%, 6) = −$411,141 With the restricted assumptions above, option 2 appears to be the best alternative.

Notes to Instructors: This problem is deceptively simple. However, it can make the problem interesting with the following embellishments. • If the required service period is changed to 5 years, what would be the best course of action? • If there are price differentials in the subcontracting option (say, $55,000 a year for a 6-year contract, $60,000 for a 5-year contract, $70,000 a year for a 4-year contract and $75,000 a year for any contract lasting less than 4 years), what would be the best option? 32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • If both processes A and B would be available in the subsequent years, but the required investment and salvage value would be increasing at the annual rate of 10%, what would be the best course of action? • If both processes A and B will be available in the subsequent years, but the required investment and salvage value (as well as the O & M costs) would be decreasing at the annual rate of 10%, what would be the best course of action?

ST 5.2

PW(8%)1 = −$680, 000 − $680, 000( P / F ,8%, 7) = −$1, 076, 773.47 PW(8%)2 = −$920, 000 ∴ Option 2 is a better choice. PW(8%)1 = −$680, 000 − $680, 000( P / F ,8%, 7) = −$1, 076, 773.47 PW(8%) 2 = −$920, 000 − $920, 000( P / F ,8%,16) = −$1,188,539.23

∴ Option 1 is a better choice.

ST 5.3 Not provided.

33 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 6 Annual Equivalence Method Identifying Cash Inflows and Outflows 6.1 AE(9%) = $300, 000( A / P ,9%, 5) = $77,127.74

6.2

AE(12%) = A( P / A,12%,3) = $250, 000 A = $104, 087.25

6.3

PW(10%) = −$300 + $100( P / F ,10%,1) − $160( P / F ,10%, 2) +  + $130( P / F ,10%, 6) = $105.54 AE(10%) = $105.44( A / P,10%, 6) = $24.23 6.4

AE(13%) = $20,000( A / P,13%,6) − $5,000 −$3,000( P / G,13%,5)( P / F ,13%,1)( A / P,13%,6) = $4,101.38 6.5  ( P / A,15%,5)( P / F ,15%,1)  AE(15%) = $100 + $50   ( A / P,15%, 6)  + ( P / A,15%,3)( P / F ,15%,3)  = $158.34

6.6

AE(10%) = [−$3, 000 − $3, 000( P / A,10%, 2) +$3, 000( P / A,10%, 4)( P / F ,10%, 2) +$1, 000( P / G,10%, 4)( P / F ,10%, 2)]( A / P,10%, 6) = $751

1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 6.7

AE(12%) = [−$8, 000 − $4, 000( P / F ,12%, 2) −$4, 000( P / F ,12%, 4) − $1, 000( P / F ,12%, 6)]( A / P,12%, 6) +$2, 000 + $1, 000( A / G,12%, 6) = $709.13

6.8

6.9

AE(10%) = −$4,500( A / P,10%, 6) + $2, 000 −[($500 + $1, 000( P / A,10%, 2))( P / F ,10%, 2) +$500( P / F ,10%,5)]( A / P,10%, 6) = $471.27

AE(8%) = −$2,154( A/ P,8%,6) $400(P / F,8%,1) + X (P / A,8%,2)(P / F,8%,1)  +  ( A/ P,8%,6) +$400(P / F,8%,4) + X (P / A,8%,2)(P / F,8%,4) = $200 200 = −465.94 + (370.36 +1.65X + 294 +1.31X )(0.2163) 924.64 = −1,489.78 + 2.96X X = $815.68

6.10.

$2, 000 ( P / F ,15%, 6)]( A / P,15%, ∞) 0.15 = $3, 000 + $864.66 = $3,864.66

AE(15%) = $3, 000 + [

6.11

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Contemporary Engineering Economics, 7th ed. ©2023

AE(15%) A = −$3,500( A / P,15%, 4) + $600 $900( P / F ,15%,1) + $1, 200( P / F ,15%, 2)  +  ( A / P,15%, 4)  +$400( P / F ,15%,3)  = $58.13 > 0 (Accept) AE(15%) B = −$4,000( A / P,15%, 4) + $1,500 +$100( P / F ,15%,1)( A / P,15%, 4) = $129.39 > 0 (Accept) AE(15%)C = −$6,500( A / P,15%, 4) + $2,000 + $500( F / A,15%, 2)( A / F ,15%, 4) = −$61.43 < 0 (Reject) 6.12

AE(10%) = [ −$600, 000 + $400, 000( P / F ,10%,1) + $300, 000( P / F ,10%, 2) +$200, 000( P / F ,10%,3) + $100, 000( P / F ,10%, 4)]( A / P,10%, 4) = $72, 600.73

6.13 AE(12%) A = −$4, 400( A / P,12%,3) + $12,500( A / F ,12%,3) = $1,872.42 > 0 (Accept) AE(12%) B = −$3, 400( A / P,12%,3) + $1, 000 +$500( A / G,12%,3) = $46.71 > 0 (Accept) AE(12%)C = −$3,300( A / P,12%,3) + $3, 000 −$1, 000( A / G ,12%,3) = $701.43 (Accept) AE(12%) D = −$6, 200( A / P,12%,3) + $3,800 = $1, 218.63 (Accept) 6.14 Since the project has the same cash flow cycle during the project life, you just can consider the first cycle.

AE(10%) = [−$2,500 + $1,800( P / F ,10%,1) + $900( P / F ,10%, 2) +$500( P / F ,10%,3)]( A / P,10%,3) = $102.87 ∴ Accept the project. 3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

6.15 PW(i ) = $20, 000( P / A, 6%,10) +

$20, 000 ( P / F , 6%,10) 0.04

= $147, 202 + $279,197 = $426,399

∴ The amount of additional funds should be $226,399.

Capital (Recovery) Cost / Annual Equivalent Cost 6.16

$1.5 $5 $9 $12 $10 + + + + = $17,835,518.55 2 3 4 1.15 1.15 1.15 1.15 1.155 AE(15%) = $17,835,518.55( A / P,15%,5) = $5,320,335.18 Yes, the project is justified.

PW(15%) = −$5 +

6.17

CR(6%)3 years = ($21, 635 − $6, 057.80)( A / P, 6%,3) + (0.06)($6, 057.80) CR(6%)5 years

= $6,191.05 = ($21, 635 − $3, 677.95)( A / P, 6%,5) + (0.06)($3, 677.95) = $4, 483.68

6.18 (a) Capital recovery cost: CR(10%) = ($40, 000 − $15, 000)( A / P,10%, 2) + $15, 000(0.10) = $15,905

(b) Annual savings: PW(10%) = $28, 000( P / F ,10%,1) + $40, 000( P / F ,10%, 2) = $58, 512.40

4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AE(10%)1 = $58, 512.40( A / P ,10%, 2) = $33, 714.84

AE(10%) = $33,714.84 − $15,905 = $17,809.84 (c)

AE(10%) = C[4, 000( P / F ,10%,1) + 6, 000( P / F ,10%, 2)]( A / P,10%, 2) = 4,952.46C C = 17,809.84 / 4,952.46 = $3.596 / hour

6.19 PW(i ) = $20, 000( P / A, 6%,10) +

$20, 000 ( P / F , 6%,10) 0.04

= $147, 202 + $279,197 = $426,399

∴ The amount of additional funds should be $226,399.

6.20 CR(10%) = ($500, 000 − 100, 000)( A / P,10%,15) + 100, 000(0.1) = $62, 600 AE(10%) = $40, 000 X − $30, 000 X CR(10%) = AE(10%) $62, 600 = $10, 000 X X = 6.26 (or rounds up to 7)

6.21

3,880 = (25, 000 − S )( A / P ,10%,10) + 0.1S S = $3, 006.35

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Contemporary Engineering Economics, 7th ed. ©2023 6.22

Given: I = $65, 000, S = $5, 000, N = 10 years, i = 9% (a) AE(9%)1 = ($65, 000 − $5, 000)( A / P,9%,10) +$5, 000(0.09) = $9, 799

(b) AE(9%) 2 = $18, 000 + $2,500( A / G ,9%,10) = $27, 495

(c) AE(9%) = $27, 495 − $9,799 = $17,696

∴ This is a good investment.

6.23 AE(6%) = ($65, 000 − $23, 000)( A / P, 6%, 6) + (0.06)($23, 000) + $13,500 = $23, 421.33

6.24 CR(12%) = ($75, 000 − $22, 000)( A / P,12%,5) + (0.12)($22, 000) = $17,342.72 O& M(12%) = [$18, 000 + $2, 000( A / G ,12%,5) = $21,549.2 AEC(12%) = $17,342.72 + $21,549.2 = $38,891.92

6.25 Given: I = $350, 000, S = $60, 000, N = 5 years, i = 15% CR(15%) = ($350, 000 − $60, 000)( A / P,15%,5) +$60, 000(0.15) = $95,512

6.26

n

Option 1 (Buy storage tank)

0

-$600-$13,760

Option 2 (Buy KS every year)

6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 1

0

-$3,784

2

0

-$4,162

3

0

-$4,579

4

+$100

-$5,037

AEC(6%)Option 1 = $14,360( A / P, 6%, 4) − $100( A / F , 6%, 4) = $4,121 AEC(6%)Option 2 = $3, 784( P / A1 ,10%, 6%, 4)( A / P, 6%, 4) = $4,360 ∴ Select Option 1

6.27 Given i = 6% compounded annually, N=20 years.

CR(6%) = ($300, 000, 000 − $20, 000, 000)( A / P, 6%, 20) +0.06($20, 000, 000) = $25, 611, 680 O&M(6%) = $1.10$15, 000, 000( P / A1 ,3%, 6%, 20)( A / P, 6%, 20) = $19, 043,120 Annual Energy Cost = ($1.10)(365 days/year) ×(50,000,000 gallons/day)/1,000 gallons/day =$20,075,000 AEC(6%) =$25, 611, 680 + $19, 043,120 + $20, 075, 000 = $64, 729,800 Amount of water to produce per year:

50, 000, 000 × 365 = 56, 007 acre-foot 325,851 Cost per acre-foot: $64, 729,800 = $1,155.74 / acre-foot 56, 007 7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 which exceeds the selling price of $950 per acre-foot. The city needs to price of water at $1,156/acre-foot, just to breakeven.

6.28

CE(10%) = $100, 000 +

$10, 000 $20, 000 + ( A / F ,10%, 4) 0.1 0.1

= $243,094.2

6.29 (a)

AE(15%) = −$4,500( A / P,15%, 4) + $1, 000 + ( X − $1,000)( P / F ,15%, 2)( A / P,15%, 4) =0 AE(15%) = −$841.21 + 0.26468 X = 0 X = $3,175.33

(b) AE(12%) B = $6,500( A / P,12%, 4) − $1, 400 = $740.02 > 0 AE(12%) A = −$744.01 + 0.262464 X > $740.02

X * > $5,654.30 6.30 •

Option 1: Purchase-annual installment option: A = $65, 000( A / P,9%,5) = $16, 711 AE(10%)1 = −$5, 000( A / P,10%,5) − $16, 711 = −$18, 030

•

Option 2: Cash payment option:

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AE(10%) 2 = −$66, 000( A / P,10%,5) = −$17, 411

∴ Option 2 is a better choice.

6.31 The total investment consists of the sum of the initial equipment cost and the installation cost, which is $145,000. Let R denote the break-even annual revenue.

AE(12%) = −$145, 000( A / P,12%,10) − $50, 000 −$5, 000 + $10, 000 + R =0 Solving for R yields

R = $70, 663

6.32 •

Capital recovery cost: CR(15%) = ($135, 000 − $6, 000)( A / P,15%,5) + $6, 000(0.15) = $39,383

•

Annual operating costs: $75, 000 ∴ Equivalent annual cost AEC(15%) = $39, 383 + $75, 000 = $114, 383

Unit-Cost Profit Calculation 6.33 Capital recovery cost: CR(15%) = ($56, 000 − $5, 000)( A / P ,15%, 5) + $5, 000(0.15) + $6, 000 = $21, 963.3

The machine cost per hour would be $8.78(= $21,963.3/2,500) 6.34. •

Option 1: Purchase units from Tompkins

9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Unit cost = $25 + ($70,000 − $35,000) / 20,000 − $3.50 = $23.25 •

Option 2: Make units in house

PW(15%)dm = $63, 000( P / A1 ,5%,15%,5) = $230, 241 PW(15%)dl = $190,800( P / A1 , 6%,15%,5) = $709, 491 PW(15%)vo = $139, 050( P / A1 ,3%,15%,5) = $490,888 AEC(15%) = ($230, 241 + $709, 491 +$490,888)( A / P,15%,5) + $70, 000 = $496, 776 Unit cost = $496, 776 / 20, 000 = $24.84 Option 1 is a better choice. 6.35 •

Annual equivalent revenue:

AERevenue = $32,000 + 40, 000X •

Annual equivalent cost:

AEC(8%)Cost = $800, 000( A / P,8%, ∞) + $133, 000 + $50, 000( A / F ,8%,5) = $64, 000 + $133, 000 + $8,525 = $205,525 AE Revenue = AECCost $32, 000 + 40, 000 X = $205,525 X = $4.34

6.36.

10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Salvage Value: $1, 200, 000( F / P,5%, 25) = $4, 063, 680 CR(12%) = ($6, 000, 000 − $4, 063, 680)( A / P,12%, 25) + (0.12)$4, 063, 680 = $734,522.4 AECO&M = $100 ×12 × 40 + $400, 000 = $448, 000 AEC(12%) = CR(12%) + AEma = $1,182,522.4 per year AEC(0.9489%) Monthly = $1,182,522.4( A / F , 0.9489%,12) = $93,506 per month

6.37

•

Pump I: (

180 )(0.746)(0.06)T = $9.368T 0.86 AEC(8%) I = $6, 000( A / P,8%,12) + $500 + $9.368T = $1, 296.2 + $9.368T

•

Pump II:

180 ( )(0.746)(0.06)T = $10.071T 0.8 AEC(8%) II = $4, 000( A / P,8%,12) + $440 + $10.071T = $970.8 + $10.071T

•

$1,296.2 + 9.368T = $970.8 + 10.071T T = 463 hours

6.38 Let C denote the green fee per round during the first year. •

Capital cost: CR(15%) = ($20, 000, 000 − $25, 000, 000( A / P,15%,10) + (0.15)($25, 000, 000) = $2, 753, 740

11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

Operating and maintenance cost: O&M(15%) = $650, 000 + $50, 000( A / G ,15%,10) = $819,160

•

Equivalent annual revenue: AE(15%) Revenue = $15 × 40, 000 +40, 000(1.15)C ( P / A1 ,5%,15%,10)( A / P,15%,10) = $600, 000 1 − (1 + 0.05)10 (1 + 0.15) −10  +$46, 000C   ( A / P,15%,10) 0.15 − 0.05   = $600, 000 + 54, 752C Breakeven green fee:

•

$600, 000 + 54, 752C = $2, 753, 740 + $819,160 54, 752C = $2,972,900 C = $54.30 6.39 •

Capital recovery cost CR(12%) = ($30, 000 − $18, 000)( A / P,12%, 2) + $18, 000(0.12) = $9, 260

•

$35, 000 $42, 000 + ]( A / P,12%, 2) 1.12 1.122 = $38,302 Net annual savings =$38,302 − $9, 260 AE savings (12%) = [

= $29, 042 •

C (6, 000) C (8, 000) ]( A / P,12%, 2) + 1.12 1.122 = $6,943.3C $29, 042 = $6,943.3C C = $4.18 per hour

AE hours (12%) = [

6.40

AE(14%) = [−$100, 000 + $35, 000( P / A1 , −3%,14%,5)]( A / P,14%,5) 12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 = $4, 095.13

C = $4,095.13 / 3,000 = $1.365 / hr 6.41 •

Capital cost: CR(5%) = ($10, 000, 000 − $850, 000)( A / P,5%,10) + (0.05)($850, 000) = $1, 227, 467

•

Operating cost: O&M(5%) = $5, 000, 000( P / A1 ,5%,5%,10)( A / P,5%,10)

•

= $6,166,885 Annual equivalent cost: AEC(5%) = $1, 227, 467 + $6,166,885

•

= $7,394,351 Suggested minimum fair per trip: $7, 394, 351 / 800, 000 = $9.24 per trip

6.42 •

Salvage value (book value): $11,000 − ($2,879 + $1,776 + $1,545) = $4,800

•

Capital recovery cost: CR(7%) = ($11, 000 − $4,800)( A / P, 7%,3) + $4,800(0.07) = $2, 698.52

•

Equivalent annual cost of operating and maintenance: (Annual O&M costs = Total of all costs – Depreciation) AEC(7%) = [$1,801( P / F , 7%,1) + $1,848( P / F , 7%, 2) + $1,876( P / F , 7%,3)]( A / P, 7%,3) = $1,839.97

•

Total annual equivalent cost: $2, 698.52 + $1,839.97 = $4, 538.49

•

Annualized cost of driven miles:

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Contemporary Engineering Economics, 7th ed. ©2023

C (14,500) C (13, 000) C (11,500) + + ]( A / P, 7%,3) 1.07 1.07 2 1.073 = 13, 068C

AECmiles (7%) = [

•

So, $4,538.49 = 13, 068C C = $0.35 per mile

6.43 Let T denote the total operating hours in full load. •

Motor I (Expensive): Annual power cost: 180 × (0.746) × (0.05) × T = $8.089T 0.83 Equivalent annual cost of operating the motor: AEC(6%) I = $5,800( A / P, 6%,10) + $870 + $8.089T = $1, 658.03 + $8.089T

•

Motor II (Less expensive): Annual power cost:

180 × (0.746) × (0.05) × T = $8.392T 0.80 Equivalent annual cost of operating the motor: AEC(6%) II = $4, 600( A / P, 6%,10) + $690 + $8.392T = $1,314.99 + $8.392T

•

Let AEC(6%)I =AEC(6%)II and solve for T . $1, 658.03 + $8.089T = $1, 314.99 + $8.392T T = 1,132.15 hours per year

6.44 •

Option 1: Purchase units from John Holland

14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Unit cost = $25 − ($35,000 / 20,000) − $3.5 = $19.75 •

Option 2: Make units in house

PW(15%) dm = $63, 000( P / A1 ,5%,15%,5) = $230, 241 PW(15%) dl = $190,800( P / A1 , 6%,15%,5) = $709, 491 PW(15%)vo = $139, 050( P / A1 ,3%,15%,5) = $490,888 AEC(15%) = ($230, 241 + $709, 491 + $490,888)( A / P,15%,5) + $70, 000 = $496, 776 Unit cost = $496, 776 / 20, 000 = $24.84 ∴ Option 1 is a better choice.

6.45 (a) Determine the unit profit of air sample test by the TEM (in-house). • Sub-contract Option: Unit profit = $400 − $300 − $0.50 − $1, 500 / 1, 000 = $98 • TEM Purchase Option: AEC(15%) = ($415, 000 + $9,500)( A / P,15%,8) + ($50, 000 +$6, 000 + $18, 000 + $20, 000) = $188, 600 Unit cost = $188, 600 /1, 000 = $188.60 Unit profit = $300 − $188.60 = $111.40

(b)Let X denote the break-even number of air samples per year. $400 − ($300 + $0.50 + $1, 500 / X ) = $300 − $188, 600 / X

Solving for X yields X = 933.16 ≈ 933 air samples per year

6.46 •

Option 1: Pay employee $0.56 per mile

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Contemporary Engineering Economics, 7th ed. ©2023

AEC(10%) total cost = $0.56(30, 000) = $16,800 •

Option 2: Provide a car to employee: CR(10%) = ($25, 000 − $8, 000)( A / P,10%,3) + (0.10)($8, 000) = $7, 636 O& M(10%) = $1, 200 + ($0.3)(30, 000) = $10, 200 AEC(10%) = $7, 636 + $10, 200 = $17,836 Cost per mile = $17,836 / 30, 000 = $0.5945

∴ Option 1 is a better choice.

6.47 •

Capital costs: AEC(7%)1 = ($25, 000 − $2, 000)( A / P, 7%,12) + (0.07)($2, 000)

•

= $3, 036 Annual battery replacement cost:

AEC(7%) 2 = $3, 000[( P / F , 7%,3) + ( P / F , 7%, 6)

•

+( P / F , 7%,9)]( A / P, 7%,12) = $765.41 Annual recharging cost: AEC(7%)3 = ($0.015)(20, 000) = $300

•

Total annual costs (including annual maintenance cost): AEC(7%) = $3, 036 + $765.41 + $300 + $700 = $4,801.41

•

Costs per mile: cost/mile = $4,801.41/ 20, 000 = $0.2401

6.48 •

Minimum operating hours:

16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AEC(10%) = ($32, 000 − $2, 000)( A / P,10%,15) + (0.10)($2, 000) + $800 = $4,944.21 Let T denote the annual operating hours. Then the total kilowatt-hours generated would be 40T . Since the value of the energy generated is $0.09 per kilowatt-hour, we can establish the following relationship:

$0.09 × 40T = $4,944.21 Solving for T yields •

T = 1,373.4 hours

Annual worth of the generator at full load operation: AE(10%) = ($0.09)(100, 000) − $4, 944.21 = $4, 055.79

•

Discounted payback period at full load of operation:

n 0 1  15

Investment

Revenue

Maintenance cost

Net Cash flow

-$32,000 $9,000  $9,000

 +$2,000

-$800  -$800

-$32,000 8,200  10,200

$32,000 = $8, 200( P / A,10%, n )

Solving for n yields n = 5.19 years 6.49 •

Capital recovery cost: CR(6%) = ($170, 000 − $12, 000)( A / P, 6%,12) + (0.06)($12, 000) = $19,565.77

•

Annual operating costs: AEC(6%)O = $70, 000 + $15, 000 + $5, 000 = $90, 000

•

Total annual system costs: AEC(6%) = $19, 565.77 + $90, 000 = $109, 565.77

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Contemporary Engineering Economics, 7th ed. ©2023

•

Number of rides required per year: Number of rides = $109,565.77 / $0.10 = 1,095,658 rides

6.50

Given: Investment cost = $10 million, plant capacity = 300,000 lbs./hour, plant operating hours = 4,000 hours per year, O&M cost = $4 million per year, useful life = 15 years, salvage value = $800,000, and MARR = 15%. (a)

PW(15%) = −$10, 000, 000 + ( R − $4, 000, 000)( P / A,15%, 6) = 3.7845R − $25,137,930.78 =0 Solving for R yields R = $6,642,370 per year

(b) Minimum processing fee per lb (after-tax):

$6,642,370 = $0.00554 per lb. (300,000)(4,000) Comment: The minimum processing fee per lb. should be higher on a before-tax basis. 6.51

Given: Investment = $5 million, plant capacity = 0.4 acre-foot, useful plant life = 20 years, salvage value = negligible, O&M cost = $250, 000 per year, MARR = 10% compounded annually (or effective monthly rate of 0.7974%) AEC(10%) = $5, 000, 000( A / P,10%, 20) + $250, 000 = $837, 298.12

Monthly water bill for each household:

$837,298.12( A / F ,0.7974%,12) = $226.33 295 6.52 •

Annual total operating hours: (0.70)(8, 760) = 6,132 hours per year

18 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

Annual electricity generated: 50, 000 × 6,132 = 306, 600, 000 kilowatt-hours

•

Equivalent annual cost: AEC(14%) = $85, 000, 000( A / P,14%, 25) + $6, 000, 000 = $18,367,364

•

Cost per kilowatt-hour: $18, 367, 364 / 306, 600, 000 = $0.06 per kilowatt-hour

6.53 Let X denote the average number of round-trip passengers per year. •

Capital recovery costs: CR(15%) = ($12, 000, 000 − $2, 000, 000)( A / P,15%,15) + (0.15)($2, 000, 000) = $2, 010,171

•

Annual crew costs: $225,000

•

Annual fuel costs for round trips: ($1.10)(3, 280)(2)(3)(52) = $1,125, 696

•

Annual landing fees: ($250)(3)(52)(2) = $78, 000

•

Annual maintenance, insurance, and catering costs: $237, 500 + $166, 000 + $75 X = $403, 500 + $75 X

•

Total equivalent annual costs: AEC(15%) = $2, 010,171 + $225, 000 + $1,125, 696 +$78, 000 + $403,500 + $75 X = $3, 400 X

19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Solving for X yields X = 1,156 Passengers per year

Or 1,156 /(52)(3) = 7.41 ≈ 8 Passengers per round trip

Comparing Mutually Exclusive Alternatives by the AE Method 6.54. (a) We suppose that the current projects are expected to continue for an indefinite period. These two projects will be available in the future, without significant changes in revenue expectations. (b) •

For A,

PW(10%)A-3year-cycle = −$10, 000 + $6, 000( P / F ,10%,1) + $5, 000( P / F ,10%, 2) +$4, 000( P / F ,10%,3) = $2,592

AE(10%) A-3-year-cycle = $2,592( A / P,10%,3) = $1, 042.24 PW(10%) A-6-year-cycle = $2,529[1 + ( P / F ,10%,3)] = $4, 429.08 AE(10%)A-6-year-cycle = $4, 429.08( A / P,10%, 6) = $1, 016.92 •

For B,

20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(10%) B-2-year-cycle = −$12, 000 + $7, 000( P / F ,10%,1) + $8, 000( P / F ,10%, 2) = $975.21 AE(10%) B-2-year-cycle = $975.21( A / P,10%, 2) PW(10%) B-6-year-cycle

= $561.89 = $975.21[1 + ( P / F ,10%, 2) + ( P / F ,10%, 4)] = $2, 447.25

AE(10%) B-6-year-cycle = $2, 447.25( A / P,10%, 6) = $561.89

Choose the Project A, which has a higher annual equivalent worth.

6.55 (a) AE(12%) A = −$40, 000( A / P,12%, 4) + [$19,120 − $1, 280( A / G ,12%, 4)] = $4, 211.59 AE(12%) B = −$40, 000( A / P,12%, 4) + $17,350 = $4,180.62

(b) Process A: $4, 211.29 / 3, 000 = $1.40 /hour Process B: $4,180.62 / 3, 000 = $1.39 /hour (c) Process A is a better choice.

6.56 AEC(10%)1 = ($95, 000 − $12, 000)( A / P,10%,3) + (0.1)$12, 000 +$3, 000 = $37,575.53 AEC(10%) 2 = ($120, 000 − $25, 000)( A / P,10%, 6) + (0.1)$25, 000 +$9, 000 = $33,312.70 AEC(10%)1 − AEC(10%) 2 = $37,575.53 − $33,312.70 = $4, 262.83

21 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

6.57 •

Capital recovery cost for both motors:

CR(12%)CV = $14, 000( A / P,12%, 20) = $1,874 CR(12%) PE = $16, 600( A / P,12%, 20) = $2, 222 •

Annual operating cost for both motors:

•

18.650kW × $0.08 / kWh × 3,120h = $3,901 0.895 18.650kW AEC PE = 0.75 × × $0.08 / kWh × 3,120h = $3, 754 0.93 Total annual cost for both motors: AECCV = 0.75 ×

AEC(12%)CV =$1,874 + $3,901 =$5,775 AEC(12%)PE =$2,222 + $3,754 =$5,976 (a) Savings per kWh: Savings =

$5,976 − $5, 775 = $0.0046 / kWh 0.75 × 3,120h ×18.650kW

(b) 18.650kW × $0.08 / kWh × T = 1.2503T 0.895 18.650kW = 0.75 × × $0.08 / kWh × T = 1.2032T 0.93

AE CV = 0.75 × AE PE

$1,874 + 1.2503T = $2, 222 + 1.2032T T = 7,393hrs

6.58 •

New lighting system cost: AEC(12%) = $50, 000( A / P,12%, 20) + ($8, 000 + $3, 000) = $17, 694

•

Old lighting system cost:

22 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AEC(12%) = $20, 000

Annual savings from installing the new lighting system = $2, 306 6.59

AEC(15%)1 = ($100, 000 − $15, 000)( A / P,15%, 7) + (0.15)$15, 000 +$47, 000 = $69, 680.63 AEC(15%) 2 = $150, 000( A / P,15%, 7) + $30, 000 + X ( P / F ,15%,5)( A / P,15%, 7) = $66, 054.05 + 0.1195 X AEC(15%)1 = AEC(15%) 2 $69, 680.63 = $66, 054.05 + 0.1195X

X * = $30,347.95 6.60

Given: i = 6% interest compounded monthly, the effective annual interest = (1.005)12 − 1 = 6.17% per year, effective semiannual interest

= (1.005)6 − 1 = 3.04% per semiannual •

Option 1: Buying a bond AE(3.04%)1 = −$2, 000( A / P,3.04%, 6) + $100 + $2, 000( A / F ,3.04%, 6) = $39.20 per semiannual AE(6.17%) = $39.20( F / A,3.04%, 2) = $79.59 per year

•

Option 2: Buying and holding a growth stock for 3 years AE(6.17%) 2 = −$2, 000( A / P, 6.17%,3) + $2, 735.26( A / F , 6.17%,3) = $107.17

•

Option 3: Receiving $150 interest per year for 3 years AE(6.17%)3 = −$2, 000( A / P, 6.17%,3) + $150 + $2, 000( A / F , 6.17%,3) = $26.60

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Contemporary Engineering Economics, 7th ed. ©2023

∴ Buying an d holding a growth stock is the best option.

6.61 •

Equivalent annual cost: AEC(13%) A = ($1,300, 000 − $60, 000)( A / P,13%, 20) +(0.13)($60, 000) + $70, 000 + $40, 000 = $294,318.70 AEC(13%) B = ($850, 000 − $30, 000)( A / P,13%,10) +(0.13)($30, 000) + $100, 000 + $30, 000 = $285, 017.44

•

Processing cost per ton: C1 = $294,318.70 / (20)(365) = $40.32 per ton C2 = $285,017.44 / (20)(365) = $39.04 per ton

∴ Incinerator B is a better choice.

6.62 Let X be the number of machines per year AEC(10%)1 = $40, 000 X AEC(10%) 2 = ($500, 000 − $100, 000)( A / P,10%,15) + (0.1)$100, 000 +$30, 000 X

AEC(10%)1 = AEC(10%) 2 $62,589.51 = $10, 000 X X = 6.26 or 7 machines per year

24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Life-Cycle Cost Analysis 6.63 •

Capital recovery cost for both systems:

CR(12%)Heat = $194, 000( A / P,12%,15) = $28, 483.90 CR(12%)Geo = $256,500( A / P,12%,15) = $37, 660.42 •

Annual operating cost for both systems:

AECHeat = 145, 000kWh × $0.06 / kWh = $8, 700 AECGeo = 25,590kWh × $0.06 / kWh = $1,535.40 •

Annual equivalent compressor replacement cost:

AEC(12%) Heat = $2,500( P / F ,12%,10)( A / P,12%,15) = $118.18 AEC(12%)Geo = $0 •

Total annual cost for both systems:

AEC(12%)Heat = $28, 483.90 + $8, 700 + $3,860 + $118.18 = $41,162.08 AEC(12%)Geo = $37, 660.42 + $1,535.40 + $3, 650 = $42,845.82 The heat pump system is more economical. 6.64

Assumption: jet fuel cost = $2.10 /gallon •

System A: Equivalent annual fuel cost: A1 = ($2.10/gal)(40,000 gals/1,000 hours)(2,000 hours) = $168,000 (assuming an end of-year convention) AEC(10%) fuel = [$168, 000( P / A1 , 6%,10%,3)]( A / P,10%,3) = $177, 623.20 AEC(10%) A = ($100, 000 − $10, 000)( A / P,10%,3) + (0.10)($10, 000) + $177, 623.20 = $214,813.53

•

System B: Equivalent annual fuel cost: A1 = ($2.10/gal) (32,000 gals/1,000 25 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 hours)(2,000 hours) = $134,400 AEC(10%) fuel = [$134, 400( P / A1 , 6%,10%,3)]( A / P,10%,3) = $142, 098.56 AEC(10%) B = ($200, 000 − $20, 000)( A / P,10%,3) + (0.10)($20, 000) + $142, 098.56 = $216, 479.23

•

Equivalent operating cost (including capital cost) per hour: System A = $214,813.53 / 2, 000 = $107.41 per hour System B = $216, 479.23 / 2, 000 = $108.24 per hour ∴ System A is a better choice.

6.65 Since the required service period is 12 years and the future replacement cost for each truck remains unchanged, we can easily find the equivalent annual cost over a 12-year period by simply finding the annual equivalent cost of the first replacement cycle for each truck. •

Truck A: Four replacements are required AEC(12%) A = ($15, 000 − $5, 000)( A / P,12%,3) + (0.12)($5, 000) + $3, 000 = $7, 763.50

•

Truck B: Three replacements are required AEC(12%) B = ($20, 000 − $8, 000)( A / P,12%, 4) + (0.12)($8, 000) + $2, 000 = $6,910.80

∴ Truck B is a more economical choice.

6.66 (a) Number of decision alternatives (required service period = 5 years): Alternative

Description

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Contemporary Engineering Economics, 7th ed. ©2023

A1

Buy Machine A and use it for 4 years. Then lease a machine for one year.

A2

Buy Machine B and use it for 5 years.

A3

Lease a machine for 5 years.

A4

Buy Machine A and use it for 4 years. Then buy another Machine A and use it for one year.

A5

Buy Machine A and use it for 4 years. Then buy Machine B and use it for one year.

There are five alternatives. Both A4 and A5 are feasible but we do not consider the alternatives because we need to know the salvage values of the machines after oneyear use. (b) With lease, the O&M costs will be paid by the leasing company: • For A1: PW(10%)1 = −$6,500 + $600( P / F ,10%, 4) −$800( P / A,10%, 4) − $200( P / F ,10%,3) −$100( P / F ,10%, 2) − ($3, 000 + $100)( P / F ,10%, 4) = −$10,976.33 AEC(10%)1 = $10,976.33( A / P,10%,5) = $2,895.53 • For A2: PW(10%) 2 = −$8,500 + $1, 000( P / F ,10%,5) −$520( P / A,10%,5) − $280( P / F ,10%, 4) = −$10, 042 AEC(10%) 2 = $10, 042( A / P,10%,5) = $2, 649

• For A3:

27 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AEC(10%)3 = [$3, 000 + $3, 000( P / A,10%, 4)]( A / P,10%,5) = $3,300

∴ A2 is a better choice.

6.67 •

Option 1:

AEC(18%)1 = $200, 000(180)( A / P,18%, 20) −(0.08)($200, 000)(180)( A / F ,18%, 20) +($0.005 + 0.215)(180, 000, 000) = $46,305,878 cost/lb = $46,305,878 /180, 000, 000 = $0.2573 per lb •

Option 2: AEC(18%)2 = ($0.05 + $0.215)(180, 000, 000) = $47, 700, 000 cost/lb = $47, 700, 000 /180, 000, 000 = $0.2650 per lb ∴ Option 1 is a better choice.

6.68

Given: Required service period = indefinite, analysis period = indefinite Plan A: Incremental investment strategy: •

Capital investment: CR(10%)1 = [$400, 000 + $400, 000( P / F ,10%,15)]( A / P,10%, ∞) = $49,576

•

Supporting equipment: AEC(10%) 2 = $75, 000( A / F ,10%,30)

•

= $456 Operating cost:

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Contemporary Engineering Economics, 7th ed. ©2023

AEC(10%)3 = [$31, 000( P / A,10%,15) +$62, 000( P / A,10%,5)( P / F ,10%,15)] $63, 000 +[ + $1, 000( P / G,10%, ∞)] 0.10 ×( P / F ,10%, 20)]( A / P,10%, ∞) = $40, 056 Note that ( P / G, i, ∞) = 1/ i 2 or ( P / G ,10%, ∞ ) = 100 •

Total equivalent annual worth:

AEC(10%) A = $49,576 + $456 + $40, 056 = $90, 088 Plan B: One time investment strategy: •

Capital investment: CR(10%)1 = $550, 000( A / P,10%, ∞)

= $55, 000 • Supporting equipment: $150, 000 AEC(10%) 2 = ( A / P,10%, ∞) 16.4494 = $912 Note that the effective interest rate for 30-year period is (1 + 0.1)30 − 1 = 16.4494 • Operating cost: AEC(10%)3 = [$35, 000( P / A,10%,15) +$55, 000( P / A,10%, ∞)( P / F ,10%,15)] ×( A / P,10%, ∞)

•

= $39, 788 Total equivalent annual worth:

AEC(10%) B = $55, 000 + $912 + $39, 788 = $95, 700 ∴ Plan A is a better choice.

Minimum Cost Analysis 6.69 29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (a) • Energy loss in kilowatt-hour:

6.516 $4, 709.11 (24 × 365)($0.0825) = A A • Material weight in pounds:

(200)(12)(555)

A = (770.83) A 123

• Total material costs: (770.83) A($6) = $4, 625 A

• Capital recovery cost: CR(11%) = [$4, 625 A − $1× 770.83 A]( A / P,11%, 25) + $1× 770.83 A × 0.11 = 542.44 A • Total equivalent annual cost:

4, 709.11 A • Optimal cross-sectional area: AEC(11%) = 542.44A +

dAEC(11%) 4, 709.11 = 542.44 − =0 dA A2 A = 2.9464 inches 2 (b) Minimum annual equivalent total cost:

AEC(11%) = 542.44(2.9464) +

4, 709.11 = $3,196.51 2.9464

(c)

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Contemporary Engineering Economics, 7th ed. ©2023

6.70 We assume the friction factor is 0.02. (a)

1 (600)3 (10,000) (8,760)($0.05/kWh) Cost ($) = (0.02) × 1,705 0.70 (6)5 = $2,038,821.39 (b)

Cost ($) =

(600)3 (10,000) (8,760)($0.05/kWh) 1 (0.02) × 0.70 1705 (10)5

= $158,538.75 (c) Not provided

Short Case Studies with Excel ST 6.1 31 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

We assume i = 10% . PW(10%)1 = $52, 740( P / A,10%,10) = $324, 064.47

PW(10%) 2 = $52, 740( P / A1 ,10%,10%,10)  10  = $52, 740    1 + 0.1  = $479, 454.55

where i = g

ST 6.2

$84,000 = $1,400 per kW. But if you consider the 60 time value of money, say 10% annual interest, the capital cost per kW without considering any salvage value at the end of its service life is as follows: Installed cost per kilowatt =

•

$84, 000( A / P,10%,10) $13, 671 = = $227.84 per kW 60 60 or $13, 671 = $0.026 per kWh 60 × 24 × 365 •

Operating cost per kilowatt-hour:

$19, 000 = $0.036 (60)(24)(365) ST 6.3 Given: annual energy requirement = 145, 000, 000, 000 BTUs, 1 ton=2,204.6 lbs, net proceeds from demolishing the old boiler unit = $1, 000 (a) Annual fuel costs for each alternative: •

Alternative 1: Weight of dry coal =

145, 000, 000, 000 BTUs (0.75)(14,300) = 13, 519,814 lb 13,519,814 lb = 2, 204.6 lb/ton

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 = 6,132.55 ton Annual fuel cost = 6,132.55 × $125

= $766,569 •

Alternative 2: Gas cost = $6.5 ×

145, 000, 000, 000(0.94) (0.78)(1, 000, 000)

= $1,135,833

145, 000, 000, 000(0.06) (0.81)(139, 400) = $225,756 Annual fuel cost = $1,135, 833 + $225, 756 Oil cost = $2.93 ×

= $1, 361, 589

(b) Unit cost per steam pound: •

Alternative 1: Assuming a zero-salvage value of the investment AEC(10%) = ($1, 770,300 + $100, 000 −$1, 000)( A / P,10%, 20) + 0.10($1, 000) +$766,569 = $986, 236 Unit cost = $986, 236 / 145, 000, 000 = $0.0068 per steam lb.

•

Alternative 2:

AEC(10%) = ($889, 200 − $1, 000)( A / P,10%, 20) + 0.10($1, 000) +$1,361,589 = $1, 466, 017 Unit cost = $1, 466, 017 / 145, 000, 000 = $0.01011 per steam lb.

(c) Select alternative 1. ST 6.4 If the cost of your drainage pipe has experienced a 4% annual inflation rate, I could estimate the cost of the pipe 20 years ago as follows.

$4, 208( P / F , 4%, 20) = $4, 208(1.04) −20 = $1,920.48

33 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 If the pipe had a 50-year service life with a zero-salvage value when it was placed in service 20 years ago, the annual capital recovery cost to the owner would be as follows. (I assumed the owner’s interest rate would be 5% per year. In other words, the owner could invest his $1,920.48 at 5% annual interest, if he did not purchase the pipe.) CR (5%) = $1, 920, 48( A / P , 5%, 50)

= $105.20 per year You can view this number as the annual amount he expects to recover from his investment considering the cost of money. With only a 20-year’s usage, he still has 30 more years to go. So, the unrecovered investment at the current point is $105.20( P / A, 5%, 30) = $1, 617.15.

On paper, the owner could claim this number, but the city’s interest rate could be different from the owner’s, so there is some room for negotiation. Assumed interest rate 0% 3% 4% 5%

Claim cost $1,152.24 $1,462.98 $1,545.89 $1,617.15

34 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 7 Rate-of-Return Analysis Concept of Rate of Return Note: Symbol convention---The symbol i* represents the break-even interest rate that makes the PW of the project equal to zero. The symbol IRR represents the internal rate of return of the investment. For a simple (or pure) investment, IRR = i*. For a non-simple investment, generally i* is not equal to IRR. 7.1 F = 2 P = P (1 + i ) 4 log 2 = 4 log (1+i ) log(1 + i ) = 0.07526 Or

∴ i ∗ = 18.92% $2,000 = $1000( F / P, i %, 4) i ∗ = 18.29%

7.2 $2529.24 = $1000( F / P , i %, 5) + $1, 000( F / P , i %, 3) i ∗ = 6%

7.3

$1,865 = $1,000(1+ i)2 i = 36.57% 7.4 $110, 500, 000 = $19, 000(1 + i )32 i = 31.11%

1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 7.5

$24, 500 = $514.55( P / A , i , 60)

i = 0.7917% per month r = 0.7917% ×12 = 9.5%

ia = (1+ 0.007917)12 −1 = 9.93% per year 7.6

$950 = $35( P / A, 4%,8) + F ( P / F , 4%,8) 0.7307 F = $714.35 F = $977.64

7.7

$15,767,500 = $1,650( F / P, i,44) 9,556.06 = (1 + i)44 i = 23.16% 7.8

$2.5 = $1.2(P / A, i,10) i = 46.98%

7.9

PW ( i ) = − $50, 000 + $100, 000( P / F , i ,1) − $40, 000( P / F , i , 2) = 0

i = 44.72% 7.10

$104, 200,000 = $30,000( F / P, i,54) 3, 473.33 = (1 + i)54 i = 16.30%

Investment Classification and Calculation of i* 7.11 PW(i ) = $6,500 + $4,500( P / F , i,1) − $13, 000( P / F , i, 2) = 0 i* = 10.9807%

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Contemporary Engineering Economics, 7th ed. ©2023 7.12

PW(25%) = −$12,000 + $2,500( P / F , 25%,1) + $5,500( P / F , 25%, 2) + X ( P / A, 25%, 2)( P / F , 25%, 2) =0 $6, 480 = 0.9216 X X = $7,031.25 7.13 Use Excel to find the rate of return: PW(i) = −$1, 000 + [$50( F / A, i,12) + $50( F / A, i,5) + $4, 000]( P / F , i,15) =0 Solving for i yields

i* = 12.08% 7.14 (a) Simple investment: Project A, D (b) Non-simple investment: Project B (c) • Project A: PW(i ) = −$32, 000 + $30, 000( P / A, i , 3) − $10, 000( P / G , i, 3) =0 i * = 49.52% • Project B: PW(i ) = −$38, 000 + $32, 000( P / A, i , 2) − $22, 000( P / F , i, 3) =0 *

i = 13.45% or − 41.69% • Project C:

PW ( i ) = $45, 000 − $18, 000( P / A , i , 3)

=0 i = 9.70% → Borrowing rate of return *

• Project D:

PW ( i ) = − $46, 500 + $2, 500( P / F , i ,1) + $6, 459( P / F , i , 2) + $78, 345( P / F , i , 3) =0

3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 i * = 24.94%

7.15 We may need to find the rate of return on the first cycle, which will be the same for the entire cycles.

$2,000 = [$900( P / F , i,1) + $1, 200( P / F , i, 2) + $1,800( P, F , i,3)] i* = 36.87%

7.16

(a) Classification of investment projects: • Simple projects: B and E • Non-simple projects: A, C and D (b)

−$250 + Let X =

$450 $120 − =0 1+ i (1+ i)2

1 , then, (1 + i)

−$250 + $450 X − $120 X 2 = 0 X1 = 0.6782, X 2 = 3.071 ∴ i * = 47.44% or -67.44%

Project

i*

A B C D E

47.44%, -67.44%% 20.01% 12.89% -39.46% 23.53%

7.17 (a) Classification of investment projects: • Simple projects: A,B, and D • Non-simple projects: C (b) • Project A: i* = −1.08% • Project B: i * = 11.84% • Project C: i * = 18.99% 4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • Project D: i * = 34.76% .

7.18 (a)

−$90,000 + ($27,000 − $8,000)( P / A, i,6) + $10,000( P / F , i,6) = 0

Solving for i yields i * = 9.4208% . (b) With the geometric expense series

−$90, 000 + $27, 000( P / A, i, 6) − $8, 000( P / A1 , 7%, i, 6) +$10, 000( P / F , i, 6) = 0 Solving for i* yields i * = 7.1745% . (c) To maintain i * = 9.4208% PW(i ) = −$90, 000 + $27, 000( P / A1 , g ,9.4208%, 6) −$8, 000( P / A1 , 7%,9.4208%, 6) + $10, 000( P / F ,9.4208%, 6) =0 Solving for g yields

g = 2.20%

7.19 (a) Rate of return calculation: • Project A: i* = 19.50% • Project B: i * = 9.18% (b)

PW Plot $40,000.00 $30,000.00

PW

$20,000.00

Project A

$10,000.00 $0.00 ($10,000.00) ($20,000.00)

0

2

4

6

8 10 12 14 16 18 20 22 24 26 28 30

Interest Rate (%)

5 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 The interest rate that makes two projects equivalent is 40%. 7.20 (a) Cash flow sign rules: Projects Number of Sign Changes A 1 B 2 C 1 D 1 E 2 F 3

Possible Number of i* 0, 1 0, 1, 2 0, 1 0, 1 0, 1, 2 0, 1, 2, 3

(b) Not provided but you may use Excel’s Chart Wizard. (c) • Project A: i * = 100% • Project B: i * = 482.92% or − 59.69% • Project C: i * = 20.04% • Project D: i * = 40.57% • Project E: i * = 265.41% • Project F: i * = 209.46% 7.21 (a) IRR = 20% (b) Use the Excel’s Chart Wizard. (c) Since MARR (15%) < IRR(20%), accept the project!

Mixed Investments, RIC and MIRR 7.22 (a) *

*

• Project 1: i 1 = 110%, i 2 = −60% • Project 2: i * = 136.22% *

*

• Project 3: i 1 = 23.85%, i 2 = −83.85% (b) Investment classification: • Project 1: non-simple and mixed investment, IRR = 75% 6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

PB(75%,12%)0 =−$1,000 PB(75%,12%)1 =−$1,000(1+ 0.75) +$2,500 = $750 PB(75%,12%)2 = $750(1+ 0.12) −840 = $0 • Project 2: simple and pure investment, IRR = 136.22% • Project 3: non-simple and mixed investment, IRR = 22.14%

PB(22.14%,12%)0 =−$1,000 PB(22.14%,12%)1 =−$1,000(1+ 0.2214) +$1,400 = $178.6 PB(22.14%,12%)2 = $178.6(1+ 0.12) − 200 = $0 (c) •

Project 1:

$1,000 + $840( P / F ,12%, 2) = $1,669.64 $2,500( F / P,12%,1) = $2,800 $2,800 = (1 + MIRR)2 $1,669.64 MIRR = 29.5% •

Project 2:

$1,000 = $1,000 $1,960( F / P,12%,1) + $950 = $3, 415.20

•

Project 3:

$1,000 = (1 + MIRR)2 $3, 415.20 MIRR = 77.35% $1,000 + $200( P / F ,12%, 2) = $1,159.44 $1, 400( F / P,12%,1) = $1,568 $1,568 = (1 + MIRR)2 $1,159.44 MIRR = 16.29%

(d) If MARR = 12%, all projects are acceptable.

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

7.23 (a)

−$150 +

$150 $36 + =0 1+ i (1+ i)2

Let X =

1 , then, (1 + i)

−$150 + $150 X + $36 X 2 = 0 X 1 = 0.8333, X 2 = −5 ∴ i * = 20%

(b) • Simple projects: A, B, and D • Non-simple projects: C and E (c) Apply the cash flow sign rule: Projects A (d) B C• P r D E o j

Number of Sign Changes 1 1 2 1 2

ect B:

Possible Number of i* 0, 1 0, 1 0, 1, 2 0, 1 0, 1, 2

Actual i* 20% 15.32% No return 17.70% 12.63%,41.42%

IRRB =15.32%

• Project C:

IRRC = None

• Project D:

IRRD =17.69%

• Project E:

IRRE =12.29%

Note that, since projects C and E are mixed investments, we need to find the IRRs for both C and E by using external interest rate of 12%.

(e) MIRR for Project E: 8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

$500(P/ F,12%, 2) + $500(P/ F,12%,3) = $754.48 $200( F / P,12%,5) +  + $600 = $1,333.82 $1,333.82 = (1 + MIRR)5 $754.48 MIRR = 12.07% (f) Apply the net investment test to Project E: Project Balances Project E (i*=12.63%) $200 $325 -$134 -$651 -$533 $0

0 1 2 3 4 5

A mixed investment 7.24

$5, 000(1 + i)2 + $5, 000(1 + i) = $11,500 i* = 9.69% 7.25 (a) • Project 1: i * = 82.21% *

*

• Project 2: i 1 = 14.64%, i 2 = 210.28% • Project 3: i1* = 100% (b) Apply the net investment test: • Project 1: PB(82.21%)0 = −$8,500 PB(82.21%)1 = −$8,500(1 + 0.8221) + $10, 000 = −$5, 487.85 PB(82.21%) 2 = −$5, 487.85(1 + 0.8221) + $10, 000 = 0 (- ,- ,0), a pure investment

9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • Project 2:

PB(14.64%)0 = −$5, 000 PB(14.64%)1 = −$5, 000(1 + 0.1464) + $10, 000 = $4, 268 PB(14.64%)2 = $4, 268(1 + 0.1464) + $30, 000 = $34,892.80 PB(14.64%)3 = $34,892.8(1 + 0.1464) − $40, 000 = 0 (-, + , + ,0), a mixed investment • Project 3:

PB(100%)0 = −$1,500 PB(100%)1 = −$1,500(1 + 1) + $6, 000 = $3, 000 PB(100%) 2 = $3, 000(1 + 1) − $6, 000 = 0

(-, +, 0), a mixed investment (c) • Project 1: IRR1 = 82.21% • Project 2: IRR 2 = RIC = −2.04% • Project 3: IRR 3 = RIC = −57.2% Sample RIC calculation at k = 12% for Project 3: MARR = RIC =

12% -57.2000%

Period 0 1 2

Cash Flow $ $ $

(1,500.00) 6,000.00 (6,000.00)

PB(RIC,MARR) $ $ $

(1,500.00) 5,358.00 0.96

(d) • • • (e)

Project 1: MIRR = 57.93% > 12% Project 2: MIRR = 11.30% < 12% Project 3: MIRR = 3.42% < 12% Project 1 is only acceptable.

7.26 10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(a) Pure investment: B (b) Mixed investments: A and C

IRRA =140.56%, Project B: IRRB = 21.86%, Project C:

(c) Project A:

IRRC =12.24% at an external rate of 12% (d) All three projects are acceptable. 7.27 (a) There are two sign changes in cash flow, indicating the possibility of multiple i* s.

i*1 = 10%, i*2 = 20 Apply the net investment test:

PB(10%)0 = −$1,000,000 PB(10%)1 =−$1,000,000(1.1) +$2,300,000 = $1,200,000 PB(10%)2 = $1,200,000(1.1) −$1,320,000 = 0 ∴ (-,+,0), mixed investment

(b) At an external interest rate of 12%, IRR (RIC) = 12.14%. (c) The project is acceptable. 7.28 *

*

(a) i 1 = 6%, i 2 = 0% (b) Project A Apply the net investment test:

PB(6%)0 = −$1,000 PB(6%)1 =−$1,000(1.06) + $2,060 = $1,000 PB(6%)2 = $1,000(1.06) −$1,060 = 0 ∴ (-,+,0), mixed investment

(c) Projects B & C are acceptable. 7.29 11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(a) *

*

•

Project A: i 1 =10%, i 2 =100%,

•

Project B: i 1 = 350.34%, i 2 = −80.83%

*

*

Pure investment: C, mixed investments: A, B, D and E (b) Project A:

IRRA = 13.57%, Project B: IRRB = 342.16%, Project C:

IRRC =18%, Project D: IRRD = 31.07%, Project E: IRRE = 19.67% (c) Sample calculation: • Project A: o Equivalent outflow at n = 0: −$1,000 − $2,200(P / F ,12%,2) = $2,753.83 o Equivalent inflow at n = 2: $3,100(F / P ,12%,1) = $3,472 o MIRR: $3,472 = $2,753.83(1 + MIRR)2

MIRR = 12.30% • • • •

Project B: MIRR = 75.43% > 12% Project C: MIRR = 16% > 12% Project D: MIRR = 20.51% > 12% Project E: MIRR = 13.54% > 12%

(d) All projects are acceptable. 7.30 * * From Excel, we find two rates of return: i 1 = 43.47%, i 2 = 77.67%

(a) Apply the net investment test:

PB(43.47%)0 = −$8,000

PB(43.47%)1 = −$8,000(1+ 0.4347) + $10,000 = −$1,477.98 PB(43.47%)2 = −$1,477.98(1+ 0.4347) + $30,000 = $27,879.5 PB(43.47%)3 = $27,879.5(1+ 0.4347) −$40,000 = 0 (b) Since RIC = -16.30 % < 18%, the project is not acceptable. (c) MIRR: •

Equivalent outlay at n = 0:

$8,000 + $40,000(P / F ,18%,3) = $32,345.23 12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • •

Equivalent inflow at n = 3:

$10,000(F / P ,18%,2) + $30,000(F / P ,18%,1) = $49,324 MIRR: $49,324 = $32,245.23(1 + MIRR)3 MIRR = 15.22% < 18% → not acceptable

7.31 (a) Apply the net investment test using i * = 10% :

PB(10%)0 =−$150,000 PB(10%)1 =−$150,000(1.1) + $465,000 = $300,000 PB(10%)2 = $300,000(1.1) −$330,000 = 0 ∴ (-,+,0), mixed investment

(b) By cash flow analyzer IRR (or RIC) = 6.30%. So, the project is not acceptable. (c) MIRR: $502,200 = $432,922(1 + MIRR)2

MIRR = 7.70% < 8% → not acceptable 7.32

(d)

IRR Analysis 7.33 PW(15%) = −$150, 000 + $120, 000( P / A,15%, 5) + $25, 000( P / F ,15%, 5) = $264, 694

7.34 PW(10%) = −$2, 000 + $1, 000( P / F ,10%,1)

X = $132

+$1, 200( P / F ,10%, 2) + $ X ( P / F ,10%,3) =0

7.35 13 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 The present worth of the project cash flow is PW ( i ) = − $10M + $1.8M ( P / A , i , 8) + $1M ( P / F , i , 8) = 0

∴ Since IRR = 10.18% > MARR, accept the project.

7.36 The present worth of the project cash flow is PW ( i ) = − $35, 000 + $15, 000( P / F , i ,1) + $14, 520( P / F , i , 2) + $13, 990( P / F , i , 3) =0

∴ Since IRR = 11.88% > 10% = MARR, the project is profitable.

7.37 (a) Since IRR = 10% and PW(10%) = 0, we have,

PW(10%) = −$12,500 + $8,000( P / F ,10%,1) + $4,000( P / F ,10%, 2) + X ( P / F ,10%,3) = 0 ∴ X = $2557.50

(b) Since IRR > 8%, the project is acceptable. 7.38 PW (13%) = − $2, 000 + $500( P / A ,13%, 3) + $ X ( P / F ,13%, 5) = 0 X = $1, 509.74

7.39 14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Let X be the annual rent per apartment unit. Then the expected net cash flows are:

N

Capital Investment

0

-14,500,000

Revenue

Maintenance

Manager

Net Cash Flow -14,500,000

1

50 X

-350,000

-85,000

50X - 435,000

2

50 X

-400,000

-85,000

50X - 485,000

3

50 X

-450,000

-85,000

50X - 535,000

4

50 X

-500,000

-85,000

50X - 585,000

50 X

-550,000

-85,000

50X +15,365,000

5

16,000,000

Through the Excel Solver function by setting PW(15%) = 0, ∴ X = $49,473.35 (or $4,122.78 per month)

7.40 Let X be the annual savings in labor PW(12%) = −$35, 000 + ( X − $4, 000)( P / A,12%, 6) +$5, 000( P / F ,12%, 6) =0 X = $11,896.82

7.41 •

Net cash flow table:

N Land Building Equipment Revenue Expenses Net Cash Flow 0 -$1.50 -$3 -$4.50 1 -$4 -$4.00 2 $3.50 -$1.40 $2.10 3 $3.68 -$1.47 $2.21 4 $3.86 -$1.54 $2.32 5 $4.05 -$1.62 $2.43 6 $4.25 -$1.70 $2.55 7 $4.47 -$1.79 $2.68 8 $4.69 -$1.88 $2.81 9 $4.92 -$1.97 $2.95 10 $5.17 -$2.07 $3.10

15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 11 12 13 14 •

$2

$1.40

$0.50

$5.43 $5.43 $5.43 $5.43

-$2.17 -$2.17 -$2.17 -$2.17

$3.26 $3.26 $3.26 $7.16

Rate of return calculation:

PW(i) = −$4.5 − $4( P / F , i,1) + $2.1( P / F , i,2) + + $7.16(P / F , i,14) = 0 ∴ i * = 24.85%

•

Since this is a simple investment, IRR = 24.85%. At MARR = 15%, the project is economically attractive.

7.42 PW(18%) = −$150, 000 + ( X − 50, 000)( P / A,18%,10) +15, 000( P / F ,18%,10) =0 X = $82, 739.26

7.43 (a)

PW(i ) = −$30 + $9( P / F , i ,1) + $18( P / F , i , 2) + $20( P / F , i , 3) + $18( P / F , i , 4) + $10( P / F , i , 5) + $5( P / F , i , 6) =0 This is a simple investment. Therefore, IRR = i * = 40.20% .

∴ Since IRR is higher than MARR, the project is acceptable.

(b) IRR = 45.47% $9( P / F , i,1) + $18( P / F , i, 2) + $20( P / F , i,3)  PW(i ) = −$30 + 1.1    +$18( P / F , i, 4) + $10( P / F , i,5) + $5( P / F , i, 6)  =0

(c) IRR = 27.30%

16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 $9( P / F , i,1) + $18( P / F , i, 2) + $20( P / F , i,3)  PW(i ) = −$35 + 0.9    +$18( P / F , i, 4) + $10( P / F , i,5) + $5( P / F , i, 6)  =0

Comparing Alternatives 7.44

(a) Project A: IRR = 18.33% Project B: IRR = 23.77% (b) Both projects are acceptable (c) n 0 1 2 3

Project A

Project B

A-B

-$100,000.00 -$100,000.00 $10,500.00 $70,000.00 $60,000.00 $50,000.00 $80,000.00 $20,500.00

$0.00 -$59,500.00 $10,000.00 $59,500.00

IRR A-B = 8.76% < 10%(MARR) Project B is better choice. 7.45 (a)

n 0 1 2 3

A1 -$16,000 $7,500 $7,500 $7,500

A2 -$20,000 $5,000 $15,000 $8,000

A2 – A1 -$4,000 -$2,500 $7,500 $500

IRR A2− A1 = 13.08% (b) Select Project A2. 7.46

Let A0 = current practice, A1 = just-in-time system, A2 = stock-less supply system. •

Comparison between A0 and A1:

n 0 1-8

A0 0 -9,000,000

A1 -$3,000,000 -3,800,000

A1 – A0 -$3,000,000 5,200,000

i*A1−A0 = IRRA1−A0 =173.28% >10% →Select A1 17 Copyright © 2023 Pearson Education, Inc.

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•

Comparison between A1 and A2:

n 0 1-8

A2 -$6,000,000 -1,900,000

A1 -$3,000,000 -3,800,000

A2 – A1 -$3,000,000 1,900,000

i*A2−A1 = IRRA2−A1 = 62% >10% →Select A2. 7.47 (c) Comments: Statement (a): Incorrect. The rate of return is still 11.32%. Statement (b): Incorrect. The rate of return is 9.42% Statement (c): Correct. F = $29, 223( F / P,9.42%,16.79) = $132, 483 Statement (d): Incorrect 7.48 (a) Project A: IRR = 16.01% Project B: IRR = 18.18% (b) Project A and B are acceptable (c) Since the incremental cash flow displays a nonsimple investment, we attempt to find the rate of return on incremental investment:

PW(i)A−B = −$15,000 + $10,000(P / F , i ,1) + $10,000(P / F , i ,2) −$10,000(P / F , i ,3) =0 A mixed investment → Need to calculate RIC on incremental investment. RIC A−B = −22.16% < 15% → accept B 7.49 Option 1: Buy a certificate. Option 2: Purchase a bond and assume that MARR = 5% .

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n 0 1 2 3 4 5

Option 1 -$10,000 0 0 0 0 15,386.24

Net Cash Flow Option 2 Option 1 – Option 2 -$10,000 $0 650 -650 650 -650 650 -650 650 -650 10,650 4736.24

The rate of return on incremental investment is

i*1−2 = 25.49% > 5% ∴ Option 1 is a better choice.

7.50 Determine the cash flow on incremental investment: Net Cash Flow

n

Project A1

Project A2

A2 – A1

0

-$4,000

-$5,000

-$1,000

1

2,600

3,600

1,000

2

2,800

3,200

400

i*A2− A1 = 30.62% > 15% → accept A2. 7.51 (a) IRR on the incremental investment: Net Cash Flow n Project A1 Project A2 0 -12,000 -14,000 1 5,000 6,200 2 5,000 6,200 3 5,000 6,200

A2 – A1 -$2,000 1,200 1,200 1,200

i*A2−A1 = 36.31% (b) Since it is an incremental simple investment, Therefore, select project A2.

IRRA2-A1 = 36.31% > 10% .

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Contemporary Engineering Economics, 7th ed. ©2023 7.52 (a) IRR on the incremental investment:

n 0 1 2 3

A1 -$12,000 7,500 7,500 7,500

Net Cash Flow A2 -$15,000 8,000 14,000 5,000

A2 – A1 -$3,000 500 6,500 -2,500

Since the incremental cash flow displays a non-simple investment, we may need to calculate RIC or MIRR, or abandon the IRR analysis and select based on the NPW criterion. (b)

RIC A2− A1 = 27.33% > 10% → Select A2. MIRR A2− A1 : $7,755 = $4,878.28(1 + MIRR)3 MIRR = 16.70% > 10% → Select A2.

7.53

(c)

7.54 (a) IRR for incremental investment:

n 0 1 2

Project A -$300 0 690

Net Cash Flow Project B -$800 1,150 40

B-A -$500 1,150 -650

i* B− A = 0% or30% Since this is a mixed incremental investment, we need to find the IRR using an external interest rate of 15%.

IRRB−A = 16.96% >15% →Select B. (b) Use the Excel’s Chart Wizard to plot a PW function.

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Contemporary Engineering Economics, 7th ed. ©2023 7.55 Incremental cash flows (Model A – Model B):

n

A–B -$2,376 0 0 0 2,500

0 1 2 3 4

IRRA−B =1.28% ∴ If MARR < 1.28%, Model A is preferred.

7.56 PW ( i ) A = − $8, 000 + ($900 − $150)( P / A, i , 20) + $500( P / F , i , 20) PW ( i ) B = − $12, 000 + ($1,100 − $100)( P / A, i , 20) + $600( P / F , i , 20) PW ( i ) B − A = − $4, 000 + $250( P / A, i , 20) + $100( P / F , i , 20) =0 * A

i = 7.11% iB* = 5.65% iB* − A = 2.39% < 12% → Select A, if no do-nothing is allowed.

7.57 (a) The least common multiple project lives = 6 years → Analysis period 6 years

n 0 1 2 3 4 5 6

Project A -$100 60 50 50-100 60 50 50

Net Cash Flow Project B -$200 120 150-200 120 150-200 120 150

B–A -$100 60 -100 170 -110 70 100

Since the incremental cash flow displays a non-simple investment, we may need to calculate RIC on the incremental investment. Or abandon the IRR analysis and make a selection based on the NPW criterion.

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IRRB− A = RICB− A = 15.98% > 15% → Select B.

PW(15%)B-A = $3.48 > 0 , select project B. (b) Incremental analysis between C and D:

n 0 1 2

Project C -$4,000 2,410 2,930

Net Cash Flow Project D -$2,000 1,400 1,720

C–D -$2,000 1,010 1,210

IRRC− D = 7.03% 10% →Select A1

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•

Comparison between A1 and A2:

n 0 1-8

A2 -$6,000,000 -1,900,000

A1 -$3,000,000 -3,800,000

A2 – A1 -$3,000,000 1,900,000

i*A2−A1 = IRRA2−A1 = 62% >10% →Select A2. 7.59 (a) • Project A vs. Project B

n 0 1 2 3 4

Project A -$1,000 900 500 100 50

Net Cash Flow Project B -$1,000 600 500 500 100

B–A $0 -300 0 400 50

i*B− A = IRRB− A = 21.27% > 12% , select B. • Project B vs. Project C

n 0 1 2 3 4

Project B -$1,000 600 500 500 100

Net Cash Flow Project C -$2,000 900 900 900 900

C–B -$1,000 300 400 400 800

i*C− B = IRRC− B = 26.36% >12% , select C. (b)

$1,000 = $300( P / A, i, 4) i = 7.71% BRR = 7.71%.

(c) Since BRR is less than MARR, project D is acceptable

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(d)

n

Project C -$2,000 900 900 900 900

0 1 2 3 4

Net Cash Flow Project E -$1,400 400 400 1,200 400

C–E -$600 500 500 -300 500

i*C−E = IRRC−E = 44.60% > 12% , select C. Note: Even though the incremental cash flows represent a non-simple investment, but it is a pure investment, indicating a unique rate of return. 7.60 (a)

i1* = 54.52%, i2* = 57.61%, and i3* = 38.41% (b) • Project 1 versus Project 2:

n 0 1 2

Project 1 -$1,500 700 2,500

Project 2 -$5,000 7,500 600

2–1 -$3,500 6,800 -1,900

$6,800(1.15) = [$3,500 + $1,900(1.15)−2 ](1 + MIRR)2 MIRR2−1 = 25.86% > 15% → Select Project 2. . RIC2−1 = 47.08% > 15% → Slect Project 2. • Project 2 versus Project 3:

n 0 1 2

Project 2 -$5,000 7,500 600

Project 3 -$2,200 1,600 2,000

2–3 -$2,800 5,900 -1,400

$5,900(1.15) = ($2,800 + $1,400(P / F ,15%,2))(1 + MIRR)2 MIRR = 32.61% > 15% → Select Project 2. RIC2−3 = 67.23% > 15% → Select Project 2. 24 Copyright © 2023 Pearson Education, Inc.

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Comments: If you want to apply the IRR decision rule to the non-simple investments, you should apply the net investment test and make the selection by calculating the return on invested capital (or true internal rate of return).

7.61 (a)

IRRB = 25.99% NPWA = $2,557.74 (b)

$5,500(P / A,15%,3) = $10,000(1+ MIRR)3 MIRR = 24.07% (c) Incremental analysis: Net Cash Flow Project A Project B -$10,000 -$20,000 5,500 0 5,500 0 5,500 40,000

n 0 1 2 3

B–A -$10,000 -5,500 -5,500 34,500

IRRB−A = 24.24% >15% →Select B. 7.62 iB*− A = 85% > 29% → Select B. ← Investment iB*−C = 30% > 29% → Select C. ← Borrowing iD* −C = 25% < 29% → Select C. ← Investment i A* −D = 50% > 29% → Select D. ← Borrowing B  A,C  D,C  B,D  A → Best project is C.

7.63

Model C

iB*− A = 5% < 12% → Select A. ← Investment iC* −B = 40% > 12% → Select C. ← Investment iC* − A = 15% > 12% → Select C. ← Investment A  B,C  A,C  B → Best project is C.

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7.64 All projects would be acceptable because individual ROR exceed the MARR. Based on the incremental analysis, we observe the following relationships:

IRRA2−A1 = 10% < 15%(Select A1) IRRA3−A1 = 18% > 15%(Select A3) IRRA3−A2 = 23% >15%(Select A3) Therefore, A3 is the best alternative. 7.65 From the incremental rate of return table, we can deduce the following relationships:

IRRA2− A1 = 8.9%15%(Select A3)

IRRA4−A3 = 0% 15%(Select A5) IRRA6−A5 = 36.3% >15%(Select A6) It is necessary to determine the preference relationship among A1, A3, and A6.

IRRA3−A1 = 16.66% > 15%(Select A3) IRRA6−A3 = 20.18% >15%(Select A6) IRRA6−A1 = 18.24% >15%(Select A6) A6 is the best alternative.

Unequal Service Lives 7.66 Assumptions are required that required service period is indefinite and both projects can be repeated at the same cost in the future.

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Net Cash Flow Machine A Machine B -$40,000 -$55,000 -$15,000 -$10,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000-$55,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000 -$15,000 -$10,000-$55,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000-$55,000

n 0 1 2 3 4 5 6 7 8 9 10 11 12

A–B $15,000 -$5,000 -$5,000 -$45,000 $50,000 -$5000 -$45000 -$5,000 $50,000 -$45,000 -$5,000 -$5,000 $10,000

IRR A− B = 42.72% > 10% (Machine A should be purchased.)

7.67 With the least common multiple of 6 project years,

n 0 1 2 3 4 5 6

Project A -$5,000 3,000 4,000 4,000 -5,000 3,000 4,000 4,000

Net Cash Flow Project B -$10,000 8,000 8,000 – 10,000 8,000 8,000 – 10,000 8,000 8,000

B–A -$5,000 5,000 -6,000 9,000 -5,000 4,000 4,000

Since the incremental cash flow series is a mixed investment, we may need to calculate the RIC (or MIRR) on incremental investment or abandon the IRR analysis and use the PW decision rule.

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•

RIC: Use the Cash Flow Analyzer to find the RIC:

RICB− A = 25.67% > 15% → Select B. •

MIRR: Equi. cash outlay at n = 0: $5,000 + 6,000(1.15)−2 + $5,000(1.15)−4 = $12,395.62 Equi. cash revenue at n = 6: $5,000(1.15)5 + $9,000(1.15)3 + $4,000(1.15) + $4,000 = $32,344.66 $29,302.90 = $9,395.62(1 + MIRR)6 MIRRB − A = 17.34% > 15% → Select B.

•

PW analysis: PW(15%) B − A = −$5,000 + $5,000( P / F ,15%,1) +  + $4, 000( P / F ,15%, 6) = $1,587.86 > 0 → Select B.

7.68 (a) Since there is not much information given regarding the future replacement options and required service period, we may assume that the required service period is 3years and project A2 can be repeated at the same cost in the future. (b) The analysis period may be chosen as the least common multiple of project lives, which is 3 years.

n 0 1 2 3

A2 – A1 -$5,000 0 0 15,000

IRRA2− A1 = 44.22% The MARR must be less than 44.22% for Project A1 to be preferred.

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Short Case Studies ST 7.1 We assume monthly service fee = $40, the number of transactions per each scanner per day = 100, and service period 10 years. Let R be additional annual revenue per scanner. PW(20%) = −$50(2) + [ R − ( 2($40) + $0.10(200)(30) ) ( F / A,

20% ,12)] 12

×( P / A, 20%,10) = −100 + [ R − 680(13.16)](4.1925) =0 R = $8, 972.65 per year

ST 7.2 (a) Analysis period of 40 years (unit: thousand $): • Without “mothballing” cost:

PW(i) = −$1,500,000 + $138,000( P / A1 ,0.05%, i,40) =0 i* = 8.95% • With “mothballing” cost of $0.75 billion: PW(i ) = −$1,500, 000 + $138, 000( P / A1 , 0.05%, i, 40) −$750, 000( P / F , i, 40) =0 i* = 8.77%

For a 40-year analysis period, the drop in IRR with the mothballing cost is only 1.9%, which is relatively insignificant. (b) Analysis period of 25 years (unit: thousand $): • Without “mothballing” cost:

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PW(i) = −$1,500,000 + $138,000( P / A1 ,0.05%, i, 25) =0 i* = 7.84%

• With “mothballing” cost of $0.75 billion:

PW(i) = −$1,500,000 + ($207,000 − $69,000)( P / A1 ,0.05%, i, 25) −$750,000( P / F , i, 25) =0 i* = 6.80% For a 25-year analysis period, the drop of IRR with the mothballing cost is about 13.24%, which is relatively significant. ST 7.3 To reach reasonable solution to the problem, we may need to lay out specific assumptions.

(a) Assumptions required: • We need to assume that there are no cash flows for the first three years if B&E Cooling decides to defer the decision. • Assume the firm will be in business for an indefinite period. • We assume that the best cooling technology will be the absorption technology that will be introduced 3 years from now. Therefore, if B&E Cooling decides to select Option 1 now, Option 2 will be adopted for an indefinite period at the end of 8 years.

(b) Investment decision: • Rate of return analysis:

We need incremental analysis for these two options and then we find that the incremental cash flow is a mixed investment case. With MARR=15%, we calculate RIC1−2 =154%.

RIC1−2 > MARR . Option 1 is a better choice. 30 Copyright © 2023 Pearson Education, Inc.

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ST 7.4

n 0 1 2

Current Pump(A) Larger Pump(B) $0 -$1,600,000 $10,000,000 $20,000,000 $10,000,000 $0

B-A -$1,600,000 $10,000,000 -$10,000,000

IRR =

25% 400%

The incremental cash flows result in multiple rates of return (25% and 400%), so we may abandon the rate of return analysis. Using the PW analysis,

PW(20%)B-A = −$1.6M + $10M ( P / F , 20%,1) − $10M ( P / F , 20%, 2) = −$0.21 < 0 Reject the larger pump. Comments: We could find the return on invested capital to be 4.17% at MARR of 20%, so we will reject the larger pump again.

ST 7.5 (a) Whenever you need to compare a set of mutually exclusive projects based on the rate of return criterion, you should perform an incremental analysis. In our example, the incremental cash flows would look like the following:

N 0 1 2

2-1 -$10,000 +$23,000 -$13,200

This is a non-simple investment with two rates of return and we calculate multiple IRRs.

i*2−1 = 10% or 20% This incremental cash flow is a mixed investment case. 31 Copyright © 2023 Pearson Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023

PB(i, MARR R)0 = −$10,, 000 PB(i, MAR RR)1 = −$10,, 000(1 + i) + $23, 000 = $13, 0000 − 10, 0000i PB(i, MAR RR)2 = ($13, 000 − 10, 0000i)(1 + MAR RR) −$13, 2200 =0 ms (Note thhat if i > 1.3, there will bbe no feasibble solution..) Rearrangging the term in

PB(i,MARR) , RR as a funnction of MA ARR. 2 gives an exprression of IR IRR = 1 .3 −

1.32 RR 1 + MAR

For exam mple, at MA ARR = 15%

IRR2−1 =15.2% >155% Select prroject 2.

(b) If you plot thee present woorth as a funnction of intterest rate, yyou will obsserve thee following:   

M < 10% % select proj oject 1. If MARR ≤ If 100% ≤ MARR R 18.1%, select projeect 2. If MARR M > 18.1%, do nothhing. 32 Copyright © 2023 Pearson E Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 8 Cost Concepts Relevant to Decision Making Classifying Cost 8.1 • Storage and material handling costs for raw materials: product cost (indirect costs) • Gains or loss on disposal of factory equipment: period costs • Lubricants for machinery and equipment used in production: product cost (mfg. Overhead) • Depreciation of a factory building: product cost (mfg. Overhead) • Depreciation of manufacturing equipment: product cost (mfg. Overhead) • Depreciation of the company president’s automobile: period cost • Leasehold costs for land on which factory buildings stand: period cost • Inspection costs of finished goods: product cost • Direct labor cost: product cost • Raw materials cost: product cost • Advertising expenses: period cost

Cost behavior 8.2 • Wages paid to temporary workers: Variable cost • Property taxes on factory building: Fixed cost • Property taxes on administrative building: Fixed cost • Sales commission: Variable cost • Electricity for machinery and equipment in the plant: Variable cost • Heat and air-conditioning for the plant: Fixed cost • Salaries paid to design engineers: Fixed cost • Regular maintenance on machinery and equipment: Fixed cost • Basic raw materials used in production: Variable cost • Factory fire insurance: Fixed cost 1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 8.3 (a)

C (301) − C (300) = 12, 519.01 − 12, 500

= $19.01 (b)

C′( x) = 0.0003x 2 − 0.16 x + 40 C ′(200) = $20 C ′′( x ) = 0.0006 x − 0.16

C ′′(200) = −0.04 < 0 → Decreasing

(c)

Average unit cost =

C ( x) 5, 000 = 0.0001x 2 − 0.08 x + 40 + x x

Average C (300) = $41.67 8.4 Output level 1,000 units 2,000 Units

Question (a) Total manufacturing cost

$98,000

$160,000

(b) Manufacturing cost per unit (c) Total variable costs

$98 $67,000

$80 $134,000

(d) Total variable costs per unit

$67

$67

(e) Total costs to be recovered

$125,000

$192,000

Cost-Volume-Profit Relationships 8.5 (a) •

RTJ Option:

Total Cost = $6,810 + ($60 + $35)N + $5, 000 + $3, 000 Total revenue = $300N 300 N = $14,810 + 95 N N * = 72.24  73 •

Wynlakes Option:

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Total Cost = $2,400 + ($100 + $40)N + $5, 000 + $3, 000 Total revenue = $300N 300 N = $10, 400 + 140 N N * = 65 (b)

14,810 + 95 N = 10, 400 + 140 N 45 N = 4, 410 Nb* = 98

8.6 (a) Break-even sales volume: $200,000 (b) Marginal contribution rate = 20% (c) Let R = break-even sales dollars; F = total fixed cost; V = variable cost per unit; Q = sales price per unit

R=

F 1−V

= Q

F V V = 0.8; ; 1 − = 0.2; MCR Q Q

V V 1 0.8 = 1− = 1− = 0.1579; Q 0.95 0.95Q 0.95 $40, 000 = $253.333 R= 0.1579

1−

(d) No change in MCR.

F = 1.1F = $44, 000 R= (e)

$44, 000 = $220, 000 0.2

V V = 0.2; = 0.8; Q Q 1.06V = 1 − 1.06(0.8) = 0.1520; 1− Q $40, 000 = $263.158 R= 0.1520 1−

(f)

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$40, 000 − $20, 000 = $100, 000 0.2 8.7 (a)

Since belt A has a maximum contribution margin, we choose to produce belt A at its maximum demand level. Then the fixed cost remains at $255, 000 − $3(20, 000) = $195, 000 . Then we calculate the breakeven $195, 000 volume for belt B: Vb = = 97,500 units. Since the demand for $2 belt B is 100,000 units, which are greater than breakeven unit, we could produce belt B up to 97,500 units. Then, the policy is to produce 20,000 units of belt A and 97,500 units of belt B, respectively to breakeven.

(b)

The total contribution margin when 200,000 units are sold is $3(20, 000) + $2(100, 000) + $1(80, 000) = $340, 000 . The operating income is $340, 000 − $255, 000 = $85, 000 .

(c)

Operating income = $3(20, 000) + $2(80, 000) + $1(100, 000) − $255, 000 = $65, 000 . Once again, we choose to produce belt A at its maximum demand level. Then the new fixed cost is $255, 000 − 20, 000($3) = $195, 000 . We choose belt B as its contribution margin comes in second. The breakeven $195, 000 = 97,500 units. However, since the new for belt B is Vb = $2 demand for belt B is 80,000 units, which is less than the breakeven unit, therefore, we can only produce belt B up to 80,000 units. Now, the new fixed cost becomes $195,000 - $2(80,000) = $35,000. Then we proceed to produce belt C. The breakeven for belt C is $35, 000 Vb = = 35, 000 units. Thus, the new breakeven is 20,000 units of $1 belt A, 80,000 units of belt B, and 35,000 units of belt C.

8.8 (a) Total fixed cost to be recovered (b) Sales volume & Profit/Loss (c) Profit = 0 (d) Total revenue (e) Break-even volume

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8.9 (a) No. 1. 2. 3. 4. 5.

Description Profit (Loss) Sales volume Total manufacturing cost Variable costs Fixed costs

No. 6. 7. 8. 9. 10.

Description Break even Loss Profit Total revenue Marginal contribution

(b) Case

Unit Sold

Sales

Variable Expenses

Contribution Margin per Unit

Fixed Expenses

Net Income (Loss)

A B C D

9,000 14,000 20,000 5,000

$270,000 $350,000 $400,000 $100,000

$162,000 $140,000 $280,000 $30,000

$12 $15 $6 $14

$90,000 $170,000 $85,000 $82,000

$18,000 $40,000 $35,000 ($12,000)

Cost Concepts Relevant to Decision Making 8.10 Additional order units = 100 Labor cost = ($12)(5)(100) = $6,000 Material cost = ($14)(100) = $1,400 Overhead cost = (50%)($6,000) = $3,000 Total cost = $6,000 + $1,400 + $3,000 = $10,400 Profit margin = (30%)($10,400) = $3,120

∴ Unit price to quote = ($10,400 + $3,120)/100 = $135.20

8.11 (a)

Product mix that must satisfy: A: B = 4:3, or 4B = 3A Break-even formula: Total revenue = Total cost: 10A + 12(0.75)A = 5A + 10(0.75)A + 2,600; 6.5A = 2,600; A = 400 units and B = 300 units

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Contemporary Engineering Economics, 7th ed. ©2023 (b)

10A + 12A = 5A + 10A + 2,600; A = 371.43 units

(c)

Compute the marginal contribution rate (MCR) for each product: Product A = $5, Product B= $2; with the assumption (A > 0 and B > 0), more weight should be given to product A.

(d) • •

Product A: MCR = $5 per unit; Production time = 0.5 hour per unit; profit per hour = $10 Product B: MCR = $2 per unit; Production time = 0.25 hour per unit; profit per hour = $8

Conclusion: Product A is more profitable, so it should be pushed first. 8.12 (a) Incremental cost In-house Option

Description Soldering operation Direct materials Direct labor Manufacturing overhead: Variable Fixed Unit cost

Outsourcing Option

$ $

7.50 5.00

$ $ $

4.80 6.00 4.25

$ $

3.80 0.20

$ $

3.23 0.20

$

16.50

$

18.48

The outsourcing option would cost $1.98 more for each unit. (b) Break-even price = $4.80 - $1.98 = $2.82 per unit

8.13 Given, F = $60,000,

v = 0.3 p

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( (a)

( (b) Break-even sales =

F 1−

v p

=

$60, 0000 = $85, 7144 1 − 0.33

( (c) p.F 0.92 p ($$60, 000) $60, 000 = = v p−v 0.922 p − v 1− 0.92 p $60, 000 $60, 000 = =  1   v  1 − (01.087)(0.3) 1−     0.92   p 

Break-even sales =

= $899, 032

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Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies with Excel ST 8.1 (a) Break-even volume: • 6-day operation: capacity → 6,000 cwt/day, 6 days, p = $22.64 / cwt

F = ($8,520) (6 days) = $51,120, v = [$0.68+($8.35) (2.35)] = $20.30 / cwt

F + Nv = Np $51,120 F Nb = = = 21,846 p − v $22.64 − $20.30 • 7-day operation: capacity → 6,000 cwt/day, 7 days, p = $22.64 / cwt

F = ($8,520) (6 days) + $850 = $51,970, v = [$0.68(6/7) + $1.36(1/7) + ($8.35) (2.35)] = $20.40 / cwt

F + Nv = Np $51,970 F Nb = = = 23, 201 p − v $22.64 − $20.40 (b) • 6-day operation:

MCR = 1 − • 7-day operation:

MCR = 1 −

v 20.30 = 1− = 0.1034 p 22.64

v 20.40 = 1− = 0.0989 p 22.64

(c) 6-day operation • Average total cost per cwt = unit F +V =

$51,120 + $20.30 = $21.72 / cwt (6000)(6)

• Net profit margin before taxes

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 MC = p - v = $22.64 - $20.30 = $2.34 / cwt, $51,120 unit F = = $1.42 / cwt (6000)(6) Net profit margin before taxes per cwt = MC - unit F = $0.92/cwt

(d) 7-day operation • Net profit margin before taxes MC = p - v = $22.64 - $20.40 = $2.24/cwt $51,970 = $1.2374 / cwt (6000)(7) Net profit margin before taxes per cwt = MC - unit F =$1.0026 /cwt unit F =

•

Total profit for each operation: • 6days operation case Total profit = $0.92(6,000) (6) = $33,120 •

7days operation case: Total profit = $1.0026(6,000) (7) = $42,109

Total profit of 7days operation is much more than the total profit of 6days operation so it would be more economical for the mill to operate on Sunday. ST 8.2 (a)

OP1 :5x1 = 0 OP2 : −5M +10x2 = 0 OP3 : −30M − 60x3 +100 = 0 x1 = 0 x2 = 0.5M x3 = 0.75M (b)

−5M +10x = −30M − 60x +100x x = 0.833M 9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c)

5x = −30M − 60x +100x x = 0.857M Select OP1 if sales volume is between 800 K and 857 K . Select OP3 if sales volume is between 857 K and 950 K .

ST 8.3 (a) Make

Variable Costs Direct materials Direct manufacturing labor Variable manufacturing overhead (power & utilities) Inspection, setup, material handling Cost to purchase chains Fixed Costs Machine lease Allocated fixed plant admin, taxes, & insurance

Buy

$40,000.00 $20,000.00 $15,000.00 $2,000.00 $0.00

$82,000.00

$3,000.00 $30,000.00

$30,000.00

$110,000.00 $112,000.00 $11.00 $11.20

Total Cost Unit Cost

Ace should not accept the offer since the unit cost is slightly more when ACE buy chains. (b) Make

Variable Costs Direct materials Direct manufacturing labor Variable manufacturing overhead (power & utilities) Inspection, setup, material handling Cost to purchase chains Cost to upgrade the bike Fixed Costs Machine lease Allocated fixed plant admin,

Buy

$40,000.00 $20,000.00 $15,000.00 $2,000.00 $0.00

$3,000.00 $30,000.00

10 Copyright © 2023 Pearson Education, Inc.

$82,000.00 $180,000.00

$30,000.00

Contemporary Engineering Economics, 7th ed. ©2023 taxes, & insurance Fixed costs of upgrades (10,000 units) Profit by upgrading the bike Total Cost

$16,000.00

$110,000.00

$200,000.00 $108,000.00

It's economical for ACE to buy from the manufacturing of the chains. (c) Make

Variable Costs Direct materials Direct manufacturing labor Variable manufacturing overhead (power & utilities) Inspection, setup, material handling (8 batches) Cost to purchase chains (6200 units)

Buy

$24,800.00 $12,400.00 $9,300.00 $1,600.00 $0.00

$50,840.00

Fixed Costs Machine lease Allocated fixed plant admin, taxes, & insurance

$3,000.00

$0.00

$30,000.00

$30,000.00

Total Cost Unit Cost

$81,100.00 $13.08

$80,840.00 $13.04

It is economical for ACE to buy.

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Contemporary Engineering Economics, 7th ed. ©2023

ST8.4 (a)

Make in-house

Unit Cost

Direct materials Direct manufacturing labor Variable manufacturing overhead Inspection, setup, material handling Testing machine lease Allocated fixed plant adm, taxes, and insurance Total cost Unit cost

$ $ $

Outsourcing Fixed plant adm, taxes, and insurance ($150,000)

20.00 10.00 6.00

Volume 10,000 $ $ $ $ $ $ $ $

200,000.00 100,000.00 60,000.00 10,000.00 15,000.00 150,000.00 535,000.00 53.50

$

40.00 15.00 55.00

$

Produce in-house. (b) Opportunity cost of producing the engines in house:

$100 − ($90 +

$80, 000 ) = $2 per unit 10, 000

Total unit cost for in-house operation: $53.50 + $2.00 = $55.50. Outsourcing option is economically more attractive.

12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(c) The in-house (make) option is more economically attractive. Volume Make in-house

Unit Cost

Direct materials Direct manufacturing labor Variable manufacturing overhead Inspection, setup, material handling Testing machine lease Allocated fixed plant adm, taxes, and insurance Total cost Unit cost

$ $ $

Outsourcing Fixed plant adm, taxes, and insurance ($150,000)

10,000

20.00 $ 200,000.00 $ 120,000.00 10.00 $ 100,000.00 $ 60,000.00 6.00 $ 60,000.00 $ 36,000.00 $ 10,000.00 $ 6,000.00 $ 15,000.00 $ 15,000.00 $ 150,000.00 $ 150,000.00 $ 535,000.00 $ 387,000.00 $ 53.50 $ 64.50 $ $

13 Copyright © 2023 Pearson Education, Inc.

6,000

40.00 $ 15.00 55.00 $

40.00 25 65.00

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 9 Depreciation and Corporate Taxes Economic Depreciation 9.1 (a), (b), (e), (f), (h) (amortization, rather than depreciation) 9.2

The loss of value is defined as the purchase price of an asset less its market value, also known as economic depreciation. Economic depreciation = $30,000 $12,000 = $18, 000

9.3

Economic depreciation = $20,000 - $14,000 = $6,000

Cost Basis 9.4

•

Total property value with the house: Original cost Add: New building Demolition expenses Property value

Land $155,000 $155,000

Building $245,000 $1,250,000 $15,000 $1,510,000

Total property value = $155,000+ $1,510,000= $1,665,000

• 9.5

Cost basis for depreciation = $15,000+ $1,250,000 = $1,265,000 (Note: The demolition expense is treated as a site preparation expense)

Cost basis for flexible manufacturing cells: Flexible Manufacturing cells ($400,000 x 3) Freight charges Handling fee Site preparation costs Start-up and testing costs Special wiring and material costs Cost basis

$1,200,000 $20,000 $15,000 $45,000 $24,000 $3,500 $1,307,500

(Note: start-up and testing costs = $20 x 40 x 6 x 5 = $24,000)

1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.6

•

Unrecognized profit Old drill press (Book value) Trade-in allowance Unrecognized loss

$46,220 $40,000 $6,220

•

Cost basis Cost of new drill Plus: unrecognized loss Cost basis of new drill

$148,000 $6,220 $154,220

Comments: If the old drill were sold on the market (instead of trade-in), there would be no unrecognized loss. In that situation, the cost basis for the new drill would be $148,000. 9.7

•

Unrecognized profit Old lift truck (Book value) Trade-in allowance Unrecognized gains

$7,808 $9,000 $1,192

•

Cost basis Cost of new truck Minus: unrecognized gains Cost basis of new truck

$38,000 $1,192 $36,808

Comments: If the old truck were sold on the market (instead of trade-in), there would be no unrecognized gains. In that situation, the cost basis for the new drill would be $38,000.

Book Depreciation Methods 9.8

n

(a) SL

(b) DDB

Dn

Bn

Dn

Bn

1

$35,000

$200,000

$94,000

$141,000

2

$35,000

$165,000

$56,400

$84,600

3

$35,000

$130,000

$24,600

$60,000

4

$35,000

$95,000

$0

$60,000

5

$35,000

$60,000

$0

$60,000

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Contemporary Engineering Economics, 7th ed. ©2023 9.9

Given: I = $125,000, n = 3 years, N = 8.

1 8 D3 = 0.25 ×125, 000 × (1 − 0.25)3−1

α = 2( ) = 0.25 = $17,578

9.10

9.11

DDB switching to SL:

n

Dn

Bn

1

$35,141

$87,859

2

$25,104

$62,755

3 4

$17,933 $12,805

$44,822 $32,017

5

$10,676

$21,341

6

$8,341

$13,000

7

$0

$13,000

Given: I = $88,000, S = $13,000, N = 6 years. (a) D1 = $29,333 , D2 = $19,556 , D3 = $13, 037 (b) DDB Switching to SL Dn n

Bn

1

$29,333

$58,667

2 3

$19,556 $13,037

$39,111 $26,074

4

$8,691

$17,383

5

$4,383

$13,000

6

$0

$13,000

Comments: If the regular DDB deduction were taken during the fifth year, B5 would be less than the salvage value. Therefore, it is necessary to adjust D5 . The number in the box represents the adjusted value. No switching is common for this type of situation whenever the salvage value is unusually large.

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Contemporary Engineering Economics, 7th ed. ©2023

9.12Given: I = $200,000, S = $32,000, N = 8 years

n

Dn

Bn

1

$50,000

$150,000

2 3

$37,500 $28,125

$112,500 $84,375

4

$21,094

$63,281

5

$15,820

$47,461

6 7 8

$11,865 $3,596 0

$35,596 $32,000 $32,000

9.13 1 (a) α =   2 = 0.3333 6 (b) D2 = (0.3333)(0.6667)(42,000) = $9,333

(c) B4 = (42, 000)(1 − 0.3333) 4 = $8, 296 9.14Given: I = $55,000, N = 5 years, S = $5,000

n

(a) SL

(b) DDB

1

$10,000

$22,000

2 3

$10,000 $10,000

$13,200 $7,920

4

$10,000

$4,752

5

$10,000

$2,128

9.15Given: I = $56,000, S = $6,500, N = 12 years $56, 000 − $6,500 = $4,125 12

(a)

D=

(b)

D3 = $6, 481

4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.16 Depreciation allowances and book values: (a) Depreciation rate = 1/6 for SL, (b) Depreciation rate = 1.5/6 = 1/4 for 150% DB SL n

Dn

0

150% DB Bn

Dn

$32,000

1 2

$4,500 $4,500

$27,500 $23,000

$32,000 $8,000 $6,000

Units-of-Production Method 9.17 Depreciation amount for each year of Limo #1 24, 000 ($32, 000 − $3, 000) = $3, 480 200, 000

D1 =

28, 000 ($32, 000 − $3, 000) = $4, 060 200, 000 The book value of Limo #1 at the end of year 2: D2 =

B2 = 32,000 − 3,480 − 4,060 = $24,460 9.18

•

•

•

•

Bn

Truck A: D=

25, 000 ($50, 000 − $5, 000) = $5, 625 200, 000

D=

12, 000 ($25, 000 − $2, 500) = $2, 250 120, 000

D=

15, 000 ($18, 500 − $1, 500) = $2,550 100, 000

Truck B:

Truck C:

Truck D:

5 Copyright © 2023 Pearson Education, Inc.

$24,000 $18,000

Contemporary Engineering Economics, 7th ed. ©2023 D=

20, 000 ($35, 600 − $3, 500) = $3, 210 200, 000

9.19 Allowed depreciation amount D=

$90, 000 − $5, 000 (30, 000) = $10, 200 250, 000    Depreciation per mile

9.20

$68, 000 − $7,500 (5,500) 50, 000 = $6, 655

D5,000 hours =

9.21 (a) Straight line: D1 = (b) UP: D1 =

257, 000 − 32, 000 = $22,500; B1 = $234,500 10

257, 000 − 32, 000 (23, 450) = $21,105; B1 = $235,895 250, 000

(c) Working hours: D1 =

257,000 − 32,000 (2, 450) = $18,375; B1 = $238, 625 30, 000

(d) DDB (without conversion to SL): D1 = 0.2($257,000) = $51, 400; B1 = $205,600 (e) DDB (with conversion to SL): D1 = 0.2(257,000) = $51, 400; B1 = $205,600 Note that: (d) and (e) are indifferent in year 1.

Tax Depreciation 9.22 Given: I = $265,000, Delivery and installment costs =$46,000, N = 12 years, and 7-year MACRS

6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (a) Cost basis = $265,000 + $46,000 = $311,000 (b)

n

MACRS Depreciation

1 2

$44,442 $76,164

3

$54,394

4

$38,844

5 6 7 8

$27,772 $27,741 $27,772 $13,871

9.23 Given: I = $35,000, S = $6,000, N = 8 years, and 5-year MACRS

n

Book Depreciation

MACRS Depreciation

1

$3,625

$7,000

2

$3,625

$11,200

3 4

$3,625 $3,625

$6,720 $4,032

5

$3,625

$4,032

6 7 8

$3,625 $3,625 $3,625

$2,016 -

9.24 (a) Cost basis: $190, 000 + $25, 000 = $215, 000 (b)

D1 = $30, 723.5, D2 = $52, 653.5, D3 = $37, 603.5, D4 = $26,853.5 D5 = D7 = $19,199.5, D6 = $19,178, D7 = $19,199.5, D8 = $9,589

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.25 Given: I = $100,000, S = $20,000, N = 7 years (a) MACRS 7-year class:

B3 = $100,000 − (0.1429 + 0.2449 + 0.1749)($100,000) = $43,730  $100, 000 − $20, 000  (b) SL: B3 = $100, 000 − 3   = $65, 714.29 7 

The Difference is $21,984.29. 9.26 Let I denote the cost basis for the equipment.

B3 = I − ( D1 + D2 + D3 ) I = I − (0.1429 + 0.2449 + 0.1749) I = I − 0.5627 I = 0.4373($185, 000) = $80,901 9.27 Given: I = $92,000, S = $12,000, N = 5 years, 7-year MACRS depreciation class

D1 = $13,147 D2 = $22,531 D3 = $16, 091 D4 = $11, 491 D5 = $8, 216 9.28 Given: I = $50,000, tax depreciation method = 6-year MACRS property class with half-year convention

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Contemporary Engineering Economics, 7th ed. ©2023

n 1

200% DB Bn −1 Dn life $50, 000 $8,333 6.5

SL Dn $7692

MACRS Dn $8,333

2 3 4

$41, 667 $27, 778 $18,519

$13,889 $9, 259 $6,173

5.5 $7,576 $13,889 4.5 $6,173 $9, 259 3.5 $5, 291 $6,173

5 6

$12,346 $7, 407

$4,115 $2, 469

2.5 $4,938 1.5 $4,938

$4,938 $4, 938

7

$2, 469

$412

0.5 $2, 469

$2, 469

: Optimal time to switch

9.29 Since the land is not depreciable, just consider the building depreciations. Given: I = $250,000, tax depreciation method = 27.5-year MACRS property Depreciation Allowed n rate depreciation 1 2 3 4

2.5758% 3.6364% 3.6364% 3.6364%

$6,439 $9,091 $9,091 $9,091

5

3.1818%

$7,955

9.30 Given: I = $53,000 and 7-year MACRS property n Dn 1 $7,574 2 $12,980 3 $9, 270 4 $6, 620 5 $4, 733 6 $4, 728 7 $4, 733 8 $2,364

9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.31 Given: I machine tool = $5,000, I CNC machine = $125,000, and I warehouse = $335,000

 Machine tools n 0 1 2 3 4

Dep. Rate 0.3333 0.4444 0.1481 0.0741

Dn $1,667 $2,222 $741 $370

Bn $5,000 $3,333 $1,111 $370 $0

 CNC machine n 0 1 2 3 4 5 6 7 8

Dep. Rate 0.1429 0.2449 0.1749 0.1249 0.0892 0.0892 0.0892 0.0446

Dn $17,857 $30,612 $21,866 $15,618 $11,156 $11,156 $11,156 $5,578

Bn $125,000 $107,143 $76,531 $54,665 $39,046 $27,890 $16,734 $5,578 $0

Dep. Rate

…….

39 40

Dn

0.0256 0.0118

$4,653 $8,590 $8,590 …….

0.0139 0.0256 0.0256

Bn $335,000 $330,347 $321,757 $313,168

…..

n 0 1 2 3 …….

 Warehouse

$8,590 $3,937

10 Copyright © 2023 Pearson Education, Inc.

$3,937 $0

Contemporary Engineering Economics, 7th ed. ©2023 9.32 Given: Residential real property (27.5-year), I = $270,000 (a)  100%  2.5 D1 =    27.5  12 = (0.00758)($270, 000) = $2, 045 (b) Total amount of depreciation over the 4-year ownership, assuming that the asset is sold at the end of 4th calendar year: n Rate Dn

1 0.7576% $2,045 2 3.6364% $9,818 3 3.6364% $9,818 4 3.4848% $9, 409 Total amount of depreciation allowed = $31,091. Note that the 4th year depreciation reflects the mid-month convention (11.5 months).

B4 = $270,000 − $31,091 + $80,000(land) = $318,909 9.33 Types of Assets Depreciating Methods End of year Initial Cost ($) Salvage value ($) Book value ($) Depreciation life Depreciable Amount ($) Accu. Dep. Amount ($)

I SL 6 30,000 6,000 12,000 8 yr 3,000 18,000

II DDB 3 25,000 5,000 5,400 5 yr 3,600 19,600

III UP 3 41,000 5,000 23.000 90,000 mi 6,000 18,000

9.34 (a) Book depreciation methods:

•

Straight-line method: n

1 2 3 4 5

Dn

Bn

Cum. Dn

$15,800

$73,200

$15,800

$15,800 $15,800

$57,400 $41,600

$31,600 $47,400

$15,800

$25,800

$63,200

$15,800

$10,000

$79,000

11 Copyright © 2023 Pearson Education, Inc.

IV MACRS 4 20,000 2,000 3,456 5 yr 2,304 16,544

Contemporary Engineering Economics, 7th ed. ©2023

•

DDB method: n

1 2 3 4 5

Dn

Bn

Cum. Dn

$35,600

$53,400

$35,600

$21,360 $12,816 $7,690 $1,534

$32,040 $19,224 $11,534 $10,000

$56,960 $69,776 $77,466 $79,000

(b) Tax depreciation: 7-year MACRS n

Dn

1 2 3 4 5 6 7 8

Bn

Cum. Dn

$12,718

$76,282

$12,718

$21,796 $15,566

$54,486 $38,920

$34,514 $50,080

$11,116

$27,804

$61,196

$7,948 $7,939

$19,856 $11,917

$69,144 $77,083

$7,948

$3,969

$85,031

$3,969

$0

$89,000

(c) Trade-in allowance Book value of the old equipment (B3)

$38,920

Less: Trade-in allowance

$20,000

Unrecognized loss Cost of new equipment Plus: Unrecognized loss on trade-in Cost basis of new lathe

($18,920) $92,000 $18,920 $110,920

Comments: If the old equipment were sold on the market (instead of trade-in), there would be no unrecognized loss. In that situation, the cost basis for the new equipment would be just $92,000. No half-year convention is assumed in this analysis.

12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.35 Given: I = $1,000,000, 39 years-MACRS real property n 0 1 2

Dep. Rate

Dn

0.016026 0.025641

$16,026 $25,641

Bn $1,000,000 $983,974 $969,017

9.36 Types of depreciation method (a) (b) (c) (d) (e)

B A D C None

Depletion 9.37 (a)

 Ore mine: Depletion rate per ton =

($8,900, 000 − $1,500, 000) = $1.85per ton 4, 000,000

 Mining equipment: Depreciation rate per ton =

$2,500, 000 = $0.625 per ton 4, 000, 000

(b)

 For tax year 2022: Depletion expense = $1.85(550,000) = $1,017,500 Depreciation expenses = $0.625(550,000) = $343,750

 For tax year 2023: Depletion expense = $1.85 (688,000) = $1,272,800 Depreciation expenses = $0.625 (688,000) = $430,000

Note: Depletion rate = (depletion base – Salvage value)/total units to be recovered

13 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 9.38

• •

$600, 000 = $88, 235.29 per MBF 6.8 Total depletion allowance = $88, 235.29(1.5) = $132, 352.94

Depletion allowed per MBF =

9.39 Percentage depletion versus cost depletion: Gross Income Depletion Computed % depletion

•

$48,365,000

× 15% $7,254,750

Percentage depletion: Gross Income Expenses Taxable income Deduction limit Maximum depletion deduction

$48,365,000 $22,250,000 $26,115,000

× 50% $13,057,500

The allowable percentage deduction is $7,254,750.

•

Cost depletion =

$80, 000, 000 (52,000) = $8,320, 000 500, 000

The cost depletion is more advantageous than the percentage depletion.

Note: If no salvage is given in the problem, we view the restoration expenses would offset any salvage value recovered after the depletion activities. 9.40 (a) Cost basis:

 Parcel A:

$39, 000, 000 = $4.33 per bbl 9, 000, 000

 Parcel B:

$24,000, 000 = $4.80 per bbl 5, 000, 000

(b) Depletion charge for parcel A:

 Cost depletion: $4.33(1, 200, 000) = $5,196, 000

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Contemporary Engineering Economics, 7th ed. ©2023

 Percentage depletion: - Gross income = $60 × 1,200,000 = $72,000,000 Gross Income $72,000,000 Depletion × 15% Computed % depletion

$10,800,000

Gross Income Expenses

$72,000,000 $3,600,000

Taxable income Deduction limit

$68,400,000

× 50%

Maximum depletion deduction

$34,200,000 The allowable percentage deduction is $10,800,000. (c) Percentage depletion versus cost depletion for parcel A in year 2022:

Cost depletion: $4.33(1, 000, 000) = $4, 330, 000  Percentage depletion: - Gross income = $75 × 1,000,000 = $75,000,000 Gross Income Depletion Computed % depletion

$75,000,000 × 15% $11,250,000

Gross Income Expenses

$75,000,000 $3,600,000

Taxable income Deduction limit

$71,400,000

× 50%

Maximum depletion deduction

$35,700,000 The allowable percentage deduction is $11,250,000. (d) Percentage depletion versus cost depletion for parcel B in year 2022

Cost depletion: $4.80(800, 000) = $3,840, 000  Percentage depletion:

15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 - Gross income = $75 × 800,000 = $60,000,000 Gross Income Depletion

$60,000,000

× 15% $9,000,000

Computed % depletion Gross Income Expenses

$60,000,000 $3,000,000

Taxable income Deduction limit

$57,000,000

Maximum depletion deduction

$28,500,000

× 50%

The allowable percentage deduction is $9,000,000. During year 2022, Oklahoma Oil claimed its depletion deduction in the amount of $9,000,000 from parcel B. Book value at the beginning of year 2023: $24,000,000-$9,000,000 = $15,000,000 The revised cost per bbl. is

$15, 000, 000 = $3.75 per bbl 4, 000, 000 Since no gross income figure is available during year 2023, we may calculate the depletion charge based on unit cost $3.75(1, 000, 000) = $3, 750, 000

9.41 (a) Cost depletion:

$30,000,000 = $4.6154 per ton 6,500,000 Depletion cost = $4.6154(1, 000, 000) = $4, 615, 400 Cost per ton =

(b) Percentage depletion: Gross Income Depletion 16 Copyright © 2023 Pearson Education, Inc.

$15,000,000

× 10%

Contemporary Engineering Economics, 7th ed. ©2023 Computed % depletion

$1,500,000

Gross Income Expenses

$15,000,000 $1,850,000

Taxable income Deduction limit

$13,150,000

Maximum depletion deduction

× 50% $6,575,000

The allowable percentage deduction is $1,500,000.

Revision of Depreciation Rates 9.42 (a) D = $800, 000 / 25 = $32, 000 (b) B = $400, 000 + $125, 000 = $525, 000 (c)

$800,000 − $400, 000 = 12.5 $32,000 Remaining years = (25 - 12.5) + 10 = 22.5 years n=

D = $525, 000 / 22.5 = $23, 333 (d) Depreciation rate for 12.5 years (June of 12th year): 2.5641%

D12.5 = ($800, 000 + $125, 000 )(0.025641) = $23, 718 9.43 (a) Book depreciation amount for 2023:

B2023 = $180, 000 − 5($18, 000) = $90, 000 revised depreciation basis = $90, 000 + $45, 000 = $135, 000 revised useful life = 10 years D2023 = $135, 000 /10 = $13,500 (b) Tax depreciation amount for 2023:

 Depreciation schedule for the original machine: 17 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

D2023 = $180,000(0.0446) = $8,028 ← (8th year) Depreciation schedule for the improvement (treated as a separate 7-year MACRS property): D2023 = $45,000(0.1749) = $7,871 ← (3rd year after improvement) Total tax depreciation for 2023: D2023 = $8,028 + $7,871 = $15,899 9.44 Given: Cost basis = $85,000 + $4,500 = $89,500 (a) Book depreciation schedule (Depreciation basis = $89,000) n

Dn

2020 2021 2022 2023

Bn

Cum. Dn

$10,688

$78,813

$10,688

$10,688 $10,688 $10,688

$68,125 $57,438 $46,750

$21,375 $32,063 $42,750

(b) Tax depreciation schedule (Depreciation basis = $89,500) n

Dn

2020 2021 2022 2023

Bn

Cum. Dn

$12,790

$76,710

$12,790

$21,919 $15,654 $11,179

$54,792 $39,138 $27,960

$34,708 $50,362 $61,540

Comments: The accessories costing $5,000 that were incurred in 2022 do not change the depreciation schedule, because these neither extended the machine’s life nor resulted in any additional salvage value. 9.45 Net income calculation:

18 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Gross income Expenses: Sarlaries Wages Depreciation Loan interest

$

34,000,000

$ $ $ $

5,000,000 4,000,000 1,000,000 210,000

Taxable income Income Taxes(21%)

$ $

23,790,000 4,995,900

Net income

$

18,794,100

a) The marginal tax rate: 21% b) The average (effective) tax rate: 21% c) The net income: $18,794,100 9.46 (a) Depreciation expenses: •

Building (39-year class, placed in service in February):  100%   10.5  Dbuilding = $400, 000     39   12  = $400, 000(2.2436%) = $8,974

•

Equipment (5-year MACRS):

Dequipment = $200, 000 (20% ) = $40, 000 •

Total depreciation allowed in year 2019:

D = $8,974 + $40,000 = $48,974 (b) Tax liability: Sales revenue Expenses: Cost of goods sold Bond interest Depreciation Taxable income Income taxes Net income

$2,500,000 $800,000 $50,000 $48,974 $1,601,026 $336,215 $1,264,810

Note: Income taxes = $1,601,026(0.21) = $336,215 19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Corporate Tax Systems Note: As of 2023, the U.S corporate tax rate remains at 21%. 9.47 Net income calculation: Gross income Expenses: Salaries Wages Depreciation Loan interest Taxable income Income taxes (21%)

$

35,000,000

$ $ $ $ $ $

6,000,000 7,000,000 800,000 150,000 21,050,000 4,420,500

Net income

$

16,629,500

9.48 (a) Taxable income = $8,500,000 - $2,280,000 - $456,000 = $5,764,000 (b) Income tax calculation using tax formula Income taxes =$5,764,000(0.21) = $1,210,440 9.49 (a) Income tax liability: Gross revenues Expenses: Manufacturing Operating Interest

$ 3,500,000

Taxable operating income Adjustment: loss

$ 2,490,000 $ 15,000

Taxable income Income taxes (21%)

$ 2,475,000 $ 519,750

Net income

$ 1,955,250

$ $ $

650,000 320,000 40,000

Note: book loss = $60,000 - $75,000 = ($15,000) (b) Operating income: (capital gains or losses are not a part of operating activities) Taxable operating income Income taxes (21%)

$ 2,490,000 $ 522,900

Net operating income

$ 1,967,100

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Gains or Losses 9.50

Allowed depreciation = $200, 000(0.1429 + 0.2449 + 0.1749 + 0.1249 / 2) = $125, 030 Book value = $200, 000 − $125, 030 = $74,970 Gain = $120, 000 − $74,970 = $45, 030 Net proceeds = $120, 000 − $45, 030(0.25) = $108, 743

9.51 (a) Disposed of in year 3:

Allowed depreciation = $80,000(0.20 + 0.32 + 0.192 / 2) = $49, 280 Book value = $80,000 − $49, 280 = $30,720 Gains = $40,000 − $30,720 = $9, 280 (b) Disposed of in year 5: Allowed depreciation = $80, 000(0.20 + 0.32 + 0.192 + 0.1152 + 0.1152 / 2) = $70, 784 Book value = $80, 000 − $70, 784 = $9, 216 Taxable gains = $30, 000 − $9, 216 = $20, 784

(c) Disposed of in year 6: Allowed depreciation = $80, 000 Book value = $0 Taxable gains = $10, 000

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Contemporary Engineering Economics, 7th ed. ©2023 9.52

Allowed depreciation = $350, 000(0.1429 + 0.2449 + 0.1749 +0.1249 + 0.0893 / 2) = $256, 288 Book value = $350, 000 − $256, 288 = $93, 713

(a) If sold at $20,000: Loss = $20, 000 − $93, 713 = ($73, 713) Loss credit = $73, 713(0.34) = $25, 062

(b) If sold at $99,000: Gains = $99, 000 − $93, 713 = $5, 287 Gains tax = $5, 287(0.34) = $1, 798

9.53 Allowed depreciation = $50, 000(0.2 + 0.32 + 0.192 + 0.1152 / 2) = $38, 480 Book value = $50, 000 − $38, 480 = $11,520

9.54 Given: I = $80,000, S = $20,000, N = 8 years. (a) Book value on December 31, 2023, using SL depreciation method n

2020 2021 2022 2023

Dn

Bn

Cum. Dn

$7,500

$72,500

$7,500

$7,500 $7,500 $7,500

$65,000 $57,500 $50,000

$15,000 $22,500 $30,000

(b) Depreciation amount for 2023 using DDB n

2020 2021 2022 2023

Dn

Bn

Cum. Dn

$20,000

$60,000

$20,000

$15,000 $11,250 $8,437.50

$45,000 $33,750 $25,312.50

$35,000 $46,250 $54,687.50

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(c) Optimal time to switch year 2024 n

Dn

2020 2021 2022 2023 2024

Bn

Cum. Dn

$20,000

$60,000

$20,000

$15,000 $11,248 $8,440

$45,000 $33,750 $25,313

$35,000 $46,250 $54,688

$1,328

$23,985

$56,016

(d) Taxable gains: Allowed depreciation = $80, 000(0.1429 + 0.2449 + 0.1749 + 0.1249 / 2) = $50, 012 Book value = $80, 000 − $50, 012 = $29,988 Taxable gains = $38, 000 − $29,988 = $8, 012

9.55 (a) Taxable operating income (Do not include ordinary gains):

Revenues: Gross income Expenses: Labor Materials Depreciation Office supplies Interest Rental

$ 4,250,000

Taxable income Income taxes (21%)

$ 1,550,000 $ 785,000 $ 332,500 $ 15,000 $ 42,200 $ , 45,000 , $ 1,480,300 $ 310,863

Net income

$$ 1,169,437

Taxable income for 2018: $1,480,300 (b) Taxable gains: $43,000 - $30,000 = $13,000 23 Copyright © 2023 Pearson Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023 (c) Totaal taxes: incoome taxes = $1, 480,3000(0.21) = $310,863 ggain taxes = (0.21)($13, 000) = $2, 730 tootal taxes = $310,863 + $2, 730 = $313,593

9.566 (a) Increemental Opeerating incoome:

Revenue Expenses: Mfg. cost O&M cossts Depreciattion Taxable inccome Income taxxes (25%) Net incomee

Operatingg Costs Year 1 Year 2 $15,000,000 $15,000,0000 $$6,000,000 $ $1,200,000 $714,500 $ $7,085,500 $ $1,771,375 $ $5,314,125

$6,000,0000 $1,200,0000 $1,224,5500 $6,575,5500 $1,643,8875 $4,931,6625

Year 3 Year 4 Yeaar 5 $15,000,0000 $15,0000,000 $15,0000,000 $6,000,000 $1,200,000 $874,500 $6,925,500 $1,731,375 $5,194,125

$6,0000,000 $1,2000,000 $624,500 $7,175,500 $1,793,875 $5,381,625

(b) Gainns or losses::

Total depreciation = $33, 661, 250 B5 = $55, 000, 000 − $3, 661, 2500 = $11,338, 750 Taxablee gains = $11, 600, 000 − $1,338, 7500 = $2261, 250

24 Copyright © 2023 Pearson E Education, Inc.

$6,0000,000 $1,2000,000 $2223,250 $7,5776,750 $1,8994,188 $5,6882,562

Contemporary Engineering Economics, 7th ed. ©2023

Marginal Tax Rate in Project Evaluation 9.57 (a) Economic depreciation for the milling machine $200, 000 − $50, 000 = $150, 000

(b) Marginal tax rates with the project:

1

$80,000

$28,580

Taxable income $51,420

2

$80,000

$48,980

$31,020

$456,020

21%

3

$80,000

$34,980

$45,020

$470,020

21%

4

$80,000

$24,980

$55,020

$480,020

21%

5

$80,000

$17,860

$62,140

$487,140

21%

6

$80,000

$8,920

$71,080

$496,080

21%

n

Revenue

Dn

Combined Marginal income rate $476,420 21%

(c) Average tax rates n 1 2 3 4 5 6

Combined income $476,420 $456,020 $470,020 $480,020 $487,140 $496,080

Combined income taxes $100,048 $95,764 $98,704 $100,804 $102,299 $99,216

Average tax rate 21% 21% 21% 21% 21% 21%

Note: The average tax rate and marginal tax rate remain at 21% under the flat tax rate system. 9.58 Incremental tax rate calculation: Year 1 Revenue Operating costs Depreciation Taxable income

$220,000 $150,000 $12,000 $58,000

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Year 2 $220,000 $150,000 $19,200 $50,800

Contemporary Engineering Economics, 7th ed. ©2023 Year 1 $650,000

Year 2 $650,000

$136,500

$136,500

Taxable income with project Income taxes

$708,000 $148,680

$700,800 $147,168

Incremental taxable income Incremental income taxes Incremental tax rate (%)

$58,000 $12,180 21%

$50,800 $10,668 21%

Taxable income without project Income taxes (21%)

Comment: Note that the marginal tax rates over the project life remain unchanged because the additional income from the new project is still taxed at a flat 21% rate. This would be a different story for a progressive tax rate system – in that case, we should make sure the additional income from the project may push the company into a higher tax bracket. 9.59

Good Taxable income Before expansion Due to expansion After expansion Income Taxes

$ $ $ $

Economic condition Fair

2,500,000 $ 2,000,000 $ 4,500,000 $ 945,000

Poor

2,500,000 $ 500,000 $ 3,000,000 $

2,500,000 (100,000) 2,400,000

630,000 $

504,000

$

(a) Marginal tax rate

21%

21%

21%

(b) Average tax rate

21%

21%

21%

9.60 Incremental tax calculations: (a) Additional taxable income due to project: Year 1

Year 2

Year 3

Annual revenue

$80,000

$80,000

$80,000

Operating cost

$20,000

$20,000

$20,000

Depreciation

$16,665

$22,225

$3,703

Taxable income

$43,335

$37,775

$56,298

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(b) Additional income tax calculation: Year 1

Year 2

Year 3

$350,000

$350,000

$350,000

$73,500

$73,500

$73,500

Taxable income with project Income taxes

$393,335 $82,600

$387,775 $81,433

$406,298 $85,323

Incremental taxable income Incremental income taxes Incremental tax rate

$43,335 $9,100 21%

$37,775 $7,933 21%

$56,298 $11,823 21%

Taxable income without project Income taxes

(c) Gains tax: Total depreciation = $42,593 Book value = $50, 000 − $42,593 = $7, 408 Taxable gains = $10, 000 − $7, 408 = $2,593 Gains tax = (0.21)($2,593) = $545

Combined Marginal Income Tax Rate 9.61 (a) Explicit calculation of state income taxes: State taxable income = $3,500, 000 − $1,800, 000 = $1, 700, 000 State income taxes = $1, 700, 000(0.05) = $85, 000 Federal taxable income = $1, 700, 000 − $85, 000 = $1, 615, 000 Federal income taxes = $1, 615, 000(0.25) = $403, 750

Combined taxes = $85,000 + 403,750 = $488,750 (b) Tax calculation based on the combined tax rate: Combined tax rate = 0.25 + 0.05 − (0.05)(0.25) = 28.75%

Combined taxes = (0.2875) ($1,700,000) = $488,750

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9.62 (a) Marginal tax rates:

State taxable income = $6,500, 000 − $3, 450, 000 − $650, 000 = $2, 400, 000 $193,120 State tax rate = = 8.05% $2, 400, 000 Federal taxable income = $2, 400, 000 − $193,120 = $2, 206,880

Federal tax rate = (b)

$463, 445 = 21% $2, 206,880

Combined marginal tax rate = 0.21 + 0.0805 − (0.0805)(0.21) = 27.36%

9.63 (a) Additional annual taxable income due to expansion = $300,000 Taxable income in year 1 = $1,700,000 + $300,000 = $2,000,000

The marginal tax rate after business expansion is 21%. (b) Average tax rate after business expansion $420,000/ $2,000,000 = 21% Note 1: Income taxes = 0.21($2,000,000) = $420,000 Note 2: The average tax rate will be the same as the marginal tax rate under the flat tax rate system in the U.S. (c) PW of income taxes:

 Depreciation schedules: depreciation base = $200,000 n 1 2 3 4

MACRS $ 66,660 $ 88,900 $ 29,620 $ 14,820

 Incremental income taxes under 3-year MACRS

Year 1

Operating Year Year 2

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Year 3

Contemporary Engineering Economics, 7th ed. ©2023 Revenue Expense Depreciation Taxable income Income taxes

$500,000 $200,000 $66,660

$500,000 $200,000 $88,900

$500,000 $200,000 $14,820

$233,340

$211,100

$285,180

$49,001

$44,331

$59,888

PW(10%) = $49,001(P/F,10%,1) + $44,331(P/F,10%,2) + $59,888(P/F,10%,3) = $126,178 9.64 (a) and (b) n 1 2 3 4 5 6 7 8

Dn

(a) Bn−1

$ 500,150 $ 857,150 $ 612,150 $ 437,150 $ 312,550 $ 312,200 $ 312,550 $ 156,100

$3,500,000 $2,999,850 $2,142,700 $1,530,550 $1,093,400 $780,850 $468,650 $156,100

(b) taxes $42,000 $35,998 $25,712 $18,367 $13,121 $9,370 $5,624 $1,873

Short Case Studies ST 9.1 Given: I = $82,000 + $3,000 = $85,000, N = 10 years, S = $3,000 •

Book depreciation expenses for 2016: D2016 =

$85, 000 − $3, 000 = $8, 200 10

n

Dn

Bn

2016 2017 2018

$8,200 $8,200 $8,200

$76,800 $68,600 $60,400

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Contemporary Engineering Economics, 7th ed. ©2023 •

Book depreciation expenses for 2019: New depreciation basis = $60, 400 + $8, 000 = $68, 400 Remaining useful life=10 years Salvage value = $3, 000

∴ D2019 =

•

$68, 400 − $3, 000 = $6,540 10

n

Dn

Bn

2019 2020 2021

$6,540 $6,540 $6,540

$61,860 $55,320 $48,780

Book depreciation expenses for 2022: New depreciation basis = $48, 780 + $5, 000 = $53, 780 Remaining useful life=7 years Salvage vlaue = $6, 000 $53, 780 − $6, 000 ∴ D2022 = = $6,825.71 7

ST 9.2 (a) Depletion basis = $32.5 million - $3 million = $29.5 million $29,500,000 = $4.54 per bbl Depletion allowance per bbl. = 6,500, 000 Cost depletion for 2022 = $4.54 / bbl × 420, 000 bbl = $1, 906,800 Cost depletion for 2023 = $4.54 / bbl × 510, 000 bbl = $2, 315, 400 (b) Depreciation basis = equipment cost + pipeline cost = $3,360,000 $3,360,000 = $0.517 per bbl Depreciation allowance per bbl. = 6,500,000 Cost depreciation for 2022 = $0.517 / bbl × 420, 000 bbl = $217,140 Cost depreciation for 2023 = $0.517 / bbl × 510, 000 bbl = $263, 670

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Contemporary Engineering Economics, 7th ed. ©2023 ST 9.3 (a) Incremental Operating income: Year

1

Revenue Mfg. Cost Depreciation O&M

2

3

4

5

$15,000,000 $15,000,000 $15,000,000 $15,000,000 $15,000,000 6,000,000 6,000,000 6,000,000 6,000,000 6,000,000 714,500 1,224,500 874,500 624,500 223,250 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000

Taxable Income Income Tax

7,085,500 1,771,375

6,575,500 1,643,875

6,925,500 1,731,375

7,175,500 1,793,875

7,576,750 1,894,188

Net Income

5,314,125

4,931,625

5,194,125

5,381,625

5,682,563

(b) Gains or losses:

Total depreciation = $3, 661, 250 B5 = $5, 000, 000 − $3, 661, 250 = $1,338, 750

Taxable gains = $1, 600, 000 − $1,338, 750 = $261, 250 ST 9.4 (a) If Diamond invests in the facilities and markets the product successfully, the expected tax rate in each year will remain at 21%. Since the local and state taxes are tax-deductible expenses for federal tax calculation purpose, the combined marginal tax rate is

tm = 0.21 + 0.05 − (0.05)(0.21) = 24.95% (b) Gains or losses •

Plant (39-year MACRS):

Total depreciation = (2.4573% + 2.5641% +  + 2.4573%)($10, 000, 000) = $2, 029,913 B8 = $10, 000, 000 − $2, 029,913 = $7,970, 087 Losses = $6, 000, 000 − $7,970, 087

•

= ($1,970, 087) Equipment (7-year MACRS):

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Total depreciation = $40, 000, 000 B8 = 0

•

Ordinary gains = $4, 000, 000 Net gains:

Net gains = $4, 000, 000 − $1,970, 087 = $2, 029,913 Gains tax(24.95%) = $2, 029,913 × 0.2495 = $506, 463 Net proceeds from sales = $10, 000, 000 − $506,963 = $9, 493,537

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(c) Net operating income (unit: $1,000) n

1 $30,000

2 $30,000

3 $30,000

4 $30,000

5 $30,000

6 $30,000

7 $30,000

8 $30,000

9,000 12000

9,000 12000

9,000 12000

9,000 12000

9,000 12000

9,000 12000

9,000 12000

9,000 12000

246 5716

256 9796

256 6996

256 4996

256 3572

256 3568

256 3572

246 1784

Taxable Income for State State Income taxes (5%) Taxable Income for Federal Federal Income taxes (21%)

$3,038 152 $2,886 606

($1,052) (53) ($999) (210)

$1,748 87 $1,661 349

$3,748 187 $3,561 748

$5,172 259 $4,913 1,032

$5,176 259 $4,917 1,033

$5,172 259 $4,913 1,032

$6,970 349 $6,622 1,391

Net Income

$2,280

($790)

$1,312

$2,813

$3,882

$3,885

$3,882

$5,231

Revenue Expenses : Mfg. cost Operating cost Depreciation Building Equipment

Corporate operating losses: Ordinary operating losses (say, year 2) can be carried back to each of the preceding 3 years and forward for the following 15 years, and can be used to offset taxable income in those years. In our example, the taxable income during the first year is large enough to offset the operating loss during the second year, so that the corporation will get a federal tax refund in the amount of $ 210,000 plus $53,000 (state tax refund) at the end of year 2.

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Chapter 10 Developing Project Cash Flows Note: The U.S. corporate tax rate is 21% flat in 2022, but considering the state and local income taxes, the combined marginal tax rate ranges between 21% and 30%. However, the US Congress has been pondering on raising the corporate income tax for some time, so a higher tax rate such as 35% or even 40% would be possible in the future.

Generating Net Cash Flows 10.1 0

1

Income Statement Revenues Expenses O&M Depreciation Taxable Income Income Taxes (28%) Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(15%)

2

3

4

5

6

$35,000

$35,000

$35,000

$35,000

$35,000

$35,000

$12,000 $13,000

$12,000 $20,800

$12,000 $12,480

$12,000 $7,488

$12,000 $7,488

$12,000 $3,744

$10,000 $2,800

$2,200 $616

$10,520 $2,946

$15,512 $4,343

$15,512 $4,343

$19,256 $5,392

$7,200

$1,584

$7,574

$11,169

$11,169

$13,864

$7,200 $13,000

$1,584 $20,800

$7,574 $12,480

$11,169 $7,488

$11,169 $7,488

$13,864 $3,744

($65,000) $0 $0 ($65,000) $10,232

$20,200

$22,384

$20,054

$18,657

Since PW(15%) > 0, This CNC machine should be purchased.

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$18,657

$17,608

Contemporary Engineering Economics, 7th ed. ©2023

10.2 Investment in industrial robot: 0

1

Income Statement Revenues Expenses O&M Depreciation Taxable Income Income Taxes (35%) Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(15 %)

2

3

4

5

6

7

$150,000

$150,000

$150,000

$150,000

$150,000

$150,000

$150,000

$42,000

$67,200

$40,320

$24,192

$24,192

$12,096

$0

$108,000 $37,800

$82,800 $28,980

$109,680 $38,388

$125,808 $44,033

$125,808 $44,033

$137,904 $48,266

$150,000 $52,500

$70,200

$53,820

$71,292

$81,775

$81,775

$89,638

$97,500

$70,200 $42,000

$53,820 $67,200

$71,292 $40,320

$81,775 $24,192

$81,775 $24,192

$89,638 $12,096

$97,500 $0

($210,000) $60,000 ($21,000) ($210,000) $261,030

$112,200

$121,020

$111,612

$105,967

$105,967

$101,734

$136,500

10.3 The income and cash flow statements are as below: 0

1

Income Statement Revenues Expenses O&M Depreciation

2

3

4

5

$3,833,542

$3,833,542

$3,833,542

$3,833,542

$3,833,542

$435,000 $294,872

$485,000 $307,692

$535,000 $307,692

$585,000 $307,692

$635,000 $294,872

Taxable Income Income Taxes (35%)

$3,103,670 $1,086,285

$3,040,850 $1,064,297

$2,990,850 $1,046,797

$2,940,850 $1,029,297

$2,903,670 $1,016,285

Net Income

$2,017,386

$1,976,552

$1,944,052

$1,911,552

$1,887,386

$2,017,386 $294,872

$1,976,552 $307,692

$1,944,052 $307,692

$1,911,552 $307,692

$1,887,386 $294,872

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment (Building) Investment(Land) Salvage (Building) Salvage (Land) Gains Tax Net Cash Flow PW(15 %) Decision Annual rate

($12,000,000) ($3,500,000) $13,500,000 $3,500,000 ($1,054,487) ($15,500,000) $2,312,257 $0.00 Reject $76,670.84

$2,284,244

$2,251,744

per apartment

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$2,219,244

$18,127,770

Contemporary Engineering Economics, 7th ed. ©2023

10.4 Net cash flow:

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Contemporary Engineering Economics, 7th ed. ©2023 10.5 Investment in an answering device: •

Depreciation: It is assumed that the building will be placed in service in January during the first project year. Then, it will be depreciated based on 39year MACRS.

•

Project cash flows 0

1

Income Statement Revenues (savings) Expenses: O&M costs Depreciation : Building Equipment Taxable Income Income Taxes (28%) Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Land Building Machines Gains Tax Land Building Equipment Net Cash Flow

2

3

4

$2,500,000

$2,500,000

$2,500,000

$2,500,000

$2,500,000

1,280,000

1,280,000

1,280,000

1,280,000

1,280,000

12,287 71,450 1,136,264 318,154

12,821 122,450 1,084,730 303,724

12,821 87,450 1,119,730 313,524

12,821 62,450 1,144,730 320,524

12,287 22,325 1,185,389 331,909

$818,110

$781,005

$806,205

$824,205

$853,480

$818,110 $83,737

$781,005 $135,271

$806,205 $100,271

$824,205 $75,271

$853,480 $34,612

($100,000) ($500,000) ($500,000)

5

115000 575000 50000 ($4,200) ($38,650) $23,484

($1,100,000) IRR =

$901,846 81%

$916,276

10.6

X = $60, 000( P / A,15%,10) = $301,128

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$906,476

$899,476

$1,608,726

Contemporary Engineering Economics, 7th ed. ©2023

10.7 0

1

Income Statement Revenues (savings) Expenses: Required annual digging (ft) Number of hours to operate Operating cost (@$10/hr) Depreciation

2

3

4

5

$50,000

$50,000

$50,000

$50,000

$50,000

8,000 500 $5,000 $60,000

8,000 500 $5,000 $96,000

8,000 500 $5,000 $57,600

8,000 500 $5,000 $34,560

8,000 500 $5,000 $17,280

Taxable Income Income Taxes (25%)

($15,000) ($3,750)

($51,000) ($12,750)

($12,600) ($3,150)

$10,440 $2,610

$27,720 $6,930

Net Income

($11,250)

($38,250)

($9,450)

$7,830

$20,790

($11,250) $60,000

($38,250) $96,000

($9,450) $57,600

$7,830 $34,560

$20,790 $17,280

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

($300,000) $100,000 ($16,360) ($300,000)

$48,750

IRR= PV(15%)= $

1.8% (97,534) < 0

$57,750

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$48,150

$42,390 BV=

Not Acceptable

$121,710 $34,560

Contemporary Engineering Economics, 7th ed. ©2023

10.8 (a) (a) 0

1

Income Statement Revenues (savings) Expenses: Operating Expenses Depreciation

2

3

4

5

6

$66,000

$70,000

$74,000

$80,000

$64,000

$50,000

29,000 10,800

28,400 17,280

32,000 10,368

38,800 6,221

31,000 6,221

25,000 3,110

Taxable Income Income Taxes (25%)

$26,200 6,550

$24,320 6,080

$31,632 7,908

$34,979 8,745

$26,779 6,695

$21,890 5,472

Net Income

$19,650

$18,240

$23,724

$26,234

$20,084

$16,417

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment $ Salvage Gains Tax Net Cash Flow

(b)

$ $

19,650 10,800

$ $

18,240 17,280

$ $

23,724 10,368

$ $

26,234 6,221

$ $

20,084 6,221

$ $

16,417 3,110

$ $

8,000 (2,000)

(54,000)

($54,000) NPV=

$30,450 $74,255

$35,520

$34,092 AE(12%)=

AE(12%) = ( $74, 255 ) × ( A / P /12%, 6) = $18, 061

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$32,455 $18,061

$26,305

$25,528

Contemporary Engineering Economics, 7th ed. ©2023

10.9 (a) Investment in Mazda automatic screw machine: Input Tax Rate(%) = MARR(%) =

Output PW(i) = IRR(%) =

25 15

0

1

2

$53,441 40.82%

3

4

5

6

Income Statement Revenues (savings) Expenses: Depreciation

$38,780

$38,780

$38,780

$38,780

$38,780

$38,780

9,817

16,825

12,016

8,581

6,135

3,064

Taxable Income Income Taxes (25%)

$28,963 7,241

$21,955 5,489

$26,764 6,691

$30,199 7,550

$32,645 8,161

$35,716 8,929

Net Income

$21,722

$16,466

$20,073

$22,649

$24,484

$26,787

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

$ $ $

21,722 9,817

$ $

16,466 16,825

$ $

20,073 12,016

$ $

22,649 8,581

$ $

24,484 6,135

$ $

26,787 3,064

(68,701) $ 3,500 $ 2,190.78 ($68,701)

$31,539

$33,291

$32,089

(b) Since PW(15%) = $53,441 > 0, accept the investment. (c) IRR = 40.82% > 15%, accept the investment.

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$31,230

$30,619

$35,542

Contemporary Engineering Economics, 7th ed. ©2023

10.10 Investment in a new trench excavator: To justify the acquisition of the trench excavator, the firm must generate an after-tax annual revenue of $64,641.43 over five years. Income Statement Revenues Expenses Required annual digging(ft) Number of hours to operate Operating cost (@50/hr) Depreciation

6400 400 $20,000 $50,000

6400 6400 6400 400 400 457 $20,000 $20,000 $22,857 $80,000 $48,000 $28,800

6400 533 $26,667 $14,400

Taxable Income Income Taxes (35%)

($70,000) ($100,000) ($68,000) ($51,657) ($41,067) ($24,500) ($35,000) ($23,800) ($18,080) ($14,373)

Net Income

($45,500) ($65,000) ($44,200) ($33,577) ($26,693)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(15 %) AEC(15%) =

($45,500) ($65,000) ($44,200) ($33,577) ($26,693) $50,000 $80,000 $48,000 $28,800 $14,400 ($250,000) $60,000 ($10,920) ($250,000) $4,500 ($216,688) $64,641.43

$15,000

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$3,800 ($4,777) $36,787

Contemporary Engineering Economics, 7th ed. ©2023

10.11

0

1

2

3

4

5

Income Statement Revenues Expenses Software development O&M Depreciation

$20,000 $15,000 $15,000 $15,000 $15,000 $15,000 $20,800 $33,280 $19,968 $11,981 $5,990

Taxable Income Income Taxes (35%)

$16,200 $23,720 $37,032 $45,019 $51,010 $5,670 $8,302 $12,961 $15,757 $17,853

Net Income

$10,530 $15,418 $24,071 $29,262 $33,156

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(13%)

$72,000 $72,000 $72,000 $72,000 $72,000

$10,530 $15,418 $24,071 $29,262 $33,156 $20,800 $33,280 $19,968 $11,981 $5,990 ($104,000) 0 $4,193 ($104,000) $31,330 $48,698 $44,039 $41,243 $43,340 $41,202.89

Conclusion – yes, it is worth doing all in-house.

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10.12 Since the net margin is 7.45%, we can easily estimate the net income each year as follows: A = $4,236(0.0745)X where X is the sales volume.

0 Operating Activities Net Income Depreciation Investing Activities Investment Salvage Value Gain Taxes Net Cash Flow + Depreciation

1

2

3

4

5

6

7

8

4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 35,725,000 61,225,000 43,725,000 31,225,000 22,325,000 22,300,000 22,325,000 11,150,000 (250,000,000) 0 0 (250,000,000) 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 4236(0.0745)X 35,725,000 61,225,000 43,725,000 31,225,000 22,325,000 22,300,000 22,325,000 11,150,000

Note that we assumed a tax depreciation here (MACRS) but a book depreciation with straight-line could be more appropriate as we used a net margin to derive the net income.

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10.13 0

1

Income Statement Revenues Expenses Labor & Mat'l Costs Depreciation

4

5

6

$130,000 $130,000 $130,000 $130,000 $130,000 $130,000 $30,000 $48,000 $28,800 $17,280 $17,280 $8,640 $160,000 $142,000 $161,200 $172,720 $172,720 $181,360 $64,000 $56,800 $64,480 $69,088 $69,088 $72,544

Net Income

Net Cash Flow PW(12%)

3

$320,000 $320,000 $320,000 $320,000 $320,000 $320,000

Taxable Income Income Taxes (40%)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax

2

$96,000

$85,200

$96,720 $103,632 $103,632 $108,816

$96,000 $30,000

$85,200 $48,000

$96,720 $103,632 $103,632 $108,816 $28,800 $17,280 $17,280 $8,640

($150,000) $0 $0 ($150,000) $126,000 $133,200 $125,520 $120,912 $120,912 $117,456 $362,986.22

Note: A tax rate between 21% and 30% may be more reasonable in 2022, but the U.S. Congress has been working on raising the corporate tax rate for some time. A few problems in this chapter retain higher tax rates such as this one as an exercise left to the readers with this possible change in the future.

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10.14

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10.15 Investment in energy management system: N = 7 years 0 Income Statement Revenues Expenses O&M Depreciation

1

2

3

4

5

6

7

$14,000

$14,000

$14,000

$14,000

$14,000

$14,000

$14,000

$0 $18,332

$0 $24,448

$0 $8,146

$0 $4,076

$0 $0

$0 $0

$0 $0

Taxable Income Income Taxes (35%)

($4,332) ($10,448) ($1,516) ($3,657)

$5,855 $2,049

$9,925 $3,474

$14,000 $4,900

$14,000 $4,900

$14,000 $4,900

Net Income

($2,815)

($6,791)

$3,805

$6,451

$9,100

$9,100

$9,100

($2,815) $18,332

($6,791) $24,448

$3,805 $8,146

$6,451 $4,076

$9,100 $0

$9,100 $0

$9,100 $0

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(12%)

($55,000) $0 $0 ($55,000) $2,015.82

$15,516

$17,657

$11,951

$10,526

$9,100

$0 $0 $9,100

PW (12%) = −$55,000 + $15,516(P / F ,12%,1) + $17,657(P / F ,12%,2) + $11,951(P / F ,12%,3) +$10,526(P / F ,12%,4) + $9,100(P / A,12%, N ')(P / F ,12%,4) = −$11,874.43 + 5,782.21(P / A,12%, N ') =0 N ' = 3 → N = 4 + 3 = 7 years (Actually, it will be a little less than 7 years.) 13 Copyright © 2023 Pearson Education, Inc.

$0 $0 $9,100

Contemporary Engineering Economics, 7th ed. ©2023

10.16 Investment decision based on IRR after-tax: Required investment = $368,808

0 Income Statement Revenues Expenses O&M Depreciation

1

2

3

4

$150,000

$150,000

$30,000 $122,924

$30,000 $163,935

Taxable Income Income Taxes (28%)

($2,924) ($819)

($43,935) ($12,302)

$65,380 $18,306

$92,671 $25,948

$120,000 $33,600

Net Income

($2,105)

($31,633)

$47,073

$66,723

$86,400

($2,105) ($31,633) $122,924 $163,935

$47,073 $54,620

$66,723 $27,329

$86,400 $0

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(15 %)

$150,000 $150,000

5

$30,000 $54,620

$30,000 $27,329

$150,000 $30,000 $0

($368,808) $0 $227 ($368,808) ($0)

$120,819

$132,302

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$101,694

$94,052

$86,627

Contemporary Engineering Economics, 7th ed. ©2023

Investment in Working Capital 10.17 Investment in working capital further reduces the desirability of an investment project. Income Statement Revenues (savings) Expenses: Required annual digging (ft) Number of hours to operate Operating cost (@$10/hr) Depreciation

$50,000

$50,000

$50,000

$50,000

$50,000

8,000 500 $5,000 $60,000

8,000 500 $5,000 $96,000

8,000 500 $5,000 $57,600

8,000 500 $5,000 $34,560

8,000 500 $5,000 $17,280

Taxable Income Income Taxes (25%)

($15,000) ($3,750)

($51,000) ($12,750)

($12,600) ($3,150)

$10,440 $2,610

$27,720 $6,930

Net Income

($11,250)

($38,250)

($9,450)

$7,830

$20,790

($11,250) $60,000

($38,250) $96,000

($9,450) $57,600

$7,830 $34,560

$20,790 $17,280

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working Capital Net Cash Flow

($300,000) $100,000 ($16,360) $50,000

($50,000) ($350,000)

$48,750

IRR= PW(15%)= $

$57,750

1.5% (122,675) < 0

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$48,150

$42,390 BV=

Not Acceptable

$171,710 $34,560

Contemporary Engineering Economics, 7th ed. ©2023

10.18 0

1

Income Statement Revenues Expenses Production cost Depreciation: Building Equipment

2

3

4

5

6

7

8

9

10

$775,000

$775,000

$775,000

$775,000

$775,000

$775,000

$775,000

$775,000

$775,000

$775,000

$465,000

$465,000

$465,000

$465,000

$465,000

$465,000

$465,000

$465,000

$465,000

$465,000

$36,860 $74,308

$38,462 $127,348

$38,462 $90,948

$38,462 $64,948

$38,462 $46,436

$38,462 $46,384

$38,462 $46,436

$38,462 $23,192

$38,462 $0

$36,860 $0

Taxable Income Income Taxes (28%)

$198,833 $55,673

$144,191 $40,373

$180,591 $50,565

$206,591 $57,845

$225,103 $63,029

$225,155 $63,043

$225,103 $63,029

$248,347 $69,537

$271,539 $76,031

$273,141 $76,479

Net Income

$143,159

$103,817

$130,025

$148,745

$162,074

$162,111

$162,074

$178,809

$195,508

$196,661

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Land Building Machines Gains Tax Land Building Machines Working Capital Net Cash Flow PW(15%) IRR

$143,159 $111,168

$103,817 $165,810

$130,025 $129,410

$148,745 $103,410

$162,074 $84,898

$162,111 $84,846

$162,074 $84,898

$178,809 $61,654

$195,508 $38,462

($350,000) ($1,500,000) ($520,000)

$500,000 $800,000 $50,000 ($42,000) $89,205 ($14,000) $250,000

($250,000) ($2,620,000) $254,327 ($951,274)

$196,661 $36,860

$269,627

$259,435

$252,155

$246,971

-9%

(a) PW(15%) with working capital = −$951, 274 < 0, not acceptable. (b) PW(15%) without working capital = −$763, 070 < 0, still not acceptable.

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$246,957

$246,971

$240,463

$233,969

$1,866,726

Contemporary Engineering Economics, 7th ed. ©2023

10.19 (a) Net cash flow: 0

1

2

3

4

5

6

Income Statement Revenues Expenses Operating cost Depreciation

$65,800

$65,800

$65,800

$65,800

$65,800

$65,800

$9,150 $17,100

$9,150 $27,360

$9,150 $16,416

$9,150 $9,850

$9,150 $9,850

$9,150 $4,925

Taxable Income Income Taxes (35%)

$39,550 $13,843

$29,290 $10,252

$40,234 $14,082

$46,800 $16,380

$46,800 $16,380

$51,725 $18,104

Net Income

$25,708

$19,039

$26,152

$30,420

$30,420

$33,621

$25,708 $17,100

$19,039 $27,360

$26,152 $16,416

$30,420 $9,850

$30,420 $9,850

$33,621 $4,925

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Working Capital Net Cash Flow PW(18%)

($85,500) $5,000 ($1,750) $15,000

($15,000) ($100,500) $42,808 $54,420.65

$46,399

$42,568

(b) IRR = 37.20% > 18%, acceptable (c) PW(18%) = $54,420.65 > 0, acceptable.

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$40,270

$40,270

$56,796

Contemporary Engineering Economics, 7th ed. ©2023

10.20 (a) Net cash flows:

(b) IRR = 47.08% > 20%, acceptable (c) AE(20%) = $3,157,266

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Effects of Borrowing 10.21 (a) Equal repayment of the principal: n 0 1 2 3 4 5 6

Loan Interest $36,000 $30,000 $24,000 $18,000 $12,000 $6,000

Repayment Loan Principal Balance $300,000 $50,000 $250,000 $50,000 $200,000 $50,000 $150,000 $50,000 $100,000 $50,000 $50,000 $50,000 0

(b) Equal repayment of the interest: n 0 1 2 3 4 5 6

Loan Interest $36,000 $36,000 $36,000 $36,000 $36,000 $36,000

Repayment Loan Principal Balance $300,000 $300,000 $300,000 $300,000 $300,000 $300,000 $300,000 0

(c) Equal annual installment:

A = $300,000( A / P,12%,6) = $72,968 n 0 1 2 3 4 5 6

Loan Interest $36,000 $31,564 $26,000 $21,031 $14,798 $7,818

Repayment Loan Principal Balance $300,000 $36,968 $263,032 $41,404 $221,628 $46,373 $175,255 $51,937 $123,318 $58,170 $65,148 $65,148 0

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Contemporary Engineering Economics, 7th ed. ©2023 10.22 (a), (b), and (c)

Income Statement 0

1

2

3

4

5

$95,000

$95,000

$95,000

$95,000

$95,000

30,000 10,800

48,000 9,100

28,800 7,196

17,280 5,063

8,640 2,675

Taxable Income Income Taxes (35%)

$54,200 $18,970

$37,900 $13,265

$59,004 $20,651

$72,657 $25,430

$83,685 $29,290

Net Income Cash Flow Statement Cash from operation: Net Income Depreciation Investment / Salvage Gains Tax Loan repayment

$35,230

$24,635

$38,353

$47,227

$54,395

35,230 $ 30,000 $

24,635 $ 48,000 $

38,353 $ 28,800 $

90,000 $ (14,167) $

(15,867) $

(17,771) $

47,227 $ 17,280 $ $ $ (19,903) $

54,395 8,640 10,000 2,548 (22,292)

$56,768

$49,382

$44,603

$53,291

Income Statement Revenue Expenses: Depreciation Interest (12%)

Net Cash Flow (actual)

$ $ $ $

(150,000)

($60,000)

$51,063

PW (20%) = IRR =

$93,479 82.19%

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10.23 0

1

2

3

4

5

Income Statement Revenues Expenses O&M Depreciation Debt interest

$3,833,542

$3,833,542

$3,833,542

$3,833,542

$3,833,542

$435,000 $294,872 $1,550,000

$485,000 $307,692 $1,296,114

$535,000 $307,692 $1,016,839

$585,000 $307,692 $709,637

$635,000 $294,872 $371,715

Taxable Income Income Taxes (35%)

$1,553,670 $543,785

$1,744,736 $610,658

$1,974,011 $690,904

$2,231,213 $780,925

$2,531,955 $886,184

Net Income

$1,009,886

$1,134,078

$1,283,107

$1,450,288

$1,645,771

$1,009,886 $294,872

$1,134,078 $307,692

$1,283,107 $307,692

$1,450,288 $307,692

$1,645,771 $294,872

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment (Building) Investment(Land) Salvage (Building) Salvage (Land) Gains Tax Financial activities Borrowed funds Principal repayment Net Cash Flow PW(15 %)

($12,000,000) ($3,500,000) $13,500,000 $3,500,000 ($1,054,487) 15,500,000

$0 $3,048,956.65

($2,538,861)

($2,792,747)

($3,072,022)

($3,379,224)

($3,717,146)

($1,234,104)

($1,350,977)

($1,481,223)

($1,621,244)

$14,169,010

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10.24 0 Income Statement Revenues Expenses Depreciation Debt interest

1

2

3

4

5

6

7

$150,000 $150,000 $150,000 $150,000 $150,000 $150,000 $150,000 $42,000 $21,000

$67,200 $18,000

$40,320 $15,000

Taxable Income Income Taxes (35%)

$87,000 $30,450

$64,800 $22,680

$94,680 $113,808 $116,808 $131,904 $147,000 $33,138 $39,833 $40,883 $46,166 $51,450

Net Income

$56,550

$42,120

$61,542

$73,975

$75,925

$85,738

$95,550

$56,550 $42,000

$42,120 $67,200

$61,542 $40,320

$73,975 $24,192

$75,925 $24,192

$85,738 $12,096

$95,550 $0

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Financing Activities Borrowed funds Principal repayment Net Cash Flow PW(15%)

$24,192 $12,000

$24,192 $9,000

$12,096 $6,000

$0 $3,000

($210,000) $60,000 ($21,000) $210,000 ($30,000) ($30,000) ($30,000) ($30,000) ($30,000) ($30,000) ($30,000) $0 $309,302.46

$68,550

$79,320

$71,862

$68,167

22 Copyright © 2023 Pearson Education, Inc.

$70,117

$67,834 $104,550

Contemporary Engineering Economics, 7th ed. ©2023 10.25 • •

Annual payment = $115,000( A / P,11%,5) = $31,116 New after-tax cash flow 0

1

Income Statement Revenues Expenses O&M Depreciation Debt interest Taxable Income Income Taxes (40%) Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Financing Activities Borrowed funds Principal repayment Net Cash Flow PW(15%)

2

3

4

5

$250,000

$250,000

$250,000

$250,000

$250,000

$50,000 $46,000 $12,650

$50,000 $73,600 $10,619

$50,000 $44,160 $8,364

$50,000 $26,496 $5,861

$50,000 $13,248 $3,084

$141,350 $56,540

$115,781 $46,312

$147,476 $58,990

$167,643 $67,057

$183,668 $73,467

$84,810

$69,469

$88,486

$100,586

$110,201

$84,810 $46,000

$69,469 $73,600

$88,486 $44,160

$100,586 $26,496

$110,201 $13,248

($230,000) $5,000 $8,598 $115,000

($115,000) $260,050.00

($18,466)

($20,497)

($22,751)

($25,254)

($28,032)

$112,344

$122,572

$109,894

$101,827

$109,015

23 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 10.26

0 Operating Activities Net income Depreciation Investment Activities Investment Salvage value Gains tax (25%) Financing Activities Borrowed funds Principal repayment Net cash flow

1 $ $

$

$

$ PW(15%) = $

2

13,000 $ 6,666 $

15,023 4,445

$ $

8,000 222

(20,000)

10,000 $

(4,762) $

(5,238)

(10,000) $

14,904 $

22,452

19,937

Note: Annual installments for the loan = $10,000( A / P,10%,2) = $5,762

24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

10.27 Income statement approach: (a) Development of after-tax cash flows Input Tax Rate(% )= MARR(%) =

Output PW(i) = IRR(%) =

28 18

$358,998 79.90%

(a) 0 Income Statement Revenues: Additional revenue Labor & materials savings Expenses: Depreciation Debt interest Taxable Income Income Taxes (28%)

1

$

$85,000 65,000

$ 31,438 $ 10,800 $ 107,762 $ 30,173

$ $ $ $

53,878 7,200 88,922 24,898

$ 38,478 $ 3,600 $ 107,922 $ 30,218

$64,024

$77,704

Net Cash Flow

$77,589

$ $ $

3

$85,000 65,000

$

Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

2

77,589 31,438

$ $

64,024 53,878

$

$ $

$85,000 65,000

77,704 38,478

4

5-7

8

9

$

$85,000 65,000

$

$85,000 65,000

$

$85,000 65,000

$

27,478

$

19,646

$

9,812

$ $

122,522 34,306

$ $

130,354 36,499

$ 140,188 $ 39,253

$ 150,000 $ 42,000

$ 150,000 $ 42,000

$93,855

$100,935

$108,000

$108,000

93,855 19,646

$ 100,935 $ 9,812

$ 108,000 $ -

$ 108,000 $ -

$88,216

$ $

88,216 27,478

$ $

$

$85,000 65,000

10

$

(220,000) $ $

$

$85,000 65,000

20,000 (5,600)

120,000 $ (40,000) $ (40,000) $ (40,000) ($100,000)

$69,027

$77,902

$76,182

$115,694

(b) IRR = 79.90% (c) Since PW(10%) = $358,998>0, accept the investment.

25 Copyright © 2023 Pearson Education, Inc.

$113,501

$110,747

$108,000

$122,400

Contemporary Engineering Economics, 7th ed. ©2023 10.28 (a) and (b) Input Tax Rate(%) = MARR(%) =

Output 35 18

0

PW(i) = AE(i)=

($1,648,462) ($527,142)

1

2

3

4

5

Income Statement Revenues (savings) Expenses: Depreciation Debt interest

357,250 100,000

612,250 83,620

437,250 65,603

312,250 45,783

111,625 23,982

Taxable Income Income Taxes (35%)

($457,250) (160,038)

($695,870) (243,555)

($502,853) (175,998)

($358,033) (125,312)

($135,607) (47,462)

Net Income

($297,213)

($452,316)

($326,854)

($232,721)

($88,144)

(297,213) $ 357,250 $

(452,316) $ 612,250 $

(326,854) $ 437,250 $

(232,721) $ 312,250 $

(88,144) 111,625

$ $

250,000 146,781

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ $ $

$

(2,500,000)

1,000,000 $

(163,797) $

(180,177) $

(198,195) $

(218,014) $

(239,816)

($1,500,000) $

(103,760) $

(20,243) $

(87,799) $

(138,486) $

180,446

AEC(18%) = $527,142, an equivalent annual revenue that must be generated each to break even.

26 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 10.29 (a) After-tax cash flow

Input Tax Rate(%) = MARR(%) = 0

Taxable Income Income Taxes (35%) Net Income

Net Cash Flow

35 15 1

Income Statement Revenues (savings) Expenses: Depreciation Debt interest

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

Output PW(i) = IRR(%) = 2

$6,718.77

3

4

5

6

$13,000

$13,000

$13,000

$13,000

$13,000

$13,000

$6,431 $5,400

$11,021 $4,735

$7,871 $3,989

$5,621 $3,155

$4,019 $2,220

$2,009 $1,173

$1,170 $409

($2,755) ($964)

$1,140 $399

$4,225 $1,479

$6,762 $2,367

$9,818 $3,436

$760

($1,791)

$741

$2,746

$4,395

$6,382

$760 $6,431

($1,791) $11,021

$741 $7,871

$2,746 $5,621

$4,395 $4,019

$6,382 $2,009

($45,000) $4,000 $1,411 $45,000 $0

($5,545)

($6,211)

($6,956)

($7,791)

($8,725)

($9,772)

$1,646

$3,019

$1,656

$576

($312)

$4,029

(b) No meaningful IRR exists. We need to use present worth analysis. Since the NPW (15%) is positive, this project is acceptable.

27 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

10.30 (a),

(b) NPW = $14,247 > 0

28 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Generalized Cash Flow Method 10.31 (a) with no borrowed funds: Input Data Tax Rate(%) = MARR(%) =

Output PW(9%) =

25 9

$1,832

Financial Data year Depreciation Book value Salvage value Gains tax Loan payment schedule Interest Principal Revenues O & M costs

0 $

1 2 3 $ 2,666 $ 3,556 $ 1,185 $ 8,000 $ 5,334 $ 1,778 $ 593 $

4

5 593 $ $ 2,000 $ (700)

$ 2,500 $ 2,500 $ 2,500 $ 2,500 $ 2,500

Cash Flow Statement

Investment Net proceeds from sale Investment in working capital Recovery of working capital (1 - 0.25) (Revenue) -(1 - 0.25) (Expenses) -(1 - 0.25) (Debt interest) + (0.25) (Depreciation) Borrowed funds Principal repayment Net Cash Flow

0 ($8,000)

1

2

3

4

5 $1,300

$1,875 $1,875 $1,875 $1,875 $ 667 $ 889 $ 296 $ 148

$1,875 -

($8,000)

$2,542

$2,764

29 Copyright © 2023 Pearson Education, Inc.

$2,171

$2,023

$3,175

Contemporary Engineering Economics, 7th ed. ©2023

(b) With borrowed funds: (b)

Input Data Tax Rate(%) = MARR(%) =

25 9

Output PW(9%) =

$2,299

Financial Data year Depreciation Book value Salvage value Gains tax Loan payment schedule Interest Principal Revenues O&M costs

0 $8,000

1 $2,666 5,334

2 $3,556 1,778

3 $1,185 593

4 $593 0

$720 $1,337 2,500

$600 $1,457 2,500

$469 $1,588 2,500

$326 $1,731 2,500

5 0 $2,000 -700 $170 $1,887 2,500

Cash Flow Statement

Investment Net proceeds from sale Investment in working capital Recovery of working capital (1 - 0.25) (Revenue) -(1 - 0.25) (Expenses) -(1 - 0.25) (Debt interest) + (0.25) (Depreciation) Borrowed funds Principal repayment Net Cash Flow

0 ($8,000)

1

2

3

4

5 $1,300

$1,875 $1,875 $1,875 $1,875 $1,875 ($540) ($450) ($351) ($244) ($127) $ 667 $ 889 $ 296 $ 148 $ $8,000

$0

($1,337)

($1,457)

($1,588)

($1,731)

($1,887)

$665

$857

$232

$48

$1,161

(c) Which alternative to choose? Equity financing option is more attractive.

30 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

10.32 Net cash flow Input Data Tax Rate(%) = MARR(%) =

Output PW(12%) =

28 12

$104,425

Financial Data year Depreciation Book value Salvage value Gains tax (28%) Loan payment schedule Interest Principal Revenues O&M costs

0 $

135,000

$

135,000

$ $

1 19,292 115,709

$ $

2 33,062 82,647

$ $

3 23,612 59,036

$ $

4 16,862 42,174

$ $ $

13,500 22,113 68,000

$ $ $

11,289 24,324 68,000

$ $ $

8,856 26,756 68,000

$ $ $

6,181 29,432 68,000

$ $ $ $

5 6,028 36,146 50,000 (3,879)

$ $ $

3,238 32,375 68,000

Cash Flow Statement 0 ($135,000)

Investment Net proceeds from sale Investment in working capital Recovery of working capital (1 - 0.28) (Revenue) -(1 - 0.28) (Expenses) -(1 - 0.28) (Debt interest) + (0.28) (Depreciation) Borrowed funds Principal repayment Net Cash Flow

1

2

3

4

5 $46,121

$

$ $

$48,960 $48,960 $48,960 $48,960 $48,960 (9,720) $ (8,128) $ (6,377) $ (4,450) $ (2,331) 5,402 $ 9,257 $ 6,611 $ 4,721 $ 1,688

$

(22,113) $ (24,324) $ (26,756) $ (29,432) $ (32,375)

135,000

$0

$22,529

$25,765

$22,438

31 Copyright © 2023 Pearson Education, Inc.

$19,799

$62,063

Contemporary Engineering Economics, 7th ed. ©2023

10.33

(a) & (b) Federal Express: (unit in million dollars) Cash Flow Statement Investment Net proceeds from sale Investment in working capital Recovery of working capital

0 $

1

2

3

4

5

6

(270)

(1 - 0.25)(Revenue) -(1 - 0.25) (Debt interest) +(0.25)(Depreciation)

$ $ $

Borrowed funds Principal repayment

$

243

Net cash flow

$

(27.00) $

Cash Flow Statement

8

33.75 $ (21.87) $ 9.65 $

21.53

$

9

33.75 $ (21.87) $ 16.53 $

28.41

$

33.75 $ (21.87) $ 11.81 $

23.69

10

$

11

33.75 $ (21.87) $ 8.44 $

20.32

$

12

33.75 $ (21.87) $ 6.02 $

17.90

$

13

33.75 $ (21.87) $ 6.03 $

17.91

$ $ $

33.75 $ (21.87) $ 3.01

33.75 $ (21.87) $

Borrowed funds Principal repayment Net cash flow

$ $

14.89

PW(18%) =

$ $

11.88

33.75 $ (21.87)

$

33.75 (21.87) 6.02

17.90

14

Investment Net proceeds from sale (1 - 0.25)(Revenue) -(1 - 0.25) (Debt interest) +(0.25)(Depreciation)

7

15

$

20.25

33.75

$

33.75

$

33.75

$

33.75

$

33.75

33.75

$

33.75

$

33.75

$

33.75

$

54.00

(243)

$ (231.12) $

41 > 0, Accept the investment.

Note: Salvage value = (0.15)× (18)×$10 million =$27 million, Book value = $0, Taxable gains = $27 million Gains tax = (0.25)($27 million) = $6.75 million, Net proceeds from sale = $27 million - $6.75 million = $20.25 million

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Comparing Mutually Exclusive Alternatives 10.34 (a) The net after-tax cash flows for each financing option: •

Option 1: Retained earnings

33 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Vermont’s PW cost of owning the equipment by borrowing: Input Tax Rate(%) = MARR(%) = Option 2: Owning the equipment by borrowing 0 Income Statement

Output PW(i) = IRR(%) =

28 18 1

2

3

$254,633 305.51% 4

5

6

Revenues (savings) Expenses: O&M costs Depreciation Debt interest

$174,000

$174,000

$174,000

$174,000

$174,000

$174,000

$22,000 $28,580 $24,000

$22,000 $48,980 $21,043

$22,000 $34,980 $17,730

$22,000 $24,980 $14,020

$22,000 $17,860 $9,866

$22,000 $8,920 $5,212

Taxable Income Income Taxes (28%)

$99,420 $27,838

$81,977 $22,954

$99,290 $27,801

$113,000 $31,640

$124,274 $34,797

$137,868 $38,603

Net Income

$71,582

$59,024

$71,489

$81,360

$89,478

$99,265

$71,582 $28,580

$59,024 $48,980

$71,489 $34,980

$81,360 $24,980

$89,478 $17,860

$99,265 $8,920

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working capital Financing Activities: Borrowed funds Principal repayment Net Cash Flow

-$200,000 $30,000 $1,596 $25,000

-$25,000 $200,000

-$25,000

-$24,645

-$27,603

-$30,915

-$34,625

-$38,780

-$43,433

$75,517

$80,401

$75,554

$71,715

$68,558

$121,348

34 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c) Vermont’s PW cost of leasing the equipment:

(d) Option 2 is the best alternative.

35 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 10.35 •

Option 1: Lease PW(12%)lease = −$144, 000(1 − 0.28)(1 + ( P / A,12%, 29))

•

= −$935,380 Option 2: Purchase

o Note 1: It is assumed that the property is placed in service during January.

D1 , D30 = (11.5 /12)(1/ 39)($650, 000) = $15,972.22 D2 to D29 = (12 /12)(1/ 39)($650, 000) = $16, 666.67 o Note 2: Property tax calculation: ($800,000) (0.05) = $40,000 Input Tax Rate(% )= MARR(%) =

Output PW(i) =

28 12

0

1

2

3

($986,595)

4

29

30

Income Statement Revenues: Expenses: Depreciation Property tax Taxable Income Income Taxes (28%)

$ $ $ $

Net Income

$ (40,300) $ (40,800) $ (40,800) $ (40,800) $

(40,800) $ (40,300)

$ (40,300) $ (40,800) $ (40,800) $ (40,800) $ $ 15,972 $ 16,667 $ 16,667 $ 16,667 $

(40,800) $ (40,300) 16,667 $ 15,972

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment(land) Investment(structure) Salvage Gains Tax Net Cash Flow

$ $

15,972 40,000 (55,972) (15,672)

$ $ $ $

16,667 40,000 (56,667) (15,867)

$ $ $ $

16,667 40,000 (56,667) (15,867)

$ 16,667 $ $ 40,000 $ $ (56,667) $ $ (15,867) $

16,667 40,000 (56,667) (15,867)

$ 15,972 $ 40,000 $ (55,972) $ (15,672)

(150,000) (650,000) $ 215,000 $ 24,189

$

(800,000) $ (24,328) $ (24,133) $ (24,133) $ (24,133) $

PW(12%) purchase = −$986, 595

36 Copyright © 2023 Pearson Education, Inc.

(24,133) $ 214,861

Contemporary Engineering Economics, 7th ed. ©2023 •

Option 3: Remodel

o Note 1: Depreciation base: Remodeling cost = $300,000

D1 , D30 = (11.5 /12)(1/ 39)($300, 000) = $7,372 D2 to D29 = (12 /12)(1/ 39)($300, 000) = $7, 692 o Note 2: Property tax calculation: $660,000 (0.05) = $33,000 Cost basis for property tax = Land + building + remodeling cost = $660,000

Input Tax Rate(% )= MARR(%) = 0 Income Statement Revenues: Expenses: Depreciation Property tax Lease fee(Parking lot) Taxable Income Income tax (28%) Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment(Remodeling)

Output PW(i) =

28 12 1

2

3

($546,110)

4

$ $

$

$

30

7,372 $ 7,692 $ 7,692 $ 7,692 $ 33,000 $ 33,000 $ 33,000 $ 33,000 $ $9,000 $9,500 $10,000 $10,500 $ (49,372) $ (50,192) $ (50,692) $ (51,192) $

7,692 $ 7,372 33,000 $ 33,000 $23,000 $23,500 (63,692) $ (63,872)

$ (13,824) $ (14,054) $ (14,194) $ $ (35,548) $ (36,138) $ (36,498) $

(14,334) $ (36,858) $

(17,834) $ (17,884) (45,858) $ (45,988)

$ (35,548) $ (36,138) $ (36,498) $ $ 7,372 $ 7,692 $ 7,692 $

(36,858) $ 7,692 $

(45,858) $ (45,988) 7,692 $ 7,372

(300,000)

Salvage Gain tax Net Cash Flow

29

(300,000) $ (28,176) $ (28,446) $ (28,806) $

PW(12%)remodel = −$546,110 Option 3 is the least cost alternative.

37 Copyright © 2023 Pearson Education, Inc.

(29,166) $

$ $

30,000 11,164

(38,166) $

2,548

Contemporary Engineering Economics, 7th ed. ©2023

10.36 Comparison by annual equivalent cost (all units in thousand dollars): A

B

Book Value (n = 20) $381 $424 Salvage Value Taxable gains

C

$471

$853 $950 $1,055 $472 $526 $584

Gains tax (28%)

$132 $147

$164

Net Proceeds from sale

$721 $803

$891

Plant A • Capital recovery cost with return:

A1 = ($8,530 − $721)( A / P,12%, 20) + $721(0.12) = $1,056

•

After-tax O&M cost:

•

Depreciation tax shield: A3 = 0.28($8,530) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20)

•

A2 = (1 − 0.28)($1,964) = $1, 414.08

= $123.65 Total equivalent annual cost:

A = $1,056 + $1,414.08 − $123.65 = $2,346.43 (unit 1,000) •

Unit cost:

$2,346, 430 = $0.0469 / kWh 50, 000, 000 kWh

Plant B • Capital recovery cost with return:

A1 = ($9, 498 − $803)( A / P,12%, 20) + $803(0.12) = $1,176

•

After-tax O&M cost:

•

Depreciation tax shield: A3 = 0.39($9, 498) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20)

•

A2 = (1 − 0.28)($1,744) = $1,255.68

= $137.67 Total equivalent annual cost: 38 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

A = $1,176 + $1,255.68 − $137.67 = $2,294.01 •

Unit cost:

$2, 294, 010 = $0.0459 / kWh 50, 000, 000 kWh

Plant C • Capital recovery cost with return:

A1 = ($10,546 − $891)( A / P,12%, 20) + $891(0.12) = $1,305 •

After-tax O&M cost:

A2 = (1 − 0.39)($1,632) = $1,175.04 •

Depreciation tax shield:

A3 = 0.39($10,546) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20) = $152.87 •

Total equivalent annual cost:

A = $1,305 + $1,175.04 − $152.87 = $2,327.17 •

Unit cost:

$2,327,170 = $0.0465 / kWh 50, 000, 000 kWh Plant B is the most economical.

Lease -Versus - Buy Decisions 10.37 (a) Prescott Welding’s cost of leasing in present worth: after-tax lease expense = (1 - 0.28)($12,000) = $8,640 PW(15%)lease = −$8, 640 − $8, 640( P / A,15%,3) = −$28,367 (b) Prescott Welding’s cost of owning in present worth:

39 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

PW of after-tax maintenance expenses:

P1 = −$1, 200(1 − 0.28)( P / A,15%, 4) = −$2, 467

•

PW of after-tax loan repayment P2 = −$14,816( P / A,15%, 4) = −$42, 299

•

PW of tax credit (shield) on depreciation and interest: Dn

In

Sum

1

$9,000

$5,400

$14,400

$4,032

2

$14,400

$4,270

$18,670

$5,228

3

$8,640

$3,005

$11,645

$3,261

4

$2,592

$1,587

$4,179

$1,170

P3=

•

Combined tax

n

savings

$10,272

PW of net proceeds from sale: Total depreciation amount=

$34,632

Book value=

$10,368

Salvage=

$10,000

Taxable Gain=

($368)

Gain tax=

($103)

Net proceeds from sale=

$10,103

P4=

$5,776

PW(15%)buy = P1 + P2 + P3 + P4 = −$28, 718 (c) Should the truck be leased or purchased? The leasing option is a better choice.

40 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 10.38 (a) PW (incremental) cost of owning the equipment: •

PW of after-tax O&M

P1 = −$50, 000(1 − 0.28)( P / A,15%, 4) = −$102, 779 •

PW of after-tax loan repayment:

P2 = −$37,857( P / A,15%, 4) = −$108, 080 •

PW of tax credit (shield) on depreciation and interest:

n

Dn

In

Combined Tax Savings

1 $24,000 $12,000 $36,000(0.28) = $10,080 2 $38,400 $9,414 $47,814(0.28) = $13,388 3 $23,040 $6,570 $29,610(0.28) = $8,291 4

$6,912

$3,441

$10,353(0.28) = $2,899

P3 = $10, 080( P / F ,15%,1) + $13,388( P / F ,15%, 2) +$8, 291( P / F ,15%,3) + $2,899( P / F ,15%, 4) = $25,997 •

PW of net proceeds from sale:

total depreciation amount = $92,352 book value = $27,648 taxable gain = $20,000 - $27,648 = -$7,648 loss credit = (0.28)(-$7,648) = -$2,141 net proceeds from sale = $20,000 -(-$2,141) = $22,141 P4 = $22,141(P / F ,15%, 4) = $12,659

41 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(15%) buy = P1 + P2 + P3 + P4 = −$172, 203 Input Tax Rate(%) = MARR(%) = 0

28 15%

Output PW(i) =

($172,202)

1

2

3

4

50000 24000 12000

50000 38400 9414

50000 23040 6570

50000 6912 3441

Income Statement Revenues (savings) Expenses: O&M cost Depreciation Debt interest Taxable Income Income Taxes (28%)

($86,000) (24,080)

($97,814) (27,388)

($79,610) (22,291)

($60,353) (16,899)

Net Income

($61,920)

($70,426)

($57,319)

($43,455)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

$ (61,920) $ (70,426) $ (57,319) $ (43,455) $ 24,000 $ 38,400 $ 23,040 $ 6,912 $

(120,000) $ $

$

20,000 2,141

120,000 $ (25,856) $ (28,442) $ (31,286) $ (34,415)

Net Cash Flow

$0

($63,776)

($60,468)

($65,566)

($48,816)

(b) PW (incremental) cost of leasing the equipment: •

PW of after-tax operating cost: common cost for both alternative, so we can ignore this item in incremental analysis.

•

PW of after-tax leasing

P = −$44,000(1 − 0.28) − $44,000(1 − 0.28)( P / A,15%,3) = −$104,013

42 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Input Tax Rate(%) = MARR(%) =

28 15%

0

1

Output PW(i) =

($104,013)

3

4

2

Income Statement Revenues (savings) Expenses: O&M cost Leasing cost Debt interest

$0

$0

$0

$0

0 $44,000

0 $44,000

0 $44,000

0

$44,000

Taxable Income Income Taxes (28%)

($44,000) (12,320)

($44,000) (12,320)

($44,000) (12,320)

($44,000) (12,320)

$0 0

Net Income

($31,680)

($31,680)

($31,680)

($31,680)

$0

Cash Flow Statement Operating Activities: Net Income

$

Net Cash Flow

(31,680) $ (31,680) $ (31,680) $ (31,680) $ ($31,680)

($31,680)

($31,680)

($31,680)

$0

(c) Should ICI buy or lease the equipment? The leasing option is a better choice.

10.39 (a) PW of after-tax cash flow of leasing:

PW(15%)lea s e = −$70, 000(1 − 0.28)( P / A,15%, 4) = −$143,891 (b) PW of after-tax cash flow of owning: •

PW of after-tax maintenance expenses:

P1 = −$10, 000(1 − 0.28)( P / A,15%, 4) = −$20,556

•

PW of after-tax loan repayment P2 = −$63, 094( P / A,15%, 4) = −$180,132

•

PW of tax credit (shield) on depreciation and interest: 43 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Dn

In

Sum

1

$66,660

$20,000

$86,660

$34,664

2

$88,900

$15,961

$104,861

$41,944

3

$29,620

$10,950

$40,570

$16,228

4

$14,820

$5,736

$20,556

$8,222

P3=

•

Combined tax

n

savings

$54,061

PW of net proceeds from sale: Total depreciation amount=

$200,000

Book value=

$0

Salvage=

$20,000

Taxable Gain=

$20,000

Gain tax=

$5,600

Net proceeds from sale=

$14,400

P4=

$8,233

PW(15%) buy = P1 + P2 + P3 + P4 = −$138,394

10.40 (a) Determine the annual cash flows for each option. •

Buy option: o Net proceeds:

44 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 S = $15, 400 Total amount of depreciation = $17,248 Book value = $10,752 Taxable gains = $15,400 - $10,752 = $4,648 Gains tax = $4,648 × 0.28 = $1,301.44 Net proceeds from sale = $15,400 - $1,301.44 = $14,099 o Annual interest payments:

n Interest payment Principal payment Total payment

1 $2,002.75

2 $1,263.49

3 $460.58

$8,572.85

$9,312.11

$10,115.02

$10,575.60

$10,575.60

$10,575.60

o Annual depreciation: D1 = $28, 000 × 0.20 = $5, 600 D2 = $28, 000 × 0.32 = $8,960 D3 = $28, 000 ×

0.192 = $2, 688 2 End of period 1 2

Cash flow elements 0 ($28,000)

Investment Net proceeds

$14,099

−(0.72) I n

($1,442)

($910)

($332)

+0.28Dn

$1,568

$2,509

$753

($8,573) ($8,447)

($9,312) ($7,713)

($10,115) 4,405

Loan repayment Net cash flow

•

3

$28,000 $0

Lease option: Cash flow elements Security deposit

0 ($500)

End of period 1 2

45 Copyright © 2023 Pearson Education, Inc.

3

Contemporary Engineering Economics, 7th ed. ©2023 Refund

$500

−(0.72) Ln Net cash flow

($500)

($6,013)

($6,013)

($6,013)

($6,013)

($6,013)

($5,513)

(b) PW(13%)buy = −$10, 462 PW(13%)lease = −$14,351 The buy option is a better choice.

Note: Our assumption here is that all cash flows within a year are lumped and placed at the end of each year. 10.41 (a) Boggs’ PW cost of leasing:

PW(15%)leasing = (0.72)[$15, 000 + $15, 000( P / A,15%, 2)] =$28,357 (b) Boggs’ PW cost of owning: •

PW of after-tax maintenance expenses: P1 = $5, 000(1 − 0.28)( P / A,15%,3) = $8, 220

•

PW cost of after-tax loan repayment: P2 = $41, 635( P / A,15%,3) = $95, 062

•

PW of tax credit (shield) on depreciation and interest: n Dn In 1 $20, 000 $12, 000

Combined Tax Savings $32, 000(0.28) = $8,960

2 $32, 000 3 $9, 600

$40, 444(0.28) = $11,324 $14, 061(0.28) = $3,937

$8, 444 $4, 461

P3 = $8,960( P / F ,15%,1) + $11,324( P / F ,15%, 2) +$3,937( P / F ,15%,3) = $18,943

46 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

•

PW of net proceeds from sale: total depreciation amount = $61,600 book value = $38,400 taxable gain = $50,000 - $38,400 = $11,600 loss credit = (0.28)($11,600) = $3,248 net proceeds from sale = $50,000 - $3,248 = $46,752 P4 = $46,752(P / F ,15%,3) = $30,740 PW(15%) buy = P1 + P2 − P3 − P4 = $53,599

(c) Leasing is much cheaper. 10.42 (a) Purchase with debt:

•

PW of after-tax revenue:

P1 = $10,000(1 − 0.30)( P / A,10%,5) = $26,536 •

PW of after-tax expenses:

P2 = −$2,500(1 − 0.3)( P / A,10%,5) = −$6,634 •

PW of after-tax loan repayment:

A = $25,000( A / P,12%,5) = −$6,935.24 P3 = −$6,935.24( P / A,10%,5) = −$26, 290 •

PW of tax credit (shield) on depreciation and interest:

47 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

n

Dn

In

Combined Tax Savings

1 $3,571 $3, 000

$6,571(0.30) = $1,971

2 $6,122 $2,528 3 $4,373 $1,999

$8, 650(0.30) = $2,595 $6,372(0.30) = $1,912

4 $3,123 $1, 407

$4,530(0.30) = $1,359

5

$1,116

$743 $1,859(0.30) =

$558

P4 = $1,971( P / F ,10%,1) + $2,595( P / F ,10%, 2) +  = $6,647 •

PW of net proceeds from sale: total depreciation amount = $18,305 book value = $6,695 taxable gain = $5,000 - $6,695=($1,695) loss credit = (0.30)($1,695) = $509 net proceeds from sale = $5,000 + $509 = $5,509 P5 = $5,509(P / F ,10%,5) = $3,421 PW(10%) purchase = P1 + P2 + P3 + P4 + P5 = $3, 680

(b) Financial lease: PW(10%)lea s e = −(0.7)[$5,500 + $5, 500( P / A,10%, 4)] +0.7 [$10, 000 − $2,500] ( P / A,10%, 5) = $3,847.65

(c) The financial lease is a better choice.

10.43 Setting the lease payment schedule: Let X denote the annual lease receipt from tractor lease. We will assume that these lease payments are received at year end.

48 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Cash Flow Elements Investment Net Proceeds: Security Deposit +(0.65)(Rn ) +(0.35)Dn

0 -$53,000

End of Period 1

2

3

0.65X $5,936 0.65X $5,936

$21,423* -$1,500 0.65X $1,781 0.65X $21,704

$1,500

Net Cash Flow

0.65X $3,710 0.65X $3,710

-$51,500

Note: total depreciation amount = $32,648 book value = $53,000 - $32,648 = $20,352 taxable gain = $22,000 - $20,352 = $1,648 gains tax = (0.35)($1,648) = $577 net proceeds from sale = $22,000 - $577 = $21,423 Now to determine the required lease receipt at an after-tax rate of return of 10%, we solve the following equation: $51,500 = 0.65 X ( P / A,10%,3) +$3, 710( P / F ,10%,1) + $5,936( P / F ,10%, 2) +$21, 704( P / F ,10%,3) X = $16, 651 per year

10.44

•

Debt Financing: Investment Net proceeds -Interest (1 – 0.28) +Depreciation (0.28) Loan Repayment Net Cash Flow

0 -$100,000

$100,000 0

1

2

-$7,200 $5,600

$45,600 -$7,200 $5,600

0 $1,600

-$100,000 -$56,000

PW(12%) = $1,600( P / F ,12%,1) − $56,000( P / F,12%,2) = −$43,214 •

Lease Financing:

49 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

-Lease payment (10.28) Net Cash Flow

0 -$30,240

1 -$30,240

-$30,240

-$30,240

2

PW(12%) = −$30,240 − $30,240(P / F ,12%,1) = −$57,240 − X (1 − 0.28) − X (1 − 0.28)( P / F ,12%,1) = −$43, 214

X = $31, 708

50 Copyright © 2023 Pearson Education, Inc.

0 0

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies with Excel ST 10.1 (a) We need to examine two situations: (1) If the firm decided not to pursue the project, then $6,000,000 can be written off as an expense in 2022, which is (0.25)($6,000,000) = $1,500,000 worth of tax savings. Therefore, if the firm decided to go with the project, it incurs an opportunity cost of $2,280,000. This should be factored into the analysis. (b) Depreciation schedules:

• • •

Land – no depreciation allowed Building – 39-year MACRS Equipment – 7-year MACRS

(c) Taxable gains:

• Land: $1,500,000 - $2,500,000 = -$1,000,000 • Building: $3,800,000 – ( $12,000,000 - $1,820,520) = -$6,379,480 • Equipment: $2,500,000 – ($9,000,000 - $7,393,500) = $893,500 (d) Working capital requirements:

• •

Year 2017: $189,400,000(0.13) = $23,452,000 Year 2018: -$23,452,00 + $189,420,000(0.13) = $1,172,600

(e) Cash flow statement (next page)

• • •

Do not undertake the project: You can write off $6,000,000 R&D to save tax in the amount of $1,500,000. Undertake the project: PW(14.5%) = $123,916,445 Incremental (net) benefit of undertaking the project: $123,916,445 $1,500,000 = $122,416,445

51 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

2022

2023

2024

2025

2026

2027

2028

2029

2030

Income Statement Unit Sales

22,000

22,000

22,000

22,000

22,000

8,200

8,610

9,041

9,493

9,967

10,466

180,400,000

189,420,000

198,891,000

208,835,550

219,277,328

230,241,194

117,260,000

123,123,000

129,279,150

135,743,108

142,530,263

149,656,776

Overhead

7,500,000

7,875,000

8,268,750

8,682,188

9,116,297

9,572,112

Amortization

1,000,000

1,000,000

1,000,000

1,000,000

1,000,000

1,000,000

294,876

307,692

307,692

307,692

307,692

294,876

1,286,100

2,204,100

1,574,100

1,124,100

803,700

401,400

Taxable Income

53,059,024

54,910,208

58,461,308

61,978,463

65,519,376

69,316,030

Income Taxes (25%)

13,264,756

13,727,552

14,615,327

15,494,616

16,379,844

17,329,008

Net Income

39,794,268

41,182,656

43,845,981

46,483,847

49,139,532

51,987,023

39,794,268

41,182,656

43,845,981

46,483,847

49,139,532

51,987,023

Amortization

1,000,000

1,000,000

1,000,000

1,000,000

1,000,000

1,000,000

Depreciation (Building & Equipment)

1,580,976

2,511,792

1,881,792

1,431,792

1,111,392

696,276

Sales Price Sales Revenue

22,000

Expenses: Variable Costs Fixed Costs:

Depreciation (Building) Depreciation (Equipment)

Cash Flow Statement Operating Activities: Net Income Add Noncash Expenses:

Investment Activities: Land

(2,500,000)

Building

(2,000,000)

1,500,000 (6,000,000)

Equipment

(4,000,000)

3,800,000

9,000,000

2,500,000

Gains Taxes: Land

250,000

Building

1,594,870

Equipment

(223,375)

Opportunity Cost (R&D)

(1,500,000)

Working Capital

(23,452,000)

(24,624,600)

(25,855,830)

(27,148,622)

(28,506,053)

(29,931,355)

23,452,000

24,624,600

25,855,830

27,148,622

28,506,053

29,931,355

(23,452,000)

(1,172,600)

(1,231,230)

(1,292,792)

(1,357,431)

(1,425,303)

29,931,355

(18,452,000)

41,202,644

43,463,218

45,434,982

47,558,208

49,825,621

93,036,149

Working capital recovery Net amount (WC) Net Cash Flow in actual dollars

(6,000,000)

(6,000,000)

PW(14.5%) = $122,416,445

52 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

ST 10.2

(a), (b) (All units are in thousand dollars) 0

Income Statement Revenue Expenses Production costs Depreciation: Building Machines

1

2

3

4

5

6

7

8

9

10

11

12

$ 51,000 $ 51,000 $ 51,000 $ 51,000 $ 85,000 $ 85,000 $ 85,000 $ 136,000 $ 136,000 $ 136,000 $ 136,000 $ 136,000 36,000

36,000

36,000

36,000

60,000

60,000

60,000

96,000

96,000

96,000

96,000

96,000

1,106 14,290

1,154 24,490

1,154 17,490

1,154 12,490

1,154 8,930

1,154 8,930

1,154 8,930

1,154 4,460

1,154

1,154

1,154

1,106

Taxable income Income taxes (28%)

(396) (111)

(10,644) (2,980)

(3,644) (1,020)

1,356 380

14,916 4,176

14,916 4,176

14,916 4,176

34,386 9,628

38,846 10,877

38,846 10,877

38,846 10,877

38,894 10,890

Net income

(285)

(7,664)

(2,624)

976

10,740

10,740

10,740

24,758

27,969

27,969

27,969

28,004

(285) 15,396

(7,664) 25,644

(2,624) 18,644

976 13,644

10,740 10,084

10,740 10,084

10,740 10,084

24,758 5,614

27,969 1,154

27,969 1,154

27,969 1,154

28,004 1,106

Cash Flow Statement Operating Activities: Net income Depreciation Investment Activities: Land Building Machines Gain Taxes Land Building Equipment Net Cash Flow

(5,000) (45,000) (100,000)

8,000 30,000 10,000 (840) 349 (2,800)

$

(150,000) $ 15,111 $ 17,980 $ 16,020 $ 14,620 $ 20,824 $ 20,824 $ 20,824 $ 30,372 $ 29,123 $ 29,123 $ 29,123 $ 73,819

PW(15%) =

$ (31,725)

IRR =

10.8%

Note: A true sense of capital gains is realized only for the sale of land. (c) Project is not acceptable.

53 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

ST 10.3

Morgantown Mining Company

(a) Unit-production method (Units are thousand dollars) 0

1

2

3

4

5

6

7

8

9

10

Revenue (Savings) Expenses: O&M costs Depreciation Taxable income Income taxes (25%)

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

2,400 1,880 5,220 1,305

Net income Cash Flow Statement Operating Activities Net income Depreciation Investment Activities Investment Salvage value Gains tax Working capital

$3,915

$3,915

$3,915

$3,915

$3,915

$3,915

$3,915

$3,915

$3,915

$3,915

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

3,915 1,880

Income Statement

Net Cash Flow

(19,300) 500 (2,500) -$21,800

2500 $5,795

$5,795

$5,795

$5,795

$5,795

54 Copyright © 2023 Pearson Education, Inc.

$5,795

$5,795

$5,795

$5,795

$8,795

Contemporary Engineering Economics, 7th ed. ©2023

(b) 7-year MACRS 0

1

2

3

4

5

6

7

8

9

10

Revenue (Savings) Expenses: O&M costs Depreciation Taxable income Income taxes (25%)

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

$9,500

2,400 2,758 4,342 1,086

2,400 4,727 2,373 593

2,400 3,376 3,724 931

2,400 2,411 4,689 1,172

2,400 1,723 5,377 1,344

2,400 1,722 5,378 1,345

2,400 1,723 5,377 1,344

2,400 861 6,239 1,560

2,400 7,100 1,775

2,400 7,100 1,775

Net income Cash Flow Statement Operating Activities Net income Depreciation Investment Activities Investment Salvage value Gains tax Working capital

$3,257

$1,780

$2,793

$3,517

$4,033

$4,034

$4,033

$4,679

$5,325

$5,325

3,257 2,758

1,780 4,727

2,793 3,376

3,517 2,411

4,033 1,723

4,034 1,722

4,033 1,723

4,679 861

5,325 -

5,325 -

Income Statement

Net Cash Flow

(19,300) 500 -125 2,500

(2,500) -$21,800

$6,015

$6,507

$6,169

$5,928

$5,756

55 Copyright © 2023 Pearson Education, Inc.

$5,756

$5,756

$5,540

$5,325

$8,200

Contemporary Engineering Economics, 7th ed. ©2023

ST 10.4

Given: • Savings = $314,000 + $35,000 = $349,000 per year • Materials (resin) = $350(400) = $140,000 per year • Cost base = $187,000 + $10,000 + $15,000 = $212,000 • Taxable gain = $30,000

(a) Equity financing (retained earnings):

56 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

0

1

2

3

4

5

6

$349,000

$349,000

$349,000

$349,000

$349,000

$349,000

36,000 140,000 20,000 42,400 110,600 27,650

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

67,840 105,160 26,290

40,704 132,296 33,074

24,422 148,578 37,145

24,422 148,578 37,145

12,211 160,789 40,197

Income Statement Revenue (Savings) Expenses: O&M costs Resin Software development Depreciation Taxable income Income taxes (25%) Net income Cash Flow Statement Operating Activities Net income Depreciation Investment Activities Investment Salvage value Gains tax Net Cash Flow

$82,950

$78,870

$99,222

$111,434

$111,434

$120,592

82,950 42,400

78,870 67,840

99,222 40,704

111,434 24,422

111,434 24,422

120,592 12,211

(212,000) 30,000 (7,500) -$212,000

$125,350

PW(20%) =

$247,441

$146,710

$139,926

$135,856

IRR =

60%

57 Copyright © 2023 Pearson Education, Inc.

$135,856

$155,303

Contemporary Engineering Economics, 7th ed. ©2023

(b) Debt financing (term loan): annual installment = $212,000(A/P, 13%, 6) = $53,032 0

1

2

3

4

5

6

Income Statement Revenue (Savings) Expenses: O&M costs Resin Software development Depreciation Debt interest Taxable income Income taxes (25%) Net income Cash Flow Statement Operating Activities Net income Depreciation Investment Activities Investment Salvage value Gains tax Financing Activities Borrowed funds Principal repayment

$349,000 $349,000 $349,000 $349,000 $349,000 $349,000 36,000 140,000 20,000 42,400 27,560 83,040 20,760

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

67,840 24,249 80,911 20,228

40,704 20,507 111,789 27,947

24,422 16,278 132,300 33,075

24,422 11,500 137,078 34,270

12,211 6,101 154,688 38,672

$62,280

$60,683

$83,842

$99,225 $102,809 $116,016

62,280 42,400

60,683 67,840

83,842 40,704

99,225 24,422

102,809 24,422

116,016 12,211

(212,000) 30,000 (7,500) 212,000

Net Cash Flow

$0 PW(20%) =

(25,472)

(28,784)

(32,526)

(36,754)

(41,532)

$79,208

$99,739

$92,020

$86,893

$85,699 $103,796

$299,628

58 Copyright © 2023 Pearson Education, Inc.

(46,931)

Contemporary Engineering Economics, 7th ed. ©2023

(c) Lease financing (financial lease): 0

1

2

3

4

5

6

Income Statement Revenue (Savings) Expenses: O&M costs Resin Software development Lease payment Taxable income Income taxes (25%)

$349,000 $349,000 $349,000 $349,000 $349,000 $349,000 36,000 140,000 20,000 62,560 62,560 (62,560) 90,440 (15,640) 22,610

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

36,000 140,000

62,560 110,440 27,610

62,560 110,440 27,610

62,560 110,440 27,610

62,560 110,440 27,610

173,000 43,250

Net income Cash Flow Statement Operating Activities Net income

(46,920)

$67,830

$82,830

$82,830

$82,830

$82,830 $129,750

(46,920)

67,830

82,830

82,830

82,830

82,830

Net Cash Flow

-$46,920

$67,830

$82,830

$82,830

$82,830

$82,830 $129,750

IRR =

157%

PW(20%) = $231,745

(d) The best financing method is the term loan option.

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129,750

Contemporary Engineering Economics, 7th ed. ©2023

ST 10.5

•

(a) The net cash flows for each alternative over 10 years:

Option 1: Leasing gas-powered lift trucks (payable at the end of each year): total annual expenses = $5,465 + $6,317 + $1,660 + $58,653 + $10,000 = $82,095 annual net cash flow = - (1 - 0.25)($82,095) = -$61,571 per year

•

Option 2: Installing AGVS:

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Input Data Tax Rate(%)= MARR(%)=

Output PW(15%)= ($193,055)

25 15

Financial Data year Depreciation Book value Interest payment Principal payment O&M costs

$159,000 159,000 $20,000

Cash Flow Statement

Investment Net proceeds from sale -(1 - 0.25) (Expenses) -(1 - 0.25) (Debt interest) + (0.25) (Depreciation) Borrowed funds Principal repayment Net Cash Flow

0

1 $22,721 136,279 15,900 (26,044) $20,000

2 $38,939 97,340 13,296 (28,648) $20,000

3 $27,809 69,531 10,431 (31,513) $20,000

4 $19,859 49,672 7,280 (34,664) $20,000

5 $14,199 35,473 3,813 (38,131) $20,000

6 $14,183 21,290

7 $14,199 7,091

8 $7,091 (0)

$20,000

$20,000

6

7

9

10 $0 (0)

$0 (0)

$20,000

$20,000

$0

8

9

(all units in thousands of dollars) 0 ($159,000)

1

2

3

4

5

($15,000) ($15,000) ($15,000) ($15,000) (11,925) (9,972) (7,823) 5,680 9,735 6,952 $159,000 (26,044) (28,648) (31,513)

($15,000) ($15,000) ($15,000) ($15,000) ($15,000) ($15,000) (5,460) (2,860) 4,965 3,550 3,546 3,550 1,773 0

($15,000) ($47,289) ($43,885) ($47,384)

($50,159) ($52,441) ($11,454) ($11,450) ($13,227) ($15,000)

(34,664)

10

$0 0

(38,131)

(b) & (c) The incremental cash flows (AGVS option – Gas truck option) PW(i )agvs - gas = −$15, 000 + $14, 282( P / F , i,1) +$17, 686( P / F , i, 2) + $14,187( P / F , i,3) + $11, 412( P / F , i, 4) + $9,130( P / F , i,5) +$50,117( P / F , i, 6) + $50,121( P / F , i, 7) + $48,344( P / F , i,8) + $46,571( P / F , i,9) +$61,571( P / F , i,10) =0 IRR = 105% > 15%,select AGVS. 61 Copyright © 2023 Pearson Education, Inc.

$0

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 11 Inflation and Its Impact on Project Cash Flows Measure of Inflation 11.1

4.12(1 + 0.04)(1 + 0.06)(1 + 0.08) = $4.91 4.91 = 4.12(1 + f )3 f = 5.99% 11.2

1.13(1 + f ) 20 = 3.49 f = 5.80% 100(1 + 0.059362) 20 = 308.85 11.3 (a) Average price index 477.65(1 + f ) 4 = 528.22 f = 2.5478% (b) 528.22(1 + 0.025478) 7 = 629.94

11.4

100(1 + 0.05)(1 + 0.08) = 113.40 100( F / P, f , 2) = 113.40 f = 6.4894%

11.5

Given : f = 8%

1(1 + 0.08) − n = 0.5 1.08− n = 0.5 ( − n) log1.08 = log 0.5 n = − log 0.5 / log1.08 = 9 years Comments: If you use the Rule of 72, you may find equal to the actual value.

1 Copyright © 2023 Pearson Education, Inc.

72 = 9 years which is 8

Contemporary Engineering Economics, 7th ed. ©2023

Actual versus Constant Dollars 11.6 538,400 − 504,000 = 6.825% 504,000 577,000 − 538,400 = 7.169% f2 = 538,400 629,500 − 577,000 f3 = = 9.099% 577,000 f1 =

 629,500  f =  504,000 

1/ 3

− 1 = 7.69%

11.7

283.72 = 26.87(1 + f )62 f = 3.87% $72,500(1 + 0.0387) −62 = $6,885.73

11.8 Given: i = 8%, f = 4% , 10 annuity payments in actual dollars P = $72, 000( P / A,8%,10) = $483,126 Comments: Since the annuity payments are made in actual dollars, we use the market interest rate to find its equivalent lump sum amount in today’s dollars.

11.9 Given: i = 13%, f = 5%, maintenance costs are given in constant dollars.

i' =

i − f 0.13 − 0.05 = = 7.62% 1 + 0.05 1+ f

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Contemporary Engineering Economics, 7th ed. ©2023 P = $25, 000( P / F , 7.62%,1) + $26, 000( P / F , 7.62%, 2) +$28, 000( P / F , 7.62%,3) + $30, 000( P / F , 7.62%, 4) +$32, 000( P / F , 7.62%,5) = $112, 672 A = $112, 672( A / P,13%,5) = $32, 034 11.10

Given: i = 14%, f = 5%

(a)

n

Actual dollars

Constant Dollars

0

$25,000

$25,000(P/F,5%,0) = $25,000

4

35,000

35,000(P/F,5%,4) = 28,795

5

45,000

45,000(P/F,5%,5) = 35,259

7

55,000

55,000(P/F,5%,7) = 39,087

(b)

P = $25, 000 + $35, 000( P / F ,14%, 4) + $45, 000( P / F ,14%,5) +$55, 000( P / F ,14%, 7) = $91, 075 11.11

Given: P = $25, 000, i = 0.75% per month, f = 0.33% per month th

• 20 payment in actual dollars:

A20 = $25, 000( A / P, 0.75%, 48) = $622.13

• 20th payment in constant dollars:

A '20 = $622.13( P / F , 0.33%, 20) = $582.07 11.12 (a) Constant-dollar analysis: we need to find the inflation-free interest rate. i− f i' = = 7.477% 1+ f Then, find the equivalent present worth of this geometric series at i ' . P = $10, 000( P / A1 ,8%, 7.477%, 4) = $37, 490

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(b) Actual-dollar analysis

1

Net Cash Flow in Constant $ $10,000

Conversion factor (1 + 0.07)1

Net Cash Flow in Actual $ $10,700

2 3

10,800 11,664

(1 + 0.07) 2

4

12,597

12,365 14,289 16,512

Period

3

(1 + 0.07) (1 + 0.07) 4

P = $10, 700( P / F ,15%,1) + $12,365( P / F ,15%, 2) +$14, 289( P / F ,15%,3) + $16,512( P / F ,15%, 4) = $37, 490 Comments: As an alternative way of finding the equivalent cash flows in actual dollars, we may use the compound growth rate (geometric growth and inflation): g = (1 + 0.08)(1 + 0.07) − 1 = 15.56% P = $10, 000(1.07)( P / A1 ,15.56%,15%, 4) = $37, 490

11.13

Given: i = 9%, f = 3.8% , we find the inflation-free interest rate as follows:

i' =

i − f 0.09 − 0.038 = = 5.01% 1 + 0.038 1+ f

First compute the equivalent present worth of the constant dollar series at i ' : P = $1, 000( P / A,5.01%, 4) = $3,545.13

Then, we compute the equivalent annual payment in actual dollars using i: A = $3,545.13( A / P,9%, 4) = $1, 094.27

11.14 Given: i = 12%, f = 6% , bond interest rate = 9% compounded semiannually, face

value = $1,000 4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

• The 15th interest payment in actual dollars:

I15 = $1, 000(0.045) = $45 • The 15th interest payment (7.5th year) in constant dollars: I '15 = $45( P / F , 6%, 7.5) = $29.07

Equivalence Calculation under Inflation 11.15

0.02 12 ) −1 12 = 2.0184%

ia = (1 +

$10, 000[( F / P, 2.0184%,5) − ( F / P,3%,5)] = −$541.97 I will lose $541.97 when the general inflation rate is 3%.

11.16

i ' = 4%, f = 6% i = 0.04 + 0.06 + (0.04)(0.06) = 10.24% No, I would not be interested in an investment opportunity with 10% of interest rate.

11.17

(a) i =4% semiannually $50( P / A, 4%,10) + $1, 000( P / F ,8%,5) = $405.54 + $680.58 = $1, 086.12

(b) i' =

i− f = 4.85% 1+ f

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Contemporary Engineering Economics, 7th ed. ©2023 $50( P / A, 2.425%,10) + $1, 000( P / F , 4.85%,5) = $439.30 + $789.15 = $1, 228.45

11.18 Given: i′ = 4%, f = 6% i = 0.04 + 0.06 + (0.04)(0.06) = 0.1024

$50,000(1+ f )5 ( P / F ,10.24%,5) = $50,000(1+0.06)5 (0.6142) = $41, 096.91

11.19 (a) i = i ′ + f + i ′f = 0.05 + 0.06 + 0.05(0.06) = 0.113

(b)

 40, 000( P / A,11.3%, 4)  A( F / A,11.3%,8) =   ( F / P,11.3%,1)  +1, 000( P / G,11.3%, 4)  11.9897 A = $141,929.68 A = $11,837.63

11.20 0.11 − 0.06 1 + 0.06 = 4.72%

i' =

11.21

Given: i = 1% per month, f = 0.5% per month, P = $25,000, N = 60 months 0.01 − 0.005 1 + 0.005 = 0.4975% A ' = $25, 000( A / P, 0.4975%, 60) i' =

= $482.97

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Contemporary Engineering Economics, 7th ed. ©2023

11.22

Given: i ' = 6%, f = 5%, N = 5 years, A = $2.2 million in constant dollars

• Market interest rate: i = 0.06 + 0.05 + (0.06)(0.05) = 11.3%

• Actual dollar analysis: Period

Net Cash Flow in Actual $

Equivalent Present Worth

1 2

Net Cash Flow in Constant $ $2,200,000 2,200,000

$2,310,000 2,425,500

$2,075,472 1,957,992

3

2,200,000

2,546,775

1,847,162

4

2,200,000

2,674,114

1,742,606

5

2,200,000

2,807,819

1,643,968 $9, 267, 200

P = $2,310, 000( P / F ,11.3%,1) +  + $2,807,819( P / F ,11.3%,5) = $9, 267, 200 11.23

Given: i = 12%, f = 5%, g = 7%, N = 5 years, A1 = $5, 000 in constant dollars • Actual dollar analysis: Period

Net Cash Flow in Actual $

Equivalent Present Worth

1

Net Cash Flow in Constant $ $5,000

$5,250

$4,688

2 3

5,350 5,725

5,898 6,627

4,702 4,717

4

6,125

7,445

4,732

5

6,554

8,365

4,746 $23,584

P = $5, 250( P / F ,12%,1) +  + $8,356( P / F ,12%,5) = $23,584

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Contemporary Engineering Economics, 7th ed. ©2023

11.24

Given: i = 0.75% per month, f = 0.5% per month, P = $8, 000, N = 24 months (a) Inflation-free interest rate:

im ' =

0.0075 − 0.005 = 0.2488% per month 1 + 0.005

i ' = (1 + 0.002488 ) − 1 = 3.0263% per year 12

(b) Equal monthly payment in constant dollars:

A ' = $8, 000( A / P, 0.2488%, 24) = $343.80 11.25

Given: i = 6% compounded monthly, f = 4% compounded annually, number of months to deposit = 240 months, number of annual withdrawals = 15, first withdrawal = 6 months after retirement. •

Effective inflation rate per semiannual: Since the first withdrawals is made after 6 months from retirement, it is necessary to calculate the effective inflation rate per semiannual.

 1.04  f =  1  •

•

1/2

−1 = 1.98% per semiannual

Annual withdrawals in actual dollars: On semiannual basis, the first withdrawal will be made after 41 semiannual periods. Then, we can calculate the equivalent amount of this withdrawals in actual dollars using this formula Actual dollar = Constant dollar (F/P, f , N ) Then, the conversion of constant dollar to actual dollar is as follows:

N 41 43 45 47 49 51 53 55

Actual dollars $134,050 $139,411 $144,986 $150,784 $156,815 $163,086 $169,608 $176,391

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Contemporary Engineering Economics, 7th ed. ©2023 57 59 61 63 65 67 69 •

$183,445 $190,782 $198,427 $206,364 $214,618 $223,203 $232,131

Equivalence calculation: To find the required equal monthly deposit amount (A), we establish the following equivalence relationship:  0.06  im =   = 0.5%  12  12

 0.06  ia =  1 +  − 1 = 6.168% 12   A( F / A,0.5%, 240)( F / P,0.5%,6) = $134, 050 + $139, 411( P / F , 6.168%,1) +$144, 986( P / F , 6.168%, 2) + ⋅⋅ ⋅ +$232,131( P / F ,6.168%,14) A = $1, 747, 271 / 476.08 = $3,670 per month. $232 131

$134,050 (240 months)

Years

2

0

21

Monthly

22

Deposits

11.26

Given : i = 0.5%per month, f = 4% per year (a) 9 Copyright © 2023 Pearson Education, Inc.

29

35

Contemporary Engineering Economics, 7th ed. ©2023 • Actual dollar analysis:

A( F / A, 0.5%, 480) = $1, 000, 000( F / P, 4%, 40) (1,991.4907) A = $4,801, 021 A = $2, 410.77 (b) • Effective annual interest rate: ia = (1 + 0.06 / 12)12 − 1 = 6.1678%

• Equivalent purchasing power of $1,000,000 today at the end of 65th birthday: $1, 000, 000( F / P, 4%, 40) = $4,801, 021 (in actual dollars)

• Conversion of gradient series to equivalent uniform series:

A = G( A / G, 6.1678%, 40) = $1, 000(12.1962) = $12,196.20 • Amount of the first deposit ( A1 ) : ( A1 + $12,196.20)( F / A, 6.1678%, 40) = $4,801, 021 ( A1 + $12,196.20)(161.4438) = $4,801, 021 A1 + $12,196.20 = $29, 738.03 A1 = $17,541.83 11.27 Given: i = 8% per year, f = 6% per year (a) Freshman-year expense in actual dollars: $40, 000( F / P , 6%,10) = $71, 634

(b) Equivalent single-sum amount at n = 0 i− f = (0.08 − 0.06) / (1 + 0.06) 1+ f = 0.01887

i' =

P = [$40, 000 + $40, 000( P / A,1.887%,3)]( P / F ,1.887%,10) = $129, 077

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(c) Required annual deposit in actual dollars: A = $129, 077( A / P,8%,10) = $19, 236

Effects of Inflation on Project Cash Flows 11.28 Consider the following project’s after-tax cash flow and the expected annual general inflation rate during the project period: End of year 0 1 2 3

Cash flow in actual dollars -$45,000 32,000 32,000 32,000

Expected general inflation rate 3.5% 4.2% 5.5%

(a) The average annual general inflation rate:

(1 + 0.035)(1 + 0.042)(1 + 0.055) = 1.1378 (1 + f )3 = 1.1378 f = 4.40% (b) Constant dollars: Actual dollars -$45,000 32,000 32,000 32,000

n 0 1 2 3

Constant dollars -$45,000 32,000(0.9662) = 30,918 32,000(0.9272) = 29,670 32,000(0.8789) = 28,125

Conversion factors:

( P / F ,3.5%,1) = 0.9662 ( P / F , 4.2%,1)( P / F ,3.5%,1) = 0.9272 ( P / F ,5.5%,1)( P / F , 4.2%,1)( P / F ,3.5%,1) = 0.8789

11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c) The project is still profitable in an inflationary economy.

P = −$45, 000 + $30,918( P / F ,5%,1) +$29, 670( P / F ,5%, 2) + $28,125( P / F ,5%,3) = $35, 653 > 0

11.29

(a) and (b) 0

1

2

Income Statement Revenue Expenses: O&M Depreciation Interest

$ $ $

56,490 11,000 5,000

$ $ $

59,315 8,800 2,619

Taxable Income Income Taxes (25%)

$ $

41,510 10,378

$ $

43,266 10,816

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Working Capital Gains Tax Loan Repayment

$

31,133

$

32,449

$ $

31,133 11,000

$ $ $ (600) $ $ (23,810) $

32,449 8,800 29,768 12,600 1,358 (26,190)

($17,000) ($17,000)

$17,723 $16,879

$58,785 $53,319

PW(18%) = $ IRR (%) =

40,238 133.57%

$114,000

Net Cash Flow (actual) Net Cash Flow (constant)

$ $ $

(55,000) (12,000) $ 50,000

$

12 Copyright © 2023 Pearson Education, Inc.

$114,000

Contemporary Engineering Economics, 7th ed. ©2023 11.30 (a) With inflation: 0

1

Income Statement Revenue Expenses: O&M Depreciation Interest

$152,250

2

3

4

5

6

$159,863 $167,856 $176,248 $185,061 $194,314

$ 86,100 $ 90,405 $ 94,925 $ 99,672 $ 104,655 $ 109,888 $ 24,000 $ 38,400 $ 23,040 $ 13,824 $ 13,824 $ 6,912 $ 10,800 $ 10,800

Taxable Income Income Taxes (25%)

$31,350 $7,838

$20,258 $5,065

$49,891 $12,473

$62,752 $15,688

$66,582 $16,646

$77,514 $19,379

Net Income

$23,513

$15,194

$37,418

$47,064

$49,937

$58,136

$23,513 $24,000

$15,194 $38,400

$37,418 $23,040

$47,064 $13,824

$49,937 $13,824

$58,136 $6,912

Cash Flow Statement Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage $ (120,000) Gains Tax Working Capital Cash from financing activities: Loan repayment $ 120,000 Net Cash Flow (actual)

$0

$ 20,101 $ (5,025)

$ (120,000) $47,513

($66,407)

$60,458

$60,888

$63,761

$80,124

PW (18%) = $118,325

Note that the interest rate to use in present value calculation is an inflation-free interest rate, which can be determined as

i′ =

0.18 − 0.05 = 12.38% 1 + 0.05

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(b) Income Statement (without inflation) 0 Income Statement Revenue Expenses: O&M Depreciation Interest

1 $145,000

2

3

4

5

6

$145,000 $145,000 $145,000 $145,000 $145,000

82,000 24,000 10,800

82,000 38,400 10,800

82,000 23,040

82,000 13,824

82,000 13,824

82,000 6,912

Taxable Income Income Taxes (25%)

$28,200 $7,050

$13,800 $3,450

$39,960 $9,990

$49,176 $12,294

$49,176 $12,294

$56,088 $14,022

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working Capital Loan repayment

$21,150

$10,350

$29,970

$36,882

$36,882

$42,066

$21,150 $24,000

$10,350 $38,400

$29,970 $23,040

$36,882 $13,824

$36,882 $13,824

$42,066 $6,912 $15,000 (3,750)

$53,010

$50,706

$50,706

$60,228

Net Cash Flow (actual)

(120,000)

120,000 $0

(120,000) $45,150

($71,250)

PW (12.38%) = $111,089

(c) Present value gain (or loss) due to inflation:

PW(12.38%) no inflation = $111, 089 PW(18%) with inflation = $118,325 Present value gains = $118,325 - $111,089 =$7,236

(d) Present value gain due to borrowing:

n 0 1 2

Principal

Financing Cost Interest (A/T)

$120,000 -$120,000

-(1-0.25) (10,800) -(1-0.25) (10,800)

14 Copyright © 2023 Pearson Education, Inc.

Net Loan Flow $120,000 -$8,100 -$128,100

Contemporary Engineering Economics, 7th ed. ©2023

PW(18%) Loan = +$120, 000 − $8,100( P / F ,18%,1) −$128,100( P / F ,18%, 2) = $21,136 Comments: Present value gain due to inflation in (c) is largely from the fact that the firm was able to finance the project at 9% interest whereas the market interest rate is 18%. In practice, it is not likely that the firm would be able to access such a cheap money during the inflationary period.

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Contemporary Engineering Economics, 7th ed. ©2023 11.31

Effects of inflation on cash flows: (a) Project Cash Flows with Inflation 0

1

2

3

4

5

$17,364

$18,233

$19,144

Income Statement

Revenue Expenses: Depreciation Interest

$23,100

$20,948

$ 4,000 $

6,400 $ 3,840 $ 2,304 $ 1,152

$ 2,000

1,396

Taxable Income Income Taxes (25%)

$17,100 $4,275

$13,152 $3,288

$12,793 $3,198

$15,929 $3,982

$17,992 $4,498

Net Income

$12,825

$9,864

$9,595

$11,947

$13,494

$9,864 $6,400

$9,595 $3,840

$11,947 $2,304

$13,494 $1,152

$

$

731

Cash Flow Statement

Cash from operation Net Income $12,825 Depreciation $4,000 Cash from investing activities: Investment / Salvage $ (20,000) Gains Tax Cash from financing activities: Loan repayment $ 20,000 $ (6,042) $ Net Cash Flow (actual $)

$0

PW (20%) =

$10,783

$ 2,553 $ (62) (6,647) $ (7,311) $9,617

$6,124

$14,251

$17,137

3

4

6

$19,000

$15,000

$15,000

$15,000

$32,967

(b) Income Statement (without inflation) 0 Income Statement Revenue Expenses: Depreciation Interest

$22,000 4,000

6,400

3,840

2,304

1,152

2,000

1,396

731

0

0

Taxable Income Income Taxes (25%)

$16,000 $4,000

$11,204 $2,801

$10,429 $2,607

$12,696 $3,174

$13,848 $3,462

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Loan repayment

$12,000

$8,403

$7,822

$9,522

$10,386

$12,000 $4,000

$8,403 $6,400

$7,822 $3,840

$9,522 $2,304

$10,386 $1,152 $2,000 76

(6,042)

(6,647)

(7,311)

$9,958

$8,156

$4,351

$11,826

$13,614

1

2

(20,000) 20,000

Net Cash Flow (constant $)

$0

PW (14.29%) =

$31,784

PW gains due to inflation = $32,967 - $31,784 = $1,183 16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

11.32

(a), (b), and (c); The project is acceptable Income Statement 0 Income Statement Revenue (Labor Savings) Expenses: O&M Depreciation Interest

1

inflation 5%

2

$84,000

$

21,435 $

3

$88,200

36,735 $

$92,610

13,118

Taxable Income Income Taxes (28%)

$62,565 $17,518

$51,465 $14,410

$79,493 $22,258

Net Income

$45,047

$37,055

$57,235

$45,047 $21,435

$37,055 $36,735

$57,235 $13,118

Cash Flow Statement Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment

$ 8%

Net Cash Flow (actual) Net Cash Flow (constant)

6%

$

(150,000) (10,000) $

$ $ (864) $

(800) $

($160,000) ($160,000)

$65,682 $61,964

PW (20%) = $ PW (13.21%) = $

38,928 38,927

$72,926 $64,904

80,000 (361) 11,664

$161,656 $135,729

Therefore, the project is acceptable.

Note: The general inflation rate is 6% and this rate should be used in converting the actual dollars to it equivalent constant dollars

i' =

0.14 − 0.06 = 0.13208 1 + 0.06

17 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Rate of Return Analysis under Inflation 11.33

(a) 0 Income Statement Revenue Expenses: O&M Depreciation

1 $22,000 $ $

2

3

4

5

6

7

8

$22,000

$22,000

$22,000

$22,000

$22,000

$22,000

$22,000

9,500 $ 9,500 $ 9,500 $ 8,860 $ 15,184 $ 10,844 $

9,500 $ 7,744 $

9,500 $ 5,537 $

9,500 $ 5,530 $

9,500 $ 5,537 $

9,500 2,765

Taxable Income Income Taxes (25%)

$3,640 910

($2,684) (671)

$1,656 414

$4,756 1,189

$6,963 1,741

$6,970 1,742

$6,963 1,741

$9,735 2,434

Net Income

$2,730

($2,013)

$1,242

$3,567

$5,223

$5,227

$5,223

$7,301

Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Net Cash Flow

$ $ $

(62,000)

$

(10,000) ($72,000)

PW (12.38%)=

2,730 $ (2,013) $ 1,242 $ 8,860 $ 15,184 $ 10,844 $

$11,590 ($10,193.04)

$13,171

$12,086 IRR' (%)

18 Copyright © 2023 Pearson Education, Inc.

3,567 $ 7,744 $

$11,311 8.40%

5,223 $ 5,537 $

$10,759

5,227 $ 5,530 $

$10,758

5,223 $ 7,301 5,537 $ 2,765 $ 5,000 $ (1,250) $ 10,000 $10,759

$23,816

Contemporary Engineering Economics, 7th ed. ©2023

i' =

0.18 − 0.05 = 12.38% 1 + 0.05

(b) Project's IRR with inflation Income Statement 0

2

3

4

5

6

7

8

$23,760

$25,661

$27,714

$29,931

$32,325

$34,911

$37,704

$40,720

10,070 8,860

10,674 15,184

11,315 10,844

11,994 7,744

12,713 5,537

13,476 5,530

14,284 5,537

15,142 2,765

Taxable Income Income Taxes (25%)

$4,830 1,208

($197) (49)

$5,555 1,389

$10,193 2,548

$14,075 3,519

$15,905 3,976

$17,883 4,471

$22,814 5,703

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Loan repayment

$3,623

($148)

$4,166

$7,645

$10,557

$11,929

$13,412

$17,110

$3,623 $8,860

($148) $15,184

$4,166 $10,844

$7,645 $7,744

$10,557 $5,537

$11,929 $5,530

$13,412 $5,537

($10,000)

($800)

($864)

($933)

($1,008)

($1,088)

($1,175)

($1,269)

$17,110 $2,765 $7,387 ($1,847) $17,138

($72,000) ($72,000)

$11,682 $11,126

$14,172 $12,854

$14,077 $12,160

$14,381 $11,831

$15,005 $11,757

$16,284 $12,151

$17,679 $12,564

$42,554 $28,802

IRR' (%) =

10.00%

Income Statement Revenue Expenses: O&M Depreciation

1

($62,000)

Net Cash Flow (actual dollars) Net Cash Flow (constant dollars

PW (18%)= ($6,474) The project is not acceptable.

19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 11.34 Income Statement 0 Income Statement Revenue Expenses: O&M Rent Depreciation

1

2

3

4

5

6

$35,000

$38,000

$55,000

$60,000

$70,000

$60,000

$16,000 $10,080 $11,000

$21,000 $10,584 $17,600

$23,000 $11,113 $10,560

$32,000 $11,669 $6,336

$33,000 $12,252 $6,336

$33,000 $12,865 $3,168

Taxable Income Income Taxes (30%)

($2,080) ($11,184) $10,327 ($624) ($3,355) $3,098

$9,995 $2,999

$18,412 $5,524

$10,967 $3,290

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital

($1,456)

($7,829)

$7,229

$6,997

$12,888

$7,677

($1,456) $11,000

($7,829) $7,229 $17,600 $10,560

$6,997 $6,336

$12,888 $6,336

$7,677 $3,168 $13,401 ($4,020)

$17,789 $15,367

$13,333 $10,969

$19,224 $15,063

$20,226 $15,093

IRR' (%) =

8.57%

Net Cash Flow (Actual Dollar) Net Cash Flow (Constant Dollar)

5%

($55,000)

($55,000) ($55,000)

$9,544 $9,090

IRR(%) =

14%

$9,771 $8,863

Since MARR (inflation-adjusted) = 10%, and IRR = 14% > 10%, accept the investment. We can calculate the MARR’ (inflation-free) as 0.10 − 0.05 = 4.76% and IRR’ = 8.57% > 4.76%, the constant dollar 1 + 0.05 analysis would reach the same decision. i' =

20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 11.35 (a) Real after-tax yield on bond investment: • Nontaxable municipal bond: 0.09 − 0.03 = 5.825% 1 + 0.03 • Taxable corporate bond: ′ imunicipal =

′ icorporate =

0.12(1 − 0.3) − 0.03 = 5.243% 1 + 0.03

(b) Given i = 6%, and f = 3% ′ isavings = 2.91% ′ ′ Since imunicipal > 2.91% and icorporate > 2.91% , both bond investments are better than the savings account. The real return is best with the municipal bond.

11.36

(a), (b), and (c) – Select Engine B.

Engine A 0

1

2

3

4

5

Income Statement Revenue Expenses: O&M Depreciation

$518,400 $12,000

$559,872 $12,000

$604,662 $12,000

$653,035 $12,000

$705,277 $12,000

Taxable Income Income Taxes (25%)

($530,400) ($132,600)

($571,872) ($142,968)

($616,662) ($154,165)

($665,035) ($166,259)

($717,277) ($179,319)

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

($397,800)

($428,904)

($462,496)

($498,776)

($537,958)

($397,800) $12,000

($428,904) $12,000

($462,496) $12,000

($498,776) $12,000

($537,958) $12,000 $40,000

($100,000)

($385,800)

($416,904)

($450,496)

($486,776)

($485,958)

PW (20%)=

($1,401,765)

AE (20%)=

($468,722)

FW (20%)=

($3,488,041)

Net Cash Flow

($100,000)

21 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Engine B 0

1

2

3

4

5

Income Statement Revenue Expenses: O&M Depreciation

$407,808 $24,000

$440,433 $24,000

$475,667 $24,000

$513,721 $24,000

$554,818 $24,000

Taxable Income Income Taxes (25%)

($431,808) ($107,952)

($464,433) ($116,108)

($499,667) ($124,917)

($537,721) ($134,430)

($578,818) ($144,705)

Net Income

($323,856)

($348,324)

($374,750)

($403,290)

($434,114)

Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

($323,856) $24,000

($348,324) $24,000

($374,750) $24,000

($403,290) $24,000

($434,114) $24,000 $80,000

($200,000)

($299,856)

($324,324)

($350,750)

($379,290)

($330,114)

PW (20%)=

($1,193,665)

AE (20%)=

($399,137)

FW (20%)=

($2,970,221)

Net Cash Flow

11.37

($200,000)

(a) & (b) Actual and constant dollar analysis: 0

1

Income Statement Revenue Expenses: O&M Depreciation

2

$126,000

$132,300

$62,400 $12,000

$64,896 $9,600

Taxable Income Income Taxes

$51,600 $15,480

$57,804 $17,341

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Working capital Gains Tax

$36,120

$40,463

$36,120 $12,000

$40,463 $9,600 $40,000 $5,200 ($480)

Net Cash Flow (actual) Net Cash Flow (constant)

($60,000) ($5,000)

($65,000) ($65,000) IRR'(%)

=

($200)

$47,920 $44,370 51.04%

(c) Given f = 8%, i = 15% 0.15 − 0.08 = 6.48% (Inflation-free MARR) 1 + 0.08 Since IRR’> 6.48%, the project is a profitable one.

i′ =

22 Copyright © 2023 Pearson Education, Inc.

$94,783 $81,261

Contemporary Engineering Economics, 7th ed. ©2023 11.38

(a) & (b) Project cash flows in actual and constant dollars: Income Statement Revenue Expenses: O&M Depreciation Interest

$84,800

$89,888

$95,281 $100,998 $107,058 $113,482

$20,000

$32,000

$19,200

$11,520

$11,520

Taxable Income Income Taxes (25%)

$64,800 $16,200

$57,888 $14,472

$76,081 $19,020

$89,478 $22,370

$95,538 $107,722 $23,885 $26,931

Net Income Cash Flow Statement Operating Activities Net Income Depreciation Investment Activities Investment/Salvage Gain taxes Working capital Financing Activities Loan principal repayment

$48,600

$43,416

$57,061

$67,109

$71,654

$80,792

$48,600 $20,000

$43,416 $32,000

$57,061 $19,200

$67,109 $11,520

$71,654 $11,520

$80,792 $5,760

Net Cash Flow (Actual Dollar) Net Cash Flow (Constant Dollar)

($100,000)

$5,760

$42,556 ($10,639)

($100,000) ($100,000)

$68,600 $64,717

PW(18%) =

$179,509

$75,416 $67,120

$76,261 $64,030

$78,629 $62,281

IRR' (%) =

61.87%

23 Copyright © 2023 Pearson Education, Inc.

$83,174 $118,469 $62,152 $83,516

Contemporary Engineering Economics, 7th ed. ©2023 (c) The effects of project financing under inflation: Income Statement Revenue Expenses: O&M Depreciation Interest

$84,800

$89,888

$95,281 $100,998 $107,058 $113,482

$20,000 $12,000

$32,000 $10,521

$19,200 $8,865

$11,520 $7,010

$11,520 $4,933

Taxable Income Income Taxes (25%)

$52,800 $13,200

$47,367 $11,842

$67,216 $16,804

$82,468 $20,617

$90,605 $105,116 $22,651 $26,279

Net Income Cash Flow Statement Operating Activities Net Income Depreciation Investment Activities Investment/Salvage Gain taxes Working capital Financing Activities Loan principal repayment

$39,600

$35,525

$50,412

$61,851

$67,954

$78,837

$39,600 $20,000

$35,525 $32,000

$50,412 $19,200

$61,851 $11,520

$67,954 $11,520

$78,837 $5,760

Net Cash Flow (Actual Dollar) Net Cash Flow (Constant Dollar)

$5,760 $2,606

($100,000)

$42,556 ($10,639)

$100,000

($12,323) ($13,801) ($15,457) ($17,312) ($19,390) ($21,717)

$0 $0

$47,277 $44,601

PW(18%) =

$201,903

$53,724 $47,814

$54,155 $45,470

24 Copyright © 2023 Pearson Education, Inc.

$56,059 $44,404

$60,084 $44,898

$94,797 $66,828

Contemporary Engineering Economics, 7th ed. ©2023 (d) The present value loss due to inflation: 0 Income Statement Revenue Expenses: O&M Depreciation Interest

1

2

3

4

5

6

$80,000

$80,000

$80,000

$80,000

$80,000

$80,000

$20,000

$32,000

$19,200

$11,520

$11,520

$5,760

Taxable Income Income Taxes (25%)

$60,000 $15,000

$48,000 $12,000

$60,800 $15,200

$68,480 $17,120

$68,480 $17,120

$74,240 $18,560

Net Income Cash Flow Statement Operating Activities Net Income Depreciation Investment Activities Investment/Salvage Gain taxes Working capital Financing Activities Loan principal repayment

$45,000

$36,000

$45,600

$51,360

$51,360

$55,680

$45,000 $20,000

$36,000 $32,000

$45,600 $19,200

$51,360 $11,520

$51,360 $11,520

$55,680 $5,760

Net Cash Flow (Actual Dollar) Net Cash Flow (Constant Dollar)

($100,000)

$30,000 ($7,500)

($100,000) ($100,000)

$65,000 $61,321

PW(11.32%) =

$134,378

$68,000 $60,520

$64,800 $54,407

$62,880 $49,807

IRR' (%) =

53.29%

$62,880 $46,988

$83,940 $59,174

Present value loss = $134,533 - $179,509 = ($44,956) (e) Required additional before-tax annual revenue in actual dollars (equal amount) to make-up the inflation loss. $4,103( A / P,18%, 6) = $1,955 1 − 0.40

25 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies ST 11.1 (a) & (b) 0

1

2

3

$

84,000 $

88,200 $

92,610

$ $ $

21,000 $ 30,000 $ 9,000 $

22,050 $ 30,000 $ 9,000 $

23,153 30,000 9,000

Taxable Income Income Taxes (25%)

$ $

24,000 $ 6,000 $

27,150 $ 6,788 $

30,457 7,614

Net Income

$

18,000 $

20,363 $

22,843

$ $

18,000 $ 30,000 $

20,363 $ 30,000 $

22,843 30,000

(500) $

$ $ (525) $

30,000 (7,500) 11,025

$

(90,000)

49,838 $ $45,204

(3,632) ($3,138)

Income Statement Revenue (Labor Savings) Expenses: O&M Depreciation Interest

Cash Flow Statement Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment Net Cash Flow (actual) Net Cash Flow (constant)

inflation 5%

5%

$

(90,000)

$

(10,000) $

$

90,000

$

(10,000) $ ($10,000)

PW (15.5%) = PW (10%) =

$ $

26 Copyright © 2023 Pearson Education, Inc.

47,500 $ $45,238 66,127 66,127

Contemporary Engineering Economics, 7th ed. ©2023

ST.11.2 (a) & (b) The project cash flows and IRR with no inflation: inflation-free interest rate = 11.32% Income Statement Revenue Expenses: O&M Labor Material Energy Depreciation : Building Milling machine Jigs & dies

1 $80,000

2 $80,000

3 $80,000

4 $80,000

5 $80,000

6 $80,000

7 $80,000

8 9 10 $80,000 $80,000 $80,000

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $15,000 $9,000 $4,500

$3,000 $3,000 $3,000 $15,000 $15,000 $15,000 $9,000 $9,000 $9,000 $4,500 $4,500 $4,500

$15,719 $3,333

$26,939 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

Taxable Income Income Taxes (28%)

$29,448 $8,245

$17,116 $4,792

$27,780 $7,778

$34,020 $9,526

$38,677 $10,830

$35,355 $9,899

$34,232 $9,585

$42,113 $47,759 $48,500 $11,792 $13,373 $13,580

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Building Milling machine Jigs & dies Investment / Salvage Building Milling machine Jigs & dies (Replacement) Gains Taxes: Building Milling machine Jigs & dies

$21,203

$12,324

$20,002

$24,494

$27,847

$25,456

$24,647

$30,321 $34,386 $34,920

$21,203

$12,324

$20,002

$24,494

$27,847

$25,456

$24,647

$30,321 $34,386 $34,920

$15,719 $3,333

$26,939 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

Net Cash Flow

0

($110,000) ($10,000)

$4,906 $1,481

$0 $741

$0 $741

$0 $0

$0 $0

$10,000 $300 ($10,000)

$300

($2,800) ($84)

($84) ($120,000)

$4,906 $1,481

$40,255

PW (11.32%) $104,731

$43,708

$40,722 IRR (%) =

$38,974

$27,886

30.50%

27 Copyright © 2023 Pearson Education, Inc.

$38,601

$38,915

$36,708 $35,127 $42,336

Contemporary Engineering Economics, 7th ed. ©2023

(c), (d) & (e): The economic gain in present worth due to inflation = $125,477 - $104,731 = $20,746. (c) and (d) Income Statement (wiht inflation) 0 1 Revenue $85,600 Expenses: O&M 3,090 Labor 15,750 Material 9,360 Energy 4,635 Depreciation : Building Milling machine 15,719 Jigs & dies 3,333

2 $91,592

3 $98,003

4 $104,864

5 $112,204

6 $120,058

7 $128,463

8 9 $137,455 $147,077

10 $157,372

3,183 16,538 9,734 4,774

3,278 17,364 10,124 4,917

3,377 18,233 10,529 5,065

3,478 19,144 10,950 5,217

3,582 20,101 11,388 5,373

3,690 21,107 11,843 5,534

3,800 22,162 12,317 5,700

3,914 23,270 12,810 5,871

4,032 24,433 13,322 6,048

26,939 4,445

19,239 1,481

13,739 741

9,823 0

9,812 3,333

9,823 4,445

4,906 1,481

0 741

0 0 $109,537 30,670

Taxable Income Income Taxes (28%)

$33,713 9,440

$25,979 7,274

$41,599 11,648

$53,181 14,891

$63,592 17,806

$66,468 18,611

$72,021 20,166

$87,088 $100,470 24,385 28,132

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Building Milling machine Jigs & dies Investment / Salvage Building Milling machine Jigs & dies (Replacement) Gains Taxes: Building Milling machine Jigs & dies

$24,273

$18,705

$29,952

$38,291

$45,786

$47,857

$51,855

$62,704

$72,339

$78,867

24,273

18,705

29,952

38,291

45,786

47,857

51,855

62,704

72,339

78,867

15,719 3,333

26,939 4,445

19,239 1,481

13,739 741

9,823 0

9,812 3,333

9,823 4,445

4,906 1,481

0 741

0 0

(110,000) (10,000)

10,000 300 (10,000)

300

(2,800) (84)

(84)

Net Cash Flow (actual $) ($120,000) Net Cash Flow (constant $ ($120,000)

$43,325 $38,647

PW (18%) = $125,477

$50,089 $42,960

$50,672 $40,100 IRR (%) =

$52,771 $38,850

$45,825 $30,899

30.53%

28 Copyright © 2023 Pearson Education, Inc.

$61,002 $39,724

$66,123 $40,620

$69,091 $39,523

$73,080 $39,093

$86,283 $43,496

Contemporary Engineering Economics, 7th ed. ©2023

ST.11.3 (a)

Assumption: The building will be placed in service in January.

(b) Inflation-free interest rate

i' =

0.20 − 0.05 = 14.29% ← inflation-free discount rate (MARR) 1 + 0.05

IRR ' = 157.45% > 14.29% ← a good investment (c) Accept the investment.

29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Income Statement

2022 -2

2023 -1

2024 0

Revenues: Sales unit Unit price Sales volume Expenses: Fixed costs Variable costs Depreciation : Building Equipment Amortization Taxable Income Income Taxes (25%) Net Income Cash Flow Statement Operating Avtivities: Net Income Depreciation Amortization Investment activities Opportunity cost* Land Building Equipment Gains Taxes Land Building Equipment Working capital Net Cash Flow (actual) Net Cash Flow (constant)

2025 1

2026 2

2027 3

2028 4

2029 5

2030 6

2,000 $95,000 $190,000,000

2,000 $99,750 $199,500,000

2,000 $104,738 $209,476,000

2,000 $109,974 $219,948,000

2,000 $115,473 $230,946,000

2,000 $121,247 $242,494,000

$5,000,000 $114,000,000

$5,250,000 $119,700,000

$5,512,500 $125,685,600

$5,788,125 $131,968,800

$6,077,531 $138,567,600

$6,381,408 $145,496,400

$122,863 $1,214,650 $250,000 $69,412,487 $17,353,122

$128,205 $2,081,650 $250,000 $72,090,145 $18,022,536

$128,205 $1,486,650 $250,000 $76,413,045 $19,103,261

$128,205 $1,061,650 $250,000 $80,751,220 $20,187,805

$128,205 $759,050 $250,000 $85,163,614 $21,290,903

$122,863 $379,100 $250,000 $89,864,229 $22,466,057

$52,059,365

$54,067,609

$57,309,784

$60,563,415

$63,872,710

$67,398,172

$52,059,365 $1,337,513 $250,000

$54,067,609 $2,209,855 $250,000

$57,309,784 $1,614,855 $250,000

$60,563,415 $1,189,855 $250,000

$63,872,710 $887,255 $250,000

$67,398,172 $501,963 $250,000

($600,000) ($1,500,000) ($1,000,000)

($2,100,000) ($1,000,000) ($2,100,000) ($952,381) IRR' = 157.45%

($4,000,000) ($8,500,000)

$2,000,000 $3,000,000 $1,500,000

($1,000,000)

($125,000) $310,363 $4,313 $8,874,100

($13,500,000) ($12,244,898)

($1,425,000) $52,221,878 $45,111,222 PW (20%) =

($1,496,400) $55,031,064 $45,274,192 $124,942,384

($1,570,800)

($1,649,700)

($1,732,200)

$57,603,839 $45,134,115

$60,353,570 $45,036,763

$63,277,765 $44,970,326

.

30 Copyright © 2023 Pearson Education, Inc.

$83,713,911 $56,660,870

Contemporary Engineering Economics, 7th ed. ©2023

ST11.4 (a),(b),(c) & (d): Assumption: The building will be placed in service in January. Income Statement

2022 -2

2023 -1

2024 0

Revenues: Sales unit Unit price Sales volume Expenses: Fixed costs Variable costs Depreciation : Building Equipment Amortization Taxable Income Income Taxes (25%) Net Income Cash Flow Statement Operating Avtivities: Net Income Depreciation Amortization Investment activities Opportunity cost* Land Building Equipment Gains Taxes Land Building Equipment Working capital

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

($3,500,000)

PW (15%,n =-2)

($3,500,000) ($3,333,333)

2025 1

2026 2

2027 3

2028

2029

4

5

2030 6

2031 7

200,000 200,000 $511 $536 $102,102,525 $107,207,651

2032 8

200,000 $400 $80,000,000

200,000 $420 $84,000,000

200,000 $441 $88,200,000

200,000 $463 $92,610,000

200,000 $486 $97,240,500

200,000 $563 $112,568,034

$8,500,000 $52,000,000

$8,925,000 $54,600,000

$9,371,250 $57,330,000

$9,839,813 $60,196,500

$10,331,803 $63,206,325

$10,848,393 $66,366,641

$11,390,813 $69,684,973

$11,960,354 $73,169,222

$258,017 $2,715,100 $1,000,000 $15,526,884 $3,881,721

$269,231 $4,653,100 $1,000,000 $14,552,670 $3,638,167

$269,231 $3,323,100 $1,000,000 $16,906,420 $4,226,605

$269,231 $2,373,100 $1,000,000 $18,931,357 $4,732,839

$269,231 $1,696,700 $1,000,000 $20,736,441 $5,184,110

$269,231 $1,694,800 $1,000,000 $21,923,460 $5,480,865

$269,231 $1,696,700 $1,000,000 $23,165,934 $5,791,484

$269,231 $847,400 $1,000,000 $25,321,828 $6,330,457

$11,645,163

$10,914,502

$12,679,815

$14,198,518

$15,552,331

$16,442,595

$17,374,451

$18,991,371

$11,645,163 $2,973,117 $1,000,000

$10,914,502 $4,922,331 $1,000,000

$12,679,815 $3,592,331 $1,000,000

$14,198,518 $2,642,331 $1,000,000

$15,552,331 $1,965,931 $1,000,000

$16,442,595 $1,964,031 $1,000,000

$17,374,451 $1,965,931 $1,000,000

$18,991,371 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($9,600,000)

($480,000)

($504,000)

($529,200)

($555,660)

($583,443)

($612,615)

($643,246)

($500,000) $1,339,343 ($875,000) $13,508,164

($35,600,000) ($32,290,249)

$15,138,279 $13,077,015

$16,332,833 $13,437,062

$16,742,945 $13,118,536

$17,285,188 $12,898,474

$17,934,819 $12,745,941

$18,794,010 $12,720,526

$19,697,135 $12,696,949

$45,580,508 $27,982,478

$29,339,006 PW (15%,n = 0)PW(F /P ,15%,2) PW(A /P ,15%,8)= AE(15%)=

$38,800,835.12 $8,646,769.58

IRR' =

25.55%

Note: If the firm decides not to invest in the project, the firm could write off the R&D expenditure. The amount of write-off will be (0.25)($8,000,000) = $2,000,000. If the firm decides to undertake this project, then an opportunity cost of $2,000,000 will be incurred.

31 Copyright © 2023 Pearson Education, Inc.

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(e) Income Statement

2017 -2

2018 -1

2019 0

Revenues: Sales unit Unit price Sales volume Expenses: Fixed costs Variable costs Depreciation : Building Equipment Amortization Taxable Income Income Taxes (25%) Net Income Cash Flow Statement Operating Avtivities: Net Income Depreciation Amortization Investment activities Opportunity cost* Land Building Equipment Gains Taxes Land Building Equipment Working capital

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

2021 2

2022 3

2023 4

2024 5

2025 6

2026 7

2027 8

132,016 $400 $52,806,477

132,016 $420 $55,446,801

132,016 $441 $58,219,141

132,016 $463 $61,130,098

132,016 $486 $64,186,603

132,016 $511 $67,395,933

132,016 $536 $70,765,729

132,016 $563 $74,304,016

$8,500,000 $34,324,210

$8,925,000 $36,040,420

$9,371,250 $37,842,441

$9,839,813 $39,734,564

$10,331,803 $41,721,292

$10,848,393 $43,807,356

$11,390,813 $45,997,724

$11,960,354 $48,297,610

$269,231 $2,715,100 $1,000,000 $5,997,936 $1,499,484

$269,231 $4,653,100 $1,000,000 $4,559,050 $1,139,762

$269,231 $3,323,100 $1,000,000 $6,413,119 $1,603,280

$269,231 $2,373,100 $1,000,000 $7,913,391 $1,978,348

$269,231 $1,696,700 $1,000,000 $9,167,577 $2,291,894

$269,231 $1,694,800 $1,000,000 $9,776,153 $2,444,038

$269,231 $1,696,700 $1,000,000 $10,411,262 $2,602,815

$269,231 $847,400 $1,000,000 $11,929,421 $2,982,355

$4,498,452

$3,419,287

$4,809,839

$5,935,043

$6,875,683

$7,332,115

$7,808,446

$8,947,066

$4,498,452 $2,984,331 $1,000,000

$3,419,287 $4,922,331 $1,000,000

$4,809,839 $3,592,331 $1,000,000

$5,935,043 $2,642,331 $1,000,000

$6,875,683 $1,965,931 $1,000,000

$7,332,115 $1,964,031 $1,000,000

$7,808,446 $1,965,931 $1,000,000

$8,947,066 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($6,336,777)

($316,839)

($332,681)

($349,315)

($366,781)

($385,120)

($404,376)

($424,594)

($500,000) $1,336,539 ($875,000) $8,916,482

($3,500,000) ($32,336,777) ($3,333,333) ($29,330,410)

$8,165,944 $7,054,049

$9,008,937 $7,411,675

$9,052,855 $7,093,149

$9,210,593 $6,873,087

$9,456,494 $6,720,554

$9,891,769 $6,695,139

$10,349,783 $6,671,562

$30,941,718 $18,995,530

($3,500,000)

PW (15%,n =-2) =

2020 1

$3,347,321 PW (15%,n =0) =PW(F /P ,15%,2)

$4,198,879.81

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(f) Income Statement

2017 -2

2018 -1

2019 0

Revenues: Sales unit Unit price Sales volume Expenses: Fixed costs Variable costs Depreciation : Building Equipment Amortization Taxable Income Income Taxes (25%) Net Income Cash Flow Statement Operating Avtivities: Net Income Depreciation Amortization Investment activities Opportunity cost* Land Building Equipment Gains Taxes Land Building Equipment Working capital

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

2020 1

2021 2

2022 3

2023

2024

4

5

2025 6

2026 7

2027 8

200,000 $400 $80,000,000

200,000 $388 $77,600,000

200,000 $376 $75,272,000

200,000 $365 $73,013,840

200,000 $354 $70,823,425

200,000 $343 $68,698,722

200,000 $333 $66,637,760

200,000 $323 $64,638,628

$8,500,000 $52,000,000

$8,925,000 $50,440,000

$9,371,250 $48,926,800

$9,839,813 $47,458,996

$10,331,803 $46,035,226

$10,848,393 $44,654,169

$11,390,813 $43,314,544

$11,960,354 $42,015,108

$269,231 $2,715,100 $1,000,000 $15,515,670 $3,878,917

$269,231 $4,653,100 $1,000,000 $12,312,670 $3,078,167

$269,231 $3,323,100 $1,000,000 $12,381,620 $3,095,405

$269,231 $2,373,100 $1,000,000 $12,072,701 $3,018,175

$269,231 $1,696,700 $1,000,000 $11,490,465 $2,872,616

$269,231 $1,694,800 $1,000,000 $10,232,129 $2,558,032

$269,231 $1,696,700 $1,000,000 $8,966,473 $2,241,618

$269,231 $847,400 $1,000,000 $8,546,536 $2,136,634

$11,636,752

$9,234,502

$9,286,215

$9,054,526

$8,617,849

$7,674,097

$6,724,855

$6,409,902

$11,636,752 $2,984,331 $1,000,000

$9,234,502 $4,922,331 $1,000,000

$9,286,215 $3,592,331 $1,000,000

$9,054,526 $2,642,331 $1,000,000

$8,617,849 $1,965,931 $1,000,000

$7,674,097 $1,964,031 $1,000,000

$6,724,855 $1,965,931 $1,000,000

$6,409,902 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($9,600,000)

$288,000

$279,360

$270,979

$262,850

$254,964

$247,315

$239,896

($500,000) $1,336,539 ($875,000) $7,756,635

($3,500,000) ($35,600,000) ($3,333,333) ($32,290,249)

$15,909,083 $13,742,864

$15,436,193 $12,699,394

$14,149,524 $11,086,523

$12,959,706 $9,670,732

$11,838,744 $8,413,574

$10,885,443 $7,367,696

$9,930,681 $6,401,405

$27,244,706 $16,725,886

($3,500,000)

PW (15%,n =-2) = $13,988,120 PW (15%) = AE(15%) =

PW(F /P ,15%,2) PW(A /P ,15%,8)

$18,499,289 $4,122,568

33 Copyright © 2023 Pearson Education, Inc.

IRR' =

18.99%

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 12 Project Risk and Uncertainty Note to Instructors: Regular MACRS depreciation is assumed in all problems unless otherwise mentioned. However, instructors may adopt the capital expensing option under the 2017 Tax Cuts and Job Acts instead. In that case, the entire capital expenditure would be claimed as depreciation amount at the end of first year. Then, any salvage value at the end of project life will be subject to taxable gains.

Sensitivity Analysis 12.1

(a)

AEC(10%) = (25,000 − 5, 000)( A / P,10%,6) + 0.1(5, 000) + 3, 000 = $8, 092

(b)

AEC(10%) = (25,000 − 5, 000)( A / P,10%,5) + 0.1(5, 000) + 3,000 = $8, 776 (c)

AEC(10%) = (25,000 − 5, 000)( A / P,10%,6) + 0.1(5, 000) + 3,300 = $8,392

1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.2 With the unit price of $40 estimate, PW(15%) = $4,553,091 > 0 0 Income Statement Revenue: Unit price Demand (units) Sales revenue Expenses: unit vairable cost Variable cost Fixed cost Depreciation Taxable income Income taxes (25%) Net income Cash Flow Statement Operating activities Net income Depreciation Investment activities Investment activities Salvage Gains tax Net cash flow

1

2

$

4

5

40 $ 100,000 $ 4,000,000 $

40 $ 40 $ 40 $ 40 100,000 100,000 100,000 100,000 4,000,000 $ 4,000,000 $ 4,000,000 $ 4,000,000

$ 18 $ 1,800,000 $ 230,000 $ 90,000 $ 1,880,000 $ 470,000 $ 1,410,000

18 1,800,000 230,000 90,000 1,880,000 470,000 1,410,000

$ $ $ $ $ $ $

$ 1,410,000 $ $ 90,000 $ $

3

$ 18 $ 18 $ 18 $ 1,800,000 $ 1,800,000 $ 1,800,000 $ 230,000 $ 230,000 $ 230,000 $ 90,000 $ 90,000 $ 90,000 $ 1,880,000 $ 1,880,000 $ 1,880,000 $ 470,000 $ 470,000 $ 470,000 $ 1,410,000 $ 1,410,000 $ 1,410,000

1,410,000 $ 1,410,000 $ 1,410,000 $ 1,410,000 90,000 $ 90,000 $ 90,000 $ 90,000

(500,000) $ $

$ (500,000) $ 1,500,000 $ PW(15%) = $ 4,553,091

50,000 -

1,500,000 $ 1,500,000 $ 1,500,000 $ 1,550,000

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Contemporary Engineering Economics, 7th ed. ©2023

0 Income Statement Revenue: Unit price Demand (units) Sales revenue Expenses: unit vairable cost Variable cost Fixed cost Depreciation Taxable income Income taxes (25%) Net income Cash Flow Statement Operating activities Net income Depreciation Investment activities Investment activities Salvage Gains tax Net cash flow

1

2

$

$ PW(15%) = $

4

5

21.89 $ 100,000 $ 2,188,989 $

21.89 $ 21.89 $ 21.89 $ 21.89 100,000 100,000 100,000 100,000 2,188,989 $ 2,188,989 $ 2,188,989 $ 2,188,989

$ 18 $ 1,800,000 $ 230,000 $ 90,000 $ 68,989 $ 17,247 $ 51,742

18 1,800,000 230,000 90,000 68,989 17,247 51,742

$ $ $

3

$ $ $ $ $ $ $

51,742 $ 90,000 $

$ 18 $ 18 $ 18 $ 1,800,000 $ 1,800,000 $ 1,800,000 $ 230,000 $ 230,000 $ 230,000 $ 90,000 $ 90,000 $ 90,000 $ 68,989 $ 68,989 $ 68,989 $ 17,247 $ 17,247 $ 17,247 $ 51,742 $ 51,742 $ 51,742

51,742 $ 90,000 $

51,742 $ 90,000 $

51,742 $ 90,000 $

51,742 90,000

$ $

50,000 -

141,742 $

191,742

(500,000)

(500,000) $ 141,742 $ -

141,742 $

141,742 $

The breakeven value of unit price is $21.89. If the unit price is greater than $21.89, the project’s rate of return will be greater than 15%. 3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.3 (a) Project cash flows based on most-likely estimates: 1

2

3

4

5

6

7

8

Income Statement Revenue: Bill savings Mile Savings Expenses: Depreciation

0

$3,000,000 1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

$3,000,000 $1,250,000

2,000,000

3,200,000

1,920,000

1,152,000

1,152,000

576,000

Taxable Income Income Tax (25%)

2,250,000 562,500

1,050,000 262,500

2,330,000 582,500

3,098,000 774,500

3,098,000 774,500

3,674,000 918,500

4,250,000 1,062,500

4,250,000 1,062,500

$1,687,500

$787,500

$1,747,500

$2,323,500

$2,323,500

$2,755,500

$3,187,500

$3,187,500

1,687,500 2,000,000

787,500 3,200,000

1,747,500 1,920,000

2,323,500 1,152,000

2,323,500 1,152,000

2,755,500 576,000

3,187,500 0

3,187,500 0

3,687,500

3,987,500

3,667,500

3,475,500

3,475,500

3,331,500

3,187,500

3,187,500

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage

(10,000,000)

Net Cash Flow

(10,000,000)

PW (18%) =

$4,615,437

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Contemporary Engineering Economics, 7th ed. ©2023

(b)

Sensitivity analysis:

Percentage deviation

Savings In T.B

-30% -20% -10% 0 (base) +10% +20% +30%

$2,100,000 2,400,000 2,700,000 3,000,000 3,300,000 3,600,000 3,900,000

PW(18%)

Savings In D.M

PW(18%)

$1,863,080 2,780,532 3,697,985 4,615.437 5,532,889 6,450,341 7,367,794

$875,000 1,000,000 1,125,000 1,250,000 1,375,000 1,500,000 1,625,000

$3,468,821 3,850,893 4,233,165 4,615,437 4,997,709 5,379,980 5,762,252

(c) Sensitivity diagrams Not provided

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Contemporary Engineering Economics, 7th ed. ©2023 12.4 (a) Project cash flows based on most-likely estimates: 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Salvage Gains Tax

1

2

3

$45,000 20,000 $25,000 6,250 $18,750

$45,000 32,000 $13,000 3,250 $9,750

$45,000 9,600 $35,400 8,850 $26,550

18,750 20,000

9,750 32,000

26,550 9,600

(100,000) 30,000 2,100

Net Cash Flow

(100,000)

PW (15%) =

38,750

41,750

68,250

$10,140.13

(b) With the required investment of $110,000 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Salvage Gains Tax Net Cash Flow PW (15%) =

1

2

3

$48,162 22,000 $26,162 6,541 $19,622

$48,162 35,200 $12,962 3,241 $9,722

$48,162 10,560 $37,602 9,401 $28,202

19,622 22,000

9,722 35,200

28,202 10,560

(110,000) 30,000 3,060 (110,000)

41,622

44,922

71,822

$7,384.12

If the required investment is $110,000, the required savings would be at least $48,162 to breakeven.

6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 12.5 (a) Project cash flows based on most-likely estimates: without the working capital 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax

1

2

3

4

$35,000 21,600 $13,400 3,350 $10,050

$35,000 34,560 $440 110 $330

$35,000 20,736 $14,264 3,566 $10,698

$35,000 6,221 $28,779 7,195 $21,584

10,050 21,600

330 34,560

10,698 20,736

21,584 6,221 30,000 (1,279)

31,650

34,890

31,434

56,526

(108,000)

Net Cash Flow

(108,000)

PW (10%) =

$11,832

The project is acceptable. (b) Project cash flows based on most-likely estimates: with the working capital 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax Working Capital Net Cash Flow PW (10%) =

1

2

3

4

$35,000 21,600 $13,400 3,350

$35,000 34,560 $440 110

$35,000 20,736 $14,264 3,566

$35,000 6,221 $28,779 7,195

$10,050

$330

$10,698

$21,584

10,050 21,600

330 34,560

10,698 20,736

21,584 6,221 30,000 (1,279) 5,000

31,650

34,890

31,434

61,526

(108,000) (5,000) (113,000) $10,247

The project is still acceptable.

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c) Required annual savings (X): $32,520 using the Goal Seek function in Excel Income Statement

0

1

2

3

4

Labor Savings Depreciation

$ $

32,520 $ 21,600 $

32,520 $ 34,560 $

32,520 $ 20,736 $

32,520 6,221

Taxable Income Income Taxes (25%)

$ $

10,920 $ 4,368 $

(2,040) $ (816) $

11,784 $ 4,714 $

26,299 10,520

Net Income Cash Flow Statement Cash from Operation Net Income Depreciation Investment & Salvage Gains tax

$

6,552 $

(1,224) $

7,070 $

15,779

$ $

6,552 $ 21,600 $

(1,224) $ 34,560 $

7,070 $ 20,736 $ $ $

15,779 6,221 30,000 (1,279)

$(108,000) $

34,704 $

32,112 $

34,877 $

66,501

$(108,000)

Net Cash Flow

PW(18%) = $

-

12.6 • Project’s IRR if the investment is made now:

PW(i ) = −$500, 000 + $200, 000( P / A, i,5) = 0

i = 28.65% • Let X denote the new after-tax annual cash flow: PW(28.65%) = −$500, 000 + X ( P / A, 28.65%, 4)( P / F , 28.65%,1) = 0

X = $290, 248

The needed additional flow is $290,248-$200,000 = $90,248. 12.7 (a) Economic building height • 5% < i < 20% : The optimal building height is 5 floors. • 20% ≤ i < 30% : The optimal building height is 2 floors.

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Contemporary Engineering Economics, 7th ed. ©2023 Net Cash Flows n 0 1 2 3 4

2 Floors ($500,000) $199,100 $199,100 $199,100 $199,100

3 Floors ($750,000) $169,200 $169,200 $169,200 $169,200

4 Floors ($1,250,000) $149,200 $149,200 $149,200 $149,200

5 Floors ($2,000,000) $378,150 $378,150 $378,150 $378,150

Sensitivity Analysis PW(i) as a Function of Interest Rate i (%) 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

2 Floors $832,115 $787,037 $744,141 $703,298 $664,388 $627,298 $591,924 $558,167 $525,937 $495,148 $465,720 $437,580 $410,657 $384,885 $360,205 $336,557 $313,889 $292,150 $271,292 $251,271 $232,044 $213,572 $195,817 $178,745 $162,323 $146,519

3 Floors $687,721 $635,264 $585,441 $538,091 $493,067 $450,230 $409,452 $370,612 $333,599 $298,309 $264,644 $232,512 $201,829 $172,516 $144,496 $117,701 $92,066 $67,527 $44,029 $21,516 ($62) ($20,753) ($40,601) ($59,650) ($77,939) ($95,505)

4 Floors $963,010 $873,011 $787,722 $706,879 $630,199 $557,428 $488,330 $422,686 $360,291 $300,953 $244,495 $190,751 $139,565 $90,792 $44,298 ($46) ($42,357) ($82,746) ($121,319) ($158,173) ($193,399) ($227,084) ($259,308) ($290,148) ($319,674) ($347,955)

5 Floors $1,987,770 $1,834,680 $1,689,448 $1,551,593 $1,420,666 $1,296,250 $1,177,957 $1,065,427 $958,321 $856,326 $759,148 $666,513 $578,166 $493,867 $413,393 $336,533 $263,091 $192,883 $125,737 $61,490 ($9) ($58,903) ($115,327) ($169,407) ($221,261) ($271,002)

Best Floor Plan 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 2 2 2 2 2 2 2 2 2 2 2

(b) Effects of overestimation on resale value: Resale value Base 10% error Difference

Present Worth as a Function of Number of Floors 2 Floors $465, 720 $435,890 $29,831

3 Floors $264, 644 $219,898 $44, 746

4 Floors $244, 495 $145, 060 $99, 435

12.8 9 Copyright © 2023 Pearson Education, Inc.

5 Floors $759,148 $609, 995 $149,153

Contemporary Engineering Economics, 7th ed. ©2023 Note: In the problem statement, the current book value for the defender is given as $13,000. This implies that the machine has been depreciated under the alternative MACRS with the half-year convention. In other words, the allowed depreciation is based on a 10-year recovery period with the straight-line method (with $0 salvage). Note that, if you decide to retain the old machine, the current book value will be $13,000 as you continue to depreciate the asset without any adjustment. This is a replacement problem which we will discuss in Chapter 14. However, we can still work on this problem without any knowledge of replacement methodology. (a) •

Cost of retaining the old machine

Keep the old machine n Financial Data Depreciation Book value Market value Gain/loss Removal cost Operating cost

4 0

5 6 7 8 9 10 1 2 3 4 5 6 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 $13,000 $11,000 $9,000 $7,000 $5,000 $3,000 $1,000 $1,000

2,000

Cash Flow Statement Removal +(.25)*(Depreciation) Net proceeds from sale -(1-0.25)*(Operating cost) Net Cash Flow

2,000

2,000

2,000

2,000

1,500 2,000

($1,125) 500 500 500 500 500 500 1,000 (1,500) (1,500) (1,500) (1,500) (1,500) (1,500) $0 ($1,000) ($1,000) ($1,000) ($1,000) ($1,000) ($1,125)

PW (10%) = ($4,426)

AEC (10%) = $1,016

With the half-year convention mandated, the book value that should be used in determining the gains tax for the defender (if sold now) is Total depreciation = $1, 000 + $2, 000 + $2, 000 + $1, 000 = $6, 000 Book value = $20, 000 − $6, 000 = $14, 000 Taxable gain (loss) = $6, 000 − $14, 000 = ($8, 000) Net proceeds from sale = $6, 000 + $8, 000 × 0.25 = $8, 000 10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

Cost of replacing the old machine:

Buy a new machine Financial Data Depreciation Book value Market value Gain/loss Operating cost

n

0 $12,000

Cash Flow Statement Sale of old equipment Investment +(.25)*(Depreciation) -(1-0.25)*(Operating cost) Net proceeds from sale

6 $691 0 2000 2000 1000

1000

1000

1000

1000

1000

600 (750)

960 (750)

576 (750)

346 (750)

346 (750)

173 (750) 1,500

($150)

$210

($174)

($404)

($404)

$923

AEC (10%) =

$941

8,000 (12,000)

Net Cash Flow

($4,000) PW (10%) = ($4,100)

•

1 2 3 4 5 $2,400 $3,840 $2,304 $1,382 $1,382 9600 5760 3456 2074 691

Incremental cash flows: Net Cash Flow Old Machine New Machine

n 0 1 2 3 4 5 6

$0 -$1,000 -$1,000 -$1,000 -$1,000 -$1,000 -$1,125

-$4,000 -$150 $210 -$174 -$404 -$404 $923

Incremental Cash Flow (New - Old) -$4,000 $850 $1,210 $826 $596 $596 $2,048

IRR new-old = 12.5% > 10% → Replace the old machine.

PW(10%) new-old = $326.5 > 0 → Replace the old machine.

11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Sensitivity analysis: The answer remains unchanged. In fact, it (an increase in O&M) will make the challenger more attractive. Keep the old machine n

4 0

5 1

$13,000

$2,000 $11,000

Financial Data

Depreciation Book value Market value Gain/loss Removal cost Operating cost

6 2

7 3

$2,000 $9,000

$2,000 $7,000

8 4 $2,000 $5,000

9 5 $2,000 $3,000

10 6 $2,000 $1,000

$1,000

2,100

2,205

2,315

2,431

2,553

1,500 2,680

Cash Flow Statement

Removal +(.25)*(Depreciation) Net proceeds from sale -(1-0.25)*(Operating cost) Net Cash Flow

(1,654)

(1,736)

(1,823)

(1,914)

($1,075)

($1,154)

($1,236)

($1,323)

($1,414)

($1,635)

500

$0

W (10%) = ($5,565)

(1,575)

($1,125) 500 1,000 (2,010)

AEC (10%) =

500

500

500

500

$1,278

Incremental cash flows: Net Cash Flow Old Machine New Machine

n 0 1 2 3 4 5 6

$0 -$1,075 -$1,154 -$1,236 -$1,323 -$1,414 -$1,635

Incremental Cash Flow (New - Old)

-$4,000 -$150 $210 -$174 -$404 -$404 $923

-$4,000 $925 $1,364 $1,062 $919 $1,010 $2,558

IRR new-old = 20.3%, and PW(10%) new-old = $1, 465 (c) Break-even salvage value: Let X denote the minimum salvage value for the 12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 old machine. Then, the net proceeds from sale of the old machine will be Total depreciation = $6, 000 (with the half-year convention) Book value = $14, 000 Salvage value = X Taxable gain = X − $14, 000 Net proceeds = X − (0.25)( X − $14, 000) = 0.75 X + $3, 500 To find the break-even salvage value, PW(10%)old = −$1, 000( P / A,10%,5) − $1,125( P / F ,10%, 6) = −$4, 426 PW(10%) new = −[$12, 000 − (0.75 X + $3,500)] +$150( P / F ,10%,1) +  + $923( P / F ,10%, 6) = 0.75 X − $8,599

Let PW(10%)old = PW(10%) new and solve for X . X = $5, 564

12.9 (a) Net cash flows: 13 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

•

Paying $0.60 per mile to a sales rep. AEC(15%) = $0.60(30, 000)(1 − 0.25) = $13, 500

•

Providing a car to a sales rep.

Provide a car to a sales rep. Financial Data Depreciation Book value Market value Gain/loss Storage cost Operating cost

n

0

1 $5,000 $25,000 20000

2400 6000

Cash Flow Statement Investment +(.25)*(Depreciation) '-(1-0.25)*(Storage cost) -(1-0.25)*(Operating cost) Net proceeds from sale

2 3 4 5 $8,000 $4,800 $2,880 $1,440 12000 7200 4320 2880 5000 2120 2400 2400 2400 2400 6000 6000 6000 6000

(25,000) 1,250 2,000 1,200 720 360 (1,800) (1,800) (1,800) (1,800) (1,800) (4,500) (4,500) (4,500) (4,500) (4,500) 4,470

Net Cash Flow

($25,000) ($5,050) ($4,300) ($5,100) ($5,580) ($1,470) PW (15%) =

($39,917)

AEC (15%) = $11,908

It is cheaper to provide a car to a sales rep. (b) Breakeven analysis: $0.60 X (1 − 0.25) = $1,1908 X b* = 26, 462 miles per year

14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.10 (a) With infinite planning horizon: We assume that both machines will be available in the future with the same cost. Model A Financial Data n

0

Depreciation Book value Market value Gains/Loss O&M cost

$6,000

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(O&M cost) Net proceeds from sale

1 $857 $5,143

2 $1,469 $3,673

3 $1,049 $2,624

4 $749 $1,874

5-7 $536 $1,339

8 $268 $0 $500 $500 $700

$700

$700

$700

$700

$700

$257 ($490)

$441 ($490)

$315 ($490)

$225 ($490)

$161 ($490)

$80 ($490) $350

($233)

($49)

($175)

($265)

($329)

($60)

($6,000)

Net Cash Flow

($6,000) PW (10%) = ($7,152)

AEC (10%) =

$1,341

Model B Financial Data n

0

Depreciation Book value Market value Gains/Loss O&M cost

$8,500

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(O&M cost) Net proceeds from sale

1 $1,215 $7,285

2 $2,082 $5,204

3 $1,487 $3,717

4 $1,062 $2,655

5-7 $759 $1,896

8 $379 ($0)

9

$520

$520

$520

$520

$520

$520

$520

$364 ($364)

$624 ($364)

$446 ($364)

$318 ($364)

$228 ($364)

$114 ($364)

$0 ($364)

$0 ($364) $700

($46)

($136)

($250)

($364)

$336

($0)

10 ($0) $1,000 $1,000 $520

($8,500)

Net Cash Flow

($8,500) PW (10%) = ($8,627)

$0

AEC (10%) =

$260

$82

$1,404

Model A is preferred.

15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Break-even annual O&M costs for machine A: Let X denotes a before-tax annual operating cost for model. PW(10%) A = −$6, 000 + ($257 − 0.7 X )( P / F ,10%,1) +  + ($430 − 0.7 X )( P / F ,10%,8) = −$4,538 − 3.734 X AEC(10%) A = $851 + 0.7 X Let AEC(10%) A = AEC(10%) B , and solve for X. $851 + 0.7 X = $1, 404

∴ X = $791 per year Model A Financial Data n Depreciation Book value Market value Gains/Loss O&M cost

0

1 2 3 4 5-7 $857 $1,469 $1,049 $749 $536 $6,000 $5,143 $3,673 $2,624 $1,874 $1,339

$791

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(O&M cost) Net proceeds from sale

$791

$791

$791

$791

8 $268 $0 $500 $500 $791

($6,000) $257 $441 $315 $225 $161 $80 ($553) ($553) ($553) ($553) ($553) ($553) $350

Net Cash Flow

($6,000) ($296) ($113) ($239) ($329) ($393) ($123) PW (10%) = ($7,490)

AEC (10%) = $1,404

(c) With a shorter service life: n 0 1 2 3 4 5 PW(10%)

Net Cash Flow Model A Model B -$6,000 -$8,500 -233 0 -49 260 -175 82 -265 -46 2,172 2,883 -$5,216 -$6,464

Model A is still preferred over Model B. 16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.11 Assuming that all old looms were fully depreciated (a) • Project cash flows: Alternative 1 Alternative 1 Financial Data n Depreciation Book value Market value Gain/Loss Annual sales Annual labor cost Annual O&M cost Cash Flow Statement Investment +(0.40)*Dn +(0.60)*Sales -(0.60)*Labor -(0.60)*O&M Net proceeds from sale Net Cash Flow

0

1 $306,669 1,839,367

2 $525,564 1,313,803

3 $375,342 938,462

4 $268,040 670,422

5 $191,641 478,781

5 $191,426 287,354

7 $191,641 95,713

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

7,915,748 261,040 1,092,000

122,667 4,749,449 (156,624) (655,200)

210,226 4,749,449 (156,624) (655,200)

150,137 4,749,449 (156,624) (655,200)

107,216 4,749,449 (156,624) (655,200)

76,656 4,749,449 (156,624) (655,200)

76,571 4,749,449 (156,624) (655,200)

76,656 4,749,449 (156,624) (655,200)

38,285 4,749,449 (156,624) (655,200) 101,400

($2,108,836) $4,060,292

$4,147,850

$4,087,761

$4,044,841

$4,014,281

$4,014,195

$4,014,281

$4,077,310

AE (18%) =

$3,549,127

$2,146,036

8 $95,713 0 169,000 169,000 7,915,748 261,040 1,092,000

($2,108,836)

PW (18%) = $14,471,800

17 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

•

Sensitivity analysis for alternative 1

Change

MARR

Labor cost

O&M

Revenue

-30%

$17,662,515

$14,511,620

$14,638,377

$13,264,313

-20%

$16,496,280

$14,498,347

$14,582,851

$13,666,809

-10%

$15,436,786

$14,485,073

$14,527,326

$14,069,305

0%

$14,471,800

$14,471,800

$14,471,800

$14,471,800

10%

$13,590,722

$14,458,527

$14,416,275

$14,874,296

20%

$12,784,336

$14,445,254

$14,360,749

$15,276,792

30%

$12,044,608

$14,431,981

$14,305,224

$15,679,287

18 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

• Project cash flows: Alternative 2 Alternative 2 Financial Data n Depreciation Book value Market value Gain/Loss Annual sales Annual labor cost Annual O&M cost Cash Flow Statement Investment +(0.40)*Dn +(0.60)*Sales -(0.60)*Labor -(0.60)*O&M Net proceeds from sale Net Cash Flow

0

1 $160,083 960,159

2 $274,347 685,812

3 $195,930 489,882

4 $139,918 349,964

5 $100,038 249,926

5 $99,926 150,000

7 $100,038 49,963

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

7,455,084 422,080 1,560,000

64,033 4,473,050 (253,248) (936,000)

109,739 4,473,050 (253,248) (936,000)

78,372 4,473,050 (253,248) (936,000)

55,967 4,473,050 (253,248) (936,000)

40,015 4,473,050 (253,248) (936,000)

39,970 4,473,050 (253,248) (936,000)

40,015 4,473,050 (253,248) (936,000)

19,985 4,473,050 (253,248) (936,000) 32,400

($1,083,042) $3,347,835

$3,393,541

$3,362,175

$3,339,770

$3,323,817

$3,323,773

$3,323,817

$3,336,188

AE (18%) =

$3,084,026

$1,120,242

8 $49,963 0 54,000 54,000 7,455,084 422,080 1,560,000

($1,083,042)

PW (18%) = $12,575,319

Note: Cost basis for the new looms = $ 1,071,240 + $ 49,002 = $ 1,120,242 Net investment required = Cost basis - Net proceeds from sale of the old looms = $ 1,120,242 - $ 62,000 (1-0.40) = $ 1,083,042

19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

• Sensitivity analysis for alternative 2 Change

MARR

Labor cost

O&M

Revenue

-30%

$15,205,898

$12,885,109

$13,720,299

$7,103,571

-20%

$14,244,385

$12,781,846

$13,338,639

$8,927,487

-10%

$13,370,886

$12,678,582

$12,956,979

$10,751,403

0%

$12,575,319

$12,575,319

$12,575,319

$12,575,319

10%

$11,848,941

$12,472,055

$12,193,658

$14,399,234

20%

$11,184,155

$12,368,792

$11,811,998

$16,223,150

30%

$10,574,337

$12,265,528

$11,430,338

$18,047,066

20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b) Sensitivity graph

$2,00,00,000

Sensitivity Graph for Alt 1

NPW

$1,50,00,000

$1,00,00,000

MARR

$50,00,000

Labor O&M Revenue

$-30%

-20%

-10%

0% Change

10%

20%

30%

Sensitivity Graph for Alt 2 $2,00,00,000

NPW

$1,50,00,000

$1,00,00,000

MARR Labor

$50,00,000

O&M Revenue

$-30%

-20%

-10%

0%

10%

Change

21 Copyright © 2023 Pearson Education, Inc.

20%

30%

Contemporary Engineering Economics, 7th ed. ©2023 12.12

Sensitivity graph $2,000,000 V = 6000

$1,800,000 $1,600,000

NPW ($)

$1,400,000

V = 5000

$1,200,000 $1,000,000

V = 4000

$800,000 V = 3000

$600,000 $400,000

V = 2000

$200,000

V = 1000

$0 20

25

30

35

40

45

Sales Price (X)

12.13 • NPW(10%)s for 200 shift: PW(10%) Electric = $38,058 PW(10%) LPG = $69,345 PW(10%)Gasoline = $54,971 PW(10%) Diesel = $49,994

∵From the below table 0

1

2

3

4

5

6

7

Electric Power O&M Initial cost

($2,025) ($2,025) ($2,025) ($2,025) ($2,025) ($2,025) ($2,025) ($29,739)

Salvage

$3,000

Net cash flow ($29,739) ($2,025) ($2,025) ($2,025) ($2,025) ($2,025) ($2,025)

$975

PW(10%) = ($38,058) LPG O&M Initial cost

($10,100) ($10,100) ($10,100) ($10,100) ($10,100) ($10,100) ($10,100) ($21,200)

Salvage

$2,000

Net cash flow ($21,200) ($10,100) ($10,100) ($10,100) ($10,100) ($10,100) ($10,100) ($8,100) 22 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 PW(10%) = ($69,345) Gasoline O&M

($7,372) ($7,372) ($7,372) ($7,372) ($7,372) ($7,372) ($7,372)

Initial cost

($20,107)

Salvage

$2,000

Net cash flow ($20,107) ($7,372) ($7,372) ($7,372) ($7,372) ($7,372) ($7,372) ($5,372) PW(10%) = ($54,971) Diesel Fuel O&M

($5,928) ($5,928) ($5,928) ($5,928) ($5,928) ($5,928) ($5,928)

Initial cost

($22,263)

Salvage

$2,200

Net cash flow ($22,263) ($5,928) ($5,928) ($5,928) ($5,928) ($5,928) ($5,928) ($3,728) PW(10%) = ($49,994)

• PW(10%)s for 260 shift: PW(10%) Electric = $40,285 PW(10%) LPG = $82,635 PW(10%)Gasoline = $64,277 PW(10%) Diesel = $57,192 • Sensitivity graph Electric Power LPG Gasoline Diesel Fuel

$90,000 $80,000 $70,000 $60,000 $50,000 $40,000 $30,000 200

23 Copyright © 2023 Pearson Education, Inc.

260

Shifts

Contemporary Engineering Economics, 7th ed. ©2023

Break-Even Analysis 12.14 (a) With the unit price of $100 and unit cost of $9.35, the PW(12%) for options A and B would be the same. Here we also assumed that the old machine is sold off and the current net after-tax salvage value in the amount of $18,750 is credited for reducing the investment cost of Option B. If the firm retains the old machines for other use, then the unit cost would be $8.93.

Option A

100 0

1

2

3

$2,500,000

$2,500,000

$2,500,000

225,000 0

225,000 0

225,000 0

2,275,000 568,750

2,275,000 568,750

2,275,000 568,750

$1,706,250

$1,706,250

$1,706,250

1,706,250 0

1,706,250 0

1,706,250 0 6,000 (1,500)

1,706,250

1,706,250

1,710,750

Income Statement Revenue:

Expenses: Operating cost Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes Net Cash Flow

0 PW (12%) =

$4,101,328

24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Option B

100 0

1

unit cost= 2

9.345227724 3

Income Statement Revenue:

Expenses: Operating cost Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes

25,000 (6,250)

Net Cash Flow

18,750 PW (12%) =

$2,500,000

$2,500,000

$2,500,000

233,631 0

233,631 0

233,631 0

2,266,369 566,592

2,266,369 566,592

2,266,369 566,592

$1,699,777

$1,699,777

$1,699,777

1,699,777 0

1,699,777 0

1,699,777 0 0

1,699,777

1,699,777

1,699,777

$4,101,328

(b) If we assume the unit price of $100, the PW(12%) for options B and C would be $4,181,646, when unit cost is set at $7.14.

25 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Option C 0

1

2

3

$2,500,000

$2,500,000

$2,500,000

168,750 7,860

168,750 13,470

168,750 4,810

2,323,391 580,848

2,317,781 579,445

2,326,440 581,610

$1,742,543

$1,738,335

$1,744,830

1,742,543 7,860

1,738,335 13,470

1,744,830 4,810 15,000 3,465

1,750,402

1,751,805

1,768,105

Income Statement Revenue: Expenses: Operating cost Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes Net Sale of old machine

(55,000) 18,750

Net Cash Flow

(36,250) PW (12%) =

$4,181,640

Option B

100 0

1

unit cost= 2

7.145517276 3

Income Statement Revenue:

Expenses: Operating cost Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes

$2,500,000

$2,500,000

178,638 0

178,638 0

178,638 0

2,321,362 580,341

2,321,362 580,341

2,321,362 580,341

$1,741,022

$1,741,022

$1,741,022

1,741,022 0

1,741,022 0

1,741,022 0 0

Net Cash Flow

0 PW (12%) =

$2,500,000

1,741,022

$4,181,640

(c). Option C is the most economical.

26 Copyright © 2023 Pearson Education, Inc.

1,741,022

1,741,022

Contemporary Engineering Economics, 7th ed. ©2023

12.15 • PW of net investment: P0 = −$2, 200, 000 − $600, 000 − $400, 000 = −$3, 200, 000 • PW of after-tax revenue: P1 = −$4, 000(365) X (1 − 0.31)( P / A,10%, 25) = $9,144, 210 X

• PW of after-tax operating costs: P2 = −($230, 000 + $170, 000 X )(1 − 0.31)( P / A,10%, 25) = −$1, 440,526 − 1, 064, 737 X

• PW of tax credit (shield) on depreciation: n 1

Depreciation Building Furniture $54,060 $57,160

Combined Tax savings $111,220(0.31) = $34,478

2

56,410

97,960

154,370(0.31) = 47,855

3

56,410

69,960

126,370(0.31) = 39,175

4

56,410

49,960

106,370(0.31) = 32,975

5

56,410

35,720

92,130(0.31) = 28,560

6

56,410

33,680

92,090(0.31) = 28,548

7

56,410

35,720

92,130(0.31) = 28,560

8

56,410

17,840

74,250(0.31) = 23,018

9-24

56,410

0

56,410(0.31) = 17,487

25

54,060

0

54,060(0.31) = 16,759

P3 = $34, 478( P / F ,10%,1) + $47,855( P / F ,10%, 2) +  + $16, 759( P / F ,10%, 25) = $247, 461

27 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • PW of net proceeds from sale: Property (asset) Furniture

Cost basis

Building

2,200,000

0

794,450

(794,450)

(246,280)

600,000

2,031,813

600,000

1,431,813

443,862

Land

$400,000

Salvage Book value value $0 $0

Gains (losses) $0

Gains Taxes $0

Net proceeds from sale = $2, 031,813 + $246, 280 − $443,862 = $1,834, 231 P4 = $1,834, 231( P / F ,10%, 25) = $169, 292

PW(10%) = P0 + P1 + P2 + P3 + P4 = −$4, 223, 772 + 8, 079, 473 X =0 X = 52.28% 12.16 Useful life of the old bulb: 14, 600 /(19 × 365) = 2.1 years

For computational simplicity, let’s assume a useful life of 2 years for the old bulb. Then, the new bulb will last 4 years. Let X denote the price for the new light bulb. With an analysis period of 4 years, we can compute the equivalent present worth cost for each option as follows:

PW(15%)old = (1 − 0.25)[$45.90 + $45.90( P / F ,15%, 2)] = $60.46 PW(15%) new = (1 − 0.25)( X + $16) The break-even price for the new bulb will be

0.75 X + 12 = $60.46 X = $64.61 Since the new light bulb costs only $60, it is worth to switch to the new light bulb 12.17 28 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • PW of net investment: P0 = −$250, 000 • PW of after-tax rental revenue: P1 = X (1 − 0.30)( P / A,15%, 20) = $4.3815 X • PW of after-tax operation costs:

P2 = −(1 − 0.30)$12, 000( P / A,15%, 20) = −$52,578 • PW of tax credit (shield) on depreciation: (In this problem, we assume that the purchasing cost of $250,000 does not include any land value. Therefore, the entire purchasing cost will be the cost basis for depreciation purpose.)

Depreciation Building n 1 $6,143 2-19 6,410 20 6,143

Combined Tax savings $6,143(0.30) = $1,843 6,410(0.30) = 1,923 6,143(0.30) = 1,843

P3 = $1,843( P / F ,15%,1) + $1,923( P / A,15%,18)( P / F ,15%,1) +$1,843( P / F ,15%, 20) = $11,962 • PW of net proceeds from sale: Total depreciation = $127, 666 Book value = $250, 000 − $127, 666 = $122, 334 Salvage value = $250, 000(1.05) 20 = $663,324 Taxable gain = $663, 324 − $122, 334 = $540, 990 Gains tax = $540, 990(0.30) = $162, 297 Net proceeds from sale = $663,324 − $162, 297 = $501, 027 P4 = $501, 027( P / F ,15%, 20) = $30, 613 • The break-even rental:

PW(15%) = P0 + P1 + P2 + P3 + P4 = −$260, 003 + 4.3815 X =0 X = $59, 341

29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.18 Let X denotes the additional annual revenue (above $14,000) for model A that is required to break even. Dep.Rate: Model A Investment Net proceeds from sale +0.75(R n )

0.2 1

0 $

0.32 2

0.192 3

0.1152 4

0.1152 5

0.0576 6

(80,000) $ 0.75X

0.75X

0.75X

0.75X

0.75X

15,000 0.75X (16,500)

-0.75(O&M n )

$

(16,500) $

(16,500) $

(16,500) $

(16,500) $

(16,500) $

+0.25(D n )

$

4,000 $

6,400 $

3,840 $

2,304 $

2,304 $

$

0.75X (12,500) $

0.75X (10,100) $

0.75X (12,660) $

0.75X (14,196) $

0.75X (14,196) $

Net cash flow

$

(80,000)

PW(20%) = Model B Investment Net proceeds from sale +0.75(R n )

1

2

3

4

5

6

(52,000) $

-0.75(O&M n ) +0.25(D n ) Net cash flow

$

0.75X (348)

-$117,424.61 + 2.4941X 0

$

1,152

$ $

(12,750) $ 2,600 $

(12,750) $ 4,160 $

(12,750) $ 2,496 $

(12,750) $ 1,498 $

(12,750) $ 1,498 $

(12,750) 749

$

0 (10,150) $

0 (8,590) $

0 (10,254) $

0 (11,252) $

0 (11,252) $

0 (751)

(52,000)

PW(20%) =

11,250

$ (82,557.81) 30 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

• By letting PW(20%) A = PW(20%) B −$117, 424.61 + 2.4941X = −$82,557.81 X = $13,980

Required additional annual revenue = $13,980 12.19

Let X denote the number of copies to break-even. • A/T annual revenue = (0.6)[$0.10 + ($1.00 − $0.20)]X = 0.54 X • A/T O&M cost = −(0.60)[$400, 000(12) + $0.15 X ] = −$2,880, 000 − 0.09 X Depreciation tax credit = (0.40)[$85, 714( P / F ,13%,1) +  +$26, 775( P / F ,13%,8)]( A / P,13%,10) = $29, 285 CR(13%) = −$600, 000( A / P,13%,10) + $60, 000( A / F ,13%,10) = −$107,316 AE(13%) = 0.54 X − $2,880, 000 − 0.09 X + $29, 285 − $107,316 = 0.45 X − $2,958, 031 = 0

∴ X = 6, 573, 402 copies per year or 6,573,402/240 =27,389 copies per day

31 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Probabilistic Analysis 12.20 (a) Base case scenario: Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

1

2

3

4

5

6

7

8

$500 $500 $500 $500 $500 $500 $500 $500 20,000 20,000 20,000 20,000 20,000 20,000 20,000 20,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $120 $120 $120 $120 $120 $120 $120 $120 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $1,143,200 $1,959,200 $1,399,200 $999,200 $714,400 $713,600 $714,400 $356,800

Taxable income Income taxes (35%)

$3,956,800 $3,140,800 $3,700,800 $4,100,800 $4,385,600 $4,386,400 $4,385,600 $4,743,200 $1,384,880 $1,099,280 $1,295,280 $1,435,280 $1,534,960 $1,535,240 $1,534,960 $1,660,120

Net income

$2,571,920 $2,041,520 $2,405,520 $2,665,520 $2,850,640 $2,851,160 $2,850,640 $3,083,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

$2,571,920 $2,041,520 $2,405,520 $2,665,520 $2,850,640 $2,851,160 $2,850,640 $3,083,080 $1,143,200 $1,959,200 $1,399,200 $999,200 $714,400 $713,600 $714,400 $356,800 ($8,000,000) $500,000 0 ($8,000,000) $3,715,120 $4,000,720 $3,804,720 $3,664,720 $3,565,040 $3,564,760 $3,565,040 $3,939,880 $10,576,354

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b) Best case scenario: (+20% deviation) Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

1

2

3

4

5

6

7

8

$600 $600 $600 $600 $600 $600 $600 $600 24,000 24,000 24,000 24,000 24,000 24,000 24,000 24,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $96 $96 $96 $96 $96 $96 $96 $96 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $914,560 $1,567,360 $1,119,360 $799,360 $571,520 $570,880 $571,520 $285,440

Taxable income Income taxes (35%)

$9,181,440 $8,528,640 $8,976,640 $9,296,640 $9,524,480 $9,525,120 $9,524,480 $9,810,560 $3,213,504 $2,985,024 $3,141,824 $3,253,824 $3,333,568 $3,333,792 $3,333,568 $3,433,696

Net income

$5,967,936 $5,543,616 $5,834,816 $6,042,816 $6,190,912 $6,191,328 $6,190,912 $6,376,864

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

$5,967,936 $5,543,616 $5,834,816 $6,042,816 $6,190,912 $6,191,328 $6,190,912 $6,376,864 $914,560 $1,567,360 $1,119,360 $799,360 $571,520 $570,880 $571,520 $285,440 ($6,400,000) $600,000 0 ($6,400,000) $6,882,496 $7,110,976 $6,954,176 $6,842,176 $6,762,432 $6,762,208 $6,762,432 $7,262,304 $27,967,318

(c) Worst case scenario: (-20% deviation) Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

1

2

3

4

5

6

7

8

$400 $400 $400 $400 $400 $400 $400 $400 16,000 16,000 16,000 16,000 16,000 16,000 16,000 16,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $144 $144 $144 $144 $144 $144 $144 $144 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $1,371,840 $2,351,040 $1,679,040 $1,199,040 $857,280 $856,320 $857,280 $428,160

Taxable income Income taxes (35%)

-$275,840 -$1,255,040 -$96,544 -$439,264

-$583,040 -$204,064

-$103,040 -$36,064

$238,720 $83,552

$239,680 $83,888

$238,720 $83,552

$667,840 $233,744

Net income

-$179,296

-$378,976

-$66,976

$155,168

$155,792

$155,168

$434,096

-$179,296 -$815,776 -$378,976 -$66,976 $1,371,840 $2,351,040 $1,679,040 $1,199,040

$155,168 $857,280

$155,792 $856,320

$155,168 $857,280

$434,096 $428,160

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

-$815,776

($9,600,000) $400,000 0 ($9,600,000) $1,192,544 $1,535,264 $1,300,064 $1,132,064 $1,012,448 $1,012,112 $1,012,448 $1,262,256 ($3,611,477) 33 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (d) We see that the NPW ranges between -$3,611,477 and $27,967,318 with the base case generating a positive NPW of $10,576,354. 12.21 (a) The breakeven sales volume PW(20%) = 0  Demand (units) = 38,388  Sales_revenue = $2, 418, 440 Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

3

4

5

6

$42 $42 $42 $42 $42 $42 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $532,000 $532,000 $532,000 $532,000 $532,000 $532,000 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Net income

Net cash flow PW(20%)

2

$63 $63 $63 $63 $63 $63 38,388 38,388 38,388 38,388 38,388 38,388 $2,418,440 $2,418,440 $2,418,440 $2,418,440 $2,418,440 $2,418,440

Taxable income Income taxes (35%)

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax

1

$114,147 $39,951

$18,147 $6,351

$120,547 $42,191

$181,987 $63,695

$181,987 $63,695

$228,067 $79,823

$74,195

$11,795

$78,355

$118,291

$118,291

$148,243

$74,195 $160,000

$11,795 $256,000

$78,355 $153,600

$118,291 $92,160

$118,291 $92,160

$148,243 $46,080

($800,000) $100,000

($800,000) $0

$234,195

$267,795

$231,955

34 Copyright © 2023 Pearson Education, Inc.

$210,451

$210,451

$294,323

Contemporary Engineering Economics, 7th ed. ©2023

(b) The base case: Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

1

2

3

4

5

6

$63 $63 $63 $63 $63 $63 65,000 65,000 65,000 65,000 65,000 65,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $42 $42 $42 $42 $42 $42 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $532,000 $532,000 $532,000 $532,000 $532,000 $532,000 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

$673,000 $235,550

$577,000 $201,950

$679,400 $237,790

$740,840 $259,294

$740,840 $259,294

$786,920 $275,422

Net income

$437,450

$375,050

$441,610

$481,546

$481,546

$511,498

$437,450 $160,000

$375,050 $256,000

$441,610 $153,600

$481,546 $92,160

$481,546 $92,160

$511,498 $46,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

($800,000) $100,000

($800,000) $1,208,007

$597,450

$631,050

$595,210

35 Copyright © 2023 Pearson Education, Inc.

$573,706

$573,706

$657,578

Contemporary Engineering Economics, 7th ed. ©2023 (c) The sales price per unit increases to $400 Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation Taxable income Income taxes (35%) Net income Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

1

2

3

4

5

6

$400 2,252 $900,723

$400 2,252 $900,723

$400 2,252 $900,723

$400 2,252 $900,723

$400 2,252 $900,723

$400 2,252 $900,723

$42 $94,576 $532,000 $160,000

$42 $94,576 $532,000 $256,000

$42 $94,576 $532,000 $153,600

$42 $94,576 $532,000 $92,160

$42 $94,576 $532,000 $92,160

$42 $94,576 $532,000 $46,080

$114,147 $39,951

$18,147 $6,351

$120,547 $42,191

$181,987 $63,695

$181,987 $63,695

$228,067 $79,823

$74,195

$11,795

$78,355

$118,291

$118,291

$148,243

$74,195 $160,000

$11,795 $256,000

$78,355 $153,600

$118,291 $92,160

$118,291 $92,160

$148,243 $46,080

($800,000) $100,000

($800,000) $0

$234,195

$267,795

$231,955

$210,451

The required breakeven sales volume is 2,252.

36 Copyright © 2023 Pearson Education, Inc.

$210,451

$294,323

Contemporary Engineering Economics, 7th ed. ©2023

(d) •

Best case:

Best Case Income statement 0

1

2

3

4

5

6

Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$35.70 $35.70 $35.70 $35.70 $35.70 $35.70 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $452,200 $452,200 $452,200 $452,200 $452,200 $452,200 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

$2,134,863 $2,038,863 $2,141,263 $2,202,703 $2,202,703 $2,248,783 $747,202 $713,602 $749,442 $770,946 $770,946 $787,074

Net income

$1,387,661 $1,325,261 $1,391,821 $1,431,757 $1,431,757 $1,461,709

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

$72.45 $72.45 $72.45 $72.45 $72.45 $72.45 74,750 74,750 74,750 74,750 74,750 74,750 $5,415,638 $5,415,638 $5,415,638 $5,415,638 $5,415,638 $5,415,638

$1,387,661 $1,325,261 $1,391,821 $1,431,757 $1,431,757 $1,461,709 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080 ($800,000) $100,000

($800,000) $1,547,661 $1,581,261 $1,545,421 $1,523,917 $1,523,917 $1,607,789 $4,367,942

37 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

•

Worst case:

Worse Case Income statement 0 Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

1

2

3

4

5

6

$53.55 $53.55 $53.55 $53.55 $53.55 $53.55 55,250 55,250 55,250 55,250 55,250 55,250 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $48.30 $48.30 $48.30 $48.30 $48.30 $48.30 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $611,800 $611,800 $611,800 $611,800 $611,800 $611,800 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

-$481,738 -$168,608

-$577,738 -$202,208

-$475,338 -$166,368

-$413,898 -$144,864

-$413,898 -$144,864

-$367,818 -$128,736

Net income

-$313,129

-$375,529

-$308,969

-$269,033

-$269,033

-$239,081

-$313,129 $160,000

-$375,529 $256,000

-$308,969 $153,600

-$269,033 $92,160

-$269,033 $92,160

-$239,081 $46,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

($800,000) $100,000

($800,000) ($153,129) ($119,529) ($155,369) ($176,873) ($176,873) ($1,288,052)

12.22

E[return] = (0.15 × 5%) + (0.25 ×15%) + (0.35 × 22%) + (0.15 × 30%) + (0.1× 40%) = 20.7%

σ 2 = (0.15 × (5 − 20.7) 2 ) + (0.25 × (15 − 20.7) 2 ) + (0.35 × (22 − 20.7) 2 ) + (0.15 × (30 − 20.7) 2 ) + (0.1× (40 − 20.7) 2 ) = 95.91

σ = 9.79%

38 Copyright © 2023 Pearson Education, Inc.

($93,001)

Contemporary Engineering Economics, 7th ed. ©2023

12.23 We can calculate the mean and variance for each period with the three- point estimate using the formula shown in Table 12.6. Period (n) 0 1 2

Pessimistic ($10,000) $5,000 $4,000

Most Likely ($8,000) $12,000 $10,000

E[PW(10%)] = −$8,167 +

Optimistic ($7,000) $15,000 $13,000

E[An]

Var[An]

($8,167) $11,333 $9,500

250,000 2,777,778 2,250,000

$11,333 $9,500 + 1.1 1.12

= $9,986.97 2, 777, 778 2, 250, 000 + 1.12 1.14 = 4, 082, 464.57 0 − 9,986.97 z= = −4.9428 4, 082, 464 NORMDIST(-4.9428,0,1,1)= 0.0000385% Var[PW(10%)] = 250, 000 +

12.24

PW(12%)light = −$8, 000, 000 + $1,300, 000( P / A,12%,3) PW(12%) moderate

= −$4,877, 619 = −$8, 000, 000 + $2,500, 000( P / A,12%, 4)

PW(12%) high

= −$406, 627 = −$8, 000, 000 + $4, 000, 000( P / A,12%, 4)

= $4,149,397 E[PW(12%)] = −$4,877, 619(0.20) − $406, 627(0.40) +$4,149,397(0.40) = $521,584 Since E[PW] is positive, it is good to invest based on the expected value criterion.

39 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 12.25 (a) 1.

A1 and A2 are mutually independent 200 500 E[PW(10%)] = −500 + + = $95.04 1 2 (1 + 0.1) (1 + 0.1) Var[PW(10%)] = 0 +

2.

502

(1 + 0.1)

2

+

502

(1 + 0.1)

4

= 3773.65

A1 and A2 are partially correlated with ρ12 = 0.3 200 500 E[PW(10%)] = −500 + + = $95.04 1 2 (1 + 0.1) (1 + 0.1) Var[PW(10%)] = 0 +

502

(1 + 0.1)

2

 0.3(50)(50)  + 2 = 4900.62 3  (1 + 0.1)  (1 + 0.1)  502

+

4

(b)

σ [PW(10%)] = Var[PW(10%)] σ [PW(10%)] = 3773.65 = 61.43 0 − 95.04 z= = 61.43 σ = −1.547 P ( z ≤ −1.547) = 0.06093 or 6.093%

X −μ

12.26 (a) 1.

Mutually independent: E[PW(10%)] = −500 +

300

(1 + 0.1)

1

Var[PW(10%)] = 1002 + 2.

+

1002

(1 + 0.1)

2

400

(1 + 0.1) +

2

= $103.30

1002

(1 + 0.1)

4

= 25, 094.60

Partial correlations: E[PW(10%)] = −500 +

300

(1 + 0.1)

1

+

400

(1 + 0.1)

40 Copyright © 2023 Pearson Education, Inc.

2

= $103.30

Contemporary Engineering Economics, 7th ed. ©2023

Var[PW(10%)] = 1002 +

1002

(1 + 0.1)

2

+

1002

(1 + 0.1)

4

 0.2(100)(100)   0.2(100)(100)  +2  + 2   1 2  (1 + 0.1)   (1 + 0.1)   0.5(100)(100)  +2   3  (1 + 0.1)  = 25, 094.60 + 14, 455.30 = 39,549.89 3.

Perfect positive correlations: E[PW(10%)] = −500 +

300

(1 + 0.1)

Var[PW(10%)] = 1002 +

1

+

1002

(1 + 0.1)

2

400

(1 + 0.1) +

2

= $103.30

1002

(1 + 0.1)

4

1(100)(100)  1(100)(100)  +2  +2 1  2   (1 + 0.1)   (1 + 0.1)  1(100)(100)  +2  3   (1 + 0.1)  = 25, 094.60 + 49, 737.04 = 74,831.64 (b)

σ [PW(10%)] = Var[PW(10%)] σ [PW(10%)] = 25, 094.60 = $158.41 X − μ 100 − 103.30 z= = = −0.0208 σ 158.41 P ( z ≥ −0.0208) = 1 − P ( z ≤ −0.0208) = 1 − 0.49170 =0.5083 (or 50.83%)

41 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 12.27 Expected value criterion: (a) • Option 1:

E[ R]1 = $2, 450(0.25) + $2, 000(0.45) + $1, 675(0.30) −$150( F / P, 7.5%,1) = $1,854 • Option 2: E[ R]2 = $25, 000(0.075) = $1,875

∴ Option 2 is the better choice based on the principle of expected value maximization. (b)

This part could be answered after covering the decision tree analysis.

Potential return

Prob.

Op.1

Op.2

Optimal choice

Opp. Loss

High

0.25

$2,288.75

$1,875

Op.1

$413.75

Medium

0.45

1,838.75

1,875

Op.2

0

Low

0.3

1,513.75

1,875

Op.2

0

1,853.75

1,875

Expected value

103.44

EPPI = 0.25($2,288.75) + 0.45($1,875) + 0.3($1,875) = $1,978.44 EV = $1,875 EVPI = EPPI – EV = $103.4

42 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.28 Let X denote the annual revenue in constant dollars and Y be the general inflation. (a) NPW as functions of X and Y: End of Period

Cash elements

0 -$9,000

Investment

1

2

Salvage value

4, 000(1+Y )2

Gains tax

−0.4 4,000(1+Y )2 - 4,000 

(0.4) Dn

1,200

800

(0.6) Rn working capital

0.6X(1+Y)

0.6 X (1 + Y )2

-2,000

2,000(-Y)

2,000(1+Y)

-$11,000

1,200-2,000Y +0.6X(1+Y)

2,400 (1+Y) 2 +2,400 +0.6X (1+Y ) 2 +2,000(1+Y)

Net cash flow(Actual) Net cash flow(Constant)

1, 200(1+Y )-1 -$11,000

−2, 000Y (1+Y )−1 + 0.6 X

2,400 + 2, 400(1+Y ) −2 +0.6 X + 2, 000(1+Y )-1

i = i '+ f + i ' f i = 0.1 + Y + 0.1Y = 0.1 + 1.1Y PW(i) = −$11,000 + [1, 200 − 2,000Y + 0.6 X (1 + Y )]( P / F , i,1) +[2, 400(1 + Y )2 + 2, 400 + 0.6 X (1 + Y )2 + 2,000(1 + Y )] ×( P / F , i, 2) or

PW(10%) = −$11, 000 + [1, 200(1 + Y )-1 − 2, 000Y (1 + Y )−1 + 0.6 X ]( P / F ,10%,1) +[2, 400 + 2, 400(1 + Y )−2 + 0.6 X + 2, 000(1 + Y )-1 ]( P / F ,10%, 2) (b) and (c) Mean and variance calculation: Note that the market interest rate is a random variable as the general inflation rate becomes a random variable. There are nine joint events for X and Y. For example, event No.2 is the joint event where X = 10, 000 and Y = 0.05 . With 43 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 these X and Y values, we calculate the market interest rate and then evaluate the PW function with this market interest rate. i = i '+ f + i ' f = 0.10 + 0.05 + (0.1)(0.05) = 15.5% PW(15.5%) = −$11,000 + [1,200 − 2,000(0.05) + 0.6(10,000)(1.05)](P / F ,15.5%,1) +[2,400(1.05)2 + 2,400 + 0.6(10,000)(1.05)2 + 2,000(1.05)] ×(P / F ,15.5%,2) = $5,722 You repeat the process for the remaining joint events to obtain the following: Event No. 1 2 3 4 5 6 7 8 9

X 10,000 10,000 10,000 20,000 20,000 20,000 30,000 30,000 30,000

Y 0.03 0.05 0.07 0.03 0.05 0.07 0.03 0.05 0.07

i

$ $ $ $ $ $ $ $ $

Event No. 1 2 3 4 5 6 7 8 9

PW(i%) $5,877 $5,722 $5,574 $16,290 $16,136 $15,988 $26,704 $26,549 $26,401

P(x) 0.3 0.3 0.3 0.4 0.4 0.4 0.3 0.3 0.3

P(y)

E[PW]=

$16,137

0.133 0.155 0.177 0.133 0.155 0.177 0.133 0.155 0.177

0.25 0.5 0.25 0.25 0.5 0.25 0.25 0.5 0.25 Var[PW] = σ[PW] =

A0 ($11,000) ($11,000) ($11,000) ($11,000) ($11,000) ($11,000) ($11,000) ($11,000) ($11,000)

A1 $7,320 $7,400 $7,480 $13,500 $13,700 $13,900 $19,680 $20,000 $20,320

A2 $13,372 $13,761 $14,157 $19,737 $20,376 $21,027 $26,102 $26,991 $27,896

PW(i%)*P(x,y) (PW(i%)-E[PW])^2*P(x,y) P(x,y) 0.075 $441 7,895,197 0.15 $858 16,270,806 0.075 $418 8,368,000 0.1 $1,629 2,345 0.2 $3,227 1 0.1 $1,599 2,238 0.075 $2,003 8,373,604 0.15 $3,982 16,259,759 0.075 $1,980 7,900,640 1 $16,137 65,072,590 65,072,590 $ 8,067

Value at Risk 44 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

12.29 (a) (b)

VaR = −1.647 × 0.6916% + 0.1769% = −0.9622% $1, 000, 000 × ( −0.9622 / 100) = −$9, 622

(c) VaR90− days = −0.9622% × 90 = −9.1282% $1, 000, 000 × (−9.1282 /100) = −$91, 282

12.30 (a) 0 − 120   P (NPW ≤ 0) = Φ  z = = −2.4  = 0.0082 50  

(b)

VaR = −1.647 × $50 + $120 = +$37.65 > 0  a very safe project

12.31 Not given

Comparing Risky Projects 12.32 NOTE: A typo error in the first printing– it should be Problem 12.25, not Problem 11.20. $500 $200 $500 + + 1.25 1.252 1.253 = ($520) < 0 ← Not acceptable

PW(25%)certainty equivalent = −$1, 000 +

No, it would not be justified.

12.33 45 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 NOTE: A typo error in the first printing – it should be Problem 12.26, not Problem 11.21. Certainty equivalent value: $500 $200 $500 + + 1.18 1.182 1.183 = ($471) < 0 ← Not acceptable

PW(25%)certainty equivalent = −$1, 000 +

No, it would not be justified, even lowering the discount rate. 12.34 The mean return for projects E[return]A = (0.1× −20%) + (0.2 × 0%) + (0.25 ×10%) + (0.3 ×15%) + (0.1× 20%) + (0.05 × 40%) = 9% E[return]B = (0.1× −35%) + (0.2 × −10%) + (0.25 × 15%) + (0.3 × 25%) + (0.1× 40%) + (0.05 × 50%) = 12.25% (b) The variance of return for projects

σ 2 A = (0.1× (−20 − 9)2 ) + (0.2 × (0 − 9) 2 ) + (0.25 × (10 − 9)2 ) + (0.3 × (15 − 9) 2 ) + (0.1× (20 − 9) 2 ) + (0.05 × (40 − 9) 2 ) = 171.50

σ 2 B = (0.1× (−35 − 12.25) 2 ) + (0.2 × (−10 − 12.25) 2 ) + (0.25 × (15 − 12.25) 2 ) + (0.3 × (25 − 12.25) 2 ) + (0.1× (40 − 12.25)2 ) + (0.05 × (50 − 12.25) 2 ) = 521.19 (c) It is not a clear case, because E[return]B > E[return]A but also VarB > VarA . If you make your investment decision solely based on the principle of maximization of expected return, you may prefer project B. However, if you consider the riskiness of the project B along with the expected return, you may focus on how much additional risk you are willing to take to gain 3.25% 46 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 additional return.

12.35 (a)

E[PW(12%)] A = (0.3) [ −$150, 000 + $35, 000( P / A,12%, 5) ]

+ (0.5) [ −$150, 000 + $40, 000( P / A,12%, 5) ]

+ (0.2) [ −$150, 000 + $50, 000( P / A,12%, 5) ] = −$4, 006.6 E[PW(12%)]B = (0.3) [ −$180, 000 + $45, 000( P / A,12%, 5) ]

+ (0.5) [ −$180, 000 + $55, 000( P / A,12%, 5) ]

+ (0.2) [ −$180, 000 + $67, 000( P / A,12%, 5) ]

= $16,100

σ

2 A

= (0.3× (−23,832 + 4,006.6)2 ) + (0.5 × (−5,808 + 4,006.6)2 ) + (0.2 × (30,240 + 4,006.6)2 ) = 354,097,713

σ

2 B

= (0.3× (−17,785 −16,100)2 ) + (0.5 × (18,263 −16,100)2 ) + (0.2 × (61,520 −16,100)2 ) = 759,392,532

(b) Project A has a higher probability of losing money. 0 + 4, 006.6 zA = = 0.2129 354, 097, 713 NORMDIST(0.2129,0,1,1) = 58.43% zB =

0 − 16,100 = −0.58428 759,392,532

NORMDIST(-0.58428,0,1,1) = 27.95%

12.36 47 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(a) The PW distribution for project 1: Event (x,y)

Joint probability

($20,$10) ($20,$20) ($40,$10) ($40,$20)

0.18 0.12 0.42 0.28

PW(10%) $400 0 2,400 1,600

(b) The mean and variance of the PW for Project 1: E[PW(10%)]1 = $400(0.18) + $0(0.12) + $2, 400(0.42) +$1, 600(0.28) = $1,528 Var[PW(10%)]1 = (400 − 1,528)2 (0.18) + (0 − 1,528)2 (0.12) +(2, 400 − 1,528) 2 (0.42) +(1, 600 − 1,528) 2 (0.28) = 830, 016 (c) The mean and variance of the PW for Project 2: E[PW(10%)]2 = $0(0.24) + $400(0.20) + $1, 600(0.36) +$2, 400(0.20) = $1,136 Var[PW(10%)]2 = (0 − 1,136) 2 (0.24) + (400 − 1,136) 2 (0.20) +(1, 600 − 1,136) 2 (0.36) +(2, 400 − 1,136) 2 (0.20) = 815,104 (d) It is not a clear case, because E1 > E2 but also Var1 > Var2 . If Juan makes the decision solely based on the principle of maximization of expected value, she may prefer contract A. However, if she considers the riskiness of the project B along with the expected return, she may focus on how much additional risk she is willing to take to gain 3.25% additional return.

12.37 48 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (a) E[PW]1 = ($2, 000)(0.20) + ($3, 000)(0.60) + ($3,500)(0.20) − $1, 000

= $1,900 E[PW]2 = ($1, 000)(0.30) + ($2,500)(0.40) + ($4,500)(0.30) − $800

= $1,850 Project 1 is preferred over Project 2.

(b) Var[PW]1 = (2, 000 − 1,900) 2 (0.20) + (3, 000 − 1,900) 2 (0.60) + (3,500 − 1,900) 2 (0.20) = 1, 240, 000 Var[PW]2 = (1, 000 − 1,850) 2 (0.30) + (2,500 − 1,850) 2 (0.40) + (4,500 − 1,850) 2 (0.30) = 2, 492,500

∴ Project 1 is still preferred, because of higher E[PW] and lower Var[PW] .

12.38 (a) Mean and variance calculations:

E[PW]1 = ($100, 000)(0.20) + ($50, 000)(0.40) + (0)(0.40) = $40, 000 E[PW]2 = ($40, 000)(0.30) + ($10, 000)(0.40) + (−$10, 000)(0.30) = $13, 000 Var[PW]1 = (100, 000 − 40, 000) 2 (0.20) + (50, 000 − 40, 000) 2 (0.40) + (0 − 40, 000) 2 (0.40) = 1, 400, 000, 000 Var[PW]2 = (40, 000 − 13, 000) 2 (0.30) + (10, 000 − 13, 000) 2 (0.40) + (−10, 000 − 13, 000) 2 (0.30) = 381, 000, 000 It is not a clear case, because E1 > E2 but also Var1 > Var2 . If she makes decision solely based on the principle of maximization of expected value, she may prefer contract A. If she is a risk-averse decision maker, then she has to ask herself whether or not $27,000 increase in expected value is worth taking an increase in risk of 1,019,000,000.

49 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Assuming that both contracts are statistically independent from each other,

Joint event (PWA >PWB )

Joint Probability

($100,000,$40,000) ($100,000,$10,000) ($100,000,-$10,000) ($50,000,$40,000) ($50,000,$10,000) ($50,000,-$10,000) ($0,-$10,000)

(0.20)(0.30) = 0.06 (0.20)(0.40) = 0.08 (0.20)(0.30) = 0.06 (0.40)(0.30) = 0.12 (0.40)(0.40) = 0.16 (0.40)(0.30) = 0.12 (0.40)(0.30) = 0.12 Σ = 0.72

P ( PWA >PWB ) = 0.72 12.39 (a) •

Machine A:

CR(10%) A = ($60, 000 − $22, 000)( A / P,10%, 6) + (0.10)($22, 000) = $10,924 E[AE(10%)]A = ($5, 000)(0.20) + ($8, 000)(0.30) + ($10, 000)(0.30) + ($12, 000)(0.20) + $10,924 = $19, 725 Var[AE(10%)]A = (15,924 − 19, 725) 2 (0.20) + (18,924 − 19, 725) 2 (0.30) + (20,924 − 19, 725) 2 (0.30) + (22,924 − 19, 725) 2 (0.20) = 5,560, 000 • Machine B: CR(10%) B = $35, 000( A / P,10%, 4) = $11, 042 E[AE(10%)]B = ($8, 000)(0.10) + ($10, 000)(0.30) + ($12, 000)(0.40) + ($14, 000)(0.20) + $11, 042 = $22, 442 Var[AE(10%)]B = (19, 042 − 22, 442) 2 (0.10) + (21, 042 − 22, 442) 2 (0.30) + (23, 042 − 22, 442) 2 (0.40) + (25, 042 − 22, 442) 2 (0.20) = 3, 240, 000

50 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Prob[ AE (10%) A > AE (10%) B ] : Joint event (O&M A ,O&M B ) (AE A >AE B )

Joint Probability

($10,000, $8,000)

($20,924, $19,042)

(0.30)(0.10) = 0.03

($12,000, $8,000)

($22,924, $19,042)

(0.20)(0.10) = 0.02

($12,000, $10,000)

($22,924, $21,042)

(0.20)(0.30) = 0.06

Σ = 0.11 12.40 (a) Mean and variance calculation (Note: For a random variable Y, which can be expressed as a linear function of another random variable X (say, Y = aX , where a is a constant) the variance of Y can be calculated as a function of variance of X, Var[Y ] = a 2Var[ X ] . E[PW]A = −$5, 000 + $4, 000( P / A,15%, 2) = $1,502.84 E[PW]B = −$10, 000 + $6, 000( P / F ,15%,1) + $8, 000( P / F ,15%, 2) = $1, 266.54 V [PW]A = 1, 0002 + ( P / F ,15%,1) 21, 0002 + ( P / F ,15%, 2) 21,5002 = 3, 042,588 V [PW]B = 2, 0002 + ( P / F ,15%,1) 21,5002 + ( P / F ,15%, 2) 2 2, 0002 = 7,988,336 (b) Comparing risky projects.

E[PW]

Project A $1,503

Project B $1,267

Var[PW]

3,042,588

7,988,336

Project A is preferred because of higher E[PW] and lower Var[PW] .

51 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Decision – Tree Analysis 12.41 • Joint & Marginal probabilities:

H L Marginal probabilities Actual

Survey M 0.12 0.35 0.47

H 0.12 0.07 0.19

L 0.06 0.28 0.34

• Conditional probabilities:

Actual

H L

H 0.632 0.368

Survey M 0.255 0.745

(a) EV0 = 0 EVIntro. = (0.3)($4M)+(0.7)(-$2M) = -$0.2M EVDon't Intro. = 0

(b) EVPI0 = EPPI-EV0 = 1.2M EPPI = (0.3)($4M)+(0.7)($0) = $1.2M EV0 = 0

52 Copyright © 2023 Pearson Education, Inc.

L 0.176 0.824

Contemporary Engineering Economics, 7th ed. ©2023 (c) Decision tree 0.3 High 4 introduce

0 0

-0.2

1 No survey

4

0.7 Low

Event 3

-2 2

0

0

0

0

-2

0 Do not 0 0

0 0.632 High 4

introduce

0 0

1.792

0.19

4

0.368 Low

S

-2

#

1

0.14048

0

0

-2

1.792 Do not 0 0

0 0.255 High 4

introduce

0 0

-0.47

0.47 Do survey

0.745 Low

I

-2 2

-0.2

4

0.34048

0

0

-2

0 Do not 0 0

0 0.176 High 4

introduce

0 0

-0.944

0.34

4

0.824 Low

F

-2 2 0

0

-2

0 Do not 0 0

0

(d) EVPIe = EPPI − EVe = $1.2M − $0.34048M = $0.85952 M (e) EVSI = EVPI0 - EVPIe = EVe − EV0 = $0.34048M (f) Optimal decision: Take the survey. If the survey says high sales (S), then introduce a new product. Otherwise, do not.

53 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 12.42 (a) Let’s define the symbols: P: Party is taking place NP: No party is planned TP: Tipster says “P” TNP: Tipster says “NP” Then, P (TP,NP) = P (NP) P (TP/NP) = (0.4)(0.2) = 0.08 (b) • Optimal decision without sample information: EMV = (0.6)(100) + (0.4)( −50) = 40 points.

Raid the dormitories. • Joint & Marginal probabilities:

P NP Marginal Probability

Actual

Tipster says TP TNP 0.24 0.36 0.08 0.32 0.32 0.68

Marginal Probability 0.6 0.4 1

• Conditional probabilities: Tipster says Actual

P NP

TP 0.750 0.250

TNP 0.529 0.471

• Optimal decision after receiving the tips: The tipster’s information has no value, even though it costs nothing. Do not reply on the tips. (c) EVPI = 60 - 40 = 20 EPPI = (0.6)(100) + (0.4)(0) = 60 points

54 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

* Decision Tree

40

R

40

NR

Do not take tips

0.4 -6

40 62.5 62.5

Take tips

39.93

0.32

R NR

TP

-7.5

0.6

29.41 29.41

R

NR

-50 -10 0

0.75

100

0.25

-50

0.75

-10 0 100

0.529 0.471

-5.3

0529 0.471

55 Copyright © 2023 Pearson Education, Inc.

100

0.4

0.25

0.68 TNP

0.6

-50 -10 0

Conntemporary E Engineeringg Economics, 7th ed. ©22023

12.443 (a)

EV V0 = (0.3)($33, 060, 763) + (0.4)($1,306,552) + (00.3)(−$728, 333) = $1, 222,349.8 56 Copyright © 2023 Pearson E Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Open the store •

EVPI0 = EPPI - EV0 = $1, 440,849.7 − $1, 222,349.8 = $218, 499.9

EPPI = $3,060,763 (.3) + $1,306,552 (.4) + $0(.3) = $1,440,849.7 You may be willing to pay up to $218,499.9 to know the true state of nature. (b) • Joint / marginal probabilities: Survey says

Actual

Marginal Probability

High

Medium

Low

High

0.21

0.075

0.015

0.3

Medium

0.08

0.24

0.08

0.4

Low

0.015

0.06

0.225

0.3

0.305

0.375

0.320

1

Marginal Probability

• Conditional probabilities: Survey says

Actual

High

Medium

Low

High

0.689

0.2

0.047

Medium

0.262

0.64

0.25

Low

0.049

0.16

0.703

o Optimal decision: Take a survey. With either “High” or “Medium” result from the survey, open the store. Otherwise, do not open the store. o Calculating the expected value of perfect information with survey.

EVPIe = EPPI-EVe = $1, 440,849.7 − $1,235,789.12= $205,060.58

EPPI= $3,060,763 (.3) + $1,306,552 (.4) + $0(.3) = $1,440,849.7 57 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

EVe = $0(0.32)+ $1,331,812.6 (0.375)+ $2,414,293.1 (0.305)=$1,235,789.12 o Expected Value of Sample Information (EVSI)

EVSI = EVPI0 - EVPIe =$218,499.9– $205,060.58= $13,439.32 EVSI = EVe - EV0 •

Decision Tree

58 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 0.3 Demand High 3,060,763 0

3060763

0.4 Demand medium

Open

1,306,552 0

1222349.8

0

1306552

0.3 Demand low

No survey 1 0

(728,333)

1222349.8

0

-728333

Do not open 0

0 0.68852459 Demand High 3,060,763 0

3060763

0.26229508 Demand medium

Open

1,306,552 0

2414293.09

0.305 Sample info High

0

1306552

0.04918033 Demand low

#

1

1234789.12

0

(728,333)

2414293.09

0

-728333

Do not open 0

0 0.2 Demand High 3,060,763 0

3060763

0.64 Demand medium

Open

1,306,552 0

1331812.6

0.375 Sample info Medium

Survey

0

1306552

0.16 Demand low 1

-1000

1235789.12

0

(728,333)

1331812.6

0

-728333

Do not open 0

0 0.046875 Demand High 3,060,763 0

3060763

0.25 Demand medium

Open

1,306,552 0

-41997.875

0.32 Sample info Low

0

1306552

0.703125 Demand low 2

0

(728,333)

0

0

-728333

Do not open 0

59 Copyright © 2023 Pearson Education, Inc.

0

Conntemporary E Engineeringg Economics, 7th ed. ©22023

12.444 ( (a) EM MVTV Network = $900 EMVMovie = $250(0.3) + $1, 000(0.6) + $3, 200(0.10) = $9995

Seleect the Moovie deal.

(b) EPPI = (0.330)($900) + (0.60)$1, 0000 + (0.10)($$3, 200) = $11,190 EVPI = $1,1190 − $995 = $195

T The maxim mum amouunt to pay is $195. (c)

R Revised proobabilities: P(S) = 0.3

P P(M) = 0.6

P(L) = 0.1

S) = 0.2 P(F|S

P P(F|M) = 0.4

P(F|L) = 0.7

P(U|S) = 0.8

P P(U|M) = 0.66

P(U|L) = 0.3



P(S) = (0.2)(00.3) = 0.06 P(F, S) = P(F|S)P



P(F, M) = P(F|M))P(M) = (0.4)(0.6) = 0.244



P(L) = (0.7)(00.1) = 0.07 P(F, L) = P(F|L)P 0 P(F)) = 0.06 + 0.224 + 0.07 = 0.37

•

P(U, S) = P(U|S)P P(S) = (0.8)((0.3) = 0.24 60 Copyright © 2023 Pearson E Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023 •

P(U, M) = P(U|M M)P(M) = (0.66)(0.6) = 0.36

•

P(U, L) = P(U|L))P(L) = (0.3)((0.1) = 0.03 P(U)) = 0.24 + 0.336 + 0.03 = 00.63

C Conditional P Probabilities •

P(L|F F) = P(F, L)/P P(F) = (0.07))/(0.37) = 0.11892

•

P(M||F) = P(F, M))/P(F) = (0.244)/(0.37) = 00.6486

•

P(S|F F) = P(F, S)/P P(F) = (0.06))/(0.37) = 0.11622

•

P(L|U U) = P(U, L))/P(U) = (0.03)/(0.63) = 00.0476

•

P(M||U) = P(U, M M)/P(U) = (0.36)/(0.63) = 0.5714

•

P(S|U U) = P(U, S)//P(U) = (0.244)/(0.63) = 0.3810

61 Copyright © 2023 Pearson E Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023 •

EVPI affter receivinng the moviee critic’s repport. EVPI afterr critic = (1055.43)(0.37) + (166.62)(0.663) = $143,9980 Favorable critic, (33,180)(0.1892)) + (980)(0.64486) + (880)(0.1622) – 1,2744.59 = 105.43 Unfaavorable critic,, (3,180)(0.04776) + (980)(0..5714) + (880))(0.3810) – 8880 = 166.62

•

What is worth of thhe movie criitic’s report?? EVSI = 1,026 – 995 = $31,000. Therefore, tthe worth of tthe movie criitic’s report iis $31,0000 + $20,000 = $51,000, w which meanss even thoughh the report costs c $20,0000, it increeases the EM MV of the deccision by $31,000.

•

What is Jay’s ultim mate strategyy for this invvestment prooblem? (4 points) S See the red aarrows in the decision treee.

45 12.4 ( (a)

( (b)

EV0 = $54 EPPI = $20(0.2)+ +$50(0.5)+$200(0.3) = $$89 EVPI0 = EPPI-EV0 = $35

62 Copyright © 2023 Pearson E Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023 ((c) • Determine the straategy that m maximizes th he expected d payoff witth the surveey.

63 Copyright © 2023 Pearson E Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

EVPI after taking the survey

EPPI = $20(0.20) + $50(0.5) + $200(0.3) = $89 EV0 = $88.25(0.485) + $44.95(0.515) = $65.95 EVPIe = EPPI - EV0 = $89 − $65.95 = $23.05

•

What is the true worth of the survey?

EVSI = EVPI0 - EVPIe = $35 − $23.05 = $11.95 or EVSI = EVe - EVe = $65.95 − $54 = $11.95 •

What is the ultimate strategy for this investment problem? Take a survey. If the result is favorable, select Action B – construct new facilities. If the result is unfavorable, select Action A – arrange for subcontracting.

12.46

a1: publish a2: do not publish (a) Prior optimal act: a1 – publish, with EMV = $5,000 (b) EVPI = $20,000 - $5,000 = $15,000 (c) Posterior optimal act: Send out for review – if he likes, publish. If he dislikes it, do not publish. EMV = $9,000 (d) EVPI after Z1: EPPI = 0.75($40,000) + 0.25(0) = $30,000, EMV = $22,500 EVPI = $30,000 - $22,500 = $7,500 (e) EVPI after Z1: EPPI = 0.333($40,000) + 0.25(0) = $13,333, EMV = $0 EVPI = $13,333 - $0 = $13,333 (f) EVSI = $9,000 - $5,000 = $4,000 (g) ENGS = $4,000 - $4,500 = -$500 (not worth it) 64 Copyright © 2023 Pearson Education, Inc.

Conntemporary E Engineeringg Economics, 7th ed. ©22023

65 Copyright © 2023 Pearson E Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies ST 12.1 • The number of possible combinations that can be obtained by taking a sample of 6 numbers from 44. n n! 44! C ( n, r ) =   = = = 7,059,052  r  r !( n − r )! 6!(38!)

Mathematically, you will be able to buy the winning ticket by spending $7,059,052. Practically it is hard to buy these many tickets in each time, unless you generate all combinations of 6-number sets on computer and buy all of them by transferring funds electronically through financial institutions. Prize First Second Third Fourth Fail

•

Probability 1 7, 059, 052 228 7, 059, 052 10,552 7, 059, 052 168, 073 7, 059, 052 178,824 7, 059, 052

Payoff

EMV

$27,007,364 - $1

$3.83

$899 - $1

$0.03

$51 - $1

$0.07

$1 - $1

$0

$0 - $1

-$0.97

Total:

$2.81

It seems the expected value is about $3.81 for each one dollar of investment.

• • •

So, a person who buys one ticket has odds of 1 in slightly more than 7 million. Holding more tickets increases the odds of winning, so that 1,000 tickets have odds of 1 in 7,000 and 1 million tickets have odds of 1 in 7. Since each ticket costs $1, it would receive at least a share in the jackpot and many of the second, third, and fourth place prizes. Together these combinations were worth $911,197 (= $27,918,561 - $17,007,364). If the jackpot is big enough, provided nobody else buys a winning ticket, it makes economic sense to buy one lottery ticket for every possible combination of numbers and be sure to win.

66 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

ST 12.2 (a) Project cash flows: (a) Project cash flows based on most-likely estimates: 0 1 Income Statement Revenue: Steam Sales $1,550,520 Tipping Fee 976,114 Expenses: O&M 832,000 Depreciation Interest(11.5%) 805,000

2

3

4

5

6

7

8-19

$1,550,520 $1,550,520 208,585 0

20

$1,550,520 895,723

$1,550,520 800,275

$1,550,520 687,153

$1,550,520 553,301

$1,550,520 395,161

$1,550,520 0

832,000

832,000

832,000

832,000

832,000

832,000

832,000

832,000

805,000

805,000

805,000

805,000

805,000

805,000

805,000

805,000

Taxable Income Income Tax (0%)

$889,634 0

$809,243 0

$713,795 0

$600,673 0

$466,821 0

$308,681 0

$122,105 0

($86,480) 0

($86,480) 0

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax Loan Repayment

$889,634

$809,243

$713,795

$600,673

$466,821

$308,681

$122,105

($86,480)

($86,480)

889,634 0

809,243 0

713,795 0

600,673 0

466,821 0

308,681 0

122,105 0

(86,480) 0

(6,688,800)

(86,480) 0 300,000

6,688,800

(7,000,000)

Net Cash Flow PW (10%) =

0 $1,639,723

889,634

809,243

713,795

600,673

466,821

308,681

122,105

Yes, enough revenue.

Note: There are no tax payments by the City of Opelika, as a municipal government.

67 Copyright © 2023 Pearson Education, Inc.

(86,480)

(6,786,480)

Contemporary Engineering Economics, 7th ed. ©2023 (b) Let X denote the steam charge per pound. Then, annual steam charge = 1, 061,962(0.001X )(365) = 387, 616 X n 1 2 3 4 5 6 7 8-19 20

Revenue $387,616X $387,616X $387,616X $387,616X $387,616X $387,616X $387,616X $387,616X $387,616X

Expenses -$660,886 -$741,277 -$836,725 -$949,847 -$1,083,699 -$1,241,839 -$1,428,415 -$1,637,000 -$8,337,000

AE(10%) = $387, 616 X − [$660,886( P / F ,10%,1) +  +$8,337, 000( P / F ,10%, 20)]( A / P,10%, 20) = $387, 616 X − $1,357,918 =0 X = $3.503 per lb. or $3,503 per thousand lbs. ST 12.3 (a) Transmission distance of 5 miles: • Option 1 (Copper wire): 5 miles = 5 × 5,280 = 26,400 feet First cost = (1.692 + 0.013 × 2,000) × 26,400 = $731,069 Annual operating cost = $731,069(0.184) = $134,517 ∴ PW(15%)1 = −$731, 069 − $134,517( P / A,15%,30) = −$1,614,305

• Option 2 (Fiber optics): Cost of ribbon = $15,000/mile × 5miles = $75,000 Cost of terminators = $30,000 × 3 × 2 = $180,000 Cost of modulating system = ($12, 092 + $21, 217)(21)(2) = $1,398,978 Cost of repeater = $15,000 Total first cost = $75,000 + $180,000 + $1,398,978 + $15,000 = $1,668,978 Annual operating costs = $1, 398, 978(0.125) +$75, 000(0.178) = $188, 222

68 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 ∴ PW(15%) 2 = −$1, 668,978 − $188, 222( P / A,15%,30) = −$2,904,840

Option 1 is the better choice. (b) Either 10 miles or 25 miles of transmission distance: • 10 miles: two repeaters need for option 2 PW(15%)1 = −$3, 228, 610 PW(15%) 2 = −$1,808,904 Option 2 is the better choice. • 25 miles: five repeaters need for option 2 PW(15%)1 = −$8, 071,512 PW(15%) 2 = −$1,853,904 Option 2 is the better choice.

69 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 ST 12.4 (a), (b), and (c). Random Variables

Low

Most

High

Likely Annual Market size (units)

5,000

8,000

10,000

Growth rate (annual)

3%

5%

8%

Unit price

$80,000

$84,000

$86,000

Unit variable cost

$56,000

$60,000

$65,000

Fixed cost (annual)

$5,000,000 $8,000,000 $9,000,000

excluding depreciation Salvage value

•

$6,000,000 $7,000,000 $8,000,000

Depreciated on a 7-year MACRS Depreciation: % 14.29 24.49 17.49 12.49 8.93 8.92 8.93 4.46

n 0 1 2 3 4 5 6 7 8

Dn $7,859,500 $13,469,500 $9,619,500 $6,869,500 $4,911,500 $4,906,000 $4,911,500 $2,453,000

70 Copyright © 2023 Pearson Education, Inc.

Bn $55,000,000 $47,140,500 $33,671,000 $24,051,500 $17,182,000 $12,270,500 $7,364,500 $2,453,000 $0

Contemporary Engineering Economics, 7th ed. ©2023

• A sample calculation: Net cash flow (Low case) (unit: 000) Income Statement Revenue Variable cost Fixed cost Depreciation Taxable income Income taxes (35%) Net income Cash Flow Statement Net income Depreciation Investment/Salvage Gains tax Net Cash Flow

0

1 400,000 280,000 5,000 7,860 107,141 37,499 69,641

2 412,000 288,400 5,000 13,470 105,131 36,796 68,335

3 424,360 297,052 5,000 9,620 112,689 39,441 73,248

4 437,091 305,964 5,000 6,870 119,258 41,740 77,518

5 450,204 315,142 5,000 4,912 125,150 43,802 81,347

6 463,710 324,597 5,000 4,906 129,207 45,222 83,984

7 477,621 334,335 5,000 4,912 133,375 46,681 86,694

8 491,950 344,365 5,000 2,453 140,132 49,046 91,086

69,641 7,860

68,335 13,470

73,248 9,620

77,518 6,870

81,347 4,912

83,984 4,906

86,694 4,912

(55,000) 77,501 PW(15%) = 324,589.09

81,804

82,867

84,387

86,259

88,890

91,605

91,086 2,453 6,000 (2,100) 97,439

(55,000)

• Result of simulation (with random variables) o Mean: $534,575,300 o Standard deviation: $97,630,260

71 Copyright © 2023 Pearson Education, Inc.

Conntemporary Engineeringg Economics, 7th ed. ©22023

72 Copyright © 2023 Pearson E Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 13 Real-Options Analysis Financial Options 13.1 • Define the option parameters for the call option. S0

K

T

r

σ

$120

$150

0.5

0.06

0.45

The value of the call option is $6.97 by Black-Scholes equation.

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K ) Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

1 Copyright © 2023 Pearson Education, Inc.

120 150 0.45 0.5 0.06 0 6.97 32.53 -0.447892596 -0.766090647 0.327115356 0.221811184 0.672884644 0.778188816

Contemporary Engineering Economics, 7th ed. ©2023 13.2 • Define the option parameters for the put option. S0

K

T

r

σ

$68.50

$70

0.167

0.05

0.6

The value of the put option is $7.20 by Black-Scholes equation.

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K ) Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

2 Copyright © 2023 Pearson Education, Inc.

68.5 70 0.6 0.167 0.05 0 6.28 7.20 0.068307205 -0.176886596 0.527229456 0.429798734 0.472770544 0.570201266

Contemporary Engineering Economics, 7th ed. ©2023

13.3 • Long call and short call

•

Long put and short put

3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

13.4 • Define the option parameters for this option. S0

K

T

r

σ

?

$52

0.75

0.05

0.40

∴ To get a value of call option of $4 by Black-Scholes equation, the current stock price should be $44.3 by Goal seek function in Excel.

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K ) Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

13.5 • u = eσ Δt = e0.3× 0.75 = 1.2967 1 1 d= = = 0.7712 u 1.2967 • Risk neutral probability

4 Copyright © 2023 Pearson Education, Inc.

44.3 52 0.4 0.75 0.05 0 4.00 9.78 -0.181169748 -0.527579909 0.42811717 0.29889547 0.57188283 0.70110453

Contemporary Engineering Economics, 7th ed. ©2023

q=

e r Δt − d e0.05×0.75 − 0.7712 = = 0.5081 u−d 1.2967 − 0.7712

• Tree valuation 100.88 Max(0, (100.88-60)) = $40.88 q 60 $9.81

77.80 $20.02 60 Max(0, (60-60)) = $0

1-q 46.27 $0.00

35.68 Max(0, (35.68-60)) = $0

European call option value = $9.81 13.6 • u = eσ

Δt

= e0.3× 1 = 1.3499 , d =

1 1 = = 0.7408 u 1.3499

• Risk neutral probability e r Δt − d e0.05×1 − 0.7408 q= = = 0.5097 u−d 1.3499 − 0.7408

98.39

• Binomial lattice valuation 72.89 Max(0,45-72.89) 0 Do not exercise

53.99

40

Max(3.34,45-53.99) 3.34 Do not exercise

8.79

29.63 Max(14.24,45-29.63) 15.37 Do not exercise

40 Max(7.17,45-40) 7.17 Do not exercise

21.95 Max(20.86,45-21.95) 23.05 Exercise

5 Copyright © 2023 Pearson Education, Inc.

Max(0,45-98.39) 0 Do not exercise

53.99 Max(0,45-53.99) 0 Do not exercise

29.63 Max(0,45-29.63) 15.37 Exercise

16.26 Max(0,45-16.26) 28.74 Exercise

Contemporary Engineering Economics, 7th ed. ©2023

American option value = $8.79 13.7 • Put-call parity:

c + Ke

− rf

= p + S0

12.00 + Ke −0.06 = 9.45 + 76.33 K = $78.34

13.8 Portfolio A long call with K = $40 A short put with K = $45 Two short calls with K = $35 Two stocks shorted at $40 Total

Premium $3 $4 $5

Payoff at stock price $60 $17 $4 ($40) ($40) ($59)

13.9 • Intrinsic value = S0 − K = $2

• Time premium = option premium – intrinsic value = $2

13.10 • Invest $10,000 in the stocks

Stock Purchase Price

Initial Cost of Stock 57.36 shares

Stock, Price at Expiration

Value of Stock at Expiration

Payoff

$174.35

$10,000

$140

$8,030

($1,968)

6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 $174.35

$10,000

$185

$10,610

$610

$174.35

$10,000

$215

$12,332

$2,332

• Invest $10,000 in the call options Option Price per Contract $1,985 $1,985 $1,985

Initial Cost of Options (5 contacts) $9,925 $9,925 $9,925

Profit Per Option at Expiration

Total Profit of Options

Payoff

$0

$0

($9,925)

$5

$2,500

($7,425)

$35

$17,500

$7,575

o Purchase just one contract of the call option and deposit the rest of money in a savings account which earns 3% annual interest. Option Price per Contract $1,985 $1,985 $1,985

Profit from Depositing $8,015 in a savings account at 3% $248 $248 $248

Profit Per Option at Expiration

Total Profit of Options

Payoff

$0

$0

($1,737)

$5

500

($1,237)

$35

$3,500

$3,748

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

13.11 (a) European call option

Discount rate per period

S0= K= T= r= u= d= σ= q= 1-q= w=

100 105 1.5 5% 1.354 0.739 35% 0.49 0.51 0.9632

183.35 135.41 36.74 100 17.23

78.35 100.00

73.85 0.00

0.00 54.54

17.23 = 0.9632*(0.49(36.74 )+ 0.51(0)) 0.00 0= 0.9632(0.49(0) + 0.51(0))

Option value=17.23

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Contemporary Engineering Economics, 7th ed. ©2023

(b) European put option S 0= K= T= r= u= d=

Discount rate per period

100 100 1 5% 1.191 0.839 35% 0.49 0.51 0.9876

σ= q= 1-q= w=

201.38

119.12 3.73 100 10.43 83.95 17.18

169.05

0.00

141.91

0.00

141.91

0.00

119.12

0.00

100.00

0.00

100.00

7.43

83.95

0.00

70.47

14.81

70.47

27.06

59.16

29.53

39.60

49.66

18.87= 0.9876(0.46(8.29) + 0.54(28.46))

50.34 Option value=

10.43

9 Copyright © 2023 Pearson Education, Inc.

MAX(0, 100 - 49.66) = 50.34

Contemporary Engineering Economics, 7th ed. ©2023 (c) American put option

Discount rate per period

S 0= K= T= r= u= d= σ= q= 1-q= w=

100 100 1 5% 1.191 0.839 35% 0.49 0.51 0.9876

201.38 169.05

0.00

141.91

0.00

141.91

119.12

0.00

119.12

0.00

100

4.04

100.00

0.00

100.00

11.36

83.95

8.05

83.95

0.00

18.73

70.47

16.05

70.47

29.53

59.16

29.53

40.84

49.66

12.42 = 0.9876*(0.46(4.51 )+ 0.54(19.54))

MAX(19.54, 100 - 83.95) = 19.54 Option value=

11.36

50.34 MAX(40.19, 100 - 59.16) = 40.84

MAX(0, 100 - 49.66) = 50.34

13.12 (a) • Define the option parameters for this call option. S0

K

T

r

σ

$40

$40

1.167

0.06

0.4

The value of call option is $8.05 by Black-Scholes equation.

10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b) • Define the option parameters for this put option. S0

K

T

r

σ

$50

$55

1.5

0.06

0.2

The value of put option is $5.02 by Black-Scholes equation. (c) • Define the option parameters for this put option. S0

K

T

r

σ

$38

$40

0.25

0.06

0.6

The value of put option is $5.35 by Black-Scholes equation. (d) • Define the option parameters for this call option. S0

K

T

r

σ

$100

$95

3

0.08

0.4

The value of call option is $38.27 by Black-Scholes equation. 13.13 o

The accumulated cost of the hedge at the end of year one is ($50, 000 − $38, 000)e0.06 = $12, 742

Let S be the market price: •

If S < $1.25, the put is in the money and the payoff is 1,000,000(1.25 – S) = $1,250,000 – 1,000,000S. The sale of the coffee beans has a payoff of 1,000,000(S – 1) - $12,742 + $1,250,000 – 1,000,000S = $237,258

•

From $1.25 to $1.40 neither option has a payoff and the profit is

11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 1,000,000(S – 1) - $12,742 = 1,000,000S -1,012,742 •

If S > $1.40, the call option is in the money and the payoff is -$1,000,000(S – 1.40) = $1,400,000 – 1,000,000S. The profit is 1,000,000(S -1) - $12,742 +$1,400,000 – 1,000,000S = $387,258

•

Therefore, the range is $237,258 to $387,258.

Real-Options Analysis 13.14 • Define the real option parameters for delaying option. V0

I

T

r

σ

$1.9 Million

$2 Million

1

0.08

0.4

The value of delaying is $0.32 Million by Black-Scholes equation. If the choice is to defer or cancel, the value of delaying is $0.32M as calculated. If the choice is to defer or upgrade now, then we have to subtract the conventional NPW and the answer is $0.32M – (-$0.1M) = $0.42M.

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Contemporary Engineering Economics, 7th ed. ©2023

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K )

1.9 2 0.4 1 0.08 0 0.32 0.27

Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

0.271766764 -0.128233236 0.607099319 0.448982199 0.392900681 0.551017801

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

13.15 • Define the real option parameters for license option. V0

I

T

r

σ

$160.5 Million

$140 Million

3

0.06

0.2

The value of license is $48.16 Million by Black-Scholes equation.

13 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K )

160.5 140 0.2 3 0.06 0 48.16 4.60

Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

1.087299282 0.740889121 0.861547717 0.770619664 0.138452283 0.229380336

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

Growth Options 13.16 • Define the real option parameters for this option. V0

T

r

$1 Million

3

0.06

14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(a) Optimal Cutting Policy – Wait until Year 3.

0 r ∆t u d w q 1-q

1.5

1.4

1

2

3

0.06 1 1.2 0.8 0.941765 0.6546 0.3454

5.80608

3.456 4.838416 4.438416

Equivalent σ = 0.1823 Rent

1.6

1.92 3.65532

3.87072 3.25532

0.4

Option value = $1.228559

1 2.628559 2.228559

2.304 3.22561

2.82561

1.28 2.311312 1.911312

2.58048

1.536 2.150407 1.750407 1.72032

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Contemporary Engineering Economics, 7th ed. ©2023

(b) Optimal Cutting Policy – Wait until Year 3.

1.6

1.5

1.4

1

2

3

0 r ∆t u d w q 1-q

0.06 1 1.2 0.8 0.941765 0.6546 0.3454

5.80608

3.456 4.838416 4.838416

Equivalent σ = 0.1823 Rent

1.92 4.032026 4.032026

3.87072

0.4

Option value = $2.36

1 3.360033 3.360033

2.304 3.22561

3.22561

1.28 2.688017 2.688017

2.58048

1.536 2.150407 2.150407 1.72032

16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Deferral Options 13.17 • Define the real option parameters for deferral option.

∴

V0

I1

I2

T

r

σ

$60,000

$38,588

$60,638

2

0.06

0.2

The value of postponing the construction decision for two years: $349,743

0 I1= I2= σ= r= T= ∆t = u= d= q= 1-q =

60,000 349,743

2 38,588 60,638 0.2 0.06 2 2 1.33 0.75 0.65 0.35

38,588=35,000(1.05)(1.05) 60,638=55,000(1.05)(1.05)

410,263=MAX(79,614-38,588,0)*10 569,289=MAX(79,614-60,638,0)*30

79,614 410,263 10 stores mall open 569,289 30 stores mall open 45,218 66,308 10 stores mall open 0 30 stores mall open

349,743=EXP(-0.06*2)*(569,289*0.65+66,308*0.35)

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Contemporary Engineering Economics, 7th ed. ©2023

Switching Options 13.18 • The NPW of project B: PW(12%) B = −$2 + $1( P / A,12%,10) = $3.65M • Define the real option parameters for switching option. V0

I

T

r

σ

$4 Million

$3.65 Million

5

0.06

0.5

value of switching is $0.85 Million by Black-Scholes equation for put option. • Therefore, the total value is: ENPW = Value of project A + Option to switch to project B

ENPW = $4 + $0.85 = $4.85M

R&D Options 13.19 • Assuming MARR = 12%, the cash flow diagram transforms to:

$73.42

0 $14.18

1

2

3

4

5

R&D Expenses

6

7

8

9

10

Manufacturing and Distribution $80

18 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • Define the real option parameters for R&D option. V0

I

T

r

σ

$46.66 Million

$80 Million

4

0.06

0.5

The value of option today is $13.70 Million by Black-Scholes equation for a call option. • Therefore, the total value is: ENPW = Cost for R&D + Option value = −$14.18 + $13.7 = −$0.48M The maximum R&D spending should not exceed $13.7M.

Abandonment Options 13.20 • Standard NPW approach 0.35 0.35 0.35 0.35

0.35

... 0

1

2

3

4

.

.

.

n

$3 PW(12%)0 = −$3 +

$0.35 = −$0.08M 0.12

• Abandon Option value through the binomial tree - Option parameters V0 I T $2.92 Million $2.2 Million 5

19 Copyright © 2023 Pearson Education, Inc.

r 0.06

σ 0.5

Contemporary Engineering Economics, 7th ed. ©2023 - Option valuations Time

0 2.92

1 4.81 1.77

2 7.94 2.92 1.07

3 13.09 4.81 1.77 0.65

4 21.58 7.94 2.92 1.07 0.40

0.50

0.23 0.77

0.06 0.42 1.13

0.00 0.12 0.69 1.55

0.00 0.00 0.23 1.13 1.80

Monetary Value

Option value

5 35.57 13.09 4.81 1.77 0.65 0.24 0.00 0.00 0.00 0.43 1.55 1.96

* It is optimal to exercise early at the shaded positions.

∴ ENPW = -0.08 + 0.50 = 0.42M

Scale-Down Options 13.21 • Scale down option parameters V0 $10 Million

I $4 Million

T 3

r 0.06

• Decision tree for a scale-down option through one-year time increment. • • •

u = eσ Δt = e0.3× 1 = 1.35 1 1 d= = = 0.74 u 1.35 erΔt − d e0.06×1 − 0.74 q= = = 0.53 u−d 1.35 − 0.74

20 Copyright © 2023 Pearson Education, Inc.

σ 0.3

Contemporary Engineering Economics, 7th ed. ©2023

0 K= σ= r= u= d= q=

1

3

4 0.3 0.06

=max(18.22*0.8+4,EXP(-0.06)*(24.6*0.53+14.8*(1-0.53))) =18.8

1.35 0.74 0.53

1.35

10 11.7671

2

1.35

13.50 14.7989 scale down

0.74

1.35

18.22 18.8003 0.74 Do not scale down

0.74

1.35

10 12.0000 scale down

1.35

7.41 9.9265 scale down

0.74

13.50 14.80 scale down

0.74 1.35

5.49 8.3905 scale down

24.60 24.60 Do not scale down

7.41 9.93 scale down

0.74

4.07 7.25 scale down

• From the result of the tree we can get the flexible NPV: ENPW = NPW + Option value = $11.77M Option value = $11.77M - $10M = $1.77M

21 Copyright © 2023 Pearson Education, Inc.

max(24.6*0.8+4,24.6) =24.6

max(13.5*0.8+4,13.5) =14.8

Contemporary Engineering Economics, 7th ed. ©2023

Expansion-Contraction Options 13.22 (a) Binomial lattice tree • u = eσ Δt = e0.15× 1 = 1.1618 1 1 = 0.8607 • d= = u 1.1618 • Tree with incremental period one year

134.99

116.18 100 100 86.07 74.08

(b) Option valuation • Risk neutral probability q=

erΔt − d e0.05×1 − 0.8607 = = 0.6329 , 1− q = 0.3671 u−d 1.1618 − 0.8607

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Contemporary Engineering Economics, 7th ed. ©2023

134.99 Max(134.99×0.9+25,134.99×1.3-20,134.99) 155.48 Expand

116.8

Max(116.8×0.9+25,116.8×1.3-20,133.76) 133.76 Keep option open

100 Max(100×0.9+25,100×1.3-20,100) 115 Contract

100 116.31

86.07 Max(86.07×0.9+25,86.07×1.3-20,101.24) 102.46 Contract

74.08 Max(74.08×0.9+25,74.08×1.3-20,74.08) 91.67 Contract

∴ Option value = ENPW – NPW = $116.31 - $100 = $16.31M

23 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Compound Options 13.23 • Compound option parameters V0

I1

I2

T1

T2

$32.43

$10

$30

1

3

r 0.06

σ 0.5

• Decision tree for a scale-down option through one-year time increment. - u = eσ Δt = e0.5× 1 = 1.6487 1 1 - d= = = 0.6065 u 1.6487 er Δt − d e0.06×1 − 0.6065 = = 0.4369 - q= u−d 1.6487 − 0.6065 145.34 Max(145.34 - 30, 0) 115.34 Invest $30

88.15 59.90 Keep option open

53.47 Max(53.47 - 30, 0) 23.47 Invest $30

53.47 Max(29.77 - 10, 0) 19.77 Invest $10

32.43 8.13

32.43 9.66 Keep option open

Max(19.67 - 30, 0) 0 Do not invest

19.67 Max(3.97 - 10, 0) 0 Do not invest

19.67

11.93 0 Do not invest

7.24 Max(7.24 -30, 0) 0 Do not invest

First Option

Second Option

ENPW is $8.13 and this exceed the initial cost $5. Initiate the phase I. 24 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

13.24 • Option parameters

∴

V0

I1

I2

$19.5M

$3M

$12M

T 3

r 0.05

The company is willing to pay $ 16,646,312 for this option.

25 Copyright © 2023 Pearson Education, Inc.

σ 0.25

Contemporary Engineering Economics, 7th ed. ©2023

Phase 1 1

0

Phase 2 2

3,000,000

I=

3

12,000,000

12,000,000

29,281,500=MAX(41,281,500-12,000,000, 0) 2,700,000: Abandon (Selling price of the land) volatility= r= T= ∆t = u= d= v= q= 1-q =

20,735,312=EXP(-0.05*1)*(29,281,500*0.54+13,038,496*0.46) 20,150,065=MAX(32,150,065-12,000,000, 0) 2,700,000: Abandon (Selling price of the land)

0.25 0.05 3.00 1.00 1.28 0.78 0.00 0.54 0.46

41,281,500

25,038,496 14,180,447 22,038,496 2,700,000 15,186,615 5,382,658 12,186,615 2,700,000

19,500,000 16,646,312

Keep Option Invest 3M Abandon Keep Option Invest 3M Abandon

5,382,658=EXP(-0.05*1)*(8,085,247*0.54+2,817,955*0.46) 12,186,615=MAX(15,186,615-3,000,000, 0) 2,700,000: Abandon (Selling price of the land) Option value=

32,150,065 20,735,312 20,150,065 2,700,000 19,500,000 8,085,247 7,500,000 2,700,000 11,827,348 2,817,955 2,700,000

Keep Option Invest 12M Abandon Keep Option Invest 12M Abandon Keep Option Invest 12M Abandon

16,646,312

26 Copyright © 2023 Pearson Education, Inc.

29,281,500 Invest 12M 2,700,000 Abandon 25,038,496 13,038,496 Invest 12M 2,700,000 Abandon 15,186,615 3,186,615 Invest 12M 2,700,000 Abandon 9,211,148 - Invest 12M 2,700,000 Abandon

Contemporary Engineering Economics, 7th ed. ©2023

13.25 (a) Contract option = $113.78 (b) Abandonment option = $105.80 (c) Expansion option = $112.82 (d) Combined options = $117.25

(a) Contract option value = $113.78 Vo K ( or I) r σ Δt u d q 1-q w

100 0.05 0.15 1 1.161834 0.860708 0.632835 0.367165 0.951229

(a) Contract option 156.8312 166.1481

134.9859 146.4873 116.1834 129.5651

100 113.7807

116.1834 129.5651

100 115 86.0708 102.4637

86.0708 102.4637

74.08182 91.67364

=MAX(H26,H26*0.9+25,(K23*B9+K31*B10)*B11

27 Copyright © 2023 Pearson Education, Inc.

63.76282 82.38653

Contemporary Engineering Economics, 7th ed. ©2023

(b) Abandonment option value = $105.89 Vo K ( or I) r σ Δt u d q 1-q w

100 100 0.05 0.15 1 1.161834 0.860708 0.632835 0.367165 0.951229

(a) Abandonment option 156.8312 156.8312

134.9859 134.9859 116.1834 117.8825

100 105.8877

116.1834 116.1834

100 104.8649 86.0708 100

86.0708 100

74.08182 100

=MAX(H26,B3,(K23*B9+K31*B10)*B11)

28 Copyright © 2023 Pearson Education, Inc.

63.76282 100

Contemporary Engineering Economics, 7th ed. ©2023

(c) Expansion option value = $112.82

Vo K ( or I) r σ Δt u d q 1-q w

100 20 0.05 0.15 1 1.161834 0.860708 0.632835 0.367165 0.951229

(c) Expansion option 156.8312 183.8806

134.9859 156.4571 116.1834 132.9417

100 112.823

116.1834 131.0385

100 110.9754 86.0708 93.90155

86.0708 91.89204

74.08182 77.58604

=MAX(H26,H26*1.3-B3,(K23*B9+K31*B10)*B11)

29 Copyright © 2023 Pearson Education, Inc.

63.76282 63.76282

Contemporary Engineering Economics, 7th ed. ©2023

(c) Combined option value = $117.25 Vo K ( or I) r σ Δt u d q 1-q w

(a) 100 25 0.05 0.15 1 1.161834 0.860708 0.632835 0.367165 0.951229

100 117.2493

(b) 20

(c) 100 (c) Combined option

Combined option value in RED

116.1834 134.3474 129.5651 132.9427 117.8825

86.0708 104.1525 102.4637 93.9015 100

134.9859 156.4571 146.4873 156.4571 134.9859

100 115 115 110.9754 104.8649

74.08182 100 91.6736 77.586 100

=MAX(H26,H28,H29,H30,(K23*B9+K31*B10)*B11)

156.8312 183.8806 166.1481 183.8806 156.8312

Equity value Combined Option value Contraction Expansion Avandonment

116.1834 131.0385 129.5651 131.0385 116.1834

Equity value Combined Option value Contraction Expansion Avandonment

86.0708 102.4637 102.4637 91.892 100

Equity value Combined Option value Contraction Expansion Avandonment

63.76282 100 82.3865 63.7628 100

Equity value Combined Option value Contraction Expansion Avandonment

30 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies ST 13.1 V0

σ T

Δt r I1

100 0.2 1 1 0.06 50

u d q 1-q e(-rt)

1.2214028 0.8187308 0.6037315 0.3962685 0.9512294

u d q 1-q e(-rt)

1.3498588 0.7408182 0.5270886 0.4729114 0.9512294

V1

122 0.3 3 1 0.06

σ T

Δt r I2

V1

σ T

Δt r I2

70

82 0.3 3 1 0.06 70

300 230 $223 $158 $165 $105 $122 $17.46

165 95 $122 $57 90 20

$90 $33

$100 8.76

67 10

50 0

$149 $84

201 131

$111 $51 $82 -$19.70 $0.00

111 41 $82 $20

$61 $10

61 0 $45 $0

31 Copyright © 2023 Pearson Education, Inc.

33 0

Contemporary Engineering Economics, 7th ed. ©2023

ST 13.2 (a) American put option value • Option parameters V0 $150 • u = eσ

K $100 Δt

Δt

T 2

= e0.3× 1 = 1.3499 , d =

1 year

r 0.05

σ 0.3

1 1 = = 0.7408 u 1.3499

• Risk neutral probability

q=

er Δt − d e0.05×1 − 0.7408 = = 0.5097 , 1 − q = 0.4903 u−d 1.3499 − 0.7408 0 v= K= σ= r= r -v = u= d= q=

1

2

0 100 0.3 0.05 0.05 1.35 0.74 0.51 1.35 1.35

150 3.84

202.48 0.00

0.74

0.74 1.35

111.12 8.24

273.32 0.00

150.00 0.00

0.74

82.32 17.68

∴ American put option value = $3.84 (In our example, dividend rate (v) = 0.)

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b) Expansion Option value • Option parameters K

V0 $400 • u = eσ

Δt

= e0.35×

1

Δt

T 2 = 1.42 , d =

1 year

r 0.07

1 1 = = 0.70 u 1.42

• Risk neutral probability

q=

erΔt − d = 0.51 , 1− q = 0.49 u−d

0

σ= r= r-v = u= d= q=

1

2

0.35 0.07 0.07 1.42 0.70 0.51 1.42 1.42

400 452.58

568.00 658.80 679.36

0.70

0.70 1.42

280.00 198.00 279.53

806.56 1040.50

397.60 397.60

0.70

∴ Expansion option value = $452.58

33 Copyright © 2023 Pearson Education, Inc.

196.00 196.00

σ 0.35

Contemporary Engineering Economics, 7th ed. ©2023

ST 13.3 (a) Since $4 M is lower than the option price, it is a good investment for Merck Co. • To give a range for the option value, first using one period lattice. V0 $36M

K $72M

Δt

T 3

3 years

r 0.06

($30×1.2M) ($60×1.2M) • u = eσ Δt = e0.5× 3 = 2.3774 1 1 • d= = = 0.4206 u 2.3774 • Risk neutral probability

er Δt − d e0.06×3 − 0.4206 q= = = 0.3969 , 1 − q = 0.6031 u−d 2.3774 − 0.4206 • One period lattice and option price: 4.50M 85.59 Max(85.59 - 72, 0) 13.59 Buy the stock 36

4.50 15.14 Max(15.14 - 72, 0) 0 Do not buy the stock • Through the B-S model, the option price should be $6.54M.

34 Copyright © 2023 Pearson Education, Inc.

σ 0.5

Contemporary Engineering Economics, 7th ed. ©2023

Note: We can think that the price from the one step binomial tree is the lower bound and B-S is the upper bound. So the option price is definitely higher than the suggested price. (b) By buying the aforesaid agreement, Merck has a chance to buy Genetics with a lower price than the market when the project is successful. Otherwise they just lose $4M. If Merck purchases the stock now, Merck also has an opportunity to earn huge money than previous position. In other case, Merck will lose more than the foresaid agreement case. To make this concept simple we may demonstrate a basic example below: • Profit/loss analysis Success Fail

Buy stock Gain $37.59M Loss $32.86M

Aforesaid agreement Gain $13.59M Loss $4M

∴ The “Buy stock” option has a higher risk than the other option.

ST 13.4

V0 = $58,490,998

0

Manufacturing Phase

Development Phase

R&D Phase

V8 = $144,821,558

4

8 9

I0 = $4,329,258

I1 = $4,435,954

I2 = $100,000,000

35 Copyright © 2023 Pearson Education, Inc.

18

Contemporary Engineering Economics, 7th ed. ©2023

Research Phase 0 Revenue Expenses: Labor Materials Overhead Depreciation Taxable inocme Taxes Net income

1

800,000 500,000

-

Development Phase

2

3

800,000 400,000

4

800,000 400,000

5

6

7

8

800,000 400,000

750,000 750,000 750,000 2,050,000 - 1,950,000 - 1,950,000 820,000 780,000 780,000 1,230,000 - 1,170,000 - 1,170,000 -

320,000 320,000 320,000 350,000 350,000 350,000 1,000,000 1,000,000 1,000,000 750,000 500,000 500,000 500,000 1,950,000 - 2,170,000 - 2,170,000 - 2,170,000 780,000 868,000 868,000 868,000 1,170,000 - 1,302,000 - 1,302,000 - 1,302,000 -

320,000 350,000 1,000,000 500,000 2,170,000 868,000 1,302,000

Operating cash Capital Exp

-

480,000 -

420,000 -

420,000 -

420,000 -

802,000 -

802,000 -

802,000 -

802,000

-

3,000,000

Net cash flow

-

3,000,000 -

480,000 -

420,000 -

420,000 -

420,000 -

802,000 -

802,000 -

802,000 -

802,000

Future Cap. Exp. I0 =

-

-

2,000,000

-

4,435,954

I2 =

4,329,258 I1 =

-

100,000,000

V8 =

144,821,558

V0 =

58,490,998

Manufacturing Phase 9

10

11

12

13

14

15

16

17

18

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

50,000,000

1,000,000 10,000,000 8,000,000 10,000,000 21,000,000 8,400,000 12,600,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

1,000,000 4,000,000 8,000,000 10,000,000 27,000,000 10,800,000 16,200,000

22,600,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

22,600,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

26,200,000

36 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Teaching Notes: •

Once students generate the cash flows in each phase of the project, ask them to determine whether it is worth investing $4,329,258 at n = 0, and $4,435,954 at n = 4. In this case, we are assuming a simple call option, not a compound option. The equivalent required investment during the R&D and Development phases is $4, 329, 258 + $4, 435, 954( P / F ,15%, 4) = $7,148, 387

which exceeds the option value calculated by the B-S Model ($8,504,641), so we may suggest that the firm may initiate both R&D and Development activities.

Black-Scholes Model Stock price (S 0) or Project Value (V0) Exercise Price or Investment Cost (K ) Volatility (σ) Time to Investment Decision (T ) Risk-free Rate (r ) Dividend Yield (ω) Call Value Put Value

d1 d2 N(d 1) N(d 2) N(-d 1) N(-d 2)

•

58490998 100000000 0.15 8 0.06 0 8504640.61 11891981.79 0.079437969 -0.344826099 0.531657866 0.365112555 0.468342134 0.634887445

However, this is not a simple option – since we have a flexibility to decide, upon completion of the R&D, whether to proceed for the Development Phase, the option value will be much higher than the simple option. To capture this additional flexibility, we may adopt an advanced real options technique, known as Geske Model, which is not covered in the text. For interested readers, the Geske model reveals that this added option value from the phased investments amounts to $9,036,605.

37 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 14 Replacement Decisions Note to Instructors: Regular MACRS depreciation is assumed in all problems unless otherwise mentioned. However, instructors may adopt the capital expensing option under the 2017 Tax Cuts and Job Acts instead. In that case, the entire capital expenditure would be claimed as depreciation amount at the end of first year. Then, any salvage value at the end of project life will be subject to taxable gains.

Sunk Costs, Opportunity Costs, and Cash Flows 14.1 Tax R ate(% ) = M A R R (% ) =

P W (i) = A E (% ) =

0.00% 10.00%

0

1

2

($20,065) ($6,329.8)

3

4

Income Statement

R evenues (savings) E xpenses: O&M D epreciation

$2,500 $0

$3,000 $0

$3,500 $0

$4,000 $0

Taxable Incom e Incom e Taxes (% )

($2,500) 0

($3,000) 0

($3,500) 0

($4,000) 0

N et Incom e

($2,500)

($3,000)

($3,500)

($4,000)

Cash Flow Statement

O perating A ctivities: N et Incom e D epreciation Investm ent A ctivities: $ Investm ent S alvage G ains Tax N et C ash Flow

$

(2,500) $ $0

(3,000) $ $0

(3,500) $ $0

(4,000) $0

(12,000) $

($12,000)

($2,500)

($3,000)

($3,500)

3,000 $0.00

($1,000)

PW(10%) = −$12, 000 − $2,500( P / F ,10%,1) ⋅⋅⋅⋅⋅ −$1, 000( P / A,10%, 4) = −$20, 065 AEC(10%) = $20, 065( A / P,10%, 4) = $6,329.8

1 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.2 A. B. C. D.

Original cost: The printing machine was purchased for $20,000 Market value: The old machine’s market value is estimated at $10,000. Book value: If now the machine sold its book value is $14,693. Trade in allowance: This amount is the same as the market value.

The market value is the most relevant information, but the defender’s current book value is also relevant as this will be the basis to determine the gains or losses related to disposal of the defender. 14.3 Option 1: Keep the defender PW(12%) D = −$8, 000( P / A,12%,3) + $2,500( P / F ,12%,3) = −$17, 434.9 AEC(12%) D = $17, 434.9( A / P,12%,3) = $7, 259.1 Option 2: Replace the defender with the challenger PW(12%)C = −$5, 000 − $6, 000( P / A,12%,3) + $5,500( P / F ,12%,3) = −$15, 495.9 AEC(12%)C = $15, 495.9( A / P,12%,3) = $6, 451.9 The replacement should be made now. 14.4 (a) Sunk cost = $12,000- $2,000 = $10,000. (b) Opportunity cost = $2,000. PW(15%) D = $2, 000 + $10, 000( P / A,15%,3) = $24,832 (c)

PW(15%)C = $15, 000 + $3, 000( P / A,15%,3) − $5, 000(P/ F,15%,3) = $18,562 The net incremental benefit is $6,270 associated with replacing the old machine tools. 2 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.5 (a) Purchase cost = $15,000, market value = $6,000, sunk cost = $15,000 - $6,000 = $9,000 (b) Opportunity cost = $6,000 (c)

PW(15%) = −$6, 000 − $1,500 − $3, 000( P / F ,15%,1) −($3,500 − $3, 000)( P / F ,15%, 2) = $10, 486.77 AEC(15%) = $10, 486.77( A / P,15%, 2) = $6, 450.58

(d) PW(15%) = −$7,500 − $3, 000( P / F ,15%,1) − $3,500( P / F ,15%, 2) −$3,800( P / F ,15%,3) − $4,500( P / F ,15%, 4) −$9,800( P / F ,15%,5) = −$22, 698.98 AEC(15%) = $22, 698.98( A / P,15%,5) = $6, 771.46

14.6 (a) Opportunity cost = $30,000 (b) Assume that the old machine’s operating cost is $30,000 per year. Then the new machine’s operating cost is zero per year. The cash flows associated with retaining the defender for two more years are n

0

1

2

Cash Flows: -$30,000 -$30,000 -$18,000

AEC D = ($30, 000 − $12, 000)( A / P,12%, 2) + $12, 000(0.12) +$30, 000 = $42, 090.57 (c) Cash flows for the challenger: Year 0: -$165,000; Years 1-7: 0; Year 8: $5,000

3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 AECC = ($165, 000 − $5, 000)( A / P,12%,8) + $5, 000(0.12) = $32,808.45

(d) Since AEC D > AECC , we should replace the defender now. 14.7 (a) Initial cash outlay for the new machine = $120,000 (b) Cash flows for the defender: Year 0: -$10,000 Years 1-5: 0 (c)

AE(15%) D = −$10, 000( A / P,15%,5) = −$2,983 AE(15%)C = − [ ($120, 000 − $30, 000)( A / P,15%, 7) + $30, 000(0.15)] +$50, 000 = $23,868 We should purchase the new machine because it has a higher annual equivalent cash flow. 14.8 (a) Cash flows Year: 0 Defender -$10K Challenger -$75K

1 0 $30K

2 0 $30K

3 0 $30K

4 0 $30K

5 $5K $30K

(b)

PW(15%) D = −$10 K + $5 K ( P / F ,15%,5) = −$7,514 PW(15%)C = −$75K + $30 K ( P / A,15%,5) = $25,565 Replace the defender. 14.9 (a) and (b) Cash flows: Year: 0 1 2 3 4 5 Defender -$5,500 $21,000 $21,000 $22,200 Challenger -$36,500 $24,000 $24,000 $24,000 $24,000 $30,300

• •

Revenue for defender = ($19 - $12) × 3,000 = $21,000 per year Revenue for challenger = ($19 - $11) × 3,000 = $24,000 per year

4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c)

AE(12%) D = − [ ($5,500 − $1, 200)( A / P,12%,3) + $1, 200(0.12) ] + $21, 000 = $19, 066 AE(12%)C = − [ ($36,500 − $6,300)( A / P,12%,5) + $6,300(0.12) ] + $24, 000 = $14,866 Keep the defender for now.

Economic Service Life 14.10 Annual changes in MV Interest rate n

Market Value 0 1 2 3 4 5 6 7 8

$200,000 $130,000 $104,000 $83,200 $66,560 $53,248 $42,598 $34,079 $27,263

20% 12% O&M Costs CR(12%)

$20,000 $20,000 $20,000 $20,000 $20,000 $22,000 $25,000 $28,000

$94,000 $69,283 $58,614 $51,920 $47,100 $43,396 $40,446 $38,044

The economic life of the asset is eight years; that is, N * = 8 years. Thus, AEC* = $59,354

14.11 Note: Missing Table P14.8 in the first printing. Year (n) 0 1 2 3 4

OC 3,200 3,700 4,800 5,850

MV 7,700 4,300 3,300 1,100 0

(a) Interest i = 12% Defender:

5 Copyright © 2023 Pearson Education, Inc.

OC(12%) AEC(12%)

$20,000 $20,000 $20,000 $20,000 $20,000 $20,246 $20,718 $21,310

$114,000 $89,283 $78,614 $71,920 $67,100 $63,642 $61,163 $59,354

Contemporary Engineering Economics, 7th ed. ©2023 Year 0 1 2 3 4

OC $3, 200 $3, 700 $4, 800 $5, 850

MV $7, 700 $4, 300 $3, 300 $1,100 0

OC

CR(12%) Total AEC (12%)

$3, 200 $3, 436 $3,840 $4, 261

$4,324 $2,999 $2,880 $2,535

$7,524 $6, 435 $6, 720 $6, 796

The defender’s remaining useful (economic) life is 2 more years with an AEC value of $6,435, i.e., N D = 2,AEC D = $6,435 . (b) N C = 10 years

AECC = $31,000( A / P,12%,10) + $1,000 − $2,500( A / F,12%,10) = $6,344 Since AEC D > AECC , the defender should be replaced now. (c) n* = 0. 14.12 (a) Economic service life = 2 years:

Interest rate n 0 1 2 3

Market Value $4,000 $3,000 $2,250 $1,688

12% O&M Costs $1,000 $3,500 $4,500 $5,500

CR(12%)

$1,480 $1,305 $1,165

OC(12%)

$4,620 $4,563 $4,841

AEC(12%)

$6,100 $5,869 $6,006

When N = 2, we have the smallest AEC value. The economic life of the system is 2 years (b) Replacement decision:

AEC D = $5,869, with N* = 2. AECC = $5,900, with N* = 5. Since AECD < AECC , do not replace the defender. 6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(c) When to replace?

if you keep the defender for one more year, the marginal cost would be AECD = $2,250(1.12)+ $5,500− $1,688 = $6,322. Since AECD > AECC , keep the defender for two years. (d) Required assumptions: The cost of challenger would remain the same. 14.13 (a) Interest i = 10% OC

MV

AE OC

CR(10%)

AEC(10%)

0 1 2

$2,500 $3, 200

$15,000 $12,800 $8,100

$2,500 $2,833

$3,700 $4,786

$6, 200 $7,619

3 4

$5,300 $6,500

$5, 200 $3,500

$3,579 $4, 208

$4, 461 $3,978

$8,039 $8,186

5

$7,800

0

$4,796

$3,957

$8,753

OC

MV

AE OC

CR(10%)

AEC(10%)

0 1

$2,500

$15, 000 $12,800

$2,500

$4, 450

$6, 950

2 3 4

$3, 200 $5, 300 $6,500

$8,100 $5, 200 $3,500

$2,826 $3,538 $4,131

$5, 459 $5, 072 $4,553

$8, 285 $8, 610 $8, 684

5

$7,800

0

$4, 675

$4, 475

$9,150

n

(b) Interest i = 15% n

∴ In both cases, the economic service life is 1 year.

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Replacement Decision with an Infinite Planning Horizon and No Technological Change 14.14 It is assumed that the required service period is very long. AECD = $6, 000( A / P,12%, 6) + $2, 000 − $1,500( A / F ,12%, 6) = $3, 274.52 AECC = $21, 000( A / P,12%,12) + $1, 000 − $500( A / F ,12%,12) = $4,369.46 We should continue to use the present machine. The economic advantage is $4,369.46-$3,274.52 = $1,094.94 per year.

14.15 (a) and (b) n Defender Challanger 0 1

-$4,000 -$3,000

-$6,000 -$2,000

2

-$4,500

-$3,000

3

-$5,000

-$2,000

AECC (15%) = ($6,000 +

$1,000 )( A / P,15%,3) + $2,000 1.152

= $4,959.04 AEC D (15%) = ($4,000 +

$3,000 $4,500 $5,000 + + )( A / P,15%,3) 1.15 1.152 1.153

= $5,824.62 Now is the time to replace the defender. 14.16 (a) Opportunity cost = $0 (b) The cash flows are:

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Year: 0 1 2 3 4 5 Defender $0 -$3K -$3K -$3K -$3K -$3K Challenger -$10K 0 0 0 0 0 C-D

-$10K

$3K

$3K

$3K

$3K

$3K

(c)

PW(i)C−D = −$10,000 + $3,000(P / A,i,5) =0 ∴ We find i* = 15.24% . Since i* > MARR, the firm should buy the challenger. 14.17 (a)

AE(15%) D = −$1,000( A / P,15%,5) + $10,000 − $7,000 = $2,701.68 AE(15%)C = −$12,000( A / P,15%,5) + $11,500 − $5,000 + $2,000( A / F,15%,5) = $3,216.84 Yes, the new machine should be purchased now. (b) Let P as the current market value of the old machine − P( A / P,15%,5) + $10,000 − $7,000 = $3,216.84

We find P = −$726.9 . Let P as the cost of new machine − P( A / P,15%,5) + $11,500 − $5,000 + $2,000( A / F,15%,5) = $2,701.68

We find P = $13,727 .

9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.18 Assume that the old system has a current market value of P.

AEC D = P( A / P,14%,5) + $20,000 AECC = $200,000( A / P,14%,10) + $5,000 − $18,000( A / F,14%,10) = $42,411.86 Let AEC D = AECC and solve for P. We find that P = $76,942. If the resale value of the defender is higher than $76,942, the installation of the system is justified. 14.19

AEC(12%) D = $60,000( A / P,12%,10) + $18,000 = $28,619 AEC(12%)C = ($200,000 − $20,000)( A / P,12%,10) + $20,000(0.12) + $4,000 = $38,257 Since AEC D < AECC , do not replace the defender for now. 14.20

AECC = $53,500( A / P,12%,5) − $12,000( A / F,12%,5) +$4,200 + $500( A / G,12%,5) = $18,039.80 AEC D = $8,500( A / P,12%,5) + $8,700 = $11,057.98 Since AECC > AEC D , don’t purchase the challenger now.

10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.21 Economic service life: n

OR

MV

AEOR

0 1 2 3 4

35,550 31,013 25,794 19,794

20,000 17,000 14,450 12,283 10,440

35,550 33,390 31,095 28,660

CR(10%)

AE(10%)

5,000 4,643 4,332 4,060

30,550 28,747 26,763 24,600

The economic life of the cab is one year. These cabs should be replaced every year. (a) Rate of return calculation:

PW(i) = −$20,000 +

$35,550 + $17,000 =0 1+ i

Solving for i yields 162.75%. The internal rate of return with one-year replacement cycle is 162.75%. 14.22 For the challenger, we have:

AECC = $50,000( A / P,14%,12) + $3,000 − $6,000 − $3,000( A / F,14%,12) = $5,724 For the defender, if sold at the end of first year, it will bring in $1,500. We assume the market values will be declined the same amount (($1,500-$1,000)/4 = $125) each year for the remaining years.

Year

OC

MV

AEOC

CR(14%) AEC(14%)

0 1

$2,000 $3,800 $1,500 $3,800

$780

$4,580

2 3 4

$3,800 $1,375 $3,800 $3,800 $1,250 $3,800 $3,800 $1,125 $3,800

$572 $498 $458

$4,372 $4,298 $4,258

5

$3,800 $1,000 $3,800

$431

$4,231

11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

N D = 5 years, AEC D = $4,231 with i = 14%. Since AECC > AEC D , the new machine should not be purchased. 14.23

AEC(15%)Option 1 = $15,000 + ($6,000 − 0)( A / P ,15%,10) +$12,000 + ($48,000 − $5,000)(A / P ,15%,10) + $5,000(0.15) = $37,513.35 AEC(15%)Option 2 = $24,000 + ($84,000 − $9,000)(A / P ,15%,10) + $9,000(0.15) = $40,293.90 Since AECOption 1 < AECOption 2 , Option 1 should be selected.

Replacement Problem with a Finite Planning Horizon 14.24 (a)

AEC D = $12,000( A / P,12%,1) + $3,750 − $9,600( A / F,12%,1) = $7,590 AECC = $18,000( A / P,12%,1) + $3,300 − $13,500( A / F,12%,1) = $9,960 Since AECC > AEC D , keep the defender for now. (b)

Best course of action:

12 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Defender

Annual changes in MV Annual increases in A/T O&M Interest rate n

Market Value 0 1 2 3

$12,000 $9,600 $7,680 $6,144

20% 35% 12% O&M Costs CR(12%) OC(12%) AEC(12%)

$3,750 $5,063 $6,834

$3,840 $3,478 $3,175

$3,750 $4,369 $5,100

$7,590 $7,847 $8,275

Challenger

Annual changes in MV Annual increases in A/T O&M Interest rate n

Market Value 0 1 2 3

$18,000 $13,500 $10,125 $7,594

AEC 1 2 3

25% 30% 12% O&M Costs CR(12%) OC(12%) AEC(12%)

$3,300 $4,290 $5,577

$6,660 $5,875 $5,244

$3,300 $3,767 $4,303

$9,960 $9,642 $9,547

Defender Challenger $7,590 $9,960 $7,847 $9,642 $8,275 $9,547

The annual equivalent cost of the defender is less than that of the challenger for 3 years, we should keep the defender for 3 years. 14.25 There are several plausible scenarios. Some of the most feasible scenarios are:

( j0 ,2),( j,3),( j,3),( j,2) :PW(12%) = $17,601 ( j0 ,1),( j,3),( j,3),( j,3) :PW(12%) = $18,081 ( j0 ,3),( j,3),( j,3),( j,1) :PW(12%) = $17,597 ( j0 ,3),( j,3),( j,4) :PW(12%) = $17,221 ( j0 ,4),( j,3),( j,3) :PW(12%) = $18,384 13 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

It appears that the ( j0 ,3), ( j ,3), ( j , 4) option becomes the best strategy. Keep the defender for 3 years; replace it with challenger j and keep it for 3 years; replace it with another challenger and keep it for 4 years. 14.26 There are several plausible scenarios. Some of the most feasible scenarios are.

( j0 ,3),( j,3),( j,2) :PW(12%) = $67,526 ( j0 ,2),( j,3),( j,3) :PW(12%) = $67,390 ( j0 ,1),( j,3),( j,3),( j,1) :PW(12%) = $66,856 ( j0 ,4),( j,3),( j,1) :PW(12%) = $65,820 ( j0 ,4),( j,4) :PW(12%) = $65,958 It appears that the ( j0 ,3), ( j ,3), ( j , 2) option becomes the best strategy. Keep the defender for 3 years; replace it with challenger j and keep it for 3 years; replace it with another challenger and keep it for 2 years. 14.27

PWD = $12,000 + $4,000(P / A,11%,6) − $2,000(P / F,11%,6) = $27,852.86 PWC = [$10,000 + $2,000(P / A,11%,3) − $4,000(P / F,11%,3)] × [1+ ( P / F,11%,3)] = $20,709.66 Since PWC < PWD , replace now. 14.28

AE D = $4,000 + $12,000( A / P,11%,6) − $2,000( A / F,11%,6) = $6,583.77 AEC = $2,000 + $10,000( A / P,11%,3) − $4,000( A / F,11%,3) = $4,895.28 Since AEC < AE D , replace now. 14 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Replacement Analysis with Tax considerations 14.29 (a) Comments: Sunk cost can be defined as either the difference between book value and market value or the cost that has already been expended ($). book value = $15,000 − ($3, 000 + $4,800 + $1, 440) = $5, 760 market value = $6, 000

In this example, if the sunk cost is defined as the difference between book value and market value, there would be no sunk cost as the market value exceeds the book value. However, the difference between the purchase cost ($15,000) and the current market value ($6,000) may be viewed as another sunk cost that should not be considered in the replacement analysis. (b) Opportunity cost of not replacing the truck: Book value = $15,000 − ($3,000 + $4,800 + $1,440)

• • •

= $5,760 Taxable gains = $6,000 - $5,760 = $240 Gains tax = $240 (0.25) = $60 Net proceeds from sale = $6,000 - $60 = $5,940

The opportunity cost for not replacing the old machine now, which is $5,940, is viewed as an investment required to keep the old machine. (c) Equivalent annual cost of operating the truck for two more years: Cash Flow Elements Investment Net proceeds -0.75(O&M) + (0.25) Dn Net cash flow

0 -$5,940

End of Period 1

2

-$1,125

-$2,250 $432

$2,746 -$2,625 $216

-$7,065

-$1,818

$337

PW (12%) = -$8,420 AEC(12%) = $4,882

15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (d) Equivalent annual cost of owning and operating the truck for 5 years: Cash Flow Elements Investment Net proceeds -0.75(O&M) + (0.25) Dn Net cash flow

0 -$5,940

1

2

-$1,125

-$2,250 $432

-$2,625 $432

-$7,065

-$1,818

-$2,193

PW (12%) = $18,627

3

4

5

-$2,850 $216

-$3,375 0

0 -$3,600 0

-$2,634

-$3,375

-$7,350

AEC (12%) = $5,167

Note: The cash flow in year 5 include the after-tax engine overhaul cost of ($5,000(0.75) = $3,750 along with the annual O&M cost of $3,600. 14.30 Book value = 0 Ordinary gains = $30,000 – 0 = $30,000 Gains tax = $30,000 (0.25) = $7,500 (a)

Opportunity cost = $30,000 - $7,500 = $22,500

(b), (c) and (d): Replace the defender now with the challenger.

16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

n

0

1

Depreciation Book value Market value

$0 0 $30,000

Cash Flow Statement +(.25)*(Depreciation) Opportunity cost

(22,500)

Net Cash Flow

($22,500) PW (10%) = ($15,062)

2 $0 0

$0 0 $12,000

0

0 9,000

$0

$9,000

AEC(10%) =

$7,427

Option 2: Replace the defender n Depreciation Book value Market value Savings in O&M cost Cash Flow Statement Investment Net proceeds from sale +(.25)*(Depreciation) (1 - 0.25)*(Savings) Net Cash Flow

0 $165,000

2 $40,409 $101,013

3 $28,859 $72,155

4 $20,609 $51,546 $30,000

5 $14,735 $36,812 $0 $30,000

6 $14,718 $22,094 $0 $30,000

7 $14,735 $7,359 $0 $30,000

8 $7,359 $0 $0 $30,000

$30,000

$30,000

$30,000

5,895 22,500

10,102 22,500

7,215 22,500

5,152 22,500

3,684 22,500

3,680 22,500

3,684 22,500

3,000 1,840 22,500

$26,180

$26,184

$27,340

(165,000)

($165,000) PW (10%) = ($13,805)

1 $23,579 $141,422

$28,395 AEC(10%) =

$32,602

$29,715

$2,316

17 Copyright © 2023 Pearson Education, Inc.

$27,652

$26,184

Contemporary Engineering Economics, 7th ed. ©2023 14.31 (a) Based on the opportunity cost approach: Cost basis = $120,000 Gains or losses at the time of disposal: • Old machine: Total depreciation = $50,000 Book value = 0 Market value = 0 Taxable gain = $0

•

New machine: Total depreciation = $109,272 Book value = $10,728 Salvage value = $30,000 Taxable gain = $19,272 Gains tax (25%) = $4,818

(b) Cash flow for the old machine: Cash Flow Elements Investment Net proceeds + (0.25) Dn Net cash flow

0 -$20,000

1

2

3

4

5

0 $2,500 $2,500 $2,500 $2,500 $2,500 -$20,000 $2,500 $2,500 $2,500 $2,500 $2,500

(c) Replacement analysis: Replace the old machine now. PW(12%)old = −$20, 000 + $2,500( P / A,12%,5) = −$10,988 AE(12%)old = −$10,988( A / P,12%,5) = −$3, 048 PW(12%)new = −$120, 000 + $41, 787( P / F ,15%,1) + $44,847( P / F ,12%, 2) +  + $64, 014( P / F ,12%, 7) = $81,813 AE(15%) new = $81,813( A / P,12%, 7) = $17,927

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Contemporary Engineering Economics, 7th ed. ©2023

Financial Data Depreciation Book value Salvage value Savings in O&M cost

n

Cash Flow Statement Investment Net proceeds from sale +(.25)*(Depreciation) (1 - 0.25)*(Savings) Net Cash Flow

0

1 $17,148 $102,852

2 $29,388 $73,464

3 $20,988 $52,476

4 $14,988 $37,488

5 $10,716 $26,772

6 $10,716 $16,056

$50,000

$50,000

$50,000

$50,000

$50,000

$50,000

7 $5,352 $10,704 $30,000 $50,000

4,287 37,500

7,347 37,500

5,247 37,500

3,747 37,500

2,679 37,500

2,679 37,500

25,176 1,338 37,500

($120,000)

$41,787

$44,847

$42,747

$41,247

$40,179

$40,179

$64,014

PW (12%) =

$81,813

$120,000

(120,000)

AE (12%) =

$17,927

14.32 (a) & (b): Decision - Replace the defender now. n

Financial Data

D epreciation B ook value M arket value O peration C ost

0 $9,600 $14,400 $10,000

1

2

3

4

5

$5,760

$5,760

$2,880

$0

$0

$8,640

$2,880

$0

$0

$0

$0

$0

$0

$0 $5,000 $0

$1,555

$1,555

$778

$0

$0

$0

$0

$0

$0 $3,650 $0

$1,555

$1,555

$778

$0

$3,650

AEC(12%) =

$568

1 $15,000 $60,000

2 $24,000 $36,000

3 $14,400 $21,600

4 $8,640 $12,960

$30,000

$30,000

$30,000

$30,000

5 $4,320 $8,640 $0 $30,000

Cash Flow Statement

+(.27)*(D epreciation) Investm ent C ost -(1-0.27)*(O peration cost)

($7,300)

Net C ash Flow

($7,300) PW (12% ) = ($2,047)

Replace the defender Financial Data

n

D epreciation B ook value M arket value Savings in O & M C osts

0 $75,000

Cash Flow Statement

Investm ent Net proceeds from sale +(.27)*(D epreciation) +(1-0.27)*(O & M Savings)

($75,000)

Net C ash Flow

($75,000) PW (12% ) = $17,948

$3,750 $21,900

$6,000 $21,900

$3,600 $21,900

$2,160 $21,900

$2,333 $1,080 $21,900

$25,650

$27,900

$25,500

$24,060

$25,313

AEC(12%) = ($4,979)

19 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.33 (a), (b) & (c) Decision: Do not replace the defender now.

(1) Keep the defender n

0

Depreciation Book value Market value (Salvage) Revenue Cash Flow Statement +(1-0.25)(Revenue) Opportunity cost Net proceeds from sale

$

$

Net Cash Flow

$ PW(10%) = $

1

2

3

$

21,000 $

$ 21,000 $

1,200 21,000

$

15,750 $

15,750 $

15,750

$

900

15,750 $

16,650

5,500

(4,125)

(4,125) $

15,750 $

35,719 AE(10%) =

$14,363

(2) Replace the defender n

0

Depreciation Book value Market value (Salvage) Revenue Cash Flow Statement Investment Net proceeds from sale +(1-0.25)(Revenue) +(0.25)(Depreciation)

$

$

Net Cash Flow

$ PW(10%) =

$

$ 36,500 $

1 7,300 $ 29,200 $

2 11,680 $ 17,520 $

3 7,008 $ 10,512 $

24,000 $

4 4,205 $ 6,307 $ $ 24,000 $ 24,000 $

5 2,102 4,205 6,300 24,000

$

24,000 $

$ $

18,000 $ 1,825 $

18,000 $ 2,920 $

$ 18,000 $ 18,000 $ 1,752 $ 1,051 $

5,776 18,000 526

(36,500) $

19,825 $

20,920 $

19,752 $ 19,051 $

24,302

(36,500)

41,754 AE(10%) =

$11,015

20 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.34 (a) Interest i = 10% Defender: Tax Rate

27%

MARR

10%

Holding

Permitted Annual Depreciation Amounts over the

Period (N )

Holding Period 1

2

3

4

5

6

7

8

Total

Book

Depreciation

Value

0

$7,810

1

$1,116

$1,116

2

$2,233

$1,115

3

$2,233

$2,230

$1,117

4

$2,233

$2,230

$2,232

Holding Period (N )

$1,115

$3,348

$4,462

$5,580

$2,230

$7,810

$0

Annual O&M Costs over the Holding Period 1

2

3

4

5

$6,694

Total PW 6

7

8

Total PW of

of A/T

O&M Costs

O&M Costs

$2,909

$2,124

0 1

$3,200

2

$3,200

$3,700

3

$3,200

$3,700

$4,800

4

$3,200

$3,700

$4,800

$5,850

Holding

Expected

Net A/T

A/T Operating Costs (in PW)

Period

Market

Taxable

Gains

Market

over the Holding Period

(N )

Value

Gains

Tax

Value

O&M Costs Tax Shield

OC(10%)

$5,967

$4,356

$9,573

$6,988

$13,569

$9,905

CR(10%)

Total AEC(10%)

Total OC

0

$7,700

$8,151

1

$4,300

($2,394)

($646)

$4,946

$2,124

$274

$1,850

$2,035

$4,020

2

$3,300

($1,162)

($314)

$3,614

$4,356

$797

$3,559

$2,051

$2,976

3

$1,100

($1,130)

($305)

$1,405

$6,988

$1,272

$5,716

$2,299

$2,853

4

$0

$0

$9,905

$1,704

$8,201

$2,587

$2,571

$0

$0

21 Copyright © 2023 Pearson Education, Inc.

$6,054 $5,026 $5,152 $5,159

Contemporary Engineering Economics, 7th ed. ©2023

After-Tax Opportunity Cost of Keeping the Defender now: Note that the cost of retaining the defender on after-tax basis is $8,151, instead of $7,700. The scheduled depreciation amount during the fourth year of ownership with a 7-year MACRS property is $3,122. Period (n) 1 2 3 4 5 6 7 8

Depreciation $3,573 $6,122 $4,373 $3,122 $2,232 $2,231 $2,231 $1,116

Book Value $21,427 $15,305 $10,932 $7,810 $5,578 $,3347 $1,116 0

Since the asset will be disposed of during the recovery period, the allowed depreciation amount will be (0.5) ($3,122) = $1,561. Then, the book value becomes $9,370, instead of $7,810. With the market value of $7,700, there will be a loss of $1,670. The tax credit on this loss will be $1,670(0.27) = $450.90. Finally, the net proceeds from sale of old asset will be $8,150.90 (= $7,700 + $450.90). The defender’s remaining useful (economic) life is 2 more years with an AEC value of $5,026.

22 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b)

NC = 10 years, AECC = $4,678 Input : Tax Rate(%) MARR(%)

27 10

Financial data n Depreciation Book value Salvage value Gains tax O&M cost

0 $

31,000

1 4,429 26,571

$ $

2 7,592 18,979

$ $

3 5,422 13,557

$ $

4 3,872 9,685

$ $

5 2,767 6,918

$ $

6 2,767 4,151

$ $

7 2,768 1,383

$ $

8 1,383 $ (0) $

9

$ $

$

1,000

$

1,000

$

1,000

$

1,000

$

1,000

$

1,000

$

1,000

$

1,000

1,000

$

1,196 $ (730)

$

-

10 $ (0) $ (0) $2,500 $ 1,000

Cash Flow Statement Investment Net proceeds from sale +(.27)*(Depreciation) (0.73)*(O&M cost) Net Cash Flow

$

(31,000)

($31,000) PW (10%) = AEC (10%) =

$466

2,050 $ (730) $1,320

1,464 $ (730)

1,045 $ (730)

$734

$315

747 $ (730) $17

($28,743) $4,678

(c) Marginal analysis: Replace the defender now as AECD = $5,026 > AECC = $4,678

23 Copyright © 2023 Pearson Education, Inc.

747 $ (730) $17

747 $ (730) $17

373 $ (730) ($357)

$ 1,825 $ (730) (730) ($730) $1,095

Contemporary Engineering Economics, 7th ed. ©2023

14.35 (a) Economic service life = 6 years with MARR of 15%

Tax Rate MARR Holding Period 0 1 2 3 4 5 6

40% 15%

1 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 0.2

2

$1,600 $3,200 $3,200 $3,200 $3,200 0.32

Investmen $10,000 Book valu $10,000 Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7 8

$960 $1,920 $1,920 $1,920

$576 $1,152 $1,152

$576 $1,152

0.192

0.1152

0.1152

$576 0.0576

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4 5 6

Holding Period 0 1 2 3 4 5 6

1

2

3

4

5

Total Book epreciatio Value $10,000 $2,000 $8,000 $3,600 $6,400 $6,160 $3,840 $7,696 $2,304 $8,848 $1,152 $10,000 $0

6

7

8

Total PW Total PW of of A/T O&M Costs O&M Costs

` $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 Expected Market Value $5,300 $3,900 $2,800 $1,800 $1,400 $600

$2,100 $2,100 $2,100 $2,100 $2,100

Taxable Gains

$2,700 $2,700 $2,700 $2,700

Gains Tax

($2,700) ($1,080) ($2,500) ($1,000) ($1,040) ($416) ($504) ($202) $248 $99 $600 $240

$3,400 $3,400 $3,400

$4,200 $4,200

$1,304 $2,892 $4,668 $6,612 $8,700 $10,818

$4,900

$783 $1,735 $2,801 $3,967 $5,220 $6,491

Net A/T A/T Operating Costs (in PW) Total Market over the Holding Period OC(15%) CR(15%) AEC(15%) Value O&M CostsTax Shield Total OC $6,380 $4,900 $3,216 $2,002 $1,301 $360

$783 $1,735 $2,801 $3,967 $5,220 $6,491

$696 $1,180 $1,916 $2,300 $2,547 $2,761

24 Copyright © 2023 Pearson Education, Inc.

$87 $556 $885 $1,667 $2,673 $3,730

$100 $342 $387 $584 $797 $986

$5,120 $3,872 $3,454 $3,102 $2,790 $2,601

$5,220 $4,214 $3,841 $3,686 $3,588 $3,587

Contemporary Engineering Economics, 7th ed. ©2023

(b) Economic service life = 5 years with MARR of 10%

Tax Rate MARR Holding Period 0 1 2 3 4 5 6

Investment $10,000 Book value $10,000 Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7

25% 10%

1 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000

2

$3,200 $3,200 $3,200 $3,200 $3,200

$1,920 $1,920 $1,920 $1,920

$1,152 $1,152 $1,152

$1,152 $1,152

8

$2,000 $5,200 $7,120 $8,272 $9,424 $10,000

$576

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4 5 6

Holding Period 0 1 2 3 4 5 6

1

2

3

4

5

6

Total Dep.

7

8

Book Value $10,000 $8,000 $4,800 $2,880 $1,728 $576 $0

Total PW Total PW of of A/T O&M Costs O&M Costs

` $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 Expected Market Value $5,300 $3,900 $2,800 $1,800 $1,400 $600

$2,100 $2,100 $2,100 $2,100 $2,100

Taxable Gains ($2,700) ($900) ($80) $72 $824 $600

$2,700 $2,700 $2,700 $2,700

Gains Tax ($675) ($225) ($20) $18 $206 $150

$3,400 $3,400 $3,400 Net A/T Market Value $5,975 $4,125 $2,820 $1,782 $1,194 $450

$4,200 $4,200

$1,023 $2,324 $3,846 $5,587 $7,543 $9,618

OC(10%)

CR(10%)

Total AEC(10%)

$625 $696 $953 $1,235 $1,501 $1,764

$5,025 $3,798 $3,169 $2,771 $2,442 $2,238

$5,650 $4,494 $4,122 $4,006 $3,944 $4,002

$4,900

A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $1,023 $2,324 $3,846 $5,587 $7,543 $9,618

$1,364 $3,099 $5,128 $7,450 $10,058 $12,824

$455 $1,116 $1,476 $1,673 $1,852 $1,933

25 Copyright © 2023 Pearson Education, Inc.

$568 $1,209 $2,369 $3,914 $5,692 $7,685

Contemporary Engineering Economics, 7th ed. ©2023

14.36 (a) At i = 12%, the economic service life = 7 years: Tax Rate MARR Holding Period 0 1 2 3 4 5 6 7 8 9 10

$30,000 Investment $30,000 Book value Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7

25% 12%

1

2

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000

8

$3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000

9

$3,000 $3,000

10

$3,000

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4 5 6 7 8 9 10

Holding Period 0 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

Total Depreciation $3,000 $6,000 $9,000 $12,000 $15,000 $18,000 $21,000 $24,000 $27,000 $30,000

Book Value $27,000 $24,000 $21,000 $18,000 $15,000 $12,000 $9,000 $6,000 $3,000 $0

Total PW Total PW of of A/T O&M Costs O&M Costs

8 `

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 Expected Market Value $20,000 $18,000 $16,000 $14,000 $12,000 $10,000 $8,000 $6,000 $4,000 $2,000

$3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450

Taxable Gains ($7,000) ($6,000) ($5,000) ($4,000) ($3,000) ($2,000) ($1,000) $0 $1,000 $2,000

$3,968 $3,968 $3,968 $3,968 $3,968 $3,968 $3,968 $3,968

$4,563 $4,563 $4,563 $4,563 $4,563 $4,563 $4,563 Net A/T Market Value

Gains Tax ($1,750) ($1,500) ($1,250) ($1,000) ($750) ($500) ($250) $0 $250 $500

$21,750 $19,500 $17,250 $15,000 $12,750 $10,500 $8,250 $6,000 $3,750 $1,500

$5,247 $5,247 $5,247 $5,247 $5,247 $5,247

$6,034 $6,034 $6,034 $6,034 $6,034

$6,939 $6,939 $6,939 $6,939

A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $2,009 $4,072 $6,190 $8,365 $10,598 $12,891 $15,245 $17,656 $20,138 $22,686

$670 $1,268 $1,801 $2,278 $2,704 $3,084 $3,423 $3,726 $3,996 $4,238

$1,339 $2,804 $4,389 $6,087 $7,894 $9,807 $11,822 $13,930 $16,142 $18,449

26 Copyright © 2023 Pearson Education, Inc.

$7,960 $7,960 $7,960

$9,177 $9,177

$10,554

OC(12%)

$1,500 $1,659 $1,827 $2,004 $2,190 $2,385 $2,590 $2,804 $3,029 $3,265

$2,679 $5,429 $8,253 $11,153 $14,130 $17,187 $20,326 $23,541 $26,850 $30,249

CR(12%)

$11,850 $8,553 $7,378 $6,739 $6,315 $6,003 $5,756 $5,551 $5,377 $5,224

$2,009 $4,072 $6,190 $8,365 $10,598 $12,891 $15,245 $17,656 $20,138 $22,686 Total AEC(12%)

$13,350 $10,212 $9,206 $8,743 $8,505 $8,388 $8,346 $8,355 $8,406 $8,489

Contemporary Engineering Economics, 7th ed. ©2023

(b)At i = 20%, the economic service life = 9 years Tax Rate MARR Holding Period 0 1 2 3 4 5 6 7 8 9 10

Investment $30,000 Book value $30,000 Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7

25% 20%

1

2

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000

8

$3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000

9

$3,000 $3,000

10

$3,000

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4 5 6 7 8 9 10

Holding Period 0 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

Total Depreciation $3,000 $6,000 $9,000 $12,000 $15,000 $18,000 $21,000 $24,000 $27,000 $30,000

Book Value $27,000 $24,000 $21,000 $18,000 $15,000 $12,000 $9,000 $6,000 $3,000 $0

Total PW Total PW of of A/T O&M Costs O&M Costs

8 `

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 Expected Market Value $20,000 $18,000 $16,000 $14,000 $12,000 $10,000 $8,000 $6,000 $4,000 $2,000

$3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450 $3,450

Taxable Gains ($7,000) ($6,000) ($5,000) ($4,000) ($3,000) ($2,000) ($1,000) $0 $1,000 $2,000

$3,968 $3,968 $3,968 $3,968 $3,968 $3,968 $3,968 $3,968

$4,563 $4,563 $4,563 $4,563 $4,563 $4,563 $4,563

Gains Tax

Net A/T Market Value

($1,750) ($1,500) ($1,250) ($1,000) ($750) ($500) ($250) $0 $250 $500

$21,750 $19,500 $17,250 $15,000 $12,750 $10,500 $8,250 $6,000 $3,750 $1,500

$5,247 $5,247 $5,247 $5,247 $5,247 $5,247

$6,034 $6,034 $6,034 $6,034 $6,034

$6,939 $6,939 $6,939 $6,939

A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $1,875 $3,672 $5,394 $7,044 $8,626 $10,142 $11,594 $12,982 $14,316 $15,595

$625 $1,146 $1,580 $1,942 $2,243 $2,494 $2,703 $2,878 $3,023 $3,144

$1,250 $2,526 $3,814 $5,103 $6,383 $7,647 $8,891 $10,105 $11,293 $12,450

27 Copyright © 2023 Pearson Education, Inc.

$7,960 $7,960 $7,960

$9,177 $9,177

$10,554

OC(20%)

$1,500 $1,653 $1,811 $1,971 $2,134 $2,300 $2,466 $2,633 $2,802 $2,970

$2,500 $4,896 $7,192 $9,393 $11,501 $13,522 $15,459 $17,310 $19,088 $20,793

CR(20%)

$14,250 $10,773 $9,503 $8,794 $8,318 $7,964 $7,684 $7,455 $7,262 $7,098

$1,875 $3,672 $5,394 $7,044 $8,626 $10,142 $11,594 $12,982 $14,316 $15,595 Total AEC(20%)

$15,750 $12,426 $11,313 $10,766 $10,452 $10,263 $10,150 $10,088 $10,064 $10,068

Contemporary Engineering Economics, 7th ed. ©2023 (c) At i = 0%, the economic service life = 5 years: •

Capital recovery cost:

gain = S n − Bn gain tax = tm ( Sn − Bn ) net proceeds = (1 − tm ) Sn + tm Bn = (1 − 0.25)(22, 000 − 2000n) + 0.25(30, 000 − 3, 000n) I − (1 − tm ) S n − tm Bn n 6, 000 + 2, 250n = n 6, 000 = + 2, 250 n

CR =

•

Equivalent annual O&M cost:

A/T O&M = (1 − tm ) 3, 000 (1.15 )  = 2, 250 (1.15 )

 =

AE O & M

n n =1

n −1

 

n −1

2, 250 (1.15 )

n −1

n

 (1.15) = 2, 250 n

n −1

n =1

n 2, 250 (1.15 − 1) (1.15n − 1) = = 15, 000 n 1.15 − 1 n n

•

Depreciation tax credit:

AE D

 =

n n =1

(tm × Dn ) n

, where Dn = $3, 000

= $750

28 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 •

Minimum total annual equivalent cost: 6, 000 (1.15n − 1) 2, 250 15, 000 + + − 750 n n 6, 000 (1.15n − 1) = + 15, 000 + 1,500 n n 1.15n 9, 000 = 15, 000 − + 1,500 n n

AEC(0%) =

n 1 2 3 4 5 6 7 8 9 10

1st 2nd 3rd term term term AEC(0%) 17250 -9000 1500 9750 9919 -4500 1500 6919 7604 -3000 1500 6104 6559 -2250 1500 5809 6034 -1800 1500 5734 5783 -1500 1500 5783 5700 -1286 1500 5914 5736 -1125 1500 6111 5863 -1000 1500 6363 6068 -900 1500 6668

29 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.37 Economic service life With i = 12% and tax rate = 25%: Economic service life = 1 year

Tax Rate MARR Holding Period 0 1 2 3 4 5

$15,000 Investment $15,000 Book value Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7

25% 12%

1

2

$3,000 $3,000 $3,000 $3,000 $3,000

$2,400 $4,800 $4,800 $4,800

$1,440 $2,880 $2,880

$864 $1,728

8

$864

Total Depreciation $3,000 $5,400 $9,240 $11,544 $13,272

$12,000 $9,600 $5,760 $3,456 $1,728

Total PW of O&M Costs

Total PW of A/T O&M Costs

$2,232 $4,783 $8,556 $12,686 $17,112

$1,674 $3,587 $6,417 $9,515 $12,834

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4 5

Holding Period 0 1 2 3 4 5

1

2

3

4

5

6

7

8

Book Value

` $2,500 $2,500 $2,500 $2,500 $2,500 Expected Market Value $12,800 $8,100 $5,200 $3,500 $0

$3,200 $3,200 $3,200 $3,200

Taxable Gains $800 ($1,500) ($560) $44 ($1,728)

$5,300 $5,300 $5,300

Gains Tax $200 ($375) ($140) $11 ($432)

$6,500 $6,500 Net A/T Market Value $12,600 $8,475 $5,340 $3,489 $432

$7,800 A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $1,674 $3,587 $6,417 $9,515 $12,834

$670 $1,148 $1,883 $2,276 $2,536

30 Copyright © 2023 Pearson Education, Inc.

$1,004 $2,439 $4,534 $7,239 $10,298

OC(12%)

$1,125 $1,443 $1,888 $2,383 $2,857

CR(12%)

$4,200 $4,878 $4,663 $4,208 $4,093

Total AEC(12%)

$5,325 $6,321 $6,551 $6,592 $6,950

Contemporary Engineering Economics, 7th ed. ©2023

14.38 (a) and (b): Quintana should purchase the new equipment. Financial Data

n

D epreciatio n B o ok value M arket value S aving s

0 $150,000 $150,000

1 $21,435 $128,565

2 $36,735 $91,830

3 $26,235 $65,595

4 $18,735 $46,860

5 $13,395 $33,465

6 $13,380 $20,085

7 $13,395 $6,690

8 $6,690 $0

$30,000

$30,000

$30,000

$30,000

$30,000

$30,000

$30,000

$30,000

$22,500 $5,359

$22,500 $9,184

$22,500 $6,559

$22,500 $4,684

$22,500 $3,349

$22,500 $3,345

$22,500 $3,349

$22,500 $1,673

$31,684

$29,059

$27,184

$25,849

$25,845

$25,849

$24,173

Cash Flow Statement

+(1-0.25)*(S avings) +(.25)*(D epreciatio n) Investm ent

($150,000)

N et C ash F low

($150,000)

$27,859

P W (10% ) =

$15,307

0

1 $12,000 $60,000

2 $12,000 $48,000

3 $12,000 $36,000

4 $12,000 $24,000

5 $12,000 $12,000

6 $12,000 $0

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

AE (10%) =

$2,491

(b) D efend er Financial Data

D epreciatio n B o ok value C urrent m arket value

n

$72,000

7

8 $0

$0

$0

$0

$3,000

$0

$0

$3,000

$0

$0

Cash Flow Statement

+(.25)*(D epreciatio n) Investm ent

($28,800)

N et C ash F low

($28,800)

31 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(c) D efend er w ith a current m arket value of $ 45,000 0 1 n $12,000 D epreciation $60,000 $72,000 B o ok value $45,000 C urrent m arket value

Financial Data

2

3

4

5

6

$12,000 $48,000

$12,000 $36,000

$12,000 $24,000

$12,000 $12,000

$12,000 $0

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

$3,000

7

8

9 - 10

$0

$0

$0

Cash Flow Statement

+(.25)*(D ep reciatio n) Investm ent

($55,800)

N et C ash Flow

($55,800) PW (10%) =

AEC(10%) =

($42,734)

(d ) C halleng er w ith an extend ed service life o f 12 years 0 1 n $21,435 D epreciation $128,565 $150,000 B o ok value $15,000 S avings

Financial Data

2

3

$6,955

4

5

6

7

8

9-12

$36,735 $91,830

$26,235 $65,595

$18,735 $46,860

$13,395 $33,465

$13,380 $20,085

$13,395 $6,690

$6,690 $0

$0 $0

$15,000

$15,000

$15,000

$15,000

$15,000

$15,000

$15,000

$15,000

$5,359 $11,250

$9,184 $11,250

$6,559 $11,250

$4,684 $11,250

$3,349 $11,250

$3,345 $11,250

$3,349 $11,250

$1,673 $11,250

$0 $11,250

$16,609

$20,434

$17,809

$15,934

$14,599

$14,595

$14,599

$12,923

$11,250

Cash Flow Statement

+(.25)*(D ep reciatio n) +(0.75)*(S avings) Investm ent

($150,000)

N et C ash Flow

($150,000) PW (10%) =

($46,292)

AEC(10%) =

$6,794

32 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.39 (a) D efend er (M od el A ) O riginal Investm ent =

$150,000 n

D epreciation B o ok value C urrent m arket value

0 $91,830

1 $26,235

2 $18,735

3 $13,395

4 $13,380

5 $13,395

6 $6,690

$65,595

$46,860

$33,465

$20,085

$6,690

$6,559

$4,684

$3,349

$3,345

$6,559

$4,684

$3,349

7

8 $0

$0

$0

$0

$0

$3,349

$1,673

$0

$0

$3,345

$3,349

$1,673

$0

$0

8 $13,380

$0

Cash Flow Statement

+(.25)*(D epreciation) Investm ent

($36,732)

N et C ash Flo w

($36,732)

P W (10% ) =

($19,075)

AEC(10%) =

$3,575

C hallenger (M od el B ) n

D epreciation B o ok value S avings

0 $300,000

1 $42,870

2 $73,470

3 $52,470

4 $37,470

5 $26,790

6 $26,760

7 $26,790

9

10

$257,130

$183,660

$131,190

$93,720

$66,930

$40,170

$13,380

$0

$75,000

$75,000

$75,000

$75,000

$75,000

$75,000

$75,000

$75,000

$10,718

$18,368

$13,118

$9,368

$6,698

$6,690

$6,698

$3,345

$0

$0

$56,250

$56,250

$56,250

$56,250

$56,250

$56,250

$56,250

$56,250

$56,250

$56,250

$66,968

$74,618

$69,368

$65,618

$62,948

$62,940

$62,948

$59,595

$56,250

$56,250

$75,000

$75,000

Cash Flow Statement

+(.25)*(D epreciation) +(0.75)*(S aving s) Investm ent

($300,000)

N et C ash Flo w

($300,000)

PW (10%) =

$99,741

AE (10%) =

$16,232

(b) It is rather difficult to predict what technological advances would be made on a typical equipment in the future. If the industrial engineer had all the relevant information available in one or two years, he or she could defer the replacement decision. Since Model A was already placed in service, the amount of $150,000 expended is a sunk cost, and it should not be considered in future replacement decisions. For now, we may say replace Model A with Model B.

33 Copyright © 2023 Pearson Education, Inc.

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14.40 Replacement Analysis (a) Keep the defender Financial Data Depreciation Book value Current market value O&M cost

n

-4 $20,000

-3 $2,858 $17,142

-2 $4,898 $12,244

-1 $3,498 $8,746

Cash Flow Statement (-0.7)*(O&M cost) +(.3)*(Depreciation) Investment Net proceeds from sale

1 $1,786 $4,462

2 $1,784 $2,678

3 $1,786 $892

4 $892 $0

5

6

$2,000

$2,000

$2,000

$2,000

$2,000

$0 $1,500 $2,000

(1,400) 536

(1,400) 535

(1,400) 536

(1,400) 268

(1,400) 0

(1,400) 0

$0

(6,074) 1,050

Net Cash Flow PW (10%) =

0 $2,498 $6,248 $6,000

$0 ($10,064)

$0

AEC(10%) =

$0

$0

($6,074)

($864)

($865)

($864)

7 $1,875 $937 $1,000

8 $937 ($0) $1,000

563 (700)

281 (700)

($1,132)

($1,400)

($350)

10-11 $0 ($0) $1,000

12

$0 ($0) $1,000

0 (700)

0 (700)

$2,311

(b) Replace the defender Financial Data Depreciation Book value O&M cost Cash Flow Statement +(.3)*(Depreciation) (-0.7)*(O&M cost) Investment Net proceeds from sale Net Cash Flow PW (10%) =

n $21,000

1 $3,001 $17,999 $1,000

900 (700)

2 $5,143 $12,856 $1,000

1,543 (700)

3 $3,673 $9,183 $1,000

1,102 (700)

4 $2,623 $6,560 $1,000

787 (700)

5 $1,875 $4,685 $1,000

563 (700)

6 $1,873 $2,812 $1,000

562 (700)

9

$0 ($0) 1000

0 (700)

(21,000) 350 ($21,000) ($21,113)

$200

$843

AEC(10%) =

$402

$87

($137)

$3,099

34 Copyright © 2023 Pearson Education, Inc.

($138)

($137)

($419)

($700)

($700)

($350)

Contemporary Engineering Economics, 7th ed. ©2023 14.41 (a) and (b): Replace defender now

Option 1 : Keep the defender n

0

Depreciation Book value Expected Market value O&M cost

$4,000

Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

2

3

$0 $0 $3,000 $3,000

$0 $0 $2,000 $4,500

$0 $0 $1,000 $6,000

(2,250) 0

(3,375) 0

(4,500) 0

(2,400) 750

Net Cash Flow PW (15%) =

1

($2,400) ($9,374)

($2,250) AEC(15%) =

($3,375)

($3,750)

$4,106

Option 2 : Replace the defender n

0

Depreciation Book value O&M cost

$6,000

Cash Flow Statement +(.25)*(Depreciation) (0.75)*(O&M cost) Investment Net proceeds from sale

2 $2,667 $1,333 $3,000

3 $444 $889 $4,000

500 (1,500)

667 (2,250)

111 (3,000)

(6,000) 1,722

Net Cash Flow PW (12%) =

1 $2,000 $4,000 $2,000

($6,000) ($8,985)

($1,000) AEC(12%) =

35 Copyright © 2023 Pearson Education, Inc.

($1,583) $3,741

($1,167)

Contemporary Engineering Economics, 7th ed. ©2023 14.42 (a), (b), and (c):

Option 1 : Keep the defender Financial Data Depreciation Book value Expected Market value O&M cost

n

0 $4,000 $0

Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

2 $800 $2,400 $0 $0

3 $800 $1,600 $0 $0

4 $800 $800 $0 $0

5 $800 $0 $0 $0

0 200

0 200

0 200

0 200

0 200

$200

$200

$200

$200

$200

(1,600)

Net Cash Flow

($1,600)

PW (10%) =

1 $800 $3,200 $0 $0

($842)

AEC(10%) =

$222

Option 2 : Replace the defender Financial Data Depreciation Book value Expected Market value Savings in O&M cost

n

0 $10,000 $0

Cash Flow Statement +(.25)*(Depreciation) (0.75)*(Savings in O&M) Investment Net proceeds from sale

2 $2,449 $6,122 $0 $3,000

3 $1,749 $4,373 $0 $3,000

4 $1,249 $3,124 $0 $3,000

5 $446 $2,678 $0 $3,000

357 2,250

612 2,250

437 2,250

312 2,250

112 2,250

(10,000) 670

Net Cash Flow

($10,000)

PW (10%) =

1 $1,429 $8,571 $0 $3,000

$387

Incremental cash flow IRR= 15.49%

$2,607

$2,862

AEC(10%) = ($8,400)

$2,407

$2,687

$2,562

$3,031

$2,487

$2,362

$2,831

($102) $2,662

>10%, replace the defender now.

36 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.43 (a) Replacement analysis Financial Data n Depreciation Book value Expected Market value O&M cost

0 $0 $1,000

Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

2

3

4

5

$0 $0 $0 $7,000

$0 $0 $0 $7,000

$0 $0 $0 $7,000

$0 $0 $0 $7,000

$0 $0 $0 $7,000

($5,250) $0

($5,250) $0

($5,250) $0

($5,250) $0

($5,250) $0

($5,250)

($5,250)

($5,250)

($5,250)

($5,250)

2 $2,939 $7,346 $5,000 $1,500

3 $2,099 $5,248 $5,000 $1,500

4 $1,499 $3,749 $5,000 $1,500

5 $535 $3,214 $5,000 $1,500

429 (3,750) 1,125

735 (3,750) 1,125

525 (3,750) 1,125

375 (3,750) 1,125

134 (3,750) 1,125

($2,196)

($1,890)

($2,100)

($2,250)

($188)

$3,150

$3,000

($600)

Net Cash Flow PW (12%) =

1

($600) ($19,525)

AEC(12%) =

$5,416

Option 2 : Replace the defender Financial Data Depreciation Book value O&M cost Increased Revenue

n

0 $12,000

Cash Flow Statement +(.25)*(Depreciation) -(0.75)*(O&M cost) (0.75)*(Revenue) Investment Net proceeds from sale

(12,000) 2,303

Net Cash Flow PW (12%) =

($12,000) ($18,500)

Incremental cash flow IRR= 15.34%

1 $1,715 $10,285 $5,000 $1,500

AEC(12%) = ($11,400)

$3,054

$5,132 $3,360

> 12%, replace the defender.

(b) Break-even market value: Let X denote the current market value of the old machine. Then, the opportunity cost for not replacing the old machine now is given by X − 0.25 X = 0.75 X PW(12%)defender = −0.75 X − 5, 250( P / A,12%,5) = −0.75 X − $18,925

This present value must be the same as the present value of the challenger.

37 Copyright © 2023 Pearson Education, Inc.

$5,062

Contemporary Engineering Economics, 7th ed. ©2023 −0.75 X − $18,925 = −18,500 X = $567

14.44

Replacement analysis: Let X denote the current market value of the old callswitching system:

AEC(10%) defender = $20, 000(0.75) + 0.75 X ( A / P,10%,5) = $14, 000 + (0.75 X )(0.2638) = $15, 000 + 0.1978 X AEC(10%)challenger = $179,174( A / P,10%,10) = $29,160 To justify the new call-switching system now, we must have

AEC(10%) defender >AEC(10%)challenger $15, 000 + 0.1978 X > $29,160 X > $71,570 •

Challenger:

38 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Challenger n

Financial Data Depreciation Book value Salvage value O&M cost

$200,000

Cash Flow Statement -(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale Net Cash Flow

n

Cash Flow Statement -(0.75)*(Savings in O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale Net Cash Flow

1 $40,000 $160,000

2 $64,000 $96,000

3 $38,400 $57,600

4 $23,040 $34,560

5 $23,040 $11,520

$5,000

$5,000

$5,000

$5,000

$5,000

(3,750) 10,000

(3,750) 16,000

(3,750) 9,600

(3,750) 5,760

(3,750) 5,760

$5,850

$2,010

$2,010

(200,000) ($200,000)

Financial Data Depreciation Book value Salvage value O&M cost

PW (10%) = AEC(10%) =

0

$6,250

$12,250

6 $11,520 $0

7

8

9

10

$0 $0

$0 $0

$0 $0

$5,000

$5,000

$5,000

$5,000

(3,750) 2,880

(3,750) 0

(3,750) 0

(3,750) 0

($870)

($3,750)

($3,750)

($3,750)

($179,174) $29,160

(a) and (b): Keep the defender for three years. Defender: Note: The opportunity cost of retaining the defender is as follows:     

(3,750) 0 13,500

14.45 Defender analysis



$0 $0 $18,000 $5,000

Current market value = $11,500 Current book value = $15,000 Losses = ($11,500 - $15,000) = ($3,500) Loss tax credit = $3,500(0.28) = $980 Cost of retaining the defender = $11,500+ $980 = $12,480

39 Copyright © 2023 Pearson Education, Inc.

$9,750

Contemporary Engineering Economics, 7th ed. ©2023

Tax Rate MARR Holding Period 0 1 2 3

28% 18%

1

$12,480 Investment $15,000 Book value Permitted Annual Depreciation Amounts over the Holding Period 2 3 4 5

$4,000 $4,000 $4,000

$4,000 $4,000

6

7

8

$4,000

Total Depreciation $4,000 $8,000 $12,000

Book Value $15,000 $11,000 $7,000 $3,000

Total PW of O&M Costs

Total PW of A/T O&M Costs

$3,814 $7,620 $11,333

$2,746 $5,486 $8,159

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3

Holding Period 0 1 2 3

1

2

3

4

5

6

7

8 `

$4,500 $4,500 $4,500 Expected Market Value $5,200 $3,500 $1,200

$5,300 $5,300

Taxable Gains ($5,800) ($3,500) ($1,800)

$6,100 Net A/T Market Value

Gains Tax ($1,624) ($980) ($504)

$6,824 $4,480 $1,704

A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $2,746 $5,486 $8,159

$949 $1,754 $2,435

40 Copyright © 2023 Pearson Education, Inc.

$1,797 $3,733 $5,724

OC(18%)

$2,120 $2,384 $2,633

CR(18%)

$7,902 $5,916 $5,263

Total AEC(18%)

$10,022 $8,300 $7,896

Contemporary Engineering Economics, 7th ed. ©2023



Challenger:

Financial Data

n

D epreciation B ook value S alvag e value O & M cost

0 $43,500

1 $6,216 $37,284

2 $10,653 $26,631

3 $7,608 $19,023

4 $5,433 $13,589

5 $3,885 $9,705

$1,500

$1,500

$1,500

$1,500

$1,500

($1,080) $1,741

($1,080) $2,983

($1,080) $2,130

($1,080) $1,521

($1,080) $1,088

$661

$1,903

$1,050

$441

6 $3,880 $5,825

7 $3,885 $1,940

8 $1,940 ($0)

9

$1,500

$1,500

$1,500

$1,500

$0 ($0) $3,500 $1,500

($1,080) $1,086

($1,080) $1,088

($1,080) $543

($1,080) $0

($1,080) $0

Cash Flow Statement

-(0.72)*(O & M cost) +(.28)*(D epreciation) Investm ent N et proceed s fro m sale

($43,500)

N et C ash F low

($43,500)

Financial Data

n

D epreciation B ook value S alvag e value O & M cost

$8

10 $0 ($0)

Cash Flow Statement

-(0.72)*(O & M cost) +(.28)*(D epreciation) Investm ent N et proceed s fro m sale

$2,520

N et C ash F low

$0 P W (18% ) = A E C (18% ) =

$6

$8

($537)

($1,080)

$1,440

($40,810) $9,081

Optimal time to replace: Since the remaining useful life for the defender is 3 years, which is the same as the physical life, keep the defender for 3 years.

41 Copyright © 2023 Pearson Education, Inc.

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14.46 Decision: Do not replace the defender now. 0 Depreciation Book value Current market value O&M cost Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Opportunity cost Net proceeds from sale Net Cash Flow

$50,000 $60,000

1 $5,000 $45,000

2 $5,000 $40,000

3 $5,000 $35,000

4 $5,000 $30,000

5 $5,000 $25,000

6 $5,000 $20,000

7 $5,000 $15,000

8 $5,000 $10,000

9 $5,000 $5,000

10 $5,000 $0

$18,000

$18,000

$18,000

$18,000

$18,000

$18,000

$18,000

$18,000

$18,000

$18,000

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) $1,250

($13,500) ($13,500) $1,250 $1,250

($12,250)

($12,250)

($12,250)

($12,250)

($12,250)

($12,250)

($12,250)

($12,250)

($12,250) ($12,250)

3 $34,980 $87,460 $4,000

4 $24,980 $62,480 $4,000

5 $17,860 $44,620 $4,000

6 $17,840 $26,780 $4,000

7 $17,860 $8,920 $4,000

8 $8,920 $0 $4,000

$0 $0 $4,000

$0 $0 $4,000

$8,745 ($3,000)

$6,245 ($3,000)

$4,465 ($3,000)

$4,460 ($3,000)

$4,465 ($3,000)

$2,230 ($3,000)

$0 ($3,000)

$0 ($3,000)

($56,000) ($56,000)

PW (10%) = ($131,271)

AEC(10%) =

$21,364

(b) Replace the defender 0 Depreciation Book value O&M cost

$200,000

Cash Flow Statement +(.25)*(Depreciation) (-0.75)*(O&M cost) Investment Net proceeds from sale Net Cash Flow

1 $28,580 $171,420 $4,000

2 $48,980 $122,440 $4,000

$7,145 ($3,000)

$12,245 ($3,000)

9

10

($200,000) $15,000 ($200,000)

PW (10%) = ($176,578)

$4,145

$9,245 AEC(10%) =

$5,745

$3,245

$28,737

Decision: Do not replace the defender now.

42 Copyright © 2023 Pearson Education, Inc.

$1,465

$1,460

$1,465

($770)

($3,000)

$12,000

Contemporary Engineering Economics, 7th ed. ©2023

14.47 Financial Data Depreciation Book value Market value Operation Cost

n

0

1

$0 $8,500

Cash Flow Statement +(.25)*(Depreciation) Opportunity cost -(1-0.25)*(Operation cost)

3

4

5

$0 $0

$0 $0

$0 $0

$0 $0

$8,700

$8,700

$8,700

$8,700

$8,700

0

0

0

0

0

(6,375)

Net Cash Flow

($6,375) PW (10%) =

2 $0 $0

($31,110)

(6,525)

(6,525)

(6,525)

(6,525)

(6,525)

($6,525)

($6,525)

($6,525)

($6,525)

($6,525)

AEC(10%) =

$8,207

Replace the defender Financial Data Depreciation Book value Market value Operation Cost

n

0 $53,500 $53,500

Cash Flow Statement Investment Net proceeds from sale +(.25)*(Depreciation) -(1-0.25)*(Operation cost) Net Cash Flow

2 $13,102 $32,753

3 $9,357 $23,396

4 $6,682 $16,713

$4,200

$4,700

$5,200

$5,700

5 $2,386 $14,327 $12,000 $6,200

1,911 (3,150)

3,276 (3,525)

2,339 (3,900)

1,671 (4,275)

12,582 597 (4,650)

($1,239)

($249)

($1,561)

($2,604)

$8,528

(53,500)

($53,500) PW (10%) = ($52,488)

1 $7,645 $45,855

AEC(10%) =

$13,846

43 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.48 • Defender analysis n Depreciation Book value Market value O&M cost Cash Flow Statement -(0.72)*(O&M cost) +(.28)*(Depreciation) Opportunity cost Net proceeds from sale Net Cash Flow

0 $2,873 $7,185 $2,000

2 $2,052 $3,080

3 $2,054 $1,026

4 $1,026 ($0)

5

6

$3,800

$3,800

$3,800

$3,800

$3,800

($2,736) $575

($2,736) $574

($2,736) $575

($2,736) $287

($2,736) $0

($2,736) $0

$0 ($0)

$0 ($0)

($3,854) $600 ($3,854)

PW (10%) = ($13,805)

1 $2,054 $5,131 $1,500 $3,800

($2,161) AEC(10%) =

($2,162)

($2,161)

($2,449)

($2,736)

($2,136)

$3,170

Note: Opportunity cost of keeping the defender is calculated as follows: • • • •

Book value (if sold now with the half-year convention) = $3,286+$5,633+$4,023 + (0.5)($2,874) = $8,622 Taxable gain (loss) = $2,000 -$8,622 = ($6,622) Tax credit (savings) = $6,622(0.28) = $1,854 Net proceeds from sale = $2,000 + $1,854 = $3,854

44 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • Challenger analysis

Financial Data Depreciation Book value Savings O&M cost

0 $50,000

Cash Flow Statement -(0.72)*(O&M cost) (0.72)*(Savings) +(.28)*(Depreciation) Investment ($50,000) Net proceeds from sale Net Cash Flow

($50,000)

Financial Data Depreciation Book value Savings O&M cost Salvage value Cash Flow Statement -(0.72)*(O&M cost) (0.72)*(Savings) +(.28)*(Depreciation) Investment Net proceeds from sale Net Cash Flow

1 $7,145 $42,855 $6,000 $3,000

2 $12,245 $30,610 $6,000 $3,000

3 $8,745 $21,865 $6,000 $3,000

4 $6,245 $15,620 $6,000 $3,000

5 $4,465 $11,155 $6,000 $3,000

6 $4,465 $6,690 $6,000 $3,000

($2,160) $4,320 $2,001

($2,160) $4,320 $3,429

($2,160) $4,320 $2,449

($2,160) $4,320 $1,749

($2,160) $4,320 $1,250

($2,160) $4,320 $1,250

$4,161

$5,589

$4,609

$3,909

$3,410

$3,410

7 $4,460 $2,230 $6,000 $3,000

8 $2,230 $0 $6,000 $3,000

9 $0 $0 $6,000 $3,000

$0 $0 $6,000 $3,000

$0 $0 $6,000 $3,000

$0 $0 $6,000 $3,000 $3,000

($2,160) $4,320 $1,249

($2,160) $4,320 $624

($2,160) $4,320 $0

($2,160) $4,320 $0

($2,160) $4,320 $0

($2,160) $4,320 $0

11

12

$2,160 $3,409

PW (10%) =

10

($24,494)

$2,784

$2,160 AEC(10%) =

Keep the defender for now.

45 Copyright © 2023 Pearson Education, Inc.

$2,160 $3,986

$2,160

$4,320

Contemporary Engineering Economics, 7th ed. ©2023

14.49 Select Option 1.

Option 1 Financial Data Depreciation Book value Current Market value O&M cost

0

n

$48,000 $6,000

Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Opportunity cost Investment Net proceeds from sale

2 $11,755 $29,386

3 $8,395 $20,990

4 $5,995 $14,995

5 $4,286 $10,709

$27,000

$27,000

$27,000

$27,000

$27,000

(20,250) 1,715

(20,250) 2,939

(20,250) 2,099

(20,250) 1,499

(20,250) 1,072

($18,535)

($17,311)

($18,151)

($18,751)

($19,178)

6 $4,282 $6,427

7 $4,286 $2,141

8 $2,141 $0

$27,000

$27,000

$27,000

$27,000

$0 $0 $5,000 $27,000

(20,250) 1,070

(20,250) 1,072

(20,250) 535

(20,250) 0

(20,250) 0

(4,500) (48,000)

Net Cash Flow

Financial Data Depreciation Book value Salvage value O&M cost

1 $6,859 $41,141

($52,500)

n

Cash Flow Statement -(.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

9

10 $0 $0

3,750

Net Cash Flow

($19,180)

PW (12%) = ($157,538)

AEC (12%) =

($19,178) $27,882

46 Copyright © 2023 Pearson Education, Inc.

($19,715)

($20,250)

($16,500)

Contemporary Engineering Economics, 7th ed. ©2023 Option 2:

Option 2 Financial Data Depreciation Book value O&M cost

n

$84,000

Cash Flow Statement (-0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

1 $12,004 $71,996 $24,000

2 $20,572 $51,425 $24,000

3 $14,692 $36,733 $24,000

4 $10,492 $26,242 $24,000

5 $7,501 $18,740 $24,000

(18,000) 3,001

(18,000) 5,143

(18,000) 3,673

(18,000) 2,623

(18,000) 1,875

($14,999)

($12,857)

($14,327)

($15,377)

($16,125)

6 $7,493 $11,248

7 $7,501 $3,746

$24,000

$24,000

$24,000

$24,000

$0 ($0) $9,000 $24,000

(18,000) 1,873

(18,000) 1,875

(18,000) 937

(18,000) 0

(18,000) 0

(84,000)

Net Cash Flow

Financial Data Depreciation Book value Salvage value O&M cost

0

($84,000)

n

Cash Flow Statement -(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

8 $3,746 ($0)

9

10 $0 ($0)

6,750

Net Cash Flow

($16,127)

PW (12%) = ($169,231)

AEC(12%) =

47 Copyright © 2023 Pearson Education, Inc.

($16,125) $29,951

($17,063)

($18,000)

($11,250)

Contemporary Engineering Economics, 7th ed. ©2023 14.50 The remaining useful life of the defender is 1 year. Its annual equivalent cost is $1,666. If the defender is replaced now by the challenger, its equivalent annual cost is $2,191, indicating that the defender should be kept for now. (a) Economic service life = one year Tax Rate MARR

30% 12%

Holding Period 0 1 2 3 4 5

1

Investment $1,050 Book value $0 Permitted Annual Depreciation Amounts over the Holding Period 3 4 5 6 7

2

Total Depreciation

8

$0 $0 $0 $0 $0 Annual O&M Costs over the Holding Period

Holding Period 0 1 2 3 4 5

Holding Period 0 1 2 3 4 5

Book Value

1

2

3

4

5

6

7

$0 $0 $0 $0 $0

Total PW Total PW of of A/T O&M Costs O&M Costs

8 `

$1,900 $1,900 $1,900 $1,900 $1,900 Expected Market Value $1,200 $1,000 $500 $0 $0

$2,300 $2,300 $2,300 $2,300

Taxable Gains $1,200 $1,000 $500 $0 $0

$2,700 $2,700 $2,700

Gains Tax $360 $300 $150 $0 $0

$3,100 $3,100 Net A/T Market Value $840 $700 $350 $0 $0

$1,696 $3,530 $5,452 $7,422 $9,351

$3,400 A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield Total OC $1,188 $2,471 $3,816 $5,195 $6,546

$0 $0 $0 $0 $0

$1,188 $2,471 $3,816 $5,195 $6,546

OC(12%)

$1,188 $2,471 $3,816 $5,195 $6,546 Total AEC(12%)

CR(12%)

$1,330 $1,462 $1,589 $1,710 $1,816

$336 $291 $333 $346 $291

$1,666 $1,753 $1,922 $2,056 $2,107

(b) Replace the defender Financial Data Depreciation Book value Salvage value Operation Cost

n

0 $6,000

Cash Flow Statement Investment Net proceeds from sale +(.30)*(Depreciation) -(1-0.30)*(Operation cost)

1 $1,200 $4,800

2 $1,920 $2,880

3 $1,152 $1,728

4

5

$691 $1,037

$1,100

$1,300

$1,500

$1,700

$346 $691 $1,000 $1,800

360 (770)

576 (910)

346 (1,050)

207 (1,190)

907 104 (1,260)

($410)

($334)

($704)

($983)

($249)

AEC(12%) =

$2,191

(6,000)

Net Cash Flow

($6,000) PW (12%) =

($7,899)

48 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 14.51 Option 1 : Keep the defender n Depreciation Book value Market value Operating cost Cash Flow Statement +(.3)*(Depreciation) -(1-0.30)*(Operating cost) Opportunity cost Net proceeds from sale Net Cash Flow PW (11%) =

0 $5,120 $7,680 $12,000

1 $3,072 $4,608

2 $1,843 $2,765

3 $1,843 $922

4 $922 $0

5

6

$4,000

$4,000

$4,000

$4,000

$4,000

$0 $0 $2,000 $4,000

$922 ($2,800)

$553 ($2,800)

$553 ($2,800)

$276 ($2,800)

$0 ($2,800)

$0 ($2,800)

$0 $0

($11,472) $1,400 ($11,472)

($20,703)

($1,878)

AEC(11%) =

($2,247)

($2,247)

($2,524)

($2,800)

($1,400)

$4,894

Option 2 : Replace the defender n Depreciation (First Cycle) Book value (First Cycle) Depreciation (Second Cycle) Book value (Second Cycle) Salvage value Operating cost Cash Flow Statement +(.3)*(Depreciation) -(1-0.30)*(Operating cost) Investment Net proceeds from sale Net Cash Flow PW (11%) =

0 $10,000

2 $3,200 $4,800

3 $960 $3,840

$2,000

$2,000

$10,000 $4,000 $2,000

$600 ($1,400)

$960 ($1,400)

($10,000)

($10,000) ($15,583)

1 $2,000 $8,000

($800)

AEC(11%) =

($440)

4

5

6

$2,000 $8,000

$3,200 $4,800

$2,000

$2,000

$960 $3,840 $4,000 $2,000

$288 ($1,400) ($10,000) $3,952

$600 ($1,400)

$960 ($1,400)

$288 ($1,400)

($7,160)

($800)

$3,952 ($440)

$2,840

$3,683

Decision: Replace the defender now with the current challenger. Then, replace the current challenger at the end of 3 years with a similar challenger.

49 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

14.52 (a) Economic service life = 4 years:

Year

OR

MV

AE OR

CR(12%) Total AE

0 1 2

$8,000 $4,000 $4,200 $4,000 $3,400 $2,940 $3,717

$4,760 $3,347

-$760 $370

3 4

$2,710 $2,058 $3,419 $1,917 $1,411 $3,104

$2,721 $2,332

$698 $772

5

$1,004 $1,008 $2,774

$2,061

$713

When N = 4, we have the largest AE value. The economic life of the system is 4 years. (Note that we want to maximize our operating revenue.)

(b) The economic service life varies with the interest rate: Interest rate Economic service life Maximum annual revenue

• •

1% 5%

3 years 4 years

$1,333 $1,127

10% 15%

4 years 4 years

$876 $613

20% 25% 30%

4 years 5 years 5 years

$341 $77 -$191

40%

6 years

-$760

Maximize the annual revenue at its economic service life varies inversely with the interest rate. Economic service life increases as the interest rate increases in our example. As the interest rate increases, the capital cost will also increase. However, the annual equivalent revenue will decrease. Thus, the net effect is that the marginal increase in the capital cost is less than the decrease in the annual equivalent revenue, resulting in extending the service life.

50 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies ST 14.1 (a)

Annual changes in MV Annual increases in O&M Interest rate Market Value $11,500 $5,200 $3,500 $1,200

18%

O&M Costs

CR(18%)

0 $4,500 $5,300 $8,000

$8,370 $5,740 $4,953

OC(18%)

AEC(18%)

$4,500 $4,867 $5,744

(b) Now is the time to replace the defender? AECD = $10,607 with N * = 2 years AECC = ($43,500 − $3,500)( A / P ,18%,10) + 0.18($3,500) + $1,500 = $11,030

Do not replace the defender now. (c) Marginal cost - Keep the defender for one more year (3rd year): $3,500(1.18)-$1,200+ $8,000 = $11,284 Replace the dender at the end of its economic service life - 2 years.

(d)

The future cost of challenger remains the same.

51 Copyright © 2023 Pearson Education, Inc.

$12,870 $10,607 $10,697

Contemporary Engineering Economics, 7th ed. ©2023

ST 14.2 (a) The current FMS manufacturing technology would prevail for several years without major cost increase and productivity improvement. Therefore, if the present system is kept for the remaining useful life, it will be replaced by the current FMS technology with the same investment and O&M costs. Option1: Keep the defender n

0

Operating cost Current market value

2 $115,000

3 $125,000

4 $135,000

5 $145,000

($75,600)

($82,800)

($90,000)

($97,200)

($104,400)

($75,600)

($82,800)

($90,000)

($97,200)

($104,400)

AE (15%) =

($118,074)

1 $185,770 $1,114,230 $664,243 $45,000

2 $318,370 $795,860 $664,243 $47,000

3 $227,370 $568,490 $664,243 $49,000

4 $162,370 $406,120 $664,243 $51,000

5 $116,090 $290,030 $664,243 $53,000

6 $115,960 $174,070 $664,243 $55,000

7 $116,090 $57,980 $664,243 $57,000

8 $57,980

9

10

$664,243 $59,000

$664,243 $61,000

$664,243 $63,000 $120,000

$52,016 $478,255 ($32,400)

$89,144 $478,255 ($33,840)

$63,664 $478,255 ($35,280)

$45,464 $478,255 ($36,720)

$32,505 $478,255 ($38,160)

$32,469 $478,255 ($39,600)

$32,505 $478,255 ($41,040)

$16,234 $478,255 ($42,480)

$478,255 ($43,920)

$478,255 ($45,360)

$497,871

$533,559

$506,639

$486,999

$472,600

$471,124

$469,720

$452,009

$434,335

$86,400 $ , $519,295

AE (15%) =

$231,683

$140,000

Cash Flow Statement -(.72)*(O&M) Opportunity cost

($100,800)

Net Cash Flow

($100,800)

PW (15%) =

1 $105,000

($395,804)

Option 2: Replace the defender n

Depreciation Book value Savings Operating cost Salvage value Cash Flow Statement +(.28)*(Depreciation) +(0.72)*(Savings) -(.72)*(O&M) Investment Net proceeds Net Cash Flow PW (15%) =

0 $1,300,000

($1,300,000) ($1,300,000)

$1,162,764

(b) Decision: The challenger should be adopted.

52 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

ST 14.3 (a) Decision: Replace the defender now. Option1: Keep the defender n

0

1 $65,000 $2,500

2 $65,000 $2,875

3 $65,000 $3,306

4 $65,000 $3,802

5 $65,000 $4,373

Cash Flow Statement -(.72)*(O&M) Opportunity cost

($48,600)

($48,870)

($49,181)

($49,538)

($49,948)

Net Cash Flow

($48,600)

($48,870)

($49,181)

($49,538)

($49,948)

Operating cost Maintenance

PW (16%) =

($160,863)

AEC(16%) =

$49,129

Option 2: Replace the defender n

Depreciation(Building) Book value (Building) Depreciation(Equipment) Book value (Equipment) Savings O&M cost Salvage value (Building) Salvage value (Equipment) Cash Flow Statement +(.25)*(Depreciation) +(0.72)*(Savings) -(.72)*(O&M) Investment Net proceeds Net Cash Flow PW (16%) =

0 $47,200 $131,400

2 $1,210 $44,830 $32,180 $80,443 $57,895 $35,000

3 $1,210 $43,620 $22,982 $57,461 $57,895 $35,000

4 $1,210 $42,410 $16,412 $41,049 $57,895 $35,000

5 $1,210 $41,200 $11,734 $29,315 $57,895 $35,000

6 $1,210 $39,990 $11,721 $17,594 $57,895 $35,000

7 $1,210 $38,780 $11,734 $5,860 $57,895 $35,000

8 $1,210 $37,570 $5,860

9 $1,210 $36,360

10 $1,160 $35,200

$57,895 $35,000

$57,895 $35,000

$57,895 $35,000 $0 $8,820

$4,984 $41,684 ($25,200)

$8,347 $41,684 ($25,200)

$6,048 $41,684 ($25,200)

$4,405 $41,684 ($25,200)

$3,236 $41,684 ($25,200)

$3,233 $41,684 ($25,200)

$3,236 $41,684 ($25,200)

$1,768 $41,684 ($25,200)

$303 $41,684 ($25,200)

$290 $41,684 ($25,200)

$21,469

$24,832

$22,532

$20,890

$19,720

$19,717

$19,720

$18,252

$16,787

$32,189

($178,600) $15,415 ($178,600)

($73,928)

1 $1,160 $46,040 $18,777 $112,623 $57,895 $35,000

AEC(16%) =

$15,296

53 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(b) The optimal time to replace the defender is “now.” If the AE cost for the defender were smaller than that of the challenger, we would need to perform the marginal analysis to see if when the right time is to replace the defender by calculating the incremental cost of operating the defender for just 1 more year.

54 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 ST 14.4 (a) Development of a combined productivity index: Combined productivity means “inherent productivity” × relative hours for challenger. Operating hours

n

Ratio Combined productivity index

Defender

Challenger

1

1,800

2,500

1.389

1.667

2

1,800

2,400

1.333

1.600

3

1,700

2,300

1.353

1.624

4

1,700

2,100

1.235

1.482

5

1,600

2,000

1.250

1.500

(b) • Adjusted annual O&M costs for Defender

n

Operating HR Maintenance

Wages

Fuel

Total

1

1,800

$46,800

$42,120

$24,408

$113,328

2

1,800

$46,800

$42,120

$24,408

$113,328

3

1,700

$46,800

$39,780

$23,052

$109,632

4

1,700

$46,800

$39,780

$23,052

$109,632

5

1,600

$46,800

$37,440

$21,696

$105,936

• Adjusted annual O&M costs for Challenger n

Operating HR

M

L

F

Combined Productivity index

M

Adjusted L

F

Total

1

2,500

$35,000 $58,500 $48,000

1.667

$21,000

$35,100 $28,800 $84,900

2

2,400

$38,400 $56,160 $46,080

1.600

$24,000

$35,100 $28,800 $87,900

3

2,300

$43,700 $53,820 $44,160

1.624

$26,917

$33,150 $27,200 $87,267

4

2,100

$48,300 $49,140 $40,320

1.482

$32,583

$33,150 $27,200 $92,933

5

2,000

$58,000 $46,800 $38,400

1.500

$38,667

$31,200 $25,600 $95,467

55 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

(c) Replacement analysis Tax Rate

28%

MARR

15%

Holding Period (N )

Permitted Annual Depreciation Amounts over the Holding Period 1

2

3

4

5

6

7

8

Total

Book

Depreciation

Value

0

$0

1

$0

2

$0

$0

3

$0

$0

$0

4

$0

$0

$0

$0

5

$0

$0

$0

$0

Holding Period (N )

$0

$0

$0

$0

$0

$0

$0

$0

$0

$0

$0

Annual O&M Costs over the Holding Period 1

2

3

4

5

Total PW 6

7

8

Total PW of

of A/T

O&M Costs

O&M Costs

0 1

$113,328

2

$113,328

$113,328

3

$113,328

$113,328 $109,632

$256,323

$184,553

4

$113,328

$113,328 $109,632 $109,632

$319,006

$229,684

5

$113,328

$113,328 $109,632 $109,632 $105,936

$371,675

$267,606

Holding

Expected

Period

Market

Taxable

Gains

Value

Gains

Tax

(N ) 0

$75,000

Net A/T

A/T Operating Costs (in PW)

Market

over the Holding Period

Value

O&M Costs Tax Shield

OC(15%)

$98,546

$70,953

$184,238

$132,652

CR(15%)

Total AEC(15%)

Total OC

$45,000

1

$60,000

$60,000

$16,800

$43,200

$70,953

$0

$70,953

$81,596

$8,550

2

$50,000

$50,000

$14,000

$36,000 $132,652

$0

$132,652

$81,596

$10,936

3

$30,000

$30,000

$8,400

$21,600 $184,553

$0

$184,553

$80,830

$13,489

4

$30,000

$30,000

$8,400

$21,600 $229,684

$0

$229,684

$80,450

$11,436

5

$10,000

$10,000

$2,800

$7,200 $267,606

$0

$267,606

$79,831

$12,356

56 Copyright © 2023 Pearson Education, Inc.

$90,146 $92,532 $94,318 $91,887 $92,187

Contemporary Engineering Economics, 7th ed. ©2023

Tax Rate

28%

MARR

15%

Holding Period (N )

Permitted Annual Depreciation Amounts over the Holding Period 1

2

3

4

Total 5

6

7

8

Depreciation

0

Value $400,000

1

$80,000

2

$80,000

$128,000

3

$80,000

$128,000

$76,800

4

$80,000

$128,000

$76,800

$46,080

5

$80,000

$128,000

$76,800

$46,080

Holding Period (N )

Book

$46,080

$80,000

$320,000

$208,000

$192,000

$284,800

$115,200

$330,880

$69,120

$376,960

$23,040

Annual O&M Costs over the Holding Period 1

2

3

4

5

Total PW 6

7

8

Total PW of

of A/T

O&M Costs

O&M Costs

0 1

$84,900

2

$84,900

$87,900

3

$84,900

$87,900

$87,267

4

$84,900

$87,900

$87,267

$92,933

5

$84,900

$87,900

$87,267

$92,933 Net A/T

$95,467

$73,826

$53,155

$140,291

$101,010

$197,671

$142,323

$250,805

$180,580

$298,269

$214,754

Holding

Expected

A/T Operating Costs (in PW)

Period

Market

Taxable

Gains

Market

(N )

Value

Gains

Tax

Value

0

$400,000

1

$300,000

($20,000)

$53,155

$19,478

$33,677

$38,728

$154,400

2

$240,000

$48,000

$13,440 $226,560 $101,010

$46,578

$54,431

$33,481

$140,670

3

$190,000

$74,800

$20,944 $169,056 $142,323

$60,718

$81,605

$35,741

$126,507

4

$150,000

$80,880

$22,646 $127,354 $180,580

$68,095

$112,485

$39,400

$114,602

5

$115,000

$91,960

$25,749

$74,509

$140,244

$41,837

$106,089

over the Holding Period O&M Costs Tax Shield

OC(15%)

CR(15%)

Total AEC(15%)

Total OC

$400,000 ($5,600) $305,600

$89,251 $214,754

57 Copyright © 2023 Pearson Education, Inc.

$193,128 $174,151 $162,248 $154,001 $147,926

Contemporary Engineering Economics, 7th ed. ©2023 ∴ Decision: The economic service life for the defender is 1 year with AEC (15%) = $90,146. On the other hand, the economic service life for the challenger is 5 years with AEC (15%) = $147,926. Even though the challenger is a better machine in terms of operating efficiency, its initial cost is too high to justify its purchase at this point. Even if the defender is kept beyond its economic service life, say even 5 years, it is still cheaper to operate the defender.

(d) Replacement analysis under inflation • Adjusted operating and maintenance cost by inflation: Sample calculation for fuel cost – Defender for Year 2:

Fuel cost before inflation = $24,408 Price index for fuel for year 2 = 120 Fuel cost after inflation = $24,408(1.2) = $29,290 Other inflation-adjusted operating and maintenance cost elements can be calculated in a manner like the fuel cost. The following table summarizes these calculations. All cost figures are in actual dollars. Defender

n

Maintenance

Wage

Fuel

Total

1

$50,544

$48,438

$26,849

$125,831

2

$54,288

$52,650

$29,290

$136,228

3

$58,032

$51,714

$29,968

$139,714

4

$58,968

$53,703

$32,273

$144,944

5

$59,904

$52,416

$32,544

$144,864

Challenger

n

Maintenance

Wage

Fuel

Total

1

$22,680

$40,365

$31,680

$94,725

2

$27,840

$43,875

$34,560

$106,275

3

$33,377

$43,095

$35,360

$111,832

4

$41,055

$44,753

$38,080

$123,888

5

$49,493

$43,680

$38,400

$131,573

58 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 The discount rate must be changed. The average general inflation rate is about 8% over five years, the inflation adjusted MARR is 24.2% (= 0.15 + 0.08 + 0.15(.08))

59 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

Tax Rate MARR

28% 24.2%

Holding Period (N )

Permitted Annual Depreciation Amounts over the Holding Period 1

2

3

4

5

6

7

8

Total

Book

Depreciation

Value

0

$0

1

$0

2

$0

$0

3

$0

$0

$0

4

$0

$0

$0

$0

5

$0

$0

$0

$0

Holding

$0

$0

$0

$0

$0

$0

$0

$0

$0

$0

Annual O&M Costs over the Holding Period

Total PW

Period (N )

$0

1

2

3

4

5

6

7

8

Total PW of

of A/T

O&M Costs

O&M Costs

0 1

$125,831

2

$125,831

$136,228

3

$125,831

$136,228 $139,714

$262,551

$189,036

4

$125,831

$136,228 $139,714 $144,944

$323,464

$232,894

5

$125,831

$136,228 $139,714 $144,944 $144,864

$372,482

$268,187

Holding

Expected

Period

Market

Taxable

Gains

Market

(N )

Value

Gains

Tax

Value

Net A/T

$101,313

$72,946

$189,626

$136,531

A/T Operating Costs (in PW) over the Holding Period O&M Costs Tax Shield

OC(24.2%) CR(24.2%)

Total AEC(24.2%)

Total OC

0

$75,000

1

$60,000

$60,000

$16,800

$43,200

$54,000 $72,946

$0

$72,946

$90,598

$23,868

2

$50,000

$50,000

$14,000

$36,000 $136,531

$0

$136,531

$93,937

$21,097

3

$30,000

$30,000

$8,400

$21,600 $189,036

$0

$189,036

$95,696

$21,629

4

$30,000

$30,000

$8,400

$21,600 $232,894

$0

$232,894

$97,216

$18,752

5

$10,000

$10,000

$2,800

$7,200 $268,187

$0

$268,187

$98,093

$18,860

60 Copyright © 2023 Pearson Education, Inc.

$114,466 $115,034 $117,325 $115,968 $116,953

Contemporary Engineering Economics, 7th ed. ©2023

Tax Rate MARR

28% 24.2%

Holding

Permitted Annual Depreciation Amounts over the

Period (N )

Holding Period 1

2

3

4

5

6

7

8

Total

Book

Depreciation

Value

0

$400,000

1

$80,000

$80,000

2

$80,000

$128,000

3

$80,000

$128,000

$76,800

4

$80,000

$128,000

$76,800

$46,080

5

$80,000

$128,000

$76,800

$46,080

Holding

$46,080

$208,000

$192,000

$284,800

$115,200

$330,880

$69,120

$376,960

$23,040

Annual O&M Costs over the Holding Period

Total PW

Period (N )

1

2

3

4

5

$320,000

6

7

8

Total PW of

of A/T

O&M Costs

O&M Costs

0 1

$94,725

2

$94,725

$106,275

3

$94,725

$106,275 $111,832

$203,535

$146,545

4

$94,725

$106,275 $111,832 $123,888

$255,599

$184,032

5

$94,725

$106,275 $111,832 $123,888 $131,573

$300,120

$216,086

Net A/T

A/T Operating Costs (in PW)

Market

over the Holding Period

$76,268

$54,913

$145,163

$104,517

Holding

Expected

Period

Market

Taxable

Gains

(N )

Value

Gains

Tax

0

$400,000

1

$300,000

($20,000)

$54,913

$18,035

$36,878

$45,802

$191,200

2

$240,000

$48,000

$13,440 $226,560 $104,517

$41,269

$63,248

$43,517

$174,160

3

$190,000

$74,800

$20,944 $169,056 $146,545

$52,494

$94,051

$47,612

$157,823

4

$150,000

$80,880

$22,646 $127,354 $184,032

$57,916

$126,116

$52,644

$144,629

5

$115,000

$91,960

$25,749

$62,282

$153,804

$56,256

$135,259

Value

O&M Costs Tax Shield

OC(24.2%) CR(24.2%)

Total AEC(24.2%)

Total OC

$400,000 ($5,600) $305,600

$89,251 $216,086

61 Copyright © 2023 Pearson Education, Inc.

$237,002 $217,676 $205,434 $197,273 $191,515

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 15 Capital-Budgeting Decision Methods of Financing 15.1 (a) Equity Financing: Let X denote the number of shares to be sold. The total flotation cost would be (0.06)($25) X = $1.5 X To net $10 million, 25 X − 1.5 X = $10, 000, 000

23.5 X = $10, 000, 000 X = 425,532 shares flotation cost = $638,298 (b) Equity Financing:

$10, 000, 000 − $10, 000, 000 = $193, 680 1 − 0.019 Number of bond = $10,193,680 / $1,000 = 10,194 units Annual interest = (10,194)($1,000)(0.12) = $1,223,280 flotation cost =

15.2 (a) Equal repayment of the principal: Repayment n Interest Principal 0 1 $45,000 $83,333 2 $37,500 $83,333 3 $30,000 $83,333 4 $22,500 $83,333 5 $15,000 $83,333 6 $7,500 $83,333 (b) Equal repayment of the interest: Repayment n Interest Principal 0 1 $45,000 $0 2 $45,000 $0 3 $45,000 $0 4 $45,000 $0 5 $45,000 $0 6 $45,000 $500,000 1 Copyright © 2023 Pearson Education, Inc.

Loan Balance $500,000 $416,667 $333,333 $250,000 $166,667 $83,333 $0

Loan Balance $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $0

Contemporary Engineering Economics, 7th ed. ©2023

(c) Equal annual installment:

A = $500, 000( A / P,9%, 6) = $111, 460 n 0 1 2 3 4 5 6

Repayment Interest Principal $45,000 $39,019 $32,499 $25,392 $17,646 $9,203

Loan Balance $500,000 $433,540 $361,099 $282,138 $196,070 $102,257 $0

$66,460 $72,441 $78,961 $86,068 $93,814 $102,257

15.3 (a) Equity Financing 0

1

2

3

4

$100,000

$100,000

$100,000

$100,000

0 $40,000

0 $64,000

0 $38,400

0 $11,520

Taxable Income Income Taxes (28%)

$60,000 $16,800

$36,000 $10,080

$61,600 $17,248

$88,480 $24,774

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

$43,200

$25,920

$44,352

$63,706

$43,200 $40,000

$25,920 $64,000

$44,352 $38,400

$63,706 $11,520 $30,000 $4,502

$83,200

$89,920

$82,752

$109,728

Income Statement Revenue Expenses: O&M Depreciation

Net Cash Flow

($200,000)

($200,000) PW (10%) =

$87,069

2 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (b) Debt Financing 1

2

3

4

$100,000

$100,000

$100,000

$100,000

20,000 $40,000

15,000 $64,000

10,000 $38,400

5,000 $11,520

Taxable Income Income Taxes (28%)

$40,000 $11,200

$21,000 $5,880

$51,600 $14,448

$83,480 $23,374

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Loan Repayment

$28,800

$15,120

$37,152

$60,106

$28,800 $40,000

$15,120 $64,000

$37,152 $38,400

$200,000

($50,000)

($50,000)

($50,000)

$60,106 $11,520 $30,000 $4,502 ($50,000)

$0

$18,800

$29,120

$25,552

$56,128

(BANK A) Income Statement Revenue Expenses: Interest Depreciation

Net Cash Flow

0

($200,000)

$98,691 1

2

3

4

$100,000

$100,000

$100,000

$100,000

20,000 $40,000

16,724 $64,000

13,121 $38,400

9,157 $11,520

Taxable Income Income Taxes (28%)

$40,000 $11,200

$19,276 $5,397

$48,479 $13,574

$79,323 $22,210

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Loan Repayment

$28,800

$13,879

$34,905

$57,113

$28,800 $40,000

$13,879 $64,000

$34,905 $38,400

$200,000

($32,759)

($36,035)

($39,638)

$57,113 $11,520 $30,000 $4,502 ($91,566)

$0

$36,041

$41,844

$33,667

$11,569

(BANK B) Income Statement Revenue Expenses: Interest Depreciation

Net Cash Flow

PW (10%) = 0

($200,000)

PW (10%) =

$100,542

(c) Best course of action: Adopt Bank B’s repayment plan. 3 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 15.4 (a) The total flotation costs to raise $65 million: • Common stock: amount of common stock = ($65, 000, 000)(0.45)

= $29, 250, 000 $29, 250, 000 Flotation cost = − $29, 250, 000 = $1, 410,377 1 − 0.046 • Preferred stock: amount of preferred stock = ($65, 000, 000)(0.10)

= $6,500, 000 $6,500, 000 Flotation cost = − $6,500, 000 = $572,905 1 − 0.081 • Bond: amount of bond = ($65, 000, 000)(0.45)

= $29, 250, 000 $29, 250, 000 Flotation cost = − $29, 250, 000 = $415,314 1 − 0.014 ∴ Total flotation costs = $2,398,596 (b) Number of shares or (bonds) to be sold to raise $65 million: • Common stock: X S (1 − 0.046)($32) = $29, 250, 000

X S = 958,137 shares • Preferred stock: X P (1 − 0.081)($55) = $6,500, 000

X P = 128,598 shares • Bond:

X B (1 − 0.014)($980) = $29, 250, 000 X B = 30, 271 units

(c) Cash requirement to meet financing costs: • Common stock:

4 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

annual cash dividends = ($2 / share)(958,137 shares) = $1,916, 274 • Preferred stock:

annual cash dividends = (0.06)($15 / share)(128,598shares) = $115, 738 • Bond:

borrowing amount = (30, 271)($1, 000) = $30, 271, 000 Annual interest = ($30,271,000)(0.12) = $3,632,520 ∴ Total annual cash requirement = $5,664,532

Cost of Capital 15.5

After-tax cost of debt: (a) (0.12)(1 − 0.25) = 0.09 or 9% (b) (0.14)(1 − 0.28) = 0.0108 or 10.08% (c) (0.15)(1 − 0.35) = 0.0975 or 9.75%

15.6 In the absence of a bond maturity date, we need to assume that the 13% yield to maturity represents the before-tax cost of debt after considering the flotation cost as well as bond discounting. Let kb = 13% . We compute the after-tax cost of debt as follows: (0.13)(1 − 0.28) = 9.36%

15.7

Cost of retaining earnings:

kr =

$1.12 + 0.12 = 18.22% $18

5 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

15.8 (a) Flotation costs in percentage:

fc = 1 − (b) Cost of new common stock:

ke =

15 = 16.67% 18

$1.10 + 0.10 = 17.33% $18(1 − 0.1667)

15.9 ie = 0.22 id = (0.13)(1 − 0.30) = 0.091 k = (0.091)(0.45) + (0.22)(0.55) = 0.1620 15.10 Given: ke = 0.30 (a) ie = (55 / 55)(0.30) = 0.3 id = (1 − 0.28)[(10 / 30)(0.14) + (20 / 30)(0.12)] = 0.0912 k = (0.0912)(0.30) + (0.3)(0.70) = 0.2374

(b)

ie = rf + β  rM − r f  = 0.06 + 1.2(0.12 − 0.06) = 0.132 (c) k = (0.0912)(0.30) + (0.132)(0.70) = 0.1198

Note: The difference between k in (a) and k in (c) is due to the expectation on future required return from equity. 15.11

Given: ie = 18%, id = (0.12)(1 − 0.28) = 0.0864 k = (0.4)(0.0864) + (0.6)(0.18) = 14.26%

6 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (a) Net equity flow method: PW(18%) = $40, 427 > 0 , accept the project. 0

2

3

4

5

$70,000

$70,000

$70,000

$70,000

$70,000

20,000 $17,148 $5,760

20,000 $29,388 $4,853

20,000 $20,988 $3,838

20,000 $15,000 $2,701

20,000 $5,352 $1,427

Taxable Income Income Taxes (28%)

$27,092 $7,586

$15,759 $4,413

$25,174 $7,049

$32,299 $9,044

$43,221 $12,102

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Loan repayment

$19,506

$11,346

$18,125

$23,255

$31,119

$19,506 $17,148

$11,346 $29,388

$18,125 $20,988

$23,255 $15,000

48,000

(7,556)

(8,462)

(9,478)

(10,615)

$31,119 $5,352 $30,000 $595 (11,889)

($72,000)

$29,098

$32,272

$29,635

$27,640

$55,177

PW (18%) =

$40,427

Income Statement Revenue Expenses: O&M Depreciation Interest (12%)

Net Cash Flow

1

($120,000)

(b) Cost of capital method: PW(14.26%) = $36, 211 > 0 , accept the project. 0

2

3

4

5

$70,000

$70,000

$70,000

$70,000

$70,000

20,000 $17,148

20,000 $29,388

20,000 $20,988

20,000 $15,000

20,000 $5,352

Taxable Income Income Taxes (28%)

$32,852 $9,199

$20,612 $5,771

$29,012 $8,123

$35,000 $9,800

$44,648 $12,501

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

$23,653

$14,841

$20,889

$25,200

$32,147

$23,653 $17,148

$14,841 $29,388

$20,889 $20,988

$25,200 $15,000

$32,147 $5,352 $30,000 $595

$40,801

$44,229

$41,877

$40,200

$68,093

Income Statement Revenue Expenses: O&M Depreciation

Net Cash Flow

1

($120,000)

($120,000) PW (14.26%)

$36,211

7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 15.12 (a) Net equity flow method: 0

Income Statement Revenue Expenses Depreciation Interest (15%)

1 2 3 4 5 $45,000 $45,000 $45,000 $45,000 $45,000 $20,000 $32,000 $19,200 $11,520 $9,000 $7,665 $6,130 $4,365

$5,760 $2,335

.

Taxable Income Income Taxes (30%)

$16,000 $5,335 $4,800 $1,600

$19,670 $29,115 $36,905 $5,901 $8,735 $11,072

Net Income

$11,200 $3,734

$13,769 $20,381 $25,834

Cash Flow Statement Cash from operation Net Income Depreciation Investment/Salvage Gains Tax Loan Repayment

$11,200 $3,734 $13,769 $20,381 $25,834 $20,000 $32,000 $19,200 $11,520 $5,760 ($100,000) $30,000 ($5,544) $60,000 ($8,899) ($10,234) ($11,769) ($13,534) ($15,564)

Net cash flow

($40,000) $22,301 $25,501 $21,200 $18,366 $40,485

PW(20%) =

$33,689

Accept

(b) Cost of capital method: ie = 20%, id = (0.15)(1 − 0.30) = 0.105 k = (0.6)(0.105) + (0.4)(0.2) = 14.3% 0

Income Statement Revenue Expenses Depreciation

1 2 3 4 5 $45,000 $45,000 $45,000 $45,000 $45,000 $20,000 $32,000 $19,200 $11,520

$5,760

.

Taxable Income Income Taxes

$25,000 $13,000 $25,800 $33,480 $39,240 $7,500 $3,900 $7,740 $10,044 $11,772

Net Income

$17,500

$9,100 $18,060 $23,436 $27,468

Cash Flow Statement Cash from operation Net Income Depreciation Investment/Salvage Gains Tax

$17,500 $9,100 $18,060 $23,436 $27,468 $20,000 $32,000 $19,200 $11,520 $5,760 ($100,000) $30,000 ($5,544)

Net cash flow

($100,000) $37,500 $41,100 $37,260 $34,956 $57,684

PW(14.3%) =

$39,268 Accept

8 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 15.13 (a) Using ie = 15% : Machine A Income Statement Revenue Expenses: O&M Depreciation Interest (10%)

$20,000

$20,000

$20,000

$20,000

$20,000

$20,000

8,000 $8,000 $1,200

8,000 $12,800 $1,044

8,000 $7,680 $873

8,000 $4,608 $685

8,000 $4,608 $478

8,000 $2,304 $250

Taxable Income Income Taxes (28%)

$2,800 $784

($1,844) ($516)

$3,447 $965

$6,707 $1,878

$6,914 $1,936

$9,446 $2,645

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Loan repayment

$2,016

($1,328)

$2,482

$4,829

$4,978

$6,801

$2,016 $8,000

($1,328) $12,800

$2,482 $7,680

$4,829 $4,608

$4,978 $4,608

12,000

(1,555)

(1,711)

(1,882)

(2,070)

(2,277)

$6,801 $2,304 $4,000 ($1,120) (2,505)

($28,000)

$8,461

$9,761

$8,280

$7,367

$7,309

$9,480

PW (15%) =

$4,127

Net Cash Flow

($40,000)

Machine B 0

1

2

3

4

5

6

$28,000

$28,000

$28,000

$28,000

$28,000

$28,000

10,000 $12,000 $1,800

10,000 $19,200 $1,567

10,000 $11,520 $1,310

10,000 $6,912 $1,028

10,000 $6,912 $717

10,000 $3,456 $376

Taxable Income Income Taxes (28%)

$4,200 $1,176

($2,767) ($775)

$5,170 $1,448

$10,060 $2,817

$10,371 $2,904

$14,168 $3,967

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

$3,024

($1,992)

$3,722

$7,243

$7,467

$10,201

$3,024 $12,000

($1,992) $19,200

$3,722 $11,520

$7,243 $6,912

$7,467 $6,912

$10,201 $3,456 $4,000 ($1,120)

Income Statement Revenue Expenses: O&M Depreciation Interest (10%)

($60,000)

9 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Loan repayment Net Cash Flow

18,000

(2,333)

(2,566)

(2,823)

(3,105)

(3,416)

(3,757)

($42,000)

$12,691

$14,642

$12,419

$11,050

$10,963

$12,780

PW (15%) =

$5,567

∴ Machine B should be better. (b) Using k = 0.7(0.15) + 0.3(0.10)(1 − 0.28) = 12.66% Machine A 0

1

2

3

4

5

6

$20,000

$20,000

$20,000

$20,000

$20,000

$20,000

8,000 $8,000

8,000 $12,800

8,000 $7,680

8,000 $4,608

8,000 $4,608

8,000 $2,304

Taxable Income Income Taxes (28%)

$4,000 $1,120

($800) ($224)

$4,320 $1,210

$7,392 $2,070

$7,392 $2,070

$9,696 $2,715

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

$2,880

($576)

$3,110

$5,322

$5,322

$6,981

$2,880 $8,000

($576) $12,800

$3,110 $7,680

$5,322 $4,608

$5,322 $4,608

$6,981 $2,304 $4,000 ($1,120)

($40,000)

$10,880

$12,224

$10,790

$9,930

$9,930

$12,165

PW(12.66%)=

$4,420

Income Statement Revenue Expenses: O&M Depreciation

Net Cash Flow

($40,000)

Machine B

0

2

3

4

5

6

$28,000

$28,000

$28,000

$28,000

$28,000

$28,000

10,000 $12,000

10,000 $19,200

10,000 $11,520

10,000 $6,912

10,000 $6,912

10,000 $3,456

Taxable Income Income Taxes (28%)

$6,000 $1,680

($1,200) ($336)

$6,480 $1,814

$11,088 $3,105

$11,088 $3,105

$14,544 $4,072

Net Income Cash Flow

$4,320

($864)

$4,666

$7,983

$7,983

$10,472

Income Statement Revenue Expenses: O&M Depreciation

1

10 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

$4,320 $12,000

($864) $19,200

$4,666 $11,520

$7,983 $6,912

$7,983 $6,912

$10,472 $3,456 $4,000 ($1,120)

($60,000)

$16,320

$18,336

$16,186

$14,895

$14,895

$16,808

PW(12.66%)=

$5,926

($60,000)

Net Cash Flow

∴ Machine B should still be better.

(c) Both methods provide the same decision.

Capital Budgeting 15.14 Based on the investment opportunity curve below, the firm’s optimal capital budget would be $177 million if there is no restriction on the firm’s debt limit. However, with a budget limit of $100 million, the firm may select projects 5 and 3 first. Since these two projects alone consume $95 million, the firm may have two choices about utilizing the remaining $5 million funds. First one is to find any projects whose rates of return exceed the cost of capital. Project 4 comes close to meeting this requirement. However, the firm’s borrowing rate is 18%, which is greater than the rate of return from project 4. Therefore, the projects that should be included in the $100 million budget would be projects 5 and 3. If money has to be raised from outside, the firm should raise only $95 million Rate of Return 90% 5

80% 3

40% 2

32% 30% 7

6

22% 15% rate

1

Borrowing

(18%)

4

Lending (12%)

Capital budget ($ Million) 11 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

15.15

(a) J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Projects Required Investment 0 A C D F AC AD BC CD CF DF ACD ABC BCF DEF CEF ABCD BCEF BCDF EFCD BCDEF

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

100.00 100.00 300.00 150.00 200.00 400.00 300.00 400.00 250.00 450.00 500.00 400.00 450.00 650.00 450.00 700.00 650.00 750.00 750.00 950.00

(b) Optimal capital budget = $750, select ALT 20 The IRR for each project exceeds its cost of capital k, all projects should be considered in the budget.

A B C D E F

0 -100 -200 -100 -300 -200 -150

1 40 150 30 200 100 50

2 60 100 80 120 100 70

3 40 30 60 150 150 80

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IRR 19% 25% 29% 28% 31% 15%

k 14.5% 16.0% 14.5% 16.0% 16.0% 14.5%

Contemporary Engineering Economics, 7th ed. ©2023

Compute the net present worth for each alternative at its cost of capital and select the alternative with the largest net present value. The optimal capital budget is found at $750 with ALT 20. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

0 0 -100 -100 -300 -150 -200 -400 -300 -400 -250 -450 -500 -400 -450 -650 -450 -700 -650 -750 -750 -950

1 0 40 30 200 50 70 240 180 230 80 250 270 220 230 350 180 420 330 430 380 530

2 0 60 80 120 70 140 180 180 200 150 190 260 240 250 290 250 360 350 370 370 470

3 0 40 60 150 80 100 190 90 210 140 230 250 130 170 380 290 280 320 320 440 470

IRR 0 19% 29% 28% 15% 24% 26% 26% 28% 20% 23% 26% 24% 22% 26% 25% 26% 25% 24% 26% 26%

k 14.5% 14.5% 16.0% 14.5% 16.0% 17.5% 16.0% 17.5% 16.0% 17.5% 17.5% 17.5% 17.5% 20.0% 17.5% 20.0% 20.0% 20.0% 20.0% 21.0%

PW(k ) $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

(c) Select ALT 20 Note: The same selection would be found by applying the incremental IRR analysis as illustrated in Example 15.12.

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7.35 27.19 57.69 0.35 28.45 51.75 46.60 70.06 20.13 42.16 72.22 41.20 31.62 62.96 63.03 62.04 53.24 50.46 78.24 74.34

Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies ST 15.1 (a) • Total market value = Present value of its expected future net cash flows + the value of current assets (5M) •

Total market value =stock price per share* number of share outstanding = $18(1M) = $18M Present value of its expected future net cash flows = $13M

(b)

Income before tax = $3.5M Earnings = $2.1M Income before tax (1-tax rate) = Earnings Tax rate = 40% (c) •

The case when the financing source is known, we use the cost of equity ( ie ) as c MARR, MARR = ie , ie = c ke , since we have only one source of equity, ce issuing common new stocks Debt to equity ratio = total debt/ total equity = 12000/18000 = 2/3 Games Inc. intends to maintain its current debt to equity ratio when financing new equipment. Therefore, they will be financing the cost of new equipment of $10M, with $4M from long-term debt and $6M from equity of stock. Therefore, cc = ce = $6M

D1 = D0 (1 + g) =1.9(1+0.225) = $2.3275, P0 = 1.8, fc = 0.11

g with the average dividend from 2016 to 2022 1.9 − 1 g= = 0.225 4 D1 ke = + g = 0.3703 P0 (1 − fc )

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Contemporary Engineering Economics, 7th ed. ©2023

MARR= ie =

•

ce ke = 37.03% cc

The case when the financing source is unknown, we use k (WACC) as MARR, MARR= k

k = id

cd ce + ie cd + ce cd + ce

cs ks (1 − t m ) , interest rate of term loan after tax= 10%. cd c ie = c ke = 0.37028 =37.03% from the first case ce

id =

Recall that Games Inc. intends to maintain its current debt to equity ratio when financing to purchase new equipment. Therefore, they will be financing the cost of new equipment, $10M, with $4M from debt of long-term debt and $6M from equity of stock. Therefore, cc = ce = $6M, cs = cd = $4M MARR = k = id

cd ce + ie = 26.22% cd + ce cd + ce

(d) The current stock price*(shares outstanding) = $18(1M) PW of increasing profit after installing and operating the new machine = $7,021K The number of shares to be sold to net $6,000K = $6,000K / {(1-fc)*stock price per share} = 374,532 shares The most likely estimate for Games’ stock price = {the current total stock price + increasing profit after installing and operating the new machine}/ total shares outstanding = {18M +7,021K}/ (1,000,000+374,532) shares = $18.2 Finally, the most likely estimate for Games’ stock price may be $ $18.2.

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Contemporary Engineering Economics, 7th ed. ©2023 Note) The income statement and cash flows of most likely case are like below: Income statement Inflation

Revenue Expense Depreciation Debt Interest rate

0

1 $48,400 $38,720 $1,445 $400 $7,835 $3,134 $4,701

(1000US$) 2 $53,240 $31,944 $1,061 $400 $19,835 $7,934 $11,901

0

1

(1000US$) 2

10% 10%

Taxable income Income taxes(40%) Net income

Cash flow statement Operating activities

Net income Depreciation

$4,701 $1,445

$11,901 $1,061

Investment activities

Investment Salvage Gain tax

-$10,000 $5,976 $572

10%

financing activities Borrowed funds Principal repayment Common stock Cash dividend Net cash flow(Actual $) PW=

$4,000 $6,000 $0

-$796 $5,350

-$4,000 -$6,742 -$880 $7,888

$7,021

NOTE: If we use the current prevailing tax rate of 25%, the net present value would be much higher. (e) Not provided

ST 15.2 (a) There are 58 alternatives. (b) Only 10 alternatives are feasible. j

Projects

j

Projects

1 2 3 4 5

1 2 1, 2 1, 4 1, 5

6 7 8 9 10

1, 7 1, 4, 7 1, 5, 7 2, 6 2, 7

16 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 (c) Without knowing the exact cash flow sequence for each project over the project life, it is not feasible to determine the optimal capital budget.

ST 15.3 (a) Select A and C with FW(10%) = $4,894. Since there are $500 left over after selecting A and C, this left-over could be lent out at 10% for 3 periods. Therefore, the total amount available for lending at the end of period 3 is calculated as follows:

F = $4,894 + $500( F / P,10%,3) = $5,559.60 (b) Select B and C. The total amount available for lending at the end of period 3 is $5,740. (c) With a budget limit of $3,500, the reasonable MARR should be the lending rate of 10%. (You select A and C and have $500 available for lending.) ST 15.4 (a) The debt repayment schedule for the loan from the equipment manufacturer: n

Loan Repayment Interest Principal

0

Loan Balance $2,000,000

1

$200,000

$125,491

$1,874,509

2

$187,451

$138,040

$1,736,469

3

$173,647

$151,844

$1,584,625

4

$158,463

$167,028

$1,417,597

5

$141,760

$183,731

$1,233,866

6

$123,387

$202,104

$1,031,762

7

$103,176

$222,315

$809,447

8

$80,945

$244,546

$564,901

9

$56,490

$269,001

$295,901

10

$29,590

$295,901

$0

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Contemporary Engineering Economics, 7th ed. ©2023

(b) The flotation costs and the number of common stocks to raise $8,500,000:

$8,500, 000 − $8,500, 000 = $749,184 1 − 0.081 $8,500, 000 = 205,537 shares number of shares = (1 − 0.081)($45)

flotation cost =

(c) The floatation costs and the number of $1,000 bonds to raise $10.5 million:

$10,500, 000 − $10,500, 000 = $203,364 1 − 0.019 $10,500, 000 = 11,893units number of bonds = (1 − 0.019)($900) flotation cost =

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Contemporary Engineering Economics, 7th ed. ©2023

ST 15.5 (a) The net cash flow the cogeneration project with bond financing Income Statement Revenue Electricity Bill Excess power Expenses O&M Misc. Standby power Overhead Overhaul Depreciation Unit Inter Equipment Interest (9%)

0

1

2

3

4

5

6

7

8

9

10

11

12

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000 $1,500,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000 $1,500,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000 $1,500,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $1,000,000 $6,400 $1,280,000

$500,000 $100,000 $1,070,370

$950,000 $160,000 $1,070,370

$855,000 $96,000 $1,070,370

$770,000 $57,600 $1,070,370

$693,000 $57,600 $1,070,370

$623,000 $28,800 $1,070,370

$590,500

$0

$0

$0

$0

$0

$1,070,370

$1,070,370

$1,070,370

$1,070,370

$1,070,370

$1,070,370

.

Taxable Income Income Taxes (28%)

$2,143,230 $600,104

$1,633,230 $457,304

$292,230 $81,824

$1,915,630 $536,376

$1,992,630 $557,936

$591,430 $165,600

$2,152,730 $602,764

$2,743,230 $768,104

$1,243,230 $348,104

$2,743,230 $768,104

$2,743,230 $768,104

$2,743,230 $768,104

Net Income

$1,543,126

$1,175,926

$210,406

$1,379,254

$1,434,694

$425,830

$1,549,966

$1,975,126

$895,126

$1,975,126

$1,975,126

$1,975,126

$1,543,126

$1,175,926

$210,406

$1,379,254

$1,434,694

$425,830

$1,549,966

$1,975,126

$895,126

$1,975,126

$1,975,126

$1,975,126

$500,000 $100,000

$950,000 $160,000

$855,000 $96,000

$770,000 $57,600

$693,000 $57,600

$623,000 $28,800

$590,500

$590,501

$590,502

$590,503

$590,504

$295,000

Cash Flow Statement Cash from operation Net Income Depreciation Unit Inter Equipment Investment/Salvage Unit Inter Equipment Gains Tax Loan Repayment Net cash flow PW(27%) =

($10,000,000) ($500,000)

$1,000,000 $381,217 ($11,893,000)

$10,500,000 $0

$2,143,126

$2,285,926

$1,161,406

$2,206,854

$2,185,294

$1,077,630

$2,140,466

$6,343,988

19 Copyright © 2023 Pearson Education, Inc.

$2,565,627

$1,485,628

$2,565,629

$2,565,630

($8,241,657)

Contemporary Engineering Economics, 7th ed. ©2023

(b) The maximum annual lease amount that ACC is willing to pay is $1,291,338. (By Excel Goal Seek) Income Statement Revenue Electricity Bill Excess power Expenses O&M Misc. Standby power Overhead Lease

0

1

2

3

4

5

6

7

8

9

10

11

12

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $6,120,000 $480,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$6,120,000 $480,000

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $500,000 $1,000,000 $1,000,000 $6,400 $6,400 $1,280,000 $1,280,000 $1,291,338 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

$500,000 $1,000,000 $6,400 $1,280,000 $1,291,338

.

Taxable Income Income Taxes (28%)

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $2,522,262 $706,233 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

$2,522,262 $706,233

Net Income

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029 $1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

Cash from operation Net Income

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029 $1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029 $1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

$1,816,029

Net cash flow PW(27%) =

$0 $6,343,988

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Contemporary Engineering Economics, 7th ed. ©2023

Chapter 16 Economic Analysis in the Service Sector Valuation of Benefits and Costs 16.1 (a) • User’s benefits: - Prevention (or retardation) of highway corrosion: resulting in lower highway maintenance cost. This lower maintenance cost implies lower users’ taxes on gasoline, and so forth. - Prevention of rust on vehicles: resulting in lower repair and maintenance costs and higher resale value of vehicles. - Prevention of corrosion to utility lines and damages to water supplies: resulting in lower utility rates. - Prevention of damages to vegetation and soil surrounding areas: increasing land values and agriculture yields. • Sponsor’s costs: - Paying a higher tax. - Unknown environmental damages due to using CMA (b) The state of Michigan may declare certain sections of highway for experimental purpose. CMA may be used exclusively for a designated area and road salts for another area for an extended period time. Then, it investigates the impact of CMA on vegetation yields, which can be compared with those of areas from road salt use. The difference in vegetation yields may be quantified in terms of market value, and so forth. 16.2 Open end question (Not provided) 16.3 Open end question (Not provided)

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Contemporary Engineering Economics, 7th ed. ©2023

Benefit-Cost Analysis 16.4

B = $117, 400( P / A, 6%,5) = $494,535.76 C = $5, 000 + $48,830( P / A, 6%,5) = $210, 691.49 $494,535.76 $210, 691.49 = 2.35 > 1 This project is justifiable based on the benefit-cost analysis. BC(6%) =

16.5 B = $250, 000( P / A, 6%, 25) + $50, 000( P / F , 6%, 25) = $3, 207,500 C = $1, 200, 000 + $100, 000( P / A, 6%, 25) = $2, 478,340 $3, 207,500 $2, 478,340 = 1.29 > 1

BC(6%) =

16.6

B = $786,000( P / A,8%,15) + $200,000( P / F,8%,15) = $6,790,799 C ′ = $233,000( P / A,8%,15) = $1,994,359 I = $2,500,000 Thus, the net savings per dollar invested is:

BC(8%) =

$6,790,799 = 1.51 $2,500, 000 + $1,994,359

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Contemporary Engineering Economics, 7th ed. ©2023 16.7 (a) BC(i) analysis: • Design A:

I = $400, 000 C ' = $50, 000( P / A,8%,15) = $427,974 B = $85, 000( P / A,8%,15) = $727,556 • Design B:

I = $300, 000 C ' = $80, 000( P / A,8%,15) = $684, 758 B = $85, 000( P / A,8%,15) = $727,556 • Incremental analysis: Fee collections in the amount of $85,000 will be the same for both alternatives. Therefore, we will not be able to compute the BC (i ) ratio. If this happens, we may select the best alternative based on either the least cost ( I + C ' ) criterion or the incremental PI(i ) criterion. Using the incremental PI(i ) criterion,

PI(8%) A− B =

ΔB − ΔC ' 0 − ($427,974 − $684, 758) = = 2.57 > 1 ΔI $100, 000

∴Select design A.

(b) Incremental analysis (A – C):

PI(8%) A−C =

ΔB − ΔC ' 0 − ($427,974 − $556,366) = = 2.57 > 1 ΔI $50, 000

∴Select design A.

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Contemporary Engineering Economics, 7th ed. ©2023

16.8

CR(5%) = ($8, 000, 000 − $750, 000)( A / P,5%,10) + $750, 000(0.05) = $976, 409 O&M(5%) = $2, 000, 000( P / A1 ,5%,5%,10)( A / P,5%,10) = $2, 466, 756 AEC(5%) = $976, 409 + $2, 466, 756 = $3, 443,165 Price for trip =$3, 443,165 / 600, 000 = $5.74 per trip 16.9

CR(10%) = ($55, 000 − $12, 000)( A / P,10%,5) + $12, 000(0.10) = $12,543 O&M(10%) = $18, 000 + $3, 000( A / G,10%,5) = $23, 430 AEC(10%) = $12,543 + $23, 430 = $35,973

Required annual collection fees = $35,973 per truck/year 16.10 (a) From BC(i ) =

B I + C′

I B C' PW(i) BC(i)

A1 $5,000 $12,000 $4,000 $3,000 1.33

Projects A2 $20,000 $35,000 $8,000 $7,000 1.25

A3 $14,000 $21,000 $1,000 $6,000 1.40

Since all projects’ NPWs are nonnegative and BC (i ) are greater than unity, all projects are good candidates for funding.

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(b) Incremental B/C ratio analysis: •

A1 versus A3

$21, 000 − $12, 000 ($14, 000 − $5, 000) + ($1, 000 − $4, 000) = 1.5 Since the ratio is greater than unity (1.5 > 1), project A3 is preferable to project A1. BC(i )3−1 =

•

A3 versus A2

$35, 000 − $21, 000 ($20, 000 − $14, 000) + ($8, 000 − $1, 000) = 1.077 Since the incremental B/C ratio is greater than unity (1.077 > 1), project A2 is preferable over project A3. Therefore, project A2 becomes our final choice. BC(i ) 2−3 =

Select project A2. 16.11 • Building X: BX = $1,960, 000( P / A,10%, 20) = $16, 686, 656 C X = $8, 000, 000 + $240, 000( P / A,10%, 20) − $4,800, 000( P / F ,10%, 20) = $9,329,984 BC(10%) X =

$16, 686, 656 = 1.79 > 1 $9,329,984

• Building Y: BY = $1,320, 000( P / A,10%, 20) = $11, 237,952 CY = $12, 000, 000 + $180, 000( P / A,10%, 20) − $7, 200, 000( P / F ,10%, 20) = $12, 462,528 BC(10%)Y =

$11, 237,904 = 0.90 < 1 $12, 462,528

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Contemporary Engineering Economics, 7th ed. ©2023 ∴ Since Building Y is not desirable at the outset, we don’t need an incremental analysis. Building X becomes the sole candidate for funding.

16.12 (a) Find the equivalent annual value of each element - I, B, and C’ , as B and C’ are already given on annual basis.

A $5.83 $20 $8 $6.17 1.45

I B C' AE(i) BC(i)

Designs B C $9.32 $12.82 $40 $60 $18 $25 $12.68 $22.18 1.46 1.59

(b)

BC(i) B-A =

$40 − $20 ($9.32 − $5.83) + ($18 − $8)

= 1.48 > 1, select B. BC(i)C-B =

$60 − $40 ($12.82 − $9.32) + ($25 − $18)

= 1.90 > 1, select C. (c)

PI(5%) B-A =

ΔB − ΔC ' ($40 − $20) − ($18 − $8) = ΔI ($9.32 − $5.83)

= 2.87 > 1, select B. PI(5%)C-B =

ΔB − ΔC ' ($60 − $40) − ($25 − $18) = ($12.82 − $9.32) ΔI

= 3.71 > 1, select C.

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16.13 Incremental BC(i ) analysis: Present Worth I B C’ BC(i )

A1 $100 $400 $100 2

Proposals A2 $300 $700 $200 1.4

A3 $200 $500 $150 1.43

Incremental A3-A1 A2-A1 $100 $200 $100 $300 $50 $100 0.67 1

C $1,600 $5,656 $2,922 1.25

Incremental C-B A-B $720 $1,560 -$1,414 $754 -$472 $471 -5.7 0.37

Select either A1 or A2. 16.14 Incremental BC(i ) analysis: Present Worth B I C’ BC(i )

A $2,440 $7,824 $3,865 1.24

Design B $880 $7,070 $3,394 1.65

Select design B. 16.15 (a) The benefit-cost ratio for each alternative: • Alternative A:

B = ($1, 000, 000 + $250, 000 + $350, 000 + $100, 000)( P / A,10%,50) = $16,855,185 C = $8, 000, 000 + $200, 000( P / A,10%,50) = $9,982,963 $16,855,185 BC(10%) A = = 1.69 > 1 $9,982,963 7 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023

• Alternative B:

B = ($1, 200, 000 + $350, 000 + $450, 000 + $200, 000)( P / A,10%,50) = $21,812,592 C = $10, 000, 000 + $250, 000( P / A,10%,50) = $12, 478, 704 $21,812,592 BC(10%) B = = 1.75 > 1 $12, 478, 704 • Alternative C:

B = ($1,800, 000 + $500, 000 + $600, 000 + $350, 000)( P / A,10%,50) = $32, 223,147 C = $15, 000, 000 + $350, 000( P / A,10%,50) = $18, 470,185 $32, 223,147 BC(10%) B = = 1.74 > 1 $18, 470,185 (b) Select the best alternative based on BC(i ) : $21,812,592 − $16,855,185 $12, 478, 704 − $9,982,963 = 1.99 > 1 (Select B)

BC(10%) B − A =

$32, 223,147 − $21,812,592 $18, 470,185 − $12, 478, 704 = 1.74 > 1 (Select C)

BC(10%)C − B =

16.16 • Option 1 – The “long” route:

user's annual cost = 22 miles × $0.25 per mile × 400, 000 cars = $2, 200, 000 sponsor's annual cost = $21, 000, 000( A / P,10%, 40) + $140, 000 = $2, 287, 448

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Contemporary Engineering Economics, 7th ed. ©2023

• Option 2 – Shortcut:

user's annual cost = 10 miles × $0.25 per mile × 400, 000 cars = $1, 000, 000 sponsor's annual cost = $45, 000, 000( A / P,10%, 40) + $165, 000 = $4, 766, 674 • Incremental analysis (Option 2 - Option 1):

Incremental user's benefit = $2, 200, 000 − $1, 000, 000 = $1, 200, 000 $1, 200, 000 BC(10%) 2−1 = $4, 766, 674 − $2, 287, 448 = 0.48 < 1 Assuming that there is no do-nothing alternative, select option 1. 16.17 • Multiple alternatives: Projects PW of Benefits A1 $40 A2 $150 A3 $70 A4 $120

PW of Costs $85 $110 $25 $73

Net PW -$45 $40 $45 $47

B/C ratio 0.47 1.36 2.80 1.64

Since the BC ratio for project A1 is less than 1, we delete it from our comparison. • Incremental analysis - A3 versus A4:

$120 − $70 $73 − $25 = 1.04 > 1

BC(10%) A4− A3 = Select A4.

- A2 versus A4:

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Contemporary Engineering Economics, 7th ed. ©2023

$150 − $120 $110 − $73 = 0.81 < 1

BC(10%) A2− A4 = Select A4.

Cost-Effective Analysis 16.18

n

Project A1

0

$

1 2 3 4 5 6

$ $ $ $ $ $

Project A2

(900,000) $ 400,000 340,000 300,000 240,000 200,000 150,000

Incremental (A2 - A1)

(1,200,000) $

(300,000)

200,000 300,000 350,000 440,000 400,000 350,000

(200,000) (40,000) 50,000 200,000 200,000 200,000

$ $ $ $ $ $

$ $ $ $ $ $

(a) •

A1:

B − C ' = $400, 000( P / F , 6%,1) + $340, 000( P / F , 6%, 2) +  +$150, 000( P / F , 6%, 6) = $1,377,141 I = $900, 000 PI A1 =

$1,377,141 = 1.530 > 1 $900, 000

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Contemporary Engineering Economics, 7th ed. ©2023 •

A2:

B − C ' = $200, 000( P / F , 6%,1) + $300, 000( P / F , 6%, 2) +  +$350, 000( P / F , 6%, 6) = $1, 643, 706 I = $1, 200, 000 PI A2 =

$1, 643, 706 = 1.370 > 1 $1, 200, 000

(b) (1) No budget limit: Select both projects as its PI > 1. (2) If A1 and A2 are mutually exclusive, select A1.

Δ( B − C ') = −$200, 000( P / F , 6%,1) − $40, 000( P / F , 6%, 2) +  +$200, 000( P / F , 6%, 6) = $266,564 ΔI = $300, 000 ΔPI A 2− A1 =

$266,564 = 0.889 < 1, Select A1 $300, 000

16.19

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Contemporary Engineering Economics, 7th ed. ©2023

n

Project A1

0

$

1 2 3 4 5 6 7 8 9 10

$ $ $ $ $ $ $ $ $ $

Project A2

(750,000) $ 100,000 100,000 100,000 240,000 200,000 180,000 180,000 180,000 180,000 180,000

Incremental (A2 - A1)

(1,000,000) $

(250,000)

200,000 200,000 200,000 300,000 300,000 250,000 250,000 150,000 100,000 50,000

100,000 100,000 100,000 60,000 100,000 70,000 70,000 (30,000) (80,000) (130,000)

$ $ $ $ $ $ $ $ $ $

$ $ $ $ $ $ $ $ $ $

(a) •

A1:

B − C ' = $100, 000( P / F , 7%,1) + $100, 000( P / F , 7%, 2) +  +$180, 000( P / F , 7%,10) = $1,114,333 I = $750, 000 PI A1 = •

$1,114,333 = 1.486 > 1 $750, 000

A2:

B − C ' = $200, 000( P / F , 7%,1) + $200, 000( P / F , 7%, 2) +  +$50, 000( P / F , 7%,10) = $1, 457, 013 I = $1, 000, 000 PI A 2 =

$1, 457, 013 = 1.457 > 1 $1, 000, 000

Project A1 has a higher PI. (b) Since A1 and A2 are mutually exclusive, select A2.

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Contemporary Engineering Economics, 7th ed. ©2023 •

A2-A1:

Δ( B − C ') = $100, 000( P / F , 6%,1) + $100, 000( P / F , 7%, 2) +  −$130, 000( P / F , 7%,10) = $342, 680 ΔI = $250, 000 ΔPI A 2− A1 =

$342, 680 = 1.371 > 1, Select A2 $2500, 000

16.20 Cost effectiveness of the alternatives Type of Treatment Antibiotic A Antibiotic B Antibiotic C

Cost Effectiveness 160 168.75 180.49

The best treatment option: Antibiotic A

16.21 • The summary of three mutually exclusive alternatives CER: Strategy Nothing Simple Complex

Cost $0 $5,000 $50,000

Effectiveness 0 years 5 years 5.5 years

Cost Effectiveness 0 1,000 9,091

Incremental CER 0 1,000 90,000

• Since there is no clear dominance, we may draw a cost-effectiveness diagram. Here the decision is based on the size budget available. If the available budget is over $50,000, all patients should do a complex strategy, whereas if the available budget is less than $50,000, it is best to do a simple strategy.

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Contemporary Engineering Economics, 7th ed. ©2023

Short Case Studies with Excel ST16.1 Capital allocation decision, assuming that the government will be able to raise the required funds at 10% interest: District I

II

III

IV

Project 1. 27 Street 2. Holden Avenue 3. Forest City Road 4. Fairbanks Avenue 5. Oak Ridge Road 6. University Blvd. 7. Hiawassee Road 8. Lake Avenue 9. Apopka-Ocoee Road 10. Kaley Avenue 11. Apopka-Vineland Road 12. Washington Street 13. Mercy Drive 14. Apopka Road 15. Old Dixie Highway 16. Old Apopka Road th

PW(10%) $1,606,431 $3,438,531 $2,682,758 $2,652,473 $1,672,473 $5,258,050 $4,130,824 $2,958,052 $552,475 $4,459,032 $1,166,557 $1,788,245 $5,066,566 $2,338,635 $1,213,846 $1,899,946

Investment $980,000 $3,500,000 $2,800,000 $1,400,000 $2,380,000 $5,040,000 $2,520,000 $4,900,000 $1,365,000 $2,100,000 $1,170,000 $1,120,000 $2,800,000 $1,690,000 $975,000 $1,462,500

(a) $6 million to each district: District I II III IV

Projects 1, 2, 4 5,7 9,10,11,12 13,14,16

NPW $7,697,488 $5,803,297 $5,966,309 $9,305,147

(b) $15 million to districts I & II and $9 million to districts III & IV: District I & II III & IV

Projects 2,4,5,6,7 10,12,13,14,15

NPW $17,152,400 $12,866,320

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Investment $14,840,000 $8,685,000

Contemporary Engineering Economics, 7th ed. ©2023

ST16.2 Given i = 8% , g = 10% , garbage amount/day = 300 tons (a) The operating cost of the current system in terms of $/ton of solid waste: • Annual garbage collection required (assuming 365 days): Total amount of garbage = 300 tons × 365 days = 109,500 tons • Equivalent annual operating and maintenance cost: PW(8%) = $905, 400( P / A1 ,10%,8%, 20) = $20, 071,500 AEC(8%) = $20, 071,500( A / P,8%, 20) = $2, 044,300 • Operating cost per ton:

cost per ton =

$2, 044,300 = $18.67 / ton 109,500

(b) The economics of each solid-waste disposal alternative in terms of $/ton: • Site 1:

AEC(8%)1 = $4, 053, 000( A / P,8%, 20) + $342, 000 − ($13, 200 + $87, 600) = $653, 000 $653, 000 cost per ton = = $5.96 / ton 109,500 • Site 2:

AEC(8%)2 = $4,384, 000( A / P,8%, 20) + $480, 000 − ($14, 700 + $99,300) = $812,520 $812,520 cost per ton = = $7.42 / ton 109,500 15 Copyright © 2023 Pearson Education, Inc.

Contemporary Engineering Economics, 7th ed. ©2023 • Site 3:

AEC(8%)3 = $4, 764, 000( A / P,8%, 20) + $414, 000 − ($15,300 + $103,500) = $780, 424 $780, 424 cost per ton = = $7.13 / ton 109,500 • Site 4:

AEC(8%)2 = $5, 454, 000( A / P,8%, 20) + $408, 000 − ($17,100 + $119, 400) = $827, 000 $827, 000 cost per ton = = $7.55 / ton 109,500 Site 1 is the most economical choice. ST 16.3 (a) Let’s define the following variables to compute the equivalent annual cost. Ala = initial land cost

Aeq = initial treatment equipment cost Ast = initial structure cost Apu = initial pumping equipment cost Aen = initial annual energy cost in today's dollars Alb = initial annual labor cost in today's dollars Aen = initial annual repair cost in today's dollars • Land: PW(10%)land = Ala − Ala (1.03 /1.1)120

= $0.99963 Ala • Equipment: Let’s define the following additional variables.

I15 n = replacement cost in year 15n S15 n = salvage value in year 15n C15 n = net replacement cost in year 15n where n = 1, 2,3, 4,5, 6, 7 and 8

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Contemporary Engineering Economics, 7th ed. ©2023 The total replacement cost over the analysis period is calculated as follows:

I15 = Aeq (1.05)15 = 2.07893 Aeq S15 = 0.5 Aeq C15 = (2.07893 − 0.5) Aeq = 1.57893 Aeq C15 n = (1.57893 Aeq )(1.05)15( n −1) S15 n = 0.5 Aeq (1.05)15( n −1) 7

PW(10%) equipment = Aeq +  C15 n − S120 n =1 7

= Aeq + 

(1.57893 Aeq )(1.05)15( n −1)

n =1

(1.1)15 n

−

0.5 Aeq (1.05)105 (1.1)120

= 1.74588 Aeq

• Structure:

 1 1  0.6 Ast PW(10%) structure = Ast + (0.40) Ast  + − 40 (1.1)80  (1.1)120  (1.1) = 1.00902 Ast • Pumping: PW(10%) pumping = 1.74588 Apu • Energy: 120

PW(10%)energy =  Aen (1.05 /1.1) j = 20.92097 Aen j =1

• Labor: 120

PW(10%)labor =  Alb (1.04 /1.1) j = 17.31264 Alb j =1

• Repair: 120

PW(10%) repair =  Are (1.02 / 1.1) j = 12.74852 Are j =1

• Present worth of the life-cycle cost:

PW(10%) = 0.99963 Ala + 1.74588 Aeq + 1.00902 Ast + 1.74588 Apu + 20.92097 Aen + 17.31264 Alb + 12.74852 Are

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Contemporary Engineering Economics, 7th ed. ©2023

Parameters

Ala Aeq Ast Apu

Aen Alb Are PW(10%) AEC(10%)

2 $2,400,000

Option 3 $49,000

4 $49,000

5 $400,000

$500,000

$500,000

$400,000

$175,000

$700,000

$2,100,000

$2,463,000

$1,750,000

$100,000

0

0

$100,000

$200,000

$125,000

$100,000

$50,000

$95,000

$65,000

$53,000

$37,000

$30,000

$20,000

$15,000

$5,000

$10,364,300 $1,036,440

$7,036,290 $703,637

$6,433,460 $643,353

$4,396,113 $439,616

Option 5 is the least cost alternative. (b) Cost / gallon = $439,616 / 2,000,000(365) = $0.0006 per gallon Monthly charge = (0.0006)(400)(30) = $7.23 per month ST 16.4 (a) Users benefits and disbenefits: • Users’ benefits (1) Reduced travel time. (2) Reduced fuel consumption. (3) Reduced air pollution. (4) Reduced number of accidents. • Users’ disbenefits: Increased automobile purchase and maintenance costs. (b) Sponsor’s cost • Development costs associated with computerized dashboard navigational systems roadside sensors, and automated steering and speed controls. • Implementation and maintenance costs. • Public promotional and educational costs.

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Contemporary Engineering Economics, 7th ed. ©2023 (c) On a national level, the sponsor’s costs are estimated to be as follows: • R&D costs = $3.5 billion • Implementation costs = $18 billion • Maintenance costs = $4 billion per year Comments: However, the users’ benefits are sketchy, except the level of reduction possible in travel time, fuel consumption, and air pollution. Ask the students to quantify these in dollar terms by consulting various government publications on public transportation. Once these figures are estimated, the benefit-cost ratio can be easily derived.

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