Some theoretical foundations of mathematical programming are expounded in the textbook: elements of convex analysis; con
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English Pages [207] Year 2011
AlFarabi Kazakh National University
S. A. Aisagaliev, Zh. Kh. Zhunussova
MATHEMATICAL PROGRAMMING Textbook
Almaty "Kаzаkh university" 2011 1
ББК 22. 1 А 36 Рекомендовано к изданию Ученым советом механикоматематического факультета и РИСО КазНУ им. альФараби (Протокол №1 от 19 октября 2010 г.) Reviewers: doctor of physical and mathematical sciences, professor M.Akhmet; candidate of philological sciences, professor A.A.Moldagalieva
Aisagaliev S.А., Zhunussova Zh.Kh. А 36 Mathematical programming: textbook. – Almaty: Kаzаkh university, 2011.  208 p. IBSN 9965296308 Some theoretical foundations of mathematical programming are expounded in the textbook: elements of convex analysis; convex, nonlinear, linear programming required for planning and production control for solution of the topical problems of the controlled processes in natural sciences, technology and economy. The tasks for independent work with concrete examples, brief theory and solution algorithms of the problems, term tasks on sections of the mathematical programming are put in the appendix. It is intended as a textbook for the students of the high schools training on specialties "applied mathematics", "mathematics", "mechanics", "economic cybernetics" and "informatics". It will be useful for the postgraduate students and scientific workers of the economic, mathematical, naturallytechnical and economic specialties. ББК 22. 1 ISBN 9965296308
© Aisagaliev S.А., Zhunussova Zh.Kh., 2011 © AlFarabi Kazakh National University, 2011
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FOREWORD The main sections of mathematical programming, the numerical methods of function minimization of the finite number variables are expounded in the textbook. It is written on the basis of the lectures on optimization methods which have been delivered by the authors in AlFarabi Kazakh National University. In connection with transition to credit technology of education the book is written in the manner of scholasticmethodical complex, containing alongside with lectures problems for independent work with solutions of the concrete examples, brief theory and algorithm on sections of the course, as well as term tasks for mastering of the main methods of the optimization problems solution. In the second half XX from necessity of practice appeared the new direction in mathematics  "Mathematical control theory" including the following sections: mathematical programming, optimal control with processes, theory of the extreme problems, differential and matrix games, controllability and observability theory, stochastic programming. Mathematical control theory was formed at period of the tempestuous development and creating of the new technology, spacecrafts, developing of the mathematical methods in economy, controlling by the different process in natural sciences. Aroused new problems could not be solved by classical methods of mathematics and required new approaches and theories. The different researchandproduction problems were solved due to mathematical control theory, in particular: production organizing to achieve maximum profit with provision for insufficiency resources, optimal control by nucleus and chemical reactors, electrical power and robotic systems, control by moving of the ballistic rockets, spacecrafts and satellites and others. Methods of the mathematical control theory were useful for developing mathematics. Classic boundary problems of the differential equations, problems of the best function approach, optimal choice of the parameters in the iterative processes, minimization of the difficulties with equations are reduced to studying of the extreme problems. Theory foundation and solution algorithms of the convex, nonlinear and linear programming are expounded on the lectures 117. Execution of the three term tasks for individual work of the students is provided for these sections. Solution methods of the extreme problems are related to one of the high developing direction of mathematics. That is why to make a textbook possessed by completion and without any shortage is very difficult. Authors will be grateful for critical notations concerning the textbook. 3
INTRODUCTION
Lecture 1 THE MAIN DETERMINATIONS. STATEMENT OF THE PROBLEM Let E n be the euclidean space of vectors u ( u1 ,u2 ,...,un ) and a scalar function J ( u ) J ( u1 ,...,un ) which is determined on a set U of the space E n . It is necessary to find the maximum ( minimum) of the function J u on set U, where U E n is the set. Production problem. The products of the five types are made on the enterprise. The cost of the unit product of the each type accordingly c1 , c2 , c3 , c4 , c5 , in particular c1 3, c2 5, c4 1,
c5 8 . For fabrication specified products the enterprise has some determined recourses expressed by the following normative data: Type of the product
Materials
Energy
Labor expenses
Recourses
1
a11 3
a21 6
a31 1
b1 50
2
a12 1
a22 3
a32 3
b2 120
3
a13 4
a23 1
a33 1
b3 20
4
a14 2
a 24 4
a34 2
5
a15 1
a25 5
a35 4
4
Here a11 is an amount of the materials required for fabricating the units of the 1st type products; a12 is a consumption of the material for fabrication of the 2nd type unit products and so on; aij ,
i 1, 2, 3,
j 1, 2, 3, 4, 5 is an expenses of the material, energy, labor expenses for fabrication of the j type unit products. It is required to find a such production plan output such that provides the maximal profit. So as the profit is proportional to total cost of marketed commodities, that hereinafter it is identified with total cost of marketed commodities. Let u1 , u 2 , u3 , u 4 , u5 be five products. The mathematical formalization of the problem function maximizations profit has the form
J (u ) J (u1 , u 2 , u 3 , u 4 , u 5 ) c1u1 c 2 u 2 c3 u 3 c 4 u 4 c5 u 5 3u1 5u 2 10u3 u 4 8u5 max (1) under the conditions (restrictions of the resources)
g1 (u ) a11u1 a12 u 2 a13u 3 a14 u 4 a15 u 5 b1 , g 2 (u ) a 21u1 a 22 u 2 a 23u 3 a 24 u 4 a 25 u 5 b2 , g 3 (u ) a31u1 a32 u 2 a33u 3 a34 u 4 a35 u 5 b3 , u1 0, u 2 0, u3 0, u 4 0, u5 0,
(2)
(3)
where aij , i 1,3, j 1,5 are normative coefficients which values presented above; b1 , b2 , b3 are recourses of the enterprise. Since amount of the products are nonnegative numbers that necessary the condition (3). If we introduce the notation
a11 a12 A a21 a22 a 31 a32
a13 a23 a33 5
a14 a24 a34
a15 a25 ; a35
c1 c2 c c3 ; c4 c5
u1 b1 u2 u u3 ; b b2 ; b u4 3 u 5
g1 g g 2 , g 3
then optimization problem (1) – (3) can be written as
J (u ) c ' u max;
(1' )
g (u ) Au b;
( 2' )
u 0,
(3' )
where c' (c1 , c 2 , c3 , c 4 , c5 ) is a vectorline; ' is a sign of transposition. Let a vectorfunction be g (u ) g (u ) b Au b , but a set
U 0 u E 5 / u 0 . Then problem (1) – (3) is written as: J (u ) c ' u max;
(4)
u U u E 5 u U 0 , g (u ) 0 E 5 .
(5)
The problem (1) – (3) (or (1' ) – (3' ) , or (4), (5)) belongs to type so called problem of the linear programming, since J (u ) is linear with respect to u function; the vectorfunction g (u ) also is linear with respect to u . In the case above vector u has five components, vectorfunction g (u ) – three and matrix A has an order 3 5 was considered. In the case vector u has dimensionality n , g (u ) – m a measured function, but matrix A has an order m n and there are restrictions of the equality type (for instance g1 (u ) A1u b1 , where A1 – a 6
m matrix of the order m1 n , b1 E 1 ) problem (4), (5) is written:
J u c' u max;
u U ,
(6)
n
(7)
U u E u U 0 , g ( u ) Au b 0 , g1( u ) A1u b1 0 ,
where set U 0 u E n u 0 , c E n . Problem (6), (7) belongs to type of the general problem of the linear programming. We suppose that J (u ) is convex function determined on the convex set U 0 (unnecessary linear); g (u ) is a convex function determined on the convex set U 0 . Now problem (6), (7) is written as J ( u ) max;
(8)
u U u E n u U 0 , g ( u ) 0 , g1( u ) A1u b1 0 .
(9)
Problem (8), (9) belongs to type so called problem of the convex programming. Let J (u ) , g (u ) , g1( u ) be arbitrary functions determined on the convex set U 0 . In this case, problem (8), (9) can be written as
J (u ) max;
(10)
u U u E n u U 0 , g (u ) 0, g 1 (u ) 0.
(11)
Optimization problem (10), (11) belongs to type so called problem of the nonlinear programming. In all presented problems of the linear, convex and nonlinear programming are specified the concrete ways of the prescription set U from E n . If it is distracted from concrete way of the prescription set U from E n , so optimization problem in finitedimensional space possible to write as
J (u ) max; u U , 7
U En.
Finally, we note that problem of the function maximization J (u ) on a set U tantamount to the problem of the function minimization – J (u ) on set U. So further it is possible to limit by consideration of the problem J (u ) min; u U , U E n , (12) for instance, in the problem (1)  (3) instead of maximization J (u ) minimization of the function 3u1 5u 2 10u3 u 4 8u5 min . Finally, the first part of the course "Methods of the optimization" is devoted to the solution methods of the convex, nonlinear, linear programming. The question arises: whether the problem statement (12) correct in general case? It is necessary the following definitions from mathematical analysis for answer. Definition 1. The point u* U called the minimum point of the function J (u ) on a set U if inequality J (u* ) J (u ) is executed under all u U . The value J (u* ) is called the least or minimum value of the function J (u ) on set U.
The set U * u* U J (u* ) min J (u ) contains all minimum uU
points of the function J (u ) on set U. It follows from the definition, that the global (or absolute) minimum of the function J (u ) on the set U is reached on the set
U * U . We remind that a point u** U is called the local minimum point of the function J (u ) , if inequality J (u** ) J (u ) is valid under all u o(u** , ) U , where set
o(u** , ) u E n  u u**  E n  an open sphere with the centre in u** and radius 0 ;  a  Euclidean norm of the vector a a1 ,..., a n E n , i.e.  a 
n
a i 1
8
2 i
.
, and U u E 1 1 2 u 1. u Then set U * 2 3 ; b) U u E 1 1 4 u 2 , then set Example 1. Let J (u ) cos 2
/ 2 u ,
U * 2 / 7, 2 / 5, 2 / 3, 2 ; c) U u E 1 then set U * , where  empty set. Example 2. The function J (u ) ln u , set U u E 1 0 u 1. The set U * . Example 3. The function J (u ) J 1 (u ) c, c const , where function
 u a , если u a; J 1 (u ) u b, если u b; c, если a u b, and set U E 1 . The set U *
a, b .
Definition 2. It is spoken, that function J (u ) is bounded below on set U if the number M such that J (u ) M under all u U exists. The function J (u ) is not bounded below on set U if the
sequence u k U such that lim J (u k ) exists. k
Function J (u ) is bounded below on set U in the examples 1, 3, but function J (u ) is not bounded on U in the example 2. Definition 3. We say, that function J (u ) is bounded from below
on set U . Then value J * inf J u is called the lower bound of the uU
function J (u ) on set U, if: 1) J * J (u ) under all u U ; 2) for arbitrary small number 0 is found the point u ( ) U such that value J u J * . When the function J (u ) is not bounded
from below on set U, lower bound J * . We notice, that for example 1 value J * 0 , but for examples 2, 3 values J * , J * c , accordingly. If set U * , so 9
J * min J u (refer to examples). Value J * always exists, but uU
min J u does not always exist. Since lower bound J * for function uU
J (u ) determined on set U always exists, independently of that, whether set U * is emptily or in emptily, problem (12) can be written in the manner of
J u inf,
u U , U E n .
(13)
Since value min J u does not exist, when set U * , that uU
correct form of the optimization problem in finitedimensional space has the form of (13). Definition 4. The sequence u k U is called minimizing for function J (u ) determined on set U, if limit lim J u k inf J u J * . k
uU
As follows from definition lower bound, in the case k 1 k ,
k 1,2,3... we have the sequences
uk U
u k
k 1,2,..., for which J u k J * 1 / k
u 1 / k
Thence follows that limit lim J u k J * . Finally, minimizing k
sequence u k U always exists.
Definition 5. It is spoken, that sequence u k U converges to
set
U * U , if limit lim u k , U * 0 , where u k ,U * k
inf  u k u*   a distance from point u k U till set U * . u*U *
It is necessary to note, if set U * , so the minimizing sequence always exists which converges to set U * . However statement about any minimizing sequence converges to set U * , in general case, untrue. Example 4. Let function J u u 4 (1 u 6 ) , U E 1 .
For the example set U * 0 , but the sequences u k 1 / k ,
k 1,2,... U , u k k , k 1,2,... U are minimizing. The first 10
sequence converges to set U * , the second infinitely is distanced from it. Finally, source optimization problem has the form of (13). We consider the following its solutions: 1st problem. Find the value J * inf J u . In this case uU
independently of that whether set U * is emptily or in emptily, problem (13) has a solution. 2nd problem. Find the value J * inf J u and the point uU
u* U * . In order to the problem (13) has a solution necessary the set U* . 3rd problem. Find the minimizing sequence u k U which converges to set U * . In this case necessary that set U * . Most often in practice it is required solution of the 2nd problem (refer to production problem).
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Lecture 2 WEIERSTRASS’S THEOREM We consider the optimization problem
J u inf,
u U , U E n .
(1)
It is necessary to find the point u* U * and value J * inf J u . We notice, if set U * , so J * J u* min J u .
uU
uU
The question arises: what requirements are imposed on the function J (u ) and on set U that the set U * u * U J (u * ) min J (u ) be in
uU
emptily? For answer it is necessary to enter the notions of compact set and halfcontinuously from below of the function J (u ) on set U.
The compact sets. Let u k E n be a certain sequence. We
remind that: a) the point v E n is called the limiting point of the sequence u k , if subsequence u k m exists for which limit
lim u km v ; b) sequence u k E n is identified bounded , if the
m
number M 0 exists, such the norm  u k  M for all k 1,2,3,... ; b) set U E n is identified bounded , if the number R 0 exists, such that norm  u  R under all u U ; c) the point v E n is identified the limiting point of the set U, if any its  set neighborhood ov, contains the points from U differenced from v ; d) for any limiting point v of the set U is found the sequence u k U for which lim u k v ; e) set U E n is identified k
closed, if it contains all their own limiting points. 12
Definition 1. The set U E n is identified compact, if any sequence u k U has at least one limiting point v, moreover v U . It is easy make sure the definition is equally to known from the course of the mathematical analysis statement about any bounded and closed set is compactly. In fact, according to Bolzano and Weierstrass theorem any bounded sequence has at least one limiting point (the set U is bounded ), but from inclusion v U follows the property of being closed of the set U . Halfcontinuously from below. Let u k E n be a sequence.
J k J uk is the number sequence. We notice that: a) numerical set J k is bounded from below, if the number exists, such that J k , k 1,2,3,... ; b) numerical set J k is not bounded
Then
from below, if J km subsequence exists such the limit lim J km . m
Definition 2. By the lower limit of the bounded from below numeric sequence J k is identified the value а denoted a lim J k if: 1) m
subsequence J km exists for which lim J km a ; 2) all other limiting m
points to sequences J k is not less the value а. If numeric sequence
J k is not bounded from below, so the value a . k Example 1. Let J k 1 1 , k 0,1,2,... . The value a 0 . Definition 3. It is spoken, the function J u determined on set
U E n , halfcontinuous from below in the point u U , if for any sequence u k U for which the limit lim u k u , the inequality k
lim J u k J u is executed. The function J (u ) is halfcontinuous k
from below on set U, if it is halfcontinuous from below in each point of the set U. Example 2. Let set U u E 1 1 u 1 , function J u
under 0  u  1, J 0 1 . The function J (u ) is uncontinuous on set 0  u  1 , consequently, it is halfcontinuous
u
2
13
from below on the set. We show, that function J (u ) is halfcontinuous from below in the point u 0 . In fact, the sequence 1/ k, k 1,2,... . belongs to the set U and the limit of the sequence
is equal to zero. Numeric sequence J u k 1 / k 2 , moreover limit
lim J u k lim J u k 0 . Consequently, lim J u k 0 1 . It k
k
k
means function J (u ) is halfcontinuous from below on set U . Similarly possible to enter the notion of the halfcontinuity from above of the function J (u ) on set U . We notice, if the function J (u ) is uncontinuous in the point u U , so it is halfcontinuous in it as from below, so and from above. The most suitable check way of the halfcontinuity from below of the function J (u ) on set U gives the following lemma. Lemma. Let function J (u ) be determined on closed set U E n . In order the function J (u ) to be halfcontinuous from below on set necessary and sufficiently that Lebesgue’s set U, n 1 M (с) u E J (u ) c be closed under all c E . Proof. Necessity. Let function J (u ) be halfcontinuous from below
on closed set U . We show, that set M (c ) is closed under all c E 1 . We notice, that empty set is considered as closed. Let v be any limiting point of the set M (c ) . From definition of the limiting point follows the existence of the sequence u k M (c ) which converges to the point v .
From inclusion
u k M (c)
follows that value
J (u k ) c,
k 1,2,... . With consideration of u k U and halfcontinuity from below J (u ) on U follows J v lim J u k c . Consequently, the k
point v M c . The property of being closed of the set M (c ) is proved. Sufficient. Let U be closed set, but set M (c ) is closed under all c E 1 . We show, that function J (u ) is halfcontinuous from
below on U. Let u k U  a sequence which converges to the 14
point u U . We consider the numeric sequence J u k . Let the
value a lim J u k . By definition of the below limit exists the k
for which lim J u a . Consequently, for any
sequence J u km
km
m
sufficiently small 0 is found the number N N , such that
J u km a under
m N . Thence with consideration of
lim u km u , u k m M a , we have J u a . Then with
m
consideration of arbitrarily 0 it is possible to write the following inequality: J u lim J u k a , i.e. function J (u ) is halfk
continuous from below in the point u U . Since set U is closed, function J (u ) is halfcontinuous from below in any point u U . Lemma is proved. We notice, in particular, that under c J * from lemma follows
that set M ( J * ) u E n u U , J (u ) J * U * is closed. Theorem 1. Let function J (u ) be determined, finite and halfcontinuous
from
below J * inf J u , set
on
compact
set
U En .
Then
uU
U * u* E n u* U , J (u* ) min J (u ) uU
is inempty, compactly and any minimizing sequence converges to set U* .
Proof. Let u k U be any minimizing sequence, i.e. the limit
of the numeric set lim J u k J * . We notice, that such minimizing k
sequence always exists. Let u*  any limiting point of the
minimizing sequence. Consequently, subsequence u km U exists for which lim u km u* . With consideration of the compactness of m
the set U all limiting points of the minimizing sequence belongs to 15
the set U . As follows from definition of lower bound and halfcontinuity from below function J (u ) on set U the following inequalities are faithful
J * J u* lim J u km lim J u k J * . m
(2)
k
Since the sequence J u k converges to value J * , that any its subsequence also converges to value J * . We have that value
J * J u* from inequality (2). Consequently, set U * and J * J u* . Since the statement faithfully for any limiting
point of the minimizing sequence, so it is possible to confirm that any minimizing sequence from U converges to set U * .
We show that set U * U is compact. Let wk  any sequence
taking from set U * . From inclusion wk U follows that wk U
. Then with consideration of compactness set U subsequence wkm
wk U ,
exists which converges to a point w* U . Since
so
J ( wk ) J * , k 1,2, ... . Consequently, the sequence wk U is minimizing. Then, with consideration of proved above,
values
limiting point to this sequences w* U * . Finally, closeness of the set
U * is proved. Restriction of the set U * follows from inclusion U * U . Compactness of the set U * is proved. Theorem is proved. Set U is not often bounded in the applied problems. In such cases the following theorems are useful. Theorem 2. Let function J (u ) be determined, finite and halfcontinuous from below on inempty closed set U E n . Let for certain point v U Lebesgue’s set
M (v ) u E n u U , J (u ) J (v ) 16
is bounded . Then J * , set U * is inempty, compact and any
minimizing sequence u k M (v) converges to set U * .
Proof. Since all condition of the lemma are executed, set M (v ) is closed. From restrictedness and closeness of the set M (v ) follows
its compactness. The set U M (v) M 1 (v) , where set M 1 (v)
u E n u U , J (u ) J (v) , moreover on set M 1 (v) function J (u ) does not reach its lower bound J * . Hereinafter proof of the theorem 1 is repeated with change set U on compact set M (v ) . Theorem is proved. Theorem 3. Let function J (u ) be determined, finite and halfcontinuous from below on inempty closed set U E n . Let for any sequence vk U , vk , under k the equality
lim J (v k ) is executed. Then J * , set U * is inempty, k
compact and any minimizing sequence u k U converges to set U * . Proof. Since lim J (v k ) , so the point v U k
such that
J (v) J * exists. We enter Lebesgue’s set M (v) u E n u U ,
J (u ) J (v ) . With consideration of lemma set M (v ) is closed. It is easy to show that set M (v ) is bounded. In fact, if set M (v ) is not
bounded, then the sequence wk M (v) , such that wk under
k exists. By condition of the theorem for such sequences the value J ( wk ) under k . Since J ( wk ) J (v) it is not possible. Finally, set M (v ) is bounded and closed, consequently, it is compact. Hereinafter proof of the theorem 1 is repeated for set M (v ) . Theorem is proved.
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Chapter I. CONVEX ROGRAMMING. ELEMENTS OF THE CONVEX ANALYSIS
Amongst methods of the optimization problems solution in finitedimensional space the most completed nature has a method of the problem solution of the convex programming. In the convex analysis developing intensive last years characteristic of the convex sets and functions are studied which allow generalizing the known methods of the problems solution on conditional extreme.
Lecture 3 CONVEX SETS In the applied researches often meet the problems of the convex programming in the following type:
J (u ) inf,
u U u E n u U 0 , g i (u ) 0, i 1, m,
(1)
g i (u ) a i , u bi 0, i m 1, s , where J (u ) , g i (u ) , i 1, s ,  convex functions determined on convex set U 0 E n . 18
Definition 1. Set U E n called by convex, if for any u U , v U and under all , 0 1 , the point u u (1 )v
v (u v) U . Example1. We show that closed sphere
S (u 0 , R ) u E n u u 0 R
Is a convex set. Let the points u S (u 0 , R ) , S (u 0 , R ) , i. e.
norms u u 0 R , v u 0 R . We take a number , 0, 1 , and define the point u u (1 )v . Norm
u u 0 u (1 )v u 0 (u u 0 ) (1 )(v u 0 ) u u 0 (1 ) v u 0 R (1 ) R R. Consequently, the point u S (u 0 , R ) , and set S (u 0 , R ) is convex. Example2. We show that hyperplane
u E n c, u
is a convex set, where с E n is the vector, is the number. Let the points u , v . Consequently, scalar products c, u ,
c, v . Let u u (1 )v , 0, 1 . Then scalar product
c, u c, u (1 )v c, u (1 ) c, v (1 ) . Thence follows that the point u and set is convex.
Example 3. We show that affine set М u E n Au b ,
where A is a constant matrix of the order m n , b E
19
n
is the
vector which is convex. For the point u u (1 )v , u М ,
v М , 0, 1 we have
Au A(u (1 )v) Au (1 ) Av b (1 )b b Thence follows that u М and set М is convex. Let u 1 , u 2 , …, u n r be linear independent solutions of the linear homogeneous system Au 0 . Then set M can be presented in the manner of
nr M u E n u u 0 v, v L , L u E n v i u i i 1
where vector u 0 E n is a partial solution of the nonhomogeneous
system Au b , L is a subspace to dimensionality n r formed by the vectors u 1 , u 2 , …, u n r , i , i 1, n r are the numbers and dimension of the affine set M is taken equal to the dimension of the space L . Definition 2. By affine cover of the arbitrary set U E n called the intersection of all affine sets containing set U and it is denoted by aff U . Dimensionality of the set U is called the dimension its affine cover and it is denoted by dim U . Since the intersection of any number convex sets is convex set, so the set aff U is convex. Instead of source problem J (u ) inf , u U , where U is an arbitrary set it is considered the approximate problem J (u ) inf , u aff U , since solution of the last problem in the many cases more simpler, than source. Definition 3. The set A called by the sum of the sets A1 , A2 ,..., Am , i.e. A A1 A2 ... Am if it contains that and only m
that points a ai , ai Ai , i 1, m. Set A is called by the set i 1
difference В and С i.e A B C , if it contains that and only that 20
points a b c, b B, c C . The set A D , where is a real number, if it contains the points a d , d D. Theorem 1. If the sets A1 , A2 ,..., Am , B, C , D are convex, so
the sets A A1 A2 ... Am , A B C , A D are convex. m
m
i 1
i 1
Proof. Let the points be a ai A, e ei A,
ai , ei Ai , i 1, m.
Then
the
point
u a 1 e
m
ai 1 ei , 0,1 . Since the sets Ai , i 1, m are i 1
convex, so ai 1 ei Ai , i 1, m . Consequently, the point m
a ai , ai ai 1 ei Ai , i 1, m. Thence follows i 1
that point a A . Convexity of the set A is proved. We show that set A B C is convex. Let the points a b c A, e b1 c1 A , where b, b1 B, c, c1 C . The point
a a 1 e b c 1 b1 c1 b
1 b1 c 1 c1 , 0,1 . Since the sets B and C c c are convex, the points b b 1 b1 B, 1 c1 C . Then the point a b c A, b B, c C . It is followed that set A is convex. From a d1 A, e d 2 A, d1 , d 2 D follows that a a 1 e d1 1 d 2 ,
d1 1 d 2 , 0,1 . Then a d A, d d 1 1 d 2 D Theorem is proved. Let U be the set from E n , v E n is a certain point. There is
one and only one of the following possibilities: a) The number 0 is found such that set o ( , ) U . In this case point is an internal point of the set U . We denote through int U the ensemble of all internal points of the set U. 21
b) Set o ( , ) does not contain nor one point of the set U . The point is called by external with respect to U . c) Set o ( , ) contains both points from set U , and points from set E n \ U . The point is called by bound point of the set U. The set of all border points of the set U is denoted by p U . d) The point U , but set o ( , ) does not contain nor one point from set U , except the points . The point is called the isolated point of the set U . Convex set does not contain the isolated points. Definition 4. The point U called comparatively internal point of the set U if the intersection o( , ) aff U U . We denote through riU the ensemble of all comparatively internal points of the set. Example 4. Let set
U u E 1 0 u 1; 2 u 3 E 1 . For the example the sets
int U u E 1 0 u 1; 2 u 3 . affU u E 1 0 u 3, riU u E 1 0 u 1; 2 u 3. the following statements are faithful: 1) If А is a convex set, so closing A is convex too. In fact, if a, b A , so subsequences ak A, bk A, such that ak a,
bk b under k exists. With consideration of convexity of the set A the point a k 1 bk A, k 1,2,3,... . Then limiting point
a lim a k 1 bk a 1 b A k
under all
0,1 . Thence follows the convexity of the set A . 2) If U is a convex set and int U , so the point
v v u 0 v int U , u 0 inf U , v U , 0 1 . 22
To prove. 3) If U is a convex set, so int U is convex too. To prove. 4) If U is a convex inempty set, so ri U and ri U is convex. To prove. 5) If U is a convex set and ri U , so the point
v v v v u 0 v riU , u 0 riU , v U , 0 1 To prove. Definition 5. The point u E n called the convex combination of the points u 1 , u 2 ,..., u m from E n , if it is represented in the manner of u
m
u , i 1
i
i
where the numbers i 0, i 1, m but their sum
1 2 ... m 1 . Theorem 2. The set U is convex if and only if, when it contains all convex combinations of any finite number their own points. Proof. Necessity. Let U be a convex set. We show, that it contains all convex combinations of the finite number of their own points. We use the method of induction to proof of the theorem. From definition of the convex set follows that statement faithfully for any two points from U. We suppose, that set U contains the convex combinations m 1 of their own points, i.e. the point m 1
v i u i U , i 0, i 1, m 1, 1 ... m1 1 . We prove, i 1
that it contains the convex combinations т their own points. In fact, m 1
expression u 1u 1 ... m u m 1 m
i
1 i 1
u i mu m . m
We denote i i / 1 m . We notice, that i 0 , i 1, m 1 , but sum 1 ... m 1 1, since 1 ... m1 1 m , i 0,
1 2 ... m 1 . Then the point u 1 m v m u m ,
v U , u m U . Thence with consideration of the set convexity U the point u U . Necessity is proved. 23
Sufficiency. Let the set U contains all convex combinations of any finite number its own points. We show that U  a convex set. In particular, for m 2 we have u u 1 1 u 2 U under any
u 1 , u 2 U under all 0,1 . The inclusion means that U  a
convex set. Theorem is proved. Definition 6. By convex cover of arbitrary set U E n called the intersection of all convex sets containing set U and it is denoted by Co U . From given definition follows that CoU is least (on remoteness from set U ) convex set containing the set U . We notice, that source problem J (u ) inf , u U , with arbitrary set U E n can be replaced on the approximate problem J (u ) inf , u Co U . We note, that if U is closed and bounded(compact) set, that CoU is also boundedand closed (compactly). Theorem 3. The set CoU contains that and only that points which are convex combination of the finite number points from U . Proof. To proof of theorem it is enough to show that CoU =W, where W is set of the points which are convex combination of any finite number points from set U . The inclusion W Co U follows from theorem 2, since W Co U and set CoU is convex. On the other
hand,
if
the
u, v W ,
points
i.е.
m
u i u i , i 1
i 0, i 1, m;
m
p
i 1
i 1
i 1; v i v i , i 0, i 1, p,
p
i 1
i
1,
m
that under all 0,1 the point u u 1 v i u i i 1
p
v i 1
i
i
, where i i 0, i 1, m, i 1 i 0, i 1, p p
m
moreover the sum
i 1
i
i 1
i
1 . Thence follows that u W ,
consequently, the set W is convex. Since set U W , so the 24
inclusion Co U W exists. From inclusion W Co U , Co U W follows that Co U W . Theorem is proved. Theorem 4. Any point u Co U can be presented as convex combination no more than n 1 points from U . Proof. As follows from theorem 3, the point u Co U is m
represented in the manner of
u , i 1
m
i 0,
i
i
i 1
i
1 . We
suppose, that number m n 1 . Then n 1 dimensional vectors
ui u i ,1 , i 1, m, m n 1 are linearly dependent. Consequently, there are the numbers 1 ,..., m , not all equal to zero, such that sum m
m
i ui 0 . Thence follows, that u
i 1
i 1
first equality, the point u m
m
m
i 1
i 1
i 1
m
i 1
i
0,
m
i 1
i
0 . Using
we present in the manner of
u i u i i u i t i u i where i i t i 0,
i
i
m
m
i 1
i 1
i t i u i i u i ,
1 under enough small t . Let
i* min i , amongst i 0 , where index i, 1 i m . We choose the number m
i 1
i
t from the condition i* / i* t . Since
0 , that such i* 0 always exists. Now the point m
m
i 1
i 1 i i*
u i u i i u i , i 0,
m
i 1 i i*
i
1.
(2)
Specified technique are used for any m n 1 . Iterating the given process, eventually get m n 1 . Theorem is proved. Definition 7. The convex cover drawing on the points 0 u , u1 ,..., u m from E n is called m dimensional simplex, if vectors 25
u
1
u 0 , i 1, m are linear independent and it is denoted by S m .
The points u 0 , u1 ,..., u m are called the tops of the simplex. Set S m is a convex polyhedron with dimension m and by theorem 3 it is represented in the manner of m S m u E n u i u i , i 0, i 0, m, i 1
26
m
i 0
i
1.
Lecture 4 CONVEX FUNCTIONS
The convex programming problem in general type is formulated such: J u inf, u U , U E n , where U  convex set from
E n ; J (u )  a convex function determined on convex set U . Definition 1. Let function J (u ) be determined on convex set U from E n . Function J (u ) is called convex on set U, if for any points u , v U and under all , 0 1 is executed the inequality
J u 1 v J u 1 J v .
(1)
If in the inequality (1) an equality possible under only 0 and 1 , then function J (u ) is called strictly convex on convex set U . Function J (u ) is concave (strongly concave) if function J (u ) concave (strongly concave) on set U . Definition 2. Let function J (u ) be determined on convex set U. Function J (u ) is called strongly convex on set U if the constant 0 exists, such that for any points u , v U and under all , 0 1 is executed the inequality
J u 1 v J u 1 J v 1  u v 2 . (2) Example 1. Let function J u c, u be determined on convex
set U from E n . We show that J u c, u  a convex function on
U . In fact, u , v U are arbitrary points, number 0,1 , then
27
the point u u 1 v U on the strength of convexity of the set U . Then
J u c, u c, u 1 c, v J u 1 J v . In this case relationship (1) is executed with sign of the equality. The symbol .,.  a scalar product. Example 2. Let function J u  u 2 be determined on convex
set U E n . We show that J (u )  strongly convex function on set U. In fact, for u , v U
and under all 0,1 the point
u u 1 v U . The value
J u u 1 v u 1 v, u 1 v 2
2  u 2 2 1 u, v 1  v 2 . (3) 2
The scalar product
u v, u v  u v  2  u  2 2 u, v  v 2 . 2 2 2 Thence we have 2 u, v  u   v   u v  . Having substituted
the equality to the right part of expression (3), we get
J u  u 2 1  v 2 1  u v 2
J u 1 J v 1  u v 2 .
Finitely, relationship (2) is executed with sign of the equality, moreover number 1 . Example 3. The function J u  u  2 determined on convex set
U E n is strongly convex. In fact, value 28
J u u 1 v  u  2 1  v  2 J u 1 J v , 2
moreover equality possible under only 0 and 1 . In general case check of convexity or strong convexity of the function J (u ) on convex set U by the definitions 1, 2 rather complex. In such events the following theorems are useful. The criteria to convexity of the even functions. Let C 1 (U ),
C 2 (U ) are accordingly spaces of the continuously differentiated and twice continuously differentiated functions J (u ) , determined on set U . We notice, that gradient
J ' u J u / u1 ,..., J u / u n E n under any fixed u U . Theorem 1. In order the function J u C 1 (U ) to be convex on the convex set U E n , necessary and sufficiently executing of the inequality
J u J v J ' v , u v ,
u, v U .
(4)
Proof. Necessity. Let function J u C 1 (U ) be convex on U. We show, that inequality (4) is executed. From inequality (1) we have
J v u v J v J u J v ,
0,1,
u, v U .
Thence on base of the formula of the finite increments we get
J ' v u v , u v J u J v ,
0 1.
Both parts of the given inequality are divided into 0 and turn to the limit under 0 , with consideration of 1 J u C (U ) we get the inequality (4). Necessity is proved. 29
Sufficiency. Let for function J u C 1 (U ) , U be a convex set inequality (4) is executed. We show, that J (u ) is convex on U.
Since the set U is convex, the point u u 1 v U ,
u , v U , 0,1 . Then from inequality (4) follows, that
J u J u J ' u , u u , J v J u J ' u , v u ,
u , u , v U . We multiply the first inequality on , but the second inequality  on the number 1 and add them. As a result we give J u 1 J v J u . Thence follows the convexity to functions J (u ) on set U. Theorem is proved. Theorem 2. In order the function J (u ) C 1 (U ) to be convex on convex set U necessary and sufficiently executing of the inequality
J ' u J ' v , u v 0, u, v U .
(5)
Proof. Necessity. Let the function J (u ) C 1 (U ) be convex on convex set U. We show, that inequality (5) is executed. Since inequality (4) faithfully for any u , v U , that, in particular, we have
J v J u J ' u , v u . We add the inequality with inequality (4),
as a result we get the inequality (5). Necessity is proved. Sufficiency. Let for the function J (u ) C 1 (U ) , U be a convex set, inequality (5) is executed. We show, that J (u ) is convex on U . For proving it is enough to show, that difference
J u 1 J v J u J u 1 J v J u 1 v 0, u, v U . Since J (u ) C 1 (U ) , the following inequality faithfully: 1
J u h J u J ' u 1h , h J ' u th , h dt , 0
u , 30
u h U .
The first equality follows from formula of the finite increments, but the second  from theorem about average value, 0 1 1 Then difference
J u 1 J u J u J u J u 1
1 J v J u J ' u t u u , u u dt 0
1
1 J ' u t v u , v u dt. 0
Let
z1 u t u u u t 1 u v ,
z 2 u
z1 z 2 t u v , but t v u u t v u . Then differences u u 1 z1 z 2 / t , v u z1 z 2 / t . Now previous inequality is written: 1
J u 1 J v J u 1 J ' z1 J ' z 2 z1 z 2 0
1 dt. t
Since the points z1 , z 2 U , that according to inequality (5) we
have J u 1 J v J u 0 . Theorem is proved. Theorem 3. In order the function J (u ) C 2 (U ) to be convex on convex set U , int U , necessary and sufficiency executing of the inequality
J " (u ) , 0,
E n , u U .
(6)
Proof. Necessity. Let function J (u ) C 2 (U ) be convex on U . We show, that inequality (6) is executed. If the point u int U , then
the number 0 0 such that points u U for any E n and 31
under all ,   0 is found. Since all conditions of theorem 2 are executed, that inequality faithfully
J ' (u ) J ' (u ), 0,
E n ,   0 .
Thence, considering that
J ' (u ) J ' (u ), J " (u ) , 2 ,
0 1,
and transferring to limit under 0 , we get the inequality (6). If u U is a border point, then the following sequence u k int U exists, such that u k 0 under k 0 . Hereinafter, by proved
J " (u k ) , 0, E n , u k inf U . Transferring
to
limit and taking into account that 2 lim J " (u k ) J " (u ) with considering of J (u ) C (U ) , we get k
the inequality (6). Necessity is proved. Sufficiency. Let for function J (u ) C 2 (U ) , U be a convex set, int U , inequality (6) is executed. We show, that J (u ) is convex on U . So as the equality faithfully
J ' u J ' v , u v J " v u v u v, u v ,
0 1,
that, denoting by u v and considering that u v u
v U , we get
J ' u J ' v , u v J " u , 0,
E n ,
u U .
Thence with considering of theorem 2, follows the convexity to function J (u ) on U . Theorem is proved. We note, that symmetrical matrix
32
2 J u / u12 2 J u / u1u 2 2 2 J u / u 22 J u / u 2 u1 J " u . . . . . . . . . . . . . . . . 2 J u / u u 2 J u / u u n n 1 2 a scalar product. J " u , , ' J " u .
... 2 J u / u1u n ... 2 J u / u 2 u n – . . . . . . . . 2 J u / u n2 ...
The theorems 1  3 for strongly convex function J (u ) on U are formulated in the manner specified below and its are proved by the similar way. Theorem 4. In order the function J (u ) C 1 (U ) to be strongly convex on the convex set U necessary and sufficiently executing of the inequality
J u J v J ' v , u v k  u v 2 , u, v U .
(7)
Theorem 5. In order the function J (u ) C 1 (U ) to be strongly convex on the convex set U, necessary and sufficiently executing of the inequality
J ' u J ' v , u v  u v  2 ,
k const 0, u, v U .
(8)
Theorem 6. In order the function J (u ) C 2 (U ) to be strongly convex on the convex set U , int U necessary and sufficiently executing of the inequality
J " u ,   2 , E n , u U ,
k const 0.
(9)
The formulas (4)  (9) can be applied for defining of the convexity and strong convexity of the even functions J (u ) determined on the convex set U E n . For studying of convergence of the successive approximation methods are useful the following lemma. 33
Definition 3. It is spoken that gradient J ' u to function
J (u ) C 1 (U ) satisfies to Lipschitz’s condition on the set U, if
 J ' u J ' v  L  u v , u, v U , L const 0.
(10)
Space of such functions is denoted by C 1,1 U .
Lemma. If function J u C 1,1 U , U inequality faithfully
be convex set, that
 J u J v J ' v , u v  L  u v 2 / 2,
(11)
Proof. From equality 1
J u J v J ' v , u v J ' v t u v J ' v , u v dt 0
follows that 1
J u J v J ' v , u v J ' v t u v J ' v u v dt. 0
Thence with consideration of inequality (10) and after integration on t we get the formula (11). Lemma is proved. The properties of the convex functions. The following statements are faithful. 1) If J i u , i 1, m are convex functions on the convex set, then function J u
m
J u , i 1
i
i
i
0, i 1, m is convex on set U .
In fact, m
m
J u i J i u i J i u 1 J i v i 1
i 1
m
m
i 1
i 1
i J i u 1 i J i v J u 1 J v , u, v U . 34
2) If J i u , i I is a certain family of the convex functions on
the convex set U, then function J u sup J i u , i I is convex on set U . In fact, by the determination of the upper boundary index i* I and
the
number
0,
such
that
i* i* , I are found. Thence we have
J (u ) J i* (u ),
J (u ) J i* (u ) 1 J i* (v ), u u 1 v. Consequently, J u J u 1 J v , under 0 .
3) Let J u be a convex function determined on the convex set
U E n . Then the inequality is correct
m m J i u i i J ui , i 0, i 1 i 1
m
i 1
i
1.
To prove by the method to mathematical induction. 4) If function f t , t a, b is convex and it is not decrease,
but function F u is convex on the convex set U E n , moreover the
values F u a, b , then function J u f F u is convex on U . In fact, the value
J u f F u f F u 1 F v f F u 1 f F v J u 1 J v . We notice, that t1 F u a, b, t 2 F v a, b and
f F u 1 F v f t1 1 t 2 f t1 1 f t 2 , t1 , t 2 [a, b] , with consideration of convexity of the function f t on segment
a, b .
35
Lecture 5 THE ASSIGMENT WAYS OF THE CONVEX SETS. THEOREM ABOUT GLOBAL MINIMUM. OPTIMALITY CRITERIA. THE POINT PROJECTION ON SET
Some properties of the convex functions and convex set required for solution of the convex programming problem and the other questions which are important for solution of the optimization problems in finitedimensional space are studied separately in the previous lectures. The assignment ways of the convex sets. At studying of the properties to convex function J (u ) and convex set U any answer to question as convex set U is assigned in space E n was not given. Definition 1. Let J (u ) be a certain function determined on set U from E n . The set
epiJ u, E n1 / u U E n , J (u ) E n1
(1)
is called the abovegraphic (or epigraph) to function J (u ) on set U. Theorem 1. In order the function J u to be convex on the convex set U necessary and sufficiently that set epi J to be convex. Proof. Necessity. Let function J (u ) be convex on the convex set U . We show, that set epiJ is convex. In order sufficiently to make sure in that for any z (u, 1 )epi J , w ( , 2 ) epi J 36
and under all , 0 1, the point z z (1 ) w epi J .
In fact, the point z u (1 ) , 1 (1 ) 2 , moreover
u (1 ) U with consideration of set convexity U, but the value 1 (1 ) 2 J (u ) (1 ) J ( w) J (u ) . Finally, the point z u , , where u u (1 ) U , but value J (u ) . Consequently, the point z epiJ. Necessity is proved. Sufficiency. Let set epi J be convex, U be a convex set. We show that function J (u ) is convex on U . If the points u , v U , so z u, J (u ) epi J , w v, J (v) epi J . For any , 0,1, the point z u ,J (u ) (1 ) J (v) epi J with consideration of convexity set epi J . From the inclusion follows that value
J (u ) (1 ) J (v) J (u ) . It means that function J (u ) is convex on U . Theorem is proved. Theorem 2. If function J (u ) is convex on the convex set
U E n , then set
M (C ) u E n u U , J (u ) C is convex under all C E 1 . Proof. Let the points и, v М(С), i.e. J (u ) C , J ( ) C , u ,
U . The point u u (1 ) U under all , 0,1
with consideration of convexity the set U. Since function J (u ) is convex on U, consequently value J (u ) J (u ) (1 ) J ( ), u ,
U . Thence with consideration of J (u ) C , J ( ) C , we get J (u ) C (1 )C C . In the end, the point u U , value J (u ) C . Consequently, for any u , U the point u M (C ) under all C E 1 . It means that set M C is convex. Theorem is proved. We consider the following optimization problem as appendix: 37
J (u ) inf ,
(2)
u U u E n / u U 0 , g i (u ) 0, i 1, m;
g i (u ) ai , u bi 0, i m 1, s ,
(3)
where J (u ) , g i (u ), i 1, m are convex functions determined on n convex set U 0 ; ai E , i m 1, s
are the given vectors;
bi , i m 1, s  the given numbers. We enter the following sets:
U i u i E n /u U 0 , g i (u ) 0 , i 1, m,
U m1 u E n / ai , u bi 0, i m 1, s . The sets U i , i 1, m are convex, since g i (u ), i 1, m are convex functions determined on convex set U 0 (refer to theorem 2,
C 0 ), but set
U m 1 u E n Au b u E n ai , u bi 0, i m 1, s where А
s m n ; the vectors are the lines of the matrix А; b bm 1 ,, bs E n s
is a matrix of the order
ai , i m 1, s
is affine set. Now the problem (2), (3) possible write as m 1
J (u ) inf, u U U i U 0 .
(4)
i 0
Thereby, the problem (2), (3) is problem of the convex programming, since the intersection of any number convex sets is a convex set. Theorem about global minimum. We consider the problem of 38
the convex programming (4), where J (u )  a convex function determined on convex set U from E n . Theorem 3. If J (u ) is a convex function determined on convex
n set U and J * inf J (u ) , U * u* E / u* U , J (u* )
uU
min J (u ) , then any point of the local minimum to function uU
J (u ) on U simultaneously is the point of its global minimum on U, moreover the set U * is convex. If J (u ) is strongly convex on U, then set U * contains not more one point. Proof. Let u* U be a point of the local function minimum J (u ) on set U, i.e. J (u* ) J (u ) under all u o(u* , ) U . Let U be an arbitrary point. Then the point w u* ( u* ) o(u* , ) U with consideration of convexity the set U, if u* . Consequently, inequality J (u* ) J ( w) is executed. On the other hand, since J (u )  a convex function on U, that the inequality J (w)
J (u* ( u* )) J ( (1 )u* ) J ( ) (1 ) J (u* ) exists and 0  sufficiently small number i.e. 0,1 . Now inequality J (u* ) J ( w) to write as J (u* ) J ( w) J ( ) (1 ) J (u* ) . This implies, that 0 J ( ) J (u* ) . Consequently, J (u* ) J (v) , U . This means that in the point u* U the global minimum to function J (u ) on U is reached. We show, that set U * is convex. In fact, for any u, U * , i.e. and J (u ) J ( ) J * under all , 0,1 the value J (u ) J (u (1 ) ) J (u ) (1 ) J ( ) J * . Thence follows that J (u ) J * , consequently, the point u U * . Convexity of the set is proved. We notice, that from U * Ø follows J * J (u* ) . If J (u ) be strongly convex function, then must be 39
J (u ) J (u ) (1 ) J ( ) J * , 0 1 . Contradiction is taken off, if u v , i.e. the set U * contains not more one of the point. Theorem is proved. Finally, in the convex programming problems any point of the local minimum to function J (u ) on U will be the point its global minimum on U, i.e. it is a solution of the problem. Optimization criteria. Once again we consider the convex programming problem (4) for case, when J (u ) C 1 (U ) . Theorem 4. If J (u ) C 1 (U ) is an arbitrary function, U is a
convex set, the set U * , then in any point u* U * necessary the inequality is executed
J (u* ), u u* 0, u U .
(5)
If J (u ) C 1 (U ) is a convex function, U is a convex set, U * , then in any point u* U * necessary and sufficiently performing of the condition (5). Proof. Necessity. Let the point u* U * . We show, that inequality (5) is executed for any function J (u ) C 1 (U ) , U  a convex set (in particular, J (u )  a convex function on U ). Let
u U be an arbitrary point and number 0,1 . Then the difference J (u ) J (u* ) 0, where u u (1 )u* U . Thence follows that 0 J ( u (1 )u* ) J (u* ) J (u* (u u* )) J (u* ) J (u* ), u u* o( ), moreover o( ) / 0 under 0 . Both parts are divided into 0 and transferring to the limit under 0 we get the inequality (5). Necessity is proved. Sufficiency. Let J (u ) C 1 (U ) be a convex function, U be a convex set, U * and inequality (5) is executed. We show that point u* U * . Let u U  an arbitrary point. Then, by theorem 2 40
(refer to section "Convexity criteria of the even functions"), we have J (u ) J (u* ) J (u* ), u u* 0 under all u U . This implies that J (u* ) J (u ), u U . Consequently, the point u* U * . Theorem is proved. Consequence. If J (u ) C 1 (U ) , U is a convex set, U *
and the point u* U * , u* int U , then the equality J (u* ) 0 necessary is executed. In fact, if u* int U , then the number 0 0 such that for any
e E n point u u* eU under all , 0 . is found. Then from (5) follows J (u* ), e 0, e E n , under all , 0 . Thence we have J (u* ) 0 . Formula (5) as precondition of the optimality for nonlinear programming problem and as necessary and sufficiently condition of the optimality for convex programming problem will find applying in the following lectures. The point projection on set. As application to theorems 1, 2 we consider the point projection on convex closed set. Definition 2. Let U be certain set from E n , but the point E n . The point w U is called the projection of the point E n on set U , if norm w inf u , and it is denoted by w PU ( ) uU
Example 1. Let the set be
U S (u0 , R) u E n / u u0 R , but the point E n . The point w PU ( ) u 0 R( u 0 ) u 0 , that follows from geometric interpretation. Example 2. If the set
U u E n / c, u , then point projection E n on U is defined by the formula 41
w PU ( ) c, c c . Norm of the second composed 2
is equal to the distance from the point v till U . Theorem 5. Any point E n has a single projection on convex closed set U E n , moreover for point w PU ( ) necessary and sufficiently executing of the condition
w v, u w 0, u U .
(6)
In particular, if U  affine set, that condition (6) to write as
w v, u w 0, u U .
(7)
Proof. The square of the distance from the point v till u U is 2
E n function J (u ) strongly convex function on convex closed set U . Any strongly convex equal to J (u ) u v . Under fixed
function is strongly convex. Then by theorem 1 function J (u ) reaches the lower boundary in the single point w , moreover w U with consideration of its completeness. Consequently, the point w PU ( ) . Since the point w U * U and gradient J ( w) 2( w v ) , that necessary and sufficient optimality condition for the point w, ( w u* ) according to formula (5) is written in the manner of 2 w v, u w 0, u U . This implies the inequality (6).
is affine set, then from
If U u E n / Au b
u0 , u U ,
u 0 u follows that 2u0 u U . In fact, A(2u 0 u ) 2 Au0 Au 2b b b In particular, if u0 w PU ( ) , then point 2w u U . Substituting to the formula (6) instead of the point u U we get (8) w v, w u 0, u U . From inequality (6) and (8) follows the relationship (7). Theorem is proved. 42
Example 3. Let the set U u E n / u 0 , the point E n .
Then projection w PU (v) (v1 ,, vn ), where vi max(0, vi ),
i 1, n . To prove. Example 4. Let the set
U u (u1 ,, u n ) E n / i u i i , i 1, n – ndimensional parallelepiped, but the point E n . Then components of the vector w PU ( ) are defined by formula
i , если vi i wi i , если vi i vi , если i vi i , i 1, n. n
In fact, from formula (6) follows that sum
(w v )(u i 1
i
i
i
wi ) 0
under all i vi i , i 1, n . If vi i , that wi i , ui i , consequently
the
product
( wi vi )(ui wi ) ( vi i )(u i
i ) 0 Similarly, if vi i , that wi i , u i i so ( wi vi )(u i wi )
(vi i )(ui i ) 0 . Finally, if i vi i , then wi vi , so ( wi vi )(ui wi ) 0 . In the end, for the point w ( w1 ,, wn ) PU ( ) inequality (6) is executed.
43
Lecture 6, 7 SEPARABILITY OF THE CONVEX SETS
Problems on conditional extreme and method of the indefinite Lagrangian coefficients as application to the theory of the implicit functions in course of the mathematical analysis were considered. At the last years the section of mathematics is reached the essential development and the new theory called "Convex analysis" appeared. The main moment in the theory is a separability theory of the convex sets. Definition 1. It is spoken, that hyperplane c, u with normal vector c, c 0 divides (separates) the sets А and В from E n if inequality are executed
sup c, b inf c, a . a A
bB
(1)
If sup c, b inf c, a , that sets А and В are strongly bB
a A
separated, but if c, b c, a, a A, b B , that they are strongly separated. We notice that if hyperplane c, u separates the sets А and В, that hyperplane c, u , where 0  any number also separates them, so under necessity possible to suppose the norm  c  1 . Theorem 1. If U  a convex set from E n , but the point v int U , that hyperplane c, u dividing set U and the point v exists. If U 
a convex set, but the point v U , that set U and the point v is strongly separable. 44
Proof. We consider the event, when the point v U . In this
case by theorem 5 the point E has a single projection w PU ( ) moreover w v, u w 0, u U . Let vector n
c wv . Then we c, u v w v, u v w v, u w w v
have
2
w v, w v w v, u w c 0 . Thence follows, that and the point v E are strongly separable. Consequently, the set U and the point v are strongly separable. We notice, that, if v int U , that v pU . Then by definition
c, u c, v , u U , i.e. set U
n
of the border point the sequence {vk } U such that vk v under
k exists. Since the point vk U , then by proved inequality ck , u ck , vk , u U , ck 1 is executed. By Bolzano–
Weierstrass’ theorem from limited sequence ck possible to select
subsequence c k m , moreover c k m c under m and c 1 . For the elements of the subsequence the previous inequality is written as: ckm , u. ckm , vkm , u U . Transferring to limit under
m with consideration of v km v, we get c, u c, v, u U . Theorem is proved. From proof of theorem 1 follows that through any border point v of the convex set U possible to conduct hyperplane for which c, u c, v, u U . c, u , Hyperplane where c , v , v U is called to be supporting to set U in the point v, but vector c E  a supporting vector of the set U in the point v U . Theorem 2. If convex sets A and B from E n have a no points in common, that hyperplane c, u separating set А and В, as n
well as their closures A and B , in the event of point y A B , 45
number c, y exists. Proof. We denote U A B . As it is proved earlier (refer to theorem 1 from lecture 3), set U is convex. Since the intersection A B , that 0 U . Then with consideration of theorem 1 hyperplane c, u , c, 0 0, separating set U and the point 0, i.e. c, u 0, u U exists. Thence with consideration of u a b U , a A, b B , we get c, a c, b, a A, b B . Finally, hyperplane c, u , where the number satisfies to inequality inf c, a sup c, b separates the sets А a A
bB
and B. Let the points a A , b B . Then the subsequences ak A, bk B such that a k a, bk b under k exist, moreover with consideration of proved above the inequalities c, ak c, bk . are executed. Thence, transferring to limit under k we get c, a c, b a A , b B . In particular, if the point y A B , that c, y . Theorem is proved. Theorem 3. If the convex closed sets A and B have a no points in common and one of them is limited, that hyperplane strongly separating sets A and B exists. Proof. Let the set U A B . We notice that point 0 U , since the intersection A B Ø, U  a convex set. We show, that set U is closed. Let и  a limiting point of the set U . Then sequence u k U such that u k u under k exists, moreover u k a k bk ,
a k A, bk B . If set А is limited, that sequence a k A is limited, consequently, subsequence a k m which converges to a certain point а exists, moreover with consideration of completeness of the set А the point a A . We consider the sequence bk B , where
bk ak u k . Since
a k m a, u k m u under m then
bk m a u b under m . With consideration of completeness of the B the point b B . Finally, the point u a b, a A, b B, 46
consequently, u U . Completeness of set U is proved. Since the point 0 U , U U , then with consideration of theorem 1 hyperplane c, u strongly separating the set U and the point 0 exists, i.e c, u 0 , u U . This implies, that c, a c, b , a A, b B . Theorem is proved. Theorem 4. If the intersection of the inempty convex sets A0 , A1 ,, Am  empty set, i.e. A0 A1 Am Ø, that necessary vectors c0 , c1 ,, cm from E n , not all equal zero and the numbers 0 , 1 ,, m such that
ci , ai i , ai Ai , i 0, m; c0 c1 cm 0;
0 1 m 0;
(2) (3) (4)
exist. Proof. We enter the set A A0 A1 Am  a direct product of the sets Ai , i 0, m with elements a (a0 , a1 ,, a m ) A , where
ai Ai , i 0, m; . We notice that set A E , where E E n ( m1) . It is easy to show, that set А is convex. In fact, if
a1
( a01 , a11 , , a1m ) A, a 2 ( a 02 , a12 , , a m2 ) A , then under all ,
0,1 , the point a a 1 (1 ) a 2 ( a 0 , a1 , , a m ) A ,
1 2 since ai ai (1 )ai Ai , i 0, m . We enter the diagonal set
B b (b0 , b1 ,, bm ) E b0 b1 bm b, bi E n , i 0, m . Set B is convex. It is not difficult to make sure, the intersection A0 A1 Am , if and only if, when intersection
A B .
Finally, under performing the condition of theorem the sets А and B are convex and A B , i.e all conditions of theorem 2 are 47
c 0
executed. Consequently, the vector c (c0 , c1 , , c m ) E ,
such that c, a c, b, a A, b B exists. The inequality can be written in the manner of m
i 0
m
ci , ai ci , bi i 0
m
c , b i 0
i
, ai A, bi E n .
As follows from inequality (5) linear function J (b)
(5)
m
c ,b i 0
i
,
b E n is limited, since in (5) a i Ai , i 0, m . The linear function
J (b) is limited if and only if, when
m
c i 0
i
0 . Thence follows
fairness of the formula (3). Now the inequality (5) is written as m
c ,a i 0
i
i
0, ai Ai , i 0, m . We fix the vectors ai ai Ai , m
i k , then ck , ak ci , ai const, a k Ak . Finally, the value i 0
ck , a k inf ck , a k k , a k Ak , k 1, m . We denote ak Ak
0 ( 1 2 m ) .
Then
c0 , a0 inf
ak Ak
m
c k 1
k
, ak
m
k 0 . k 1
Thereby,
values
ci , ai i
a i Ai ,
i 0, m ,
0 1 m 0 . Faithfulness of the relationships (2), (4) is proved. Theorem is proved. Convex cones. Theorems 1  5 are formulated and proved for convex sets from E n . Separability theorems are often applied in the extreme problem theories for events when convex sets are convex cones. 48
Definition 2. Set K from E n is called by cone with top in zero, if it together with any its point u K contains the points u K under all 0 . If the set K is convex, that it is called the convex cone, if K is closed, that it is called by closed cone, if K is open, that it is called by open cone. Example 1. The set K u E n / u 0 a convex closed cone.
In fact if u K , that point u K under all 0, u 0 .
Example 2. The set K u E n / a, u 0  a convex closed
cone. Since for any 0 , scalar product a, u a, u 0, u K , consequently, vector u K . Example 3. The set K u E n / a, u 0  an open convex
cone, K u E / a, u 0  closed convex cone, K u E n E n
E  a convex cone. We enter the set n
K * c E n / c, u 0, u K .
(6)
We notice, that set K K * , since from c u K follows that 2
c, u c 0 . The set K * , since it contains the element
c 0 . The condition c, u 0 , u K means that vector c K * forms an acute angle (including / 2 ) with vectors u K . To give geometric interpretation of the set K * . * Finally, the set K * from E n is the cone, if c K , i.e. c, u 0 , u K , then for vector c, 0 we have c, u c, u 0 . Consequently, vector c K * . Definition 3. The set K * determined by formula (6) is called dual (or reciprocal) cone to cone K . It is easy to make sure, that for cone K from example 1 the dual cone K * c E n / c 0, for cone K from example 2 the reciprocal cone K K * c E n c , number, for example 3 the 49
K * c E n / c , 0, K * c E n / c , 0, K * 0 accordingly.
dual cones
Theorem 5. If the intersection of the inempty convex with vertex in zero cones K 0 , K1 ,, K m  empty set, that necessary the vectors
ci K i* , i 0, m not all are equal to zero and such that c0 c1 cm 0 exist. Proof. Since all condition of theorem 4 are executed, that the
relationship (2)  (4) are faithful. We notice, that K i , i 0, m cones,
then
from
ai K i , i 0, m
follows
that
vectors
ai K i , i 0, m under all 0 . Then inequality (2) is written so: ci , ai i , a i K i , i 0, m , for any 0 . Thence we have ci , ai i / , i 0, m . In particular, when , we get
ci , ai 0 ,
i 0, m, ai Ai . On the other hand,
i inf ci , ai 0 ,
i 0, m .
ai Ai , i 0, m,
follows
ai A
From that
condition
vectors
ci , ai 0,
ci K i* , i 0, m .
Theorem is proved. Theorem 6. (Dubovicky–Milyutin’s theorem). In order the intersection of the inempty convex with vertex in zero cones K 0 , K1 ,, K m to be an empty set, when all these cones, except, can be one open, necessary and sufficiently existence of the vectors
ci K i* , i 0, m , not all are equal to zero and such that c0 c1 cm 0 . Proof. Sufficiency. We suppose inverse, i.e. that vectors
ci K i* , i 0, m , c0 c1 cm 0 , though K 0 K 1 K K m . Then the point w K 0 K 1 K m exists. Since * vectors ci K i , i 0, m , that inequalities ci , w 0, i 0, m
50
are executed. Since sum c0 c1 cm 0
and
ci , w 0,
i 0, m , then from equality c0 c1 cm , w 0 follows that
ci , w 0, i 0, m . By condition of the theorem not all ci , i 0, m are equal to zero, consequently, at least, two vectors ci , c j , i j differenced from zero exist. Since all cones are open, except can be one, that in particular, it is possible to suppose that cone K i  an open set. Then K i contains the set o( w, 2 )
u K i / u w 2 , 0. In particular, the point u w w ci / ci K i . Since ci , u 0, u K i , ci K i* , that we have
c , w c / c c , w c c 0, i
ci 0 .
i
It
i
is
i
impossible.
i
i
Consequently,
the
where 0, intersection
K 0 K1 K m . Necessity follows from theorem 5 directly. Theorem is proved.
51
Lecture 8 LAGRANGE’S FUNCTION. SADDLE POINT Searching for the least value (the global minimum) to function J(u) determined on set U from E n is reduced to determination of the saddle point to Lagrange’s function. By such approach to solution of the extreme problems the necessity to proving of the saddle point existence of the Lagrange’s function appears. We consider the next problem of the nonlinear programming:
J (u) inf ,
(1)
u U u E n u U 0 , g i (u ) 0, i 1, m;
g i (u ) 0, i m 1, s ,
(2)
where U 0 is the given convex set from E n ; J (u ), g i (u ), i 1, m are the functions determined on set U 0 . In particular, J (u ),
g i (u ), i 1, m are convex functions determined on convex set U 0 , but functions g i (u ) ai , u bi , a i E n ,
i m 1, s,
bi ,
i m 1, s, are the given numbers. In this case problem (1), (2) is related to the convex programming problems. In the beginning we consider a particular type of the Lagrange’s function for problem (1), (2). Lagrange’s Function. Saddle point. Function s
L (u , ) J (u ) i g i (u ), u U 0 , i 1
0 E / 1 0, , m 0 s
is called by Lagrange’s function to problem (1), (2). 52
(3)
Definition 1. Pair u* , * U 0 0 , i.e. u* U 0 , * 0 is called by saddle point to Lagrange’s function (3) if the inequalities are executed
L u* , L u* , * L u , * , u U 0 , 0 .
(4)
We notice that in the point u* U 0 minimum to function
L u, * is reached on set U 0 , but in point * 0 the maximum
to function L u* , is reached on set 0 . By range of definition of the Lagrange’s function is the set U 0 0 .
The main lemma. In order the pair u* , * U 0 0 to be a saddle point to Lagrange’s function (3) necessary and sufficiently performing of the following conditions:
L u* , * L u , * , u U 0 ;
(5)
*i g i (u* ) 0, i 1, s, u* U .
(6)
Proof. Necessity. Let pair u* , * U 0 0  a saddle point. We show, that the condition (5). (6) are executed. Since pair u* , * U 0 0  saddle point to Lagrange’s function (3), that inequalities (4) are executed. Then from the right inequality follows the condition (5). It is enough to prove fairness of the equality (6). The left inequality from (4) is written so:
s
s
i 1
i 1
J (u* ) i g i (u* ) J (u* ) *i g i (u* ) ,
(1 ,, s ) 0 E s . Consequently, the inequality exists s
( i 1
* i
i ) g i (u* ) 0, 0 . 53
(7)
At first, we show that u* U , i.e. g i (u* ) 0, i 1, m,
g i (u* ) 0, i m 1, s . It is easy to make sure in that vector i *i , i 1, s, i j , (1 , , s ) * j j 1, for j of 1 j m,
(8)
belongs to the set E s / 1 0,, m 0 .
Substituting value 0 from (8) to the inequality (7) we get
(1) g i (u* ) 0 . Thence follows that g i (u* ) 0, i 1, m (since it is faithful for any j of 1 j m ). Similarly the vector
i *i , i 1, s, i j , (1 ,, s ) * j j g j (u* ), для j из m 1 j s, also belongs to the set 0 . Then from inequality (7) we have 2
g i (u* ) 0, j m 1, s . Consequently, the values g j (u* ) 0,
j m 1, s . From relationship g j (u* ) 0, j 1, m , g j (u* ) 0, j m 1, s follows that point u* U . We choose the vector 0 as follows:
*i , i 1, s, i j , (1 ,, s ) i j 0, для j из 1 j m. In this case inequality (7) is written so: *j g j (u* ) 0, j 1, m . Since the value *j 0, j 1, m under proved above value g j (u* ) 0,
j 1, m , consequently, the equality *j g j (u* ) 0, j 1, m
exists.
The equalities *j g j (u* ) 0, j m 1, s follow from equality
g j (u * ) 0, j m 1, s . Necessity is proved. 54
Sufficiency. We suppose, that the conditions (5), (6) for a certain pair u* , * U 0 0 are executed. We show, that u* , *  saddle point to Lagrange’s function (3). It is easy to make sure in that product (*i i ) g i (u* ) 0 for any i , 1 i m . In fact, from
condition u* U follows that g i (u* ) 0, for any i , 1 i m . If g i (u* ) 0 , that (*i i ) g i (u* ) 0 . In the event of g i (u* ) 0 the value *i 0 , since the product *i g i (u* ) 0 , consequently, m
(
(*i i ) g i (u* ) i g i (u* ) 0, i 0 . Then sum
i 1
i ) g i (u * )
s
(
i m 1
* i
i ) g i (u * )
s
( i 1
* i
* i
i ) g i (u* ) 0, since
g i (u* ) 0, i m 1, s, u* U . Thence follows that s
s
i 1
i 1
(*i g i (u* ) (i g i (u* ), s
s
i 1
i 1
J (u* ) i g i (u* ) J (u* ) i g i (u* ) . The second inequality can be written in the manner of L u* , L u* , * . The inequality in the ensemble with (5) defines the condition
L u* , L u* , * L u , * , u U 0 , 0 .
This means that pair u* , * U 0 0  saddle point to Lagrange’s function (3). Theorem is proved. The main theorem. If pair u* , * U 0 0  a saddle point to
Lagrange’s function (3), that vector u* U is a solution of the problem (1), (2) i.e.
u* U * u* E n u* U , J (u* ) min J (u ) . uU
55
follows from the main lemma for pair u* , U 0 0 the conditions (5), (6) are executed. Then value
Proof. *
As
J (u* , * ) J (u* )
s
g (u ) J (u ) , since the point i 1
* i
i
*
*
u* U .
Now inequality (5) is written so: s
J (u* ) J (u ) *i g i (u ), u U 0 .
(9)
i 1
Since the set U U 0 , that inequality (9), in particular, for any
u U U 0 faithfully, i.e. s
J (u* ) J (u ) *i g i (u ), u U , * 0 .
(10)
i 1
As follows from condition (2) if u U , that g i (u ) 0, i 1, m , and g i (u ) 0, i m 1, s , consequently, the product *i g i (u* ) 0 for any i, 1 i s . We notice, that * (1* ,, *s ) 0 , where
1* 0,, *m 0 . Then from (10) follows J (u* ) J (u ), u U . It means that in the point u* U the global (or absolute) minimum of the function J (u ) is reached on set U. Theorem is proved. We note the following: 1) The main lemma and the main theorem for the nonlinear programming problem (1), (2) were proved, in particular, its are true and for the convex programming problem. 2) Relationship between solutions of the problem (1), (2) and saddle point to Lagrange’s function in type (3) is established. In general event for problem (1), (2) Lagrange’s function is defined by the formula
56
s
L (u,λ ) 0 J (u ) i g i (u* ), u U 0 , i 1
0 (1 ,, s ) E s 1 0 0,, m 0.
(11)
If the value 0 0 , that Lagrange’s function (11) possible to
present in the manner of L (u , ) 0 L u, , where i i / 0 ,
function L u, is defined by formula (3). In this case the main lemma and the main theorem are true for Lagrange’s function in type (11). 3) We consider the following optimization problems: J (u ) inf, u U , J 1 (u ) inf, u U , where function s
J 1 (u ) J (u ) *i g i (u ). As follows from proof of the main i 1
theorem the conditions J1 (u ) J (u ), u U ; J 1 (u* ) J (u* ),
u* U are executed. 4) For existence of the saddle point to Lagrange’s function necessary that set U * . However this condition does not guarantee existence of the saddle point to Lagrange’s function for problem J (u ) inf, u U . It is required to impose the additional requirements on function J (u ) and set U that reduces efficiency of the Lagrange’s coefficients method. We notice, that Lagrange’s coefficients method is an artificial technique solution of the optimization problems requiring spare additional condition to problem. KuhnTucker’s theorem. The question arises: what additional requirements are imposed on function J (u ) and on set U that Lagrange’s function (3) would have a saddle point? Theorems in which becomes firmly established under performing of which conditions Lagrange’s function has the saddle point are called KuhnTacker’s theorems (Kuhn and Tucker  American mathematicians). 57
It can turn out to be that source problem J (u ) inf, u U ,
U * Ø , has a solution, i.e. the point u* U * exists, however Lagrange’s function for the given problem has not saddle point. Example. Let function J (u ) u 1 , but set U u E 1 / 0
/ 0 u 1, moreover
0 u 1; (u 1) 2 0 . The function g (u ) (u 1) 2 , set U 0
uE
1
functions
J (u ) and g (u ) are
convex on set U 0 . Since set U consists of single element, i.e.
U {1} , that set U * {1} . Consequently, the point u* 1 . Lagrange’s function L u , (u 1) (u 1) 2 , 0, u U 0 for the problem has not saddle point. In fact, from formula (4) follows
L u* , 0 L u* , * 0 L u, * (u 1) * (u 1) 2 ,
0, 0 u 1. * 2 Thence we have 0 , where u 1 , 0 1 .
Under sufficiently small 0 does not exist the number 0 such the given inequality is executed. 1st case. We consider the next problem of the convex programming: *
J (u ) inf;
(12)
u U u E n u U 0 , g i (u ) 0, i 1, m ,
(13)
where J (u ), g i (u ), i 1, m  convex functions determined on convex set U 0 . Definition 2. If the point u i U 0 such that value g i (u i ) 0
exists, that it is spoken that restriction g i (u ) 0 on set is regularly. It is spoken that set U
is regularly, if all restrictions
g i (u ) 0, i 1, m from the condition (13) are regular on set U 0 . 58
We suppose, that point u U 0 exists such that
g i (u ) 0, i 1, m .
(14)
The condition (14) is called by Sleyter’s condition. Let in the points u 1 , u 2 ,, u k U 0 all restriction are regularly, i.e. k
g j (u i ) 0, j 1, m, i 1, k . Then in the point u i u i U 0 , i 1
i 0, i 1, k , 1 k 1 all restrictions g j (u ) 0, j 1, m, are regular. In fact,
k k g j (u ) g j i u i i g j (u i ) 0, j 1, m. i 1 i 1 Theorem 1. If
J (u ), g i (u ), i 1, m  convex functions
determined on convex set U 0 , the set U is regularly and
U* Ø ,
then for each point u* U * necessary the Lagrangian multipliers
* (1* ,, *m ) 0 E m / 1 0,, m 0 such that pair
u , U *
*
0
0 forms a saddle point to Lagrange’s function m
L u , J (u ) i g i (u ), u U 0 , 0 exist. i 1
As follows from condition of the theorem, for the convex programming problem (12), (13) Lagrange’s function has a saddle point, if set U is regularly (under performing the condition that set U * Ø ). We note, that in the example is represented above set U is not regular, since in the single point u 1U 0 restriction
g (1) 0 . We notice, that if set is regularly, so as follows from Sleyter’s condition (14), the point u U U 0 . Proof of Theorem 1 is provided in the following lectures.
59
Lectures 9, 10 KUHNTUCKER’S THEOREM For problem of the convex programming the additional conditions imposed on convex set U under performing of which Lagrange’s function has a saddle point are received. We remind that for solving of the optimization problems in finitedimensional space by method of the Lagrangian multipliers necessary, except performing the condition U * , existence of the saddle point to Lagrange’s function. In this case, applying of the main theorem is correct. Proof of the theorem 1. Proof is conducted on base of the separability theorem of the convex sets А and В in space E m 1 . We define as follows the sets А and B:
A a ( a 0 , a1 , , a m ) E m 1 a 0 J (u ),
a1 g1 (u ) , am g m (u ), u U 0 ;
(15)
B b (b0 , b1 , , bm ) E m 1 b0 J * , b1 0 , bm 0 , (16) where J * inf J (u ) min J (u ) since set U * uU
uU
.
a) We show that sets А and B have a no points in common. In fact, from the inclusion a A follows that a0 J (u ),
a1 g1 (u ), , am g m (u ), u U 0 . If the point u U U 0 , that inequality a0 J (u ) J * is executed, consequently, the point a B . If the point u U 0 \ U , then for certain number i, from 1 i m is g i (u ) 0 the inequality is executed. Then ai g i (u ) 0 and once again the point a B . Finally, the intersection A B . 60
b) We show that sets А and В are convex. We take two arbitrary points a A, d A and the number , 0,1 . From inclusion
a A
follows that the point u U 0 such that a 0 J (u ),
ai g i (u ), i 1, m, a (a0 , a1 ,, am ) exists. Similarly from
d A we have d 0 J (v), d i g i (v), i 1, m, where v U 0 , d (d 0 , d1 , , d m ) . Then the point a a (1 )d a0 (1 )d 0 , a1 (1 )d1 , , a m (1 )d m , 0,1, moreover a 0 (1 )d 0 a 0 J (u ) (1 ) J (v) J (u ), i 1, m with consideration of convexity J (u ) on U 0 , where
u u (1 )v . Similarly
a (1 )d a g (u ) (1 ) g (v ) g i (u ), i
i
i
i
i
i 1, m , with consideration of function convexity g i (u ), i 1, m on set U 0 . Finally, the point
a (a0 , a1 , , a m ), a0 J (u ), ai
g i (u ), i 1, m, moreover u U 0 . This implies that point a A under all , 0 1 . Consequently set А is convex. By the similar way it is possible to prove the convexity of the set B. c) Since sets А and В are convex and A B Ø , that, by theorem 2 (the lecture 6), hyperplane c, w , where normal m 1 vector c (*0 , 1* , *m ) E m 1 , c 0, w E , separating sets
А
and
В,
b1 , , bm ) E
as
its
closures
A A , B b (b0 ,
as
well
m 1
b0 J * , b1 0 , bm 0 exists. Consequently,
inequality are executed m
m
i 0
i 0
c, b *i bi c, a *i ai , a A, b B . 61
(17)
We notice that if the point u* U * , that
J (u* ) J * ,
g i (u* ) 0, i 1, m , consequently, the vector y ( J * ,0,,0) A B and the value c, y *0 J * . Now inequalities (17) are written as m
m
i 0
i 0
*0 b0 *i bi *0 J * *0 a 0 *i ai , a A, b B .
(18)
Thence, in particular, when vector b ( J * 1, 0, ,0) B ,
from the left inequality we have *0 ( J * 1) *0 J * . Consequently, value *0 0 . Similarly, choosing vector b ( J * ,0,,0,1, 0,0) from the left inequality we get *0 J * *i *0 J * . This implies that
*i 0, i 1, m . d) We take any point u* U * . It is easy to make sure in that point b ( J * ,0, ,0, g i (u* ), 0 ,0) A B , since value g i (u* ) 0 . Substituting the point to the left and right inequalities (18), we get *0 J * *i g i (u * ) *0 J * *0 J * *i g i (u* ) . Thence we have
*i g i (u* ) 0 *i g i (u* ) . Consequently, the values *i g i (u* ) 0, i 1, m . f) We show, that value *0 0 . In item c) it is shown that *0 0 . By condition of the theorem set U is regular, i.e. the Sleyter’s condition is executed (14). Consequently, the point u U U 0 such that
g i (u ) 0, i 1, m exists. We notice, that point a ( J (u ), g1 (u ),, g m (u )) A . Then from the right inequality (18) we have
*0 J * *0 J * (u ) 1* g 1 (u ) *m g m (u ) . We
suppose
opposite
i.e. 62
*0 0 .
Since
vector
c (*0 , 1* ,, *m ) 0 , that at least one of the numbers
*i , 1 i m are different from zero. Then from the previous inequality we have 0 *i g i (u ) 0 . It is impossible, since
*i 0, *i 0 . Finally, the number *0 is not equal to zero, consequently, *0 0 . Not derogating generalities, it is possible to put *0 1 . g) Let u U 0 be an arbitrary point. Then vector a ( J (u ),
g1 (u ),, g m (u )) A , from the right inequality (18) we get J * J (u ) *i g i (u ) L u , * , u U 0 . Since the product m
i 1
m
*i g i (u* ) 0, i 1, m ,
J * *i g i (u ) L u* , *
that
i 1
L u , * , L u , * , u U 0 . From the inequality and from equality g i (u* ) 0, i 1, m, u* U , where *i 0, i 1, m * i
follows that pair u* , * U 0 0  saddle point to Lagrange’s function. Theorem is proved. 2nd case. Now we consider the next problem of the convex programming:
J (u ) inf ,
(19)
u U u E n u j 0, j I , g i (u ) ai , u bi 0, i 1, m;
g i (u ) ai , u bi 0, i m 1, s ,
(20)
where I 1,2, , n  a subset; the set U 0 u E n / u j
0, j I ,
J (u)  a convex function determined on convex set
U 0 , a i E n , i 1, s  the vectors; bi , i 1, s  the numbers. For problem (19), (20) Lagrange’s function 63
m
L u , J (u ) i g i (u ), u U 0 ,
(21)
i 1
(1 , , s ) 0 E s 1 0, , m 0. We notice that, if function J (u ) cu is linear, the task (19), (20) is called the general problem of the linear programming. We suppose, that set U * . It is appeared that for task of the convex programming (19), (20) Lagrange’s function (21) always has saddle point without some additional requirements on convex set U . For proving of this it is necessary the following lemma. n Lemma. If a1 , , a p  a finite number of the vectors from E and the numbers i 0, i 1, p , that set p Q a E n a i ai , i 0, i 1, p i 1
(22)
Is convex closed cone. Proof. We show, that set Q is a cone. In fact, if a Q and p
p
i 1
i 1
0  an arbitrary number, that a i ai i ai , where i i i 0, i 1, p . This implies that vector
a Q .
Consequently, set Q is the cone. We show, that Q is a convex cone. In fact, from a Q, b Q follows that
a (1 )b i ai (1 ) i ai p
i 1
p
p
i 1
i 1 p
( i (1 ) i )ai i ai , i i (1 ) i 0, i 1, m , where 0,1,
b i ai , i 0, i 1, p . i 1
64
Thence we have a (1 )b Q under all , 0 1 . Consequently, set Q  a convex cone. We show, that Q is convex closed cone. We prove this by the method of mathematical induction. For values p 1 we have
Q a E n a a1 , 0  a halfline. Therefore, Q  closed p set. We suppose, that set Q p 1 a E n a i ai , i 0 is i 1 closed. We prove, that set Q Q p 1 a p , 0 is closed. Let
c E n is limiting point of the set Q. Consequently, the sequence cm Q exists, moreover cm c, m . The sequence cm is represented in the manner of cm bm m a p , where
bm Q p 1 , m
 a numeric sequences. As far as set Q p 1 is
closed, that bm b, m , under b Q p 1 . It is remains to
prove that numeric sequence m is bounded. We suppose opposite
i.e. mk under k . Since bmk / mk c mk / mk a p and
b
mk
/ mk Q p 1 the sequence {c mk } is bounded, so far as
c mk c, k , then under k we have a p Q p 1 . Since set Q p 1 Q
(in the event of 0, Q p 1 Q ), that vector
a p Q . It is impossible. Therefore, the sequence bounded.
Consequently,
vector
c b a p Q ,
m
is
where
mk , 0 under k . Theorem is proved. We formulate and prove Farkas’s theorem having important significant in theories of the convex analysis previously than prove the theorem about existence of saddle point to Lagrange’s function for task (19), (20). Farkas’ theorem. If cone K with vertex in zero is determined by inequalities
65
K e E n ci , e 0, i 1, m;
ci , e 0, i m 1, p;
ci , e 0, i p 1, s, ci E n , i 1, s ,
the dual to it cone K * c E n c, e 0, e K K * c E n
(23)
has the type
s c i c i , 1 0, , p 0 . i 1
(24)
Proof. Let cone K is defined by formula (23). We show, that set * of the vectors c K for which c, e 0, e K is defined by
formula (24), i.e. set s Q c E n c i ci , 1 0, , p 0 i 1
(25)
complies with K * c E n c, e 0, e K . We represent
i i i , i 0, i 0, i p 1, s . Then set Q from (25) is written as p s s Q с E n c i (ci ) i (ci ) i (ci ), i 1 i p 1 i p 1 i 0, i 1, p, i 0, i p 1, s, i 0, i p 1, s .
(26)
As follows from expressions (26) and (22) the set Q is convex closed cone generated by the vectors c1 , , c p , c p 1 ,....,
cs , c p 1 , , cs (refer to the lemma). We show, that Q K * . In fact, if c Q , i.e. c
s
c , i 1
66
i i
s
1 0,, p 0 , that product c, e i ci , e 0 with i 1
consideration of correlations (23). Consequently, vector c Q belongs to set
K * c E n c, e 0, e K . This implies that Q K * . We show, that K * Q . We suppose opposite, i.e. vector a K , however a Q . Since set Q – convex closed cone and the point *
a Q , that by theorem 1 (the lecture 6) the point a E n is strongly separable from set Q i.e. d , c d , a , c Q , where  d , d 0 is a normal vector to the hyperplane d , u d , a . Thence we have s
d , c i d , ci d , a , i , i 1, s, i 1
1 0,, p 0 .
(27)
We choose the vector ( 1 0,, p 0, p 1 ,, s ) so:
0, i j , i 1, s ( 1 , , s ) i j t , t 0, j из 1 j p. For the vector inequality (27) is written in the form t d , ci d , a . . We divide into t 0 and t , as a result we get c j , d 0, j 1, p. Hereinafter, we take the vector
i 0, ( 1 ,, s ) j t c j , d , t 0, 67
i j , i 1, s, j из p 1 j s.
From inequality (27) we have t c j , d
2
d , a . We divide into
t 0 and, directing t , we get c j , d 0, j p 1, s . Finally, for the vector d E inequalities c j , d 0, n
j 1, p, c j , d 0,
j p 1, s are executed. Then from (23) follows that d K , where K  a closure of the set K . Since the vector a K * , that the inequality a , e 0, e K exists. Thence, in particular, for e d K follows
a, d 0 .
However from (27) under i 0, i 1, s , we have a, d 0 . We have got the contradiction. Consequently, vector a K * belongs to the set Q. From inclusions Q K * , K * Q follows that K * Q . Theorem is proved. Now we formulate theorem about existence of the saddle point to Lagrange’s function (21) for problem of the convex programming (19), (20). Theorem 2. If J (u ) is a convex function on convex set
U 0 , J (u ) C 1 (U 0 ) and set U * for problems (19), (20), then for each point u * U * necessary the Lagrangian multipliers * (1* , , *s ) 0 E s / 1 0, , m 0 exist, such
that pair u* , U 0 0 forms a saddle point to Lagrange’s *
function (21) on set U 0 0 . Proof. Let the condition of the theorem is executed. We show the pair u* , * U 0 0  saddle point to Lagrange’s function (21).
Let u* U *  an arbitrary point. We define the feasible directions coming from point u* for convex set U . We notice that the vector
e (e1 ,, en ) E n is called by the feasible direction in the point
u* , if the number 0 0 such that u u * e U under all
, 0 0
exists. From inclusion 68
u u* e U
with
consideration of the expressions (20) we have
u *j e j 0, j I ;
ai , u* e bi 0, i 1, m ;
ai , u* e bi 0, i m 1, p, , 0 0 .
If the sets of the indexes I 1 j u *j 0, j I ,
(28)
I 2 j ai ,
u * bi , 1 i m are entered, that conditions (28) are written so:
e j 0, j I1 ; ai , e 0, i I 2 ; ai , e 0, i m 1, s.
(29)
Finally, the set of the feasible directions in the point u* according to correlation (29) is a cone
K e E n e j e j , e 0, j I 1 ;
ai , e 0, i m 1, s ,
ai , e 0, i I 2 , (30)
where e j 0,,0,1,0,,0 E n  a single vector. And inverse statement faithfully, i.e. if e K , that е  feasible direction. Since J (u ) is a convex function on convex set U U 0 and
J (u ) C 1 (U ) , that by theorem 4 (the lecture 5) in the point u * U * , necessary and sufficiently executing of the inequality J (u* ), u u* 0, u U . Thence with consideration of that
u u* e, 0 0 , e K , we have J (u* ), e 0, e K . Consequently, the vector J (u* ) K * . By Farkas’s theorem dual cone K * to cone (30) is defined by the formula (24), so the numbers *j 0, j I 1 ; *i 0, i I 2 ; *m 1 , , *s such that
69
J (u* ) *j e j *i ai jI1
iI 2
s
a
i m 1
* i i
(31)
exist. Let *i 0 for values i 1,2, , m \ I 2 . Then expression (31) is written as s
J (u* ) *i ai *j e j . i 1
(32)
jI1
We notice that *i g i (u* ) 0, i 1, s , since g i (u* ) 0, i I 2 and As follows from the expression (21) u U 0
i m 1, s .
the
difference
s
L u , * L u* , * J (u ) J (u* ) *i ai , u u* .
(33)
i 1
J (u ) C 1 (U ) , then according to the theorem 1 (the lecture 4) the difference J (u ) J (u* ) J (u* ), u Since convex function
u u * , u U 0 . Now inequality (33) with consideration of correlations (32) can be written in the manner of
s
L u , * L u* , * J (u* ) *i a i , u u* i 1
jI1
since
* j
e , u u* (u j u *j ) j
jI1
* j
u jI1
* j
j
0,
e j , u u* u j u *j , u *j 0, j I 1 . This implies that
L u * , * L u , * , u U 0 . From the inequality and equality
*i g i (u* ) 0, i 1, s follows that pair u* , * U 0 0  saddle point to Lagrange’s function (21). Theorem is proved. 3rd case. We consider more general problem of the convex 70
programming:
J (u ) inf ,
(34)
u U u E n u U 0 , g i (u ) 0, i 1, m; g i (u ) ai , u bi 0 , i m 1, p,
g i (u ) ai , u bi 0 , i p 1, s , (35)
where J (u ), g i (u ), i 1, m are convex functions determined on convex set U 0 ; a i E n , i m 1, s are the vectors;
bi , i
m 1, s are the numbers. Lagrange’s function for tasks (34), (35) s
L u, J (u ) i g i (u ), u U 0 ,
0 E
i 1
s
1 0,, m 0.
(36)
Theorem 3. If J (u ), g i (u ), i 1, m are convex functions
determined on convex set U 0 , set U * Ø for tasks (34), (35) and the points u riU 0 U such that g i (u ) 0, i 1, m exists, then for each
u* U *
point
necessary
the
Lagrangian
multiplier
( , , ) 0 E 1 0, , p 0 exist, such *
* 1
* s
s
that the pair u* , * U 0 0 forms a saddle point to Lagrange’s function (36) on set U 0 0 . Proof of theorem requires of using more fine separability theorems of the convex set, than theorems 1  6 (the lectures 6, 7). For studying more general separability theorems of the convex set and other KuhnTucker’s theorems is recommended the book: Rocafellar R. Convex analysis. Moskow, 1973. We note the following: 10. The theorems 13 give the sufficient conditions of existence of the saddle point for the convex programming problem. Example 1. Let function be J (u) 1 u , but set U u
71
E 1 0 u 1; (1 u ) 2 0. For this example the functions J (u ), g (u ) (1 u ) 2 are convex on convex set U 0 u E 1 0
u 1 . The set U 1 , consequently U * 1, set U * 1. The point u* 1 and the value J (u * ) 0 are solution of the problem J (u ) inf, u U . Lagrange’s function L u, (1 u ) (1
L u , 0 L u, (1 u)
1 u ) 2 , u U 0 , 0 . The pair u* 1, * 0  saddle point to Lagrange’s function, since
*
*
*
* (1 u ) 2 , 0 u 1, * g (u * ) 0 . We notice, that for the example neither Sleyter’s theorem from theorem 1, nor condition of theorem 3 is not executed. 2°. In general event Lagrangian multipliers for the point u* U *
are defined ambiguous. In specified example the pair u* 1, * 0
is the saddle point to Lagrange’s function under any 0 . 3°. Performing of the theorems 1  3 condition guarantees the existence of the saddle point to Lagrange’s function. Probably, this circumstance for problem as convex, so and nonlinear programming is a slight learned area in theories of the extreme tasks. Solution algorithm of the convex programming problem. The problems of the convex programming written in the manner of (34), (35) often are occurred in the applied researches. On base of the theories stated above we briefly give a solution sequence of the convex programming problem. 10. To make sure in that for task (34), (35) set U * . In order to use the theorems 13 from lecture 2 (Weierstrass’ theorem). 20. To check performing the conditions of the theorems 13 in depending on type of the convex programming problem to be a warranty of existence saddle points to Lagrange’s function. For instance, if task has the form of (12), (13), then to show that set U is regularly; if task has the form of (19), (20), that necessary J (u ) C 1 (U 0 ) , but for task (34), (35) necessary existence of the *
point u riU 0 U for which g i (u ) 0, i 1, m .
72
30. To form Lagrange’s function L u , J (u )
s
g (u ) i
i 1
i
with definitional domain U 0 0 , where
0 E s 1 0, , m 0 .
40. To find a saddle point u* , * U 0 0 to Lagrange’s function from the conditions (the main lemma from lecture 8):
L u* , * L u , * , u U 0 ,
*i g i (u* ) 0, i 1, s,
(37)
1* 0, , *m 0.
a) As follows from the first condition, function L u , * reaches the minimum on set U 0 in the point u* U * U 0 . Since functions
J (u ), g i (u ), i 1, m are convex on convex set U 0 , that function
L u , * is convex on U 0 . If functions J (u ) C 1 (U 0 ), i 1, m , then according to the optimality criterion (the lecture 5) and theorems about global minimum (the lecture 5) the first condition from (37) can be replaced on L u u * , * , u u * 0, u U 0 ,
where L u u * , * J (u * )
s
g (u ) . i 1
* i
i
*
Now condition (37) are
written as
L u u* , * , u u* 0 , u U 0 ,
*i g i (u* ) 0, i 1, s, 1* 0, , *m 0.
(38)
b) If we except J (u ) C 1 (U 0 ), g i (u ) C 1 (U 0 ), i 1, m , set U 0 E n , then according to optimality criteria (lecture 5) the conditions (38) possible present in the manner of 73
L u u* , * 0, *i g i (u* ) 0, i 1, s,
1* 0, , *m 0 .
(39)
The conditions (39) are represented by the system n s of the algebraic equations comparatively n s unknowns u* u1* ,, u n* ,
,, . We notice, that if Lagrange’s function has the *
* 1
* n
saddle point, that system of the algebraic equations (39) has a solution, moreover must be 1* 0,, *m 0. Conditions (39) are used and in that events, when U 0 E n , however in this case necessary to make sure the point u* U 0 .
5°. We suppose, that pair u* , * U 0 0 is determined. Then the point u * U and value J * J (u* ) is the solution of the problem (34), (35) (refer to the main theorem from lecture 8).
74
Chapter II. NONLINEAR PROGRAMMING
There are not similar theorems for problem of the nonlinear programming as for problems of the convex programming guaranteed existence of the saddle point to Lagrange’s function. It is necessary to note that if some way is installed that pair (u* , * ) U 0 0 is the saddle point to Lagrange’s function then according to the main theorem the point u* U * is the point of the global minimum in the problems of the nonlinear programming. We formulate the sufficient conditions of the optimality for the nonlinear programming problem by means of generalized Lagrange’s function. We notice, that point u* U determined from sufficient optimality conditions, in general event is not a solution of the problem, but only "suspected" point. It is required to conduct some additional researches; at least, to answer on the question: will be the point u* U by point of the local minimum to function J (u ) on ensemble U ?
75
Lectures 11, 12. STATEMENT OF THE PROBLEM. NECESSARY CONDITIONS OF THE OPTIMALITY Statement of the problem. The following problem often is occurred in practice:
J (u ) inf ,
(1)
u U u E n / u U 0 , g i (u ) 0, i 1, m; g i (u ) 0
0, i m 1, s , where J ( u ), g i ( u ), i 1 , s
(2)
 the functions determined on
n
convex ensemble U 0 from E . Entering notations
U i u E n g i (u ) 0 , i 1, m,
(3)
(3)
U m 1 u E n g i (u ) 0 , i m 1, s , ensemble U possible to present in the manner of
U i u E n g i (u ) 0 , i 1, m,
U m 1 u E n g i (u ) 0 , i m 1, s , Now problem (1), (2) can be written in such form:
J (u ) inf, u U We suppose that J * inf J (u ) , ensemble 76
U * u* E n u* U , J (u* ) min J (u ) . We notice that uU
if ensemble U * , so J (u* ) min J (u ) . It is necessary to find uU
the point u* U * and value J * J (u* ) . For problem (1), (2) generalized Lagrange’s function has the form s
L (u, ) 0 J (u ) i g i (u ), u U 0 , (0 , 1 , , s ), i 1
0 E
s 1
0 0, 1 0, , m 0.
(5)
Let U 01  an open ensemble containing set U 0 U 01 . Theorem 1 (the necessary optimality conditions). If functions J (u ) C 1 (U 01 ), g i (u ) C 1 (U 01 ), int U 0 0, U 0 is a convex ensemble, but ensemble U* , then for each point u* U * the Lagrange’s multipliers * (*0 , 1* ,, *s ) 0 necessary exist, such that the following condition is executed:
* 0, *0 0, 1* 0, , *m 0 ,
(6)
L u (u * , * ), u u* s
*0 J (u* ) *i g i(u* ), u u* 0, u U 0 ,
(7)
i 1
*i g i (u* ) 0, i 1, s.
(8)
Proof of the theorem is released on theorems 5, 6 (the lecture 7) about condition of the emptiness intersection of the convex cones (DubovickyMilutin’s theorem) and it is represented below. We comment the conditions of the theorem 1. We note the following: a) In contrast to similar theorems in the convex programming 77
problems
is
not become firmly established that pair (u* , ) U 0 0 is an saddle point to Lagrange’s function, i.e. the conditions of the main theorem are not executed. b) Since pair (u* , * ) in general event is not a saddle point, then *
from the condition (6)  (8) does not follow that point u* U  a solution of the problem (1), (2). c) If the value *0 0 , that problem (1), (2) is identified nondegenerate. In this case it is possible to take *0 1 , since Lagrange’s function is linear function comparatively . g) If *0 0 , that problem (1), (2) is called degenerate. Number of the unknown Lagrangian multipliers, independently the problem (1), (2) is degenerate or nondegenerate, possible to reduce on unit by entering the normalization condition, i.e. condition (6) can be changed on
* , *0 0, 1* 0, , *m 0 ,
(9)
where 0  any given number, in particular, 1 . Example 1. Let J (u ) u 1,
U u E 1 0 u 1; g (u ) (u 1) 2 0
Generalized Lagrange’s function for the problem has the form L (u , ) 0 J (u ) (u 1) 2 , u U 0 u E 1 0 u 1, 0 0,
0 . Ensemble U 1 , consequently, ensemble U * 1
moreover J (u* ) 0, g (u* ) 0 . The condition (7) is written as
(*0 J (u* ) * g (u* )) (u u* ) 0, u U 0 . Thence we have (*0 )(u 1) 0, u U 0 , or ( *0 )(1 u ) 0, 0 u 1 . Since expression (1 u ) 0 under 0 u 1 , then for executing the inequality necessary that value *0 0 , so as *0 0 by the condition (6). Finally, the source problem is degenerate. The 78
* condition (8) is executed for any 0 . Thereby, all conditions of
the theorem 1 are executed in the point (u* 1, *0 0, * 0) . We notice, the ordinary Lagrange’s function L (u, ) (u 1)
(u 1) 2 for the problem has not saddle point. Example 2. Let J (u ) J (u1 , u 2 ) u1 cos u 2 ,
U (u1 , u2 ) E 2 g (u1 ) u1 0 . Ensemble U 0 E 2 . Generalized Lagrange’s function
L (u1 , u 2 , 0 , ) 0 (u1 cos u 2 ) u1 , 0 0, 0,
u (u1 , u2 ) E 2 . Since ensemble U 0 E 2 , that condition (7) is written as L u (u * , * ) 0 . Thence follows that
g i (u * ), e 0,
i m 1, s . Consequently, *0 0, u 2* k , k 0,1,2, , * * . From condition (9) follows that it is possible to take 0 1 ,
where 2 . From condition (8) we have u1* 0 . Finally, necessary conditions of the optimality (6)(8) are executed in the point u * (u1* , u 2* ) (0, k ), *0 1, * 1 . To define in which
points from (0, k ) is reached minimum J (u ) on U the additional researches is required.. It is easy to make sure in the points u* 0, (2m 1) 0, k minimum J (u ) on U is reached, where m 0,1, . It is required to construct of the cones to prove the theorem 1: K y  directions of the function J (u ) decrease in the point u* ; K bi  internal directions of the ensembles U i , i 0, m in point u* ;
K m1 K k  tangent directions for ensemble U m1 in the point u* . Cone construction. We define the cones K y , K bi , i 1, m,
K k in the point u* U . 79
Definition 1. By direction of the function decrease J (u ) in the n point u* U is identified vector e E , e 0 if the numbers
0 0, 0 exist such that under all e o( , e) e E n
e e the following inequality J (u* e ) J (u* ), , 0 0
(10)
is executed. We denote through K y K y (u* ) the ensemble of all directions of the function decrease J (u ) in the point u* . Finally, the ensemble
K y e E n e e , J (u* e ) J (u* ),
0 0 .
(11)
As follows from the expression (11), ensemble K y contains together with the point е and its a neighborhood, consequently, K y  open ensemble. The point e K y . Since function
J (u ) C 1 (U 01 ) , that difference [refer to the formula (10), e K y ]:
J (u* e) J (u* ) J (u* e), e 0, , 0 0 . We notice, that open ensemble U 01 contains the neighborhood of the point u* U . Thence, dividing in 0 and directing 0 , we get J (u* ), e 0 . Consequently, ensemble
K y e E n J (u* ), e 0
(12)
 open convex cone. By Farkas’s theorem the dual cone to cone (12) is defined on formula 80
K y* c y E n c y *0 J (u* ), *0 0 .
(13)
Definition 2. By internal direction of the ensemble U i , in the n point u* U is identified the vector e E , e 0 , if there are
the numbers 0 0, 0 ,
such that under all e o( , e)
inclusion u* e U i , , 0 0 exists. We denote through K bi K bi (u* ) the ensemble of all internal ensemble U i directions in the point u* U . Finally, the ensemble
K bi e E n e e , u* e U i ,
, 0 0 .
(14)
We notice, that K bi  the open ensemble moreover the point
e K bi . For problem (1), (2) ensemble U i , i 1, m is defined by the formula (3), i.e.
U i u E n g i (u ) 0 , i 1, m Then ensemble K bi defined by formula (14) can be written as
K bi e E n g i ( u* e ) 0, , 0 0 .
(15)
Since the point u* U , that it is possible two events: 1) g i (u* ) 0 ; 2) g i (u* ) 0 . We consider the event, when g i (u* ) 0 . In this case, on the strength of function continuity g i (u ) on ensemble U 0 the number 0 0 such that value g i ( u* e) 0 under all
, 0 0 and for any vector e E n is found. Then ensemble K bi E n  open cone, but dual to it cone K bi 0 .
In the event g i (u* ) 0 we have g i ( u* e) g i ( u* ) 0 81
[refer to formula (15)]. Thence with consideration of that function g i (u ) C 1 (U 01 ) , we get g i (u* e), e 0, , 0 0 . We divide into 0 and direct 0 , as a result the given inequality is written in the manner of g i (u* ), e 0 . Then open ensemble K bi is defined by formula
K bi e E n g i (u* ), e 0 , i 1, m.
(16)
By Farkas’s theorem the dual cone to cone (16) is written as:
K b*i ci E n ci *i g i (u* ), *i 0 , i 1, m.
(17)
We notice, that K bi , i 1, m  opened convex cones. Definition 3. By tangent direction to ensemble U m1 in the point
u* U is identified the vector e E n , e 0 , if the number
0 0 and function r ( ) (r1 ( ),, rn ( )), 0 0 such that r (0) 0; r ( ) / 0 under 0 and u* e r ( ) U m 1 , , 0 0 exist. We denote
through K m 1 K m1 (u* ) the ensemble of all
tangent directions of the ensemble U m 1 in the point u* U . From given determinations follows that ensemble
K m 1 e E n u* e r ( ) U m 1 , , 0 0 ;
r (0) 0, r ( ) / 0 при 0.
According to formula (3) the ensemble
U m1 u E n g i (u ) 0, i m 1, s , consequently, 82
K m1 e E n g i (u* e r ( )) 0, i m 1, s, , 0 0 ,
(18)
where vectorfunction r ( ) possesses by the properties r ( 0 ) 0; r ( ) / 0 under
0 . Since functions g i (u ) C 1 (U 01 ),
i m 1, s and g i (u* ) 0, i m 1, s, u* U , that differences
g i (u* e r ( )) g i (u* ) g i (u* ), e r ( ) oi ( , u* ) 0, i m 1, s .. Thence, dividing in 0 and directing 0 , with consideration of that r ( ) / 0, oi ( , u* ) / 0 under 0 , we get g i (u* ), e 0, i m 1, s . Consequently, ensemble (18)
K m 1 e E n g i (u* ), e 0, i m 1, s .
(19)
 closed convex cone. Dual cone to cone (19) by Farkash’s theorem is defined by formula s K m* 1 c E n c *i g i (u* ) . i m 1
(20)
Finally, we define K 0  an ensemble internal directions of the convex ensemble U 0 in the point u* U U 0 . It is easy to make
sure in that, if u* intU 0 , that K 0 E n , consequently, K * 0. If u* pU 0 , that
K 0 e E n e (u u* ), u int U 0 , 0
(21)
 an open convex cone, but dual to it cone
K 0 с 0 E n с 0 , u с 0 , u * при всех u U 0 .
(22)
Cone’s constructions are stated without strict proofs. Full details 83
of these facts with proofs reader can find in the book: Vasiliev F. P. Numerical solution methods of the extreme problems. M.: Nauka, 1980. Hereinafter, We denote by K i , i 1, m the cones K bi , i 1, m . Lemma. If u* U is a point of the function minimum J (u ) on ensemble U , that necessary the intersection of the convex cones
K y K 0 K 1 K m K m 1 .
(23)
Proof. Let the point be u* U * U . We show, that correlation (23) is executed. We suppose opposite, i.e. existence of the vector
e K y K 0 K 1 K m K m 1 . From inclusion e K y follows that J (u* e ) J (u* ) under all , 0 y , e e y , but from e K i , i 0, m follows that u* e U i under all , 0 i , e e i , i 0, m . Let numbers be min( y , 0 ,, m ), min( y , 0 ,, m ) . Then
inequality
J (u* e ) J (u* )
is
true
and
inclusion
m
u* e U i exists under all , 0 and e e . i 0
u ( ) u* e r ( ) U m 1 under all , 0 m 1 , r (0) 0; r ( ) / 0 under 0 . We choose the vector e e r ( ) / . If m1 and m 1 0  sufficiently small number, that norm e e Since
vector
e K m1 ,
the
point
r ( ) / . Then the point u* e u* e r ( ) U 0 U 1 U m U m 1 U . Finally, the point u* e U and J (u* e ) J (u* ) . It is 84
impossible, since the point u* U U 0 is solution of the problem (1), (2). The lemma is proved. Previously than transfer to proof of the theorem, we note the following: 1) If J (u* ) 0, that under *0 1, 1* *2 ... *s 0 all condition of the theorem 1 are fulfilled. In fact, norm * 1 0 is a scalar product
L u (u * , * ), u u * *0 J (u * ), u u * 0,
u U 0 , *i g i (u* ) 0, i 1, s. 2) If g i (u* ) 0 under a certain i, 1 i m , then for values
*i 1, *j 0, j 1, s, j i , all conditions of the theorems 1 are also executed. In fact, * 1 ,
L u (u* , * ), u u* *i g i (u* ),
(u* ), u u* 0, u U , *i g i (u* ) 0, i 1, s .
3) Finally if the vectors g i (u* ), i m 1, s are linearly dependent, also the conditions (6)  (8) of the theorem 1 are occurred. In fact, in this case the numbers *m 1 , *m 2 , ., *s not all equal to zero, such that
*m 1 g m 1 (u* ) *m 2 g m 2 (u* ) *s g s (u* ) 0 . exist. We take *0 1* *m 0 . Then norm, * 0 , L u ( u * , * ), u u *
s
i m 1
* i
g i ( u * ) , u u * 0 ,
u U 0 , *i g i ( u * ) 0 , i 1, s ,
since g i (u* ), i m 1, s . From items 1  3 results that theorem 1 should prove for event, 85
J (u* ) 0, g i (u* ) 0 under all i 1, m , and vectors
when
g (u ), i m 1, s are linearly independent. i
*
Proof of the theorem 1. By the data of the theorem the ensemble u* U . Then, as follows from proved lemma, the intersection of the convex cones
K y K 0 K 1 K m K m 1
(24)
in the point u* U * , moreover all cones, except K m 1 , are open. We
J (u* ) 0, g i (u* ) 0 , i 1, m ;
notice, that in the case of
g (u ), i m 1, s i
*
are linearly independent, all cones K y ,
K 0 , , K m 1 are inempty [refer to the formulas (12), (16), (19),
(21)]. Then by DubovickyMilutin’s theorem, for occurring of the correlation (24) necessary and sufficiently existence of the vectors c y K *y , c0 K 0* , c1 K1* ,, cm1 K m* 1 , not all equal to zero and such that
c y c0 c1 cm1 0 .
(25)
As follows from formulas (13), (17), (20) the vectors
c y *0 J (u* ), *0 0, ci *i g i (u* ), *i 0, i 1, m, s
cm1 *i g i (u* ) . i m 1
We have c 0 c y c1 c m c m 1 *0 J (u * )
s
g (u ) i 1
* i
i
from equality (25). Since cone K 0* is defined by formula (22), that s
c0 , u u 0 *0 J (u* ) *i g i (u* ), u u 0 i 1
L u (u* , * ), u u 0 0, u U 0 . 86
*
Thence follows fairness of the correlations (7). The condition (6) follows from that not all c y , c 0 , c1 , , c m , are equal to zero. We notice, that if g i (u* ) 0 for a certain i from 1 i m , that cone
Ki E n ,
consequently,
K i* 0.
It
means
that
ci *i g i (u* ) 0, g i (u* ) 0 . Thence follows that *i 0 . Thereby, the products i g i (u* ) 0, i 1, s , i.e. the condition (8) is occurred. Theorem is proved. *
87
Lecture 13 SOLUTION ALGORITHM OF THE NONLINEAR PROGRAMMING PROBLEM We show a sequence of the solution of the nonlinear programming problem in the following type:
J (u ) inf ,
(1)
u U u E n u U 0 , g i (u ) 0, i 1, m;
g i (u ) 0, i m 1, s ,
(2)
1 1 where functions J (u ) C (U 01 ), g i (u ) C (U 01 ), i 1, s, U 01
 an open ensemble containing convex ensemble U 0 from E n , in particular, U 01 E n on base of the correlations (6)  (8) from theorem 1 of the previous lecture. We notice, that these correlations are true not only for the point u* U * , but also for points of the local function minimum J (u ) on ensemble U. The question arises: under performing which conditions of the problem (1), (2) will be nondegenerate and the point u* U will be a point of the local minimum J (u) on U? 10. It is necessary
U * u* E
n
to
that ensemble u* U , J (u* ) min J (u ) for which to use
the theorems 13 (the lecture 2). 20. To form the generalized (1), (2):
make uU
sure
in
Lagrange’s function for problems
88
s
L (u, ) 0 J (u ) i g i (u ), u U 0 ; i 1
(0 , 1 , , s ) 0 E s 1 / 0 0, 1 0, , m 0. 30. To find the points u* (u1* , , u n* ) U ,
* (*0 , ,
*s ) 0 from the following conditions: * , *0 0, 1* 0, , *m 0 ,
(3)
L u (u * , * ), u u * s
*0 J (u* ) *i g i(u* ), u u* 0, u U 0 ,
(4)
i 1
*i g i (u* ) 0, i 1, s,
(5)
where 0  the number, in particular 1 . a) If the point u* int U 0 or U 0 E n , that condition (4) can be replaced on s
L u (u* , * ) *0 J (u * ) *i g i (u* ) 0 .
(6)
i 1
In this case we have a system n 1 s of the algebraic equations (3), (5), (6) to determinate n 1 s of the unknowns u1* ,, u n* ; *0 ,, *s . b) If after solution of the algebraic equations system (3), (5), (6) [or formulas (3)  (5)] it is turned out that value *0 0 , that problem (1), (2) is identified by nondegenerate. The conditions (3) in it can be changed by more simpler condition *0 1, 1* 0, *2 0,, *m 0 . If in the nondegenerate problem a pair (u* , * ), s* (1* , , *s ) forms saddle point of the Lagrange’s function 89
s
L (u, ) J (u ) i g i (u ), u U 0 , i 1
0 E s / 1 0, , m 0, that u* U is the point of the global minimum. 4°. We consider the necessary problem of the nonlinear programming:
J (u ) inf
(7)
u U u E n g i (u ) 0, g i (u ) C 1 ( E n ), i 1, m .
(8)
The problem (7), (8) is a particular event of the problem (1), (2). The ensemble U 0 E n . The point in the problem (7), (8) is named
by normal point of the minimum, if the vectors g i (u* ), i 1, m are linearly independent. We notice, that if u* U  normal point, that problem (7), (8) are nondegenerate. In fact, for problem (7), (8) the equality m
*0 J (u * ) *i g i (u * ) 0
(9)
i 1
exists. If *0 0 , that on the strength of linearly independence of the
* vectors g i (u* ), i 1, m we would get i 0, i 1, m . Then the
vector 0 that contradicts to (3). *
Let u* U  a normal point for problem (7), (8). Then it is possible to take *0 1 , and Lagrange’s function has the form m
L (u, ) J (u ) i g i (u ), u E n , (1 , , m ) E m i 1
90
Theorem. Let functions J (u ), g i (u ), i 1, m are determined and
twice continuously differentiable in neighborhoods of the point u* U . In order the point u* U to be a point of the local minimum J (u ) on
ensemble U , i.e. J (u* ) J (u ), u, u o(u* , ) U sufficiently
2 L (u* , * ) that quadratic form y y to be positively determined on u 2 hyperplane
g1 u* / u1 g u* g 2 u* / u1 y u g u / u 1 m * *
g1 u* / u n y1 g 2 u* / u n y 2 0. (10) g m u* / u n y n
Proof. Let quadratic form
y
2 L (u* , * ) y y 2
2 L (u* , * ) / u12 2 L (u* , * ) / u2 u1 ( y1 , , yn ) 2 L (u , * ) / u u * 1 n
2 L (u* , * ) / u1un 2 L (u* , * ) / u2 un 2 L (u* , * ) / un2
y1 y 2 0, y 0, y n on hyperplane (10). We show that u* U  a point of the local 91
function minimum J (u ) on ensemble U. We notice, that u*  a normal point and it is determined from the conditions
L u (u* , * ) 0, *i g i (u* ) 0, i 1, m , i.e. g i (u* ) 0, i 1, m . Finally, pair (u* , * ) is known. Since functions
J (u ) C 2 (o(u* , )), g i (u ) C 2 (o(u* , )) , 2 * 2 that continuous function y ( L (u* , ) / u ) y comparatively
variable у reaches the lower bound on compact ensemble V y E n y 1, (g u* / u ) * y 0 . Let the number
min y
2 L (u* , * ) y, y V u 2
We enter the ensemble
A u E n u u* y, y 1, (g u* / u )* y 0 ,
(11)
where *  a transposition sign for matrixes; 0  sufficiently small number. If the point u A , that quadratic form
(u u* )L uu (u* , * )(u u* ) 2 y L uu (u* , * ) y 2 , where
L uu (u * , * ) 2 L (u * , * ) / u 2
on
the
strength
correlation (11). For the points u A difference
1 L (u, * ) L (u* , * ) L u (u* , * ), u u* (u u* ) 2
L uu (u* , )(u u* ) o u u* *
92
2
2 2
o( 2 )
(12) of
o( 2 ) 2 2 2 , 0 1 , 4 2 and
since
L u (u* , * ) 0
inequality
(12)
faithfully
(13)
and
o( ) / 0 under 0 . 2
2
We enter the ensemble
B u E n u u* , g i (u ) 0, i 1, m U .
(14)
Since hyperplane (10) is tangent to manifold g i (u ) 0, i 1, m
in the point u* , then for each point u~ B the point u A such
2 that norm u~ u K , K const 0 is found. In fact, if
u A : g i (u ) g i (u * ) g i (u ) g i (u * ), u u *
1 2 (u u* ) g iuu (u* ) (u u* ) oi u u* 2 1 2 y g iuu (u * ) y oi ( ), y 1; i 1, m; 2
(15)
1 u~ B : g i (u~ ) g i (u* ) 0 g i(u* ), u~ u* (u~ u* ) 2 2 g iuu (u * )(u~ u * ) oi u~ u * g i (u* ), u~ u u u *
1 ~ (u u u u * ) g iuu (u* )(u~ u u u * ) oi ( 2 ) 2 1 1 g i (u * ), u~ u (u~ u ) g iuu (u * )(u~ u * ) (u u * ) 2 2 ~ g (u )(u u ) (u u ) g (u )(u u )
iuu
*
*
iuu
*
*
oi ( 2 ), i 1, m;
93
(16)
2 From (11), (14) and (15), (16) follows that norm u~ u K ,
u~ B , u A . Since the function L u (u , * ) continuously differentiable by u in neighborhoods of the point u* , and derivative L u (u* , * ) 0 , * that difference L u (u, ) L u (u* , ) L u (u , * ) L uu (u* , )(×
*
×(u u* ) o u u*
*
in neighborhoods of the point u* . This
L (u , * ) K 1 , if
implies that, in particular, u A , norm
0 1 , 1 0  sufficiently small number. ~, * ) L For the points u A , u~ B the difference L (u 1 L (u, * ) L u (u , * ), u~ u (u~ u )L uu (u* , * )(u~ u ) 2 2 consequently, norm L (u~, * ) L (u, * ) o u~ u ,
*
KK 1 3 o1 3 K 2 3
under sufficiently small
1 0,
0 1 . Then the difference (u A , u~ B ) L (u~, * ) L (u* , * ) L (u, * ) L (u* , * ) L (u~, * ) L (u, * ) L (u, * ) L (u* , * ) L (u~, * ) L (u, * )
4
2 K 2 3 0, 0 1 ,
(17)
on the strength of correlations (13). So as values L (u~, * ) J (u~ ),
L (u* , * ) J (u* ) , since g i (u~ ) g i (u* ) 0, i 1, m then from (17) we have J (u ) J (u~ ) . Consequently, u U  the point of the *
*
local function minimum J (u ) on ensemble U . The Theorem is proved. Example. Let function be J u u12 5u1u 2 u 22 , ensemble 94
U u (u1 , u 2 ) E 2 u12 u 22 1 . To find the function minimum J (u ) on ensemble U . For the example the ensemble U 0 E 2 , functions J (u ) C 2 ( E 2 ), g i (u ) u12 u22 1 C 2 ( E 2 ) , ensemble U * , since U E 2  compact ensemble. Necessary optimality conditions
2u1* 5u 2* 2*u1* 0, L u (u , ) 0 : 2u 2* 5u1* 2*u 2* 0, *
*
u
g (u* ) 0 : u1*
2
* 2 2
1,
where function L (u, ) J (u ) g (u ), u E 2 , E 1 . Thence we find the points u1* , u 2* , * :
1) * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 ; 2) * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 ; 3) * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 ; 4) * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 . It is necessary to find in which point ( , u1 , u2 ) from 14 the minimum J (u ) on U is reached. First of all we select the points where local minimum J (u ) on U is reached. We note, that problem *
*
*
is nondegenerate and matrix L uu (u * , * ) and vector equal to
2 2* L uu (u* , * ) 5
g u ( u* )
are
2u * 5 , g u (u* ) g (u* ) 1* . 2u 2 2* 2
For the first point * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 quadric form is
y L uu (u* , * ) y y12 2 5 y1 y 2 5 y 22 , a hyperplane 95
equation 2 5 / 6 y1 2 1 / 6 y 2 0 . Thence we have y 2 5y1 . Substituting the values y 2 5 y1 to the quadric form we get
y L uu (u * , * ) y 36 y12 0, y1 0 .
(* 3 / 2,
Consequently,
u1* 5 / 6 , u 2* 1 / 6 )  point of the local minimum J (u ) on U . By the similar way it is possible to make sure in that * 3 / 2, u1* 5 / 6 , u 2* 1 / 6  a point of the local
minimum, but the points * 3 / 2, u1* 5 / 6 , u 2* 1 / 6 and
3 / 2, u 5 / 6 , u 1 / 6 are not the points of the local minimum J (u ) on U . In order to find minimum J (u ) on U necessary to calculate the values of the function J (u ) in the points *
* 1
* 2
1) and 2). It can be shown, that. J ( 5 / 6 , 1 / 6 ) J ( 5 / 6 ,
, 1 / 6 ) 3 / 2 . Consequently, in the points 1) and 2) the global minimum J (u ) on U is reached. 50. Now we consider the problem (1), (2) in the case
U 0 E n , J (u ) C 2 ( E n ), g i (u ) C 2 ( E n ), i 1, s . We suppose,
that problem (1), (2) is nondegenerate and the points u* , *0 0, * are determined by algorithm 10  30. We select amongst constraints
g i (u ) 0, i 1, m, g i (u ) 0, i m 1, s for which g i (u* ) 0, i I , where index ensembles
I i i m 1, s, g i (u* ) 0, 1 i m . If the problem (1), (2) is nondegenerate, that vectors g i (u* ), i I are linearly independent. According to specified above theorem from item 40 the point u* U is the point of the local minimum to functions J (u ) on U , if quadratic form y L uu (u* , ) y 0, y 0 on hyperplane *
g i u* / u *
y 0, i I . 96
Lecture 14 DUALITY THEORY On base of the Lagrange’s function the main and dual problem are formulated and relationship between its solutions is established. The dual problems for the main, general and canonical forms writing of the linear programming problem are determined. Since dual problem  a problem of the convex programming regardless of that whether the main problem is the problem of the convex programming or not, then in many events reasonable to study the dual problem and using the relationship between its solutions to return to the source problem. Such acceptance often is used at solution of the linear programming problem. We consider the nondegenerate nonlinear programming problem in the following type:
J (u ) inf ,
(1)
u U u E n u U 0 , g i (u ) 0, i 1, m;
g i (u ) 0, i m 1, s ,
(2)
For problem (1), (2) Lagrange’s function s
L (u, ) J (u) i gi (u), u U 0 ; i 1
(0 , 1 ,, s ) 0 E 1 0,, m 0.
(3)
s
The main task. We enter the function [refer to. the formula (3)]: 97
X (u ) sup L (u, ), u U 0 .
(4)
0
We show, that function
J (u ) при всех u U , X (u ) при всех u U 0 \ U . In fact, if
u U , that
g i (u ) 0,
(5)
i 1, m, g i (u ) 0,
i m 1, s , consequently, s X (u ) sup J (u ) i g i (u ) 0 i 1 m sup J (u ) i g i (u ) J (u ) , 0 i 1
since 1 0,, m 0,
m
g (u ) 0 i 1
i
i
under all u U , moreover
0 0 . If u U 0 \ U , that possible for a certain number i 1 i m, g i (u ) 0 and for a certain j from m 1 j s, g j (u ) 0 . In the both events by choice a sufficient
from
big i 0 or j kg j (u ), k 0  sufficiently large number, the value X (u ) can be done by arbitrary large number. Now the source problem (1), (2) can be written in the equal type:
X (u ) inf, u U 0 ,
(6)
on the strength of correlations (4), (5). We notice that inf X (u ) uU 0
inf J (u ) J * , consequently, if the ensemble U * , that uU
98
U * u* U min J (u ) J (u* ) J * uU
u* U 0 X (u* ) J (u* ) J * min X (u ) . uU
The source problem (1), (2) or tantamount its problem (6) is named by the main task. Dual problem. On base of the Lagrange’s function (3) we enter function
( ) inf L (u, ), 0 .
(7)
uU
Optimization problem of the following type:
( ) sup, 0 ,
(8)
is called by dual problem to problem (1), (2) or tantamount its problem (6), but Lagrange’s multipliers (1 ,, s ) 0  dual variables with respect to variables u (u1 ,, u n ) U 0 . We denote through
* * E s * 0 , (* ) max ( ) 0 We notice, that if Ø , that (* ) sup ( ) * . *
0
Lemma. The values J * inf X (u ), * sup ( ) for main uU 0
0
(6) and dual (8) problems accordingly satisfy to the inequalities
* J * X (u ), u U 0 , 0 . Proof. As follows from formula (7), function
(9)
( )
inf L (u, ) L (u, ), u U 0 , 0 . Thence we have uU 0
* sup ( ) sup L (u, ) X (u ), u U 0 , 0
0
99
(10)
on the strength of correlations (4). From correlations (10) transferring to lower bound on u we get * inf X (u ) J * . uU 0
Thence and from determinations of the lower bound the inequalities (9) follow. Lemma is proved. Theorem 1. In order to execute the correlations
U * , * , X (u* ) J * * * ,
(11)
necessary and sufficiently that Lagrange’s function (3) has saddle point on ensemble U 0 0 . The ensemble of the saddle points to function L (u , ) on U 0 0 coincides with ensemble U * * .
u* U * , * * correlations (11) are complied. We show, that pair u* , *  saddle point to Lagrange’s function (3) on ensemble U 0 0 . Since Proof. Necessity. Let for the points
* * inf L (u, * ) L (u* , * ) sup L (u* , ) X (u* ) uU 0
0
J * , that on the strength of correlations (11) we have L (u* , * ) inf L (u , * ) sup L (u* , ) , uU 0
0
u* U * , * 0 .
(12)
From inequality (12) follows that
L (u* , ) L (u* , * ) L (u, * ), u U 0 , 0 .
(13)
It means that pair u* , * U * *  saddle point. Moreover ensemble U * belongs to the ensemble of the saddle points to *
Lagrange’s function, since u* U * , * *  arbitrary taken points from ensemble U * , * accordingly. Necessity is proved.
Sufficiency. Let pair u* , * U 0 0 be a saddle point to 100
Lagrange’s function (3). We show that correlations (11) are fulfilled. As follows from determination of the saddle point which has the form (13) inequality L (u* , ) L (u* , * ), 0 faithfully. Consequently,
X (u* ) sup L (u* , ) L (u* , * ) .
(14)
0
Similarly from the right inequality (13) we have
(* ) inf L (u, * ) L (u* , * ).
(15)
uU 0
From inequalities (14), (15) with consideration of correlation (9) we get
L (u* , * ) (* ) * J * X (u* ) L (u* , * ) . Thence we have (* ) * J * X (u* ) . Consequently, ensemble U * , and moreover ensemble of the saddle *
points to function (3) belongs to the ensemble U * * . The theorem is proved. The following conclusions can be made on base of the lemma and theorem 1: 1°. The following four statements are equivalent: a) or u* , * U 0 0  the saddle point to Lagrange’s function (3) on
ensemble U 0 0 ; b) or correlations (11) are executed; c) or the points u* U 0 , * 0 such that X (u* ) (* ) exist; d) or equality
max inf L (u, ) min sup L (u, ) 0 uU 0
uU 0 0
is equitable. 2°. If u* , * , a* , b * U 0 0 are saddle points of the
Lagrange’s function (3) on U 0 0 , that u* , b* , a* , *
 also
saddle points to function (3) on U 0 0 , moreover L (u* , b* ) L 101
L (a* , * ) L (u* , * ) L (a* , b* ) (* ) X (u* ) J * * . However inverse statement, i.e. that from L (u* , * ) L (a, b) follows a, b U 0 0  saddle point in general event untrue. 3°. Dual problem (8) it is possible to write as
( ) inf, 0 .
(16)
Since function L (u , ) is linear by on convex ensemble 0 , that optimization problem (16) is the convex programming problem, ( ) is convex on 0 regardless of that, the main task (1), (2) would be convex or no. We notice, that in general event the dual problem to dual problem does not comply with source, i.e. with the main problem. There is such coincidence only for tasks of the linear programming. We consider the problem of the linear programming as applications to duality theories. The main task of the linear programming has the form
J (u ) c, u inf,
(17)
u U u E n u 0, Au b 0 ,
where c E n , b E m are the vectors; А is the matrix of the order m n ; the ensemble
U 0 u E n u u1 0, , u n 0 0 Lagrange’s function for task (17) is written as
L (u, ) c, u , Au b c A* , u b, ,
u U 0 , (1 , , m ) 0 E m 1 0, , m 0 . As follows from formulas (17), (18), function
102
(18)
b, , если c A* 0 inf L (u, ) uU 0 , если ( c A* ) i 0, 0 . We notice, that under c A* 0 lower bound which is equal to b,
is got under u 0 U 0 . If ( c A* ) i 0 , that it is
possible to choose ui , but all u j 0, j 1, n, i j ,
herewith , * 0 . Finally, the dual task to task (18) has the form
b, inf,
(19)
E m 0, c A* 0.
The dual task to task (19) complies with source task (17). By introducing of the additional variables u n i 0, i 1, m , optimization problem (17) can be written in the following type:
J (u ) inf
Au i bi u ni 0,
u 0, u n i
. 0, i 1, m
(20)
General task of the linear programming has the form
J (u ) c, u inf,
u U u E n u j 0, j I , Au b 0, A u b 0 ,
(21)
where c E n , b E m , b E s are the vectors; A, A are the matrixes accordingly to the orders m n, s n ; index ensemble
I 1,2,, n. The ensemble
U 0 u u1 ,, u n E n u j 0, j I Lagrange’s function for task (21) is written as 103
L (u , ) c, u , Au b , A u b c A* A * , u b, b , ,
u U 0 , ( , ) 0 ( , ) E m E s 0 . Function
inf L(u, ) uU 0
b, b , , if c A* A * i 0, i I ; c A* A * j 0, j I ; under rest.
The dual task to task (21) is written as
b, b , inf; c A* A *
c A A *
*
j
i
0, i I ;
0, j I ; ( , ) E n E s , 0.
(22)
It is possible to show that dual task to task (22) complies with (21). By introducing of the additional variables u n i 0, i 1, m and representations u i qi vi , q i 0, vi 0, i I (21) possible write as
J (u ) c, u inf,
Au i bi u ni
the problem
0, A u b 0,
u j 0, j I , u i q i vi , qi 0, vi 0, i I .
(23)
Canonical task of the linear programming has the form
J (u ) c, u inf,
u U u E n u 0, Au b 0 , 104
(24)
where c E n , b E s are the vectors; А is the matrix of the order s n ; the ensemble
U 0 u E n u u1 , , u n 0
Lagrange’s function for task (24) is written so:
L (u , ) c, u , Au b c A* , u b, ,
u U 0 , 0 E s . Function
b, , если c A* 0, , если c A* 0.
( ) inf L (u, ) uU 0
Then the dual problem to the problem (24) has the form
( ) b, inf;
c A* 0, E s .
(25)
It is easy to make sure in the dual task to task (25) complies with task (24). Finally, we note that main and the general task of the linear programming by the way of introducing some additional variables are reduced to the canonical tasks of the linear programming [refer to formulas (20), (23)].
105
Chapter III. LINEAR PROGRAMMING
As it is shown above, the main and the general problem of the linear programming are reduced to the canonical problems of the linear programming. So it is reasonable to develop the general solution method of the canonical linear programming problem. Such general method is a simplexmethod. Simplexmethod for nondegenerate problems of the linear programming in canonical form is stated below.
Lectures 15, 16 STATEMENT OF THE PROBLEM. SIMPLEXMETHOD We consider the linear programming problem in the canonical form
J (u ) c, u inf, u U u E n u 0, Au b 0 , (1) where c E n , b E m are the vectors; А is the matrix of the order
m n . Matrix A aij , i 1, m, j 1, n can be represented in the manner of
106
a1 j a2 j n j 1 2 A a , a ,..., a , a , j 1, n . ... a mj
The vectors a j , j 1, n are identified by the condition vectors, but vector b E m  a vector of the restrictions. Now the equation Au b can be written in the manner of a1u1 a 2 u 2 ... a n u n b
. Since ensemble U 0 u E n u 0 and U u E n Au b  affine ensemble which are convex, that (1) is a problem of the convex programming. We notice that if ensemble
U * u* E n u* U , J (u* ) min c, u , uU
that Lagrange’s function for problem (1) always has saddle point, any point of the local minimum simultaneously is the point of the global minimum and necessary and sufficiently condition of the optimality is written as J ' (u* ), u u* c, u u* 0, u U 0 . We suppose, that ensemble U * . It is necessary to find the point
u* U * and value J * inf J (u ) J (u* ) min J (u ) . uU
uU
Simplexmethod. For the first time solution of the problem (1) was considered on simplex n U u E n u 0, ui 1 i 1
so solution method of such linear programming problem was called by simplexmethod. Then method was generalized for event of ensemble U specified in the problem (1), although initial name of the method is kept. Definition 1. The point u U is identified by extreme (or angular), if it is not represented in the manner of 107
u u 1 1 u 2 , 0 1, u 1 , u 2 U . From given definition follows that extreme point is not an internal point of any segment belonging to ensemble U. Lemma 1. Extreme point u U has not more m positive coordinates. Proof. Not derogating generalities, hereinafter we consider that first components k , k m of the extreme point are positive, since by the way of recalling the variables always possible to provide the given condition. We suppose opposite, i.e. that extreme point u U has m 1 positive coordinates
u u1 0, u 2 0, ..., u m1 0,
0,...,0 .
We compose the matrix A1 (a 1 , a 2 ,..., a m 1 ) of the order
m m 1 from condition vectors corresponding to positive coordinates of the extreme point. We consider the homogeneous linear equation A1 z 0 for vector z E m 1 . The equation has a
z ,0 and nonzero solution ~ z, ~ z 0 . We define nvector u~ ~
consider two vectors: u 1 u u~, u 2 u u~ , where 0 
sufficiently small number. We notice that vectors u 1 , u 2 U under
0 1 , where 1 0 is sufficiently small number. In fact, Au 1 Au A u~ Au A1 ~ z Au b, u 1 u u~ 0 under sufficiently small 1 0 similarly Au 2 b, u 2 0 . Then the extreme point u 1 / 2 u 1 1 / 2 u 2 , 1 / 2 . It opposites to the definition of the extreme point. The lemma is proved. Lemma 2. Condition vectors corresponding to positive coordinates of the extreme point are linear independent Proof. Let u u1 0, u 2 0, ..., u k 0, 0,...,0 U , k m be an extreme point. We show that vectors a1 , a 2 ,..., a k , k m are linear independent. We suppose opposite, i.e. that there are the numbers 1 ,..., k not all equal to zero such that 108
1a1 2 a 2 ...
k ak 0 (the vectors a1 ,..., a k are linear dependent). From inclusion
u U
follows
a1u1 a 2 u 2 ... a k u k b,
that
ui 0, i 1, k . We multiply the first equality on 0 and add (subtract) from the second equality as a result we get a 1 u1 1 a 2 u2 2 ... a k u k k b . We denote by
u 1 u1 1 ,..., u k k ,0,...,0 E n , u 2 u1 1 ,...,
k ,0,...,0 E n . There is a number
1 0
such that u 1 U ,
u 2 U under all , 0 1 . Then vector u 1 / 2u 1 1 / 2u 2 U , u 1 U , u 2 U . We obtained the contradiction in that u U  extreme point. Lemma is proved. From lemmas 1, 2 follows that: a) The number of the extreme points of the ensemble U is finite m
and it does not exceed the sum
C k 1
k n
, where C nk  a combinations
number from nelements on k. In fact, the number of the positive coordinates of the extreme points is equal to k , k m (on the strength of lemma 1), but the number of the linear independent vectors corresponding to positive coordinates of the extreme point is equal to C nk (on the strength of lemma 2). Adding on k within from 1 till m we get the maximum possible number of the extreme points. b) The ensemble U u E n u 0, Au b 0 is convex polyhedron with final number of the extreme points under any matrix А of the order m n . Definition 2. The problem of the linear programming in canonical form (1) is identified nondegenerate if the number of the positive coordinates of he feasible vectors not less than rank of the matrix А, i.e. in the equation a1u1 a 2 u 2 ... a n u n b under
ui 0, i 1, n the number differenced from zero summands not less, than rank of the matrix A. Lemma 3. Let rangA m, m n . If in the nondegenerate 109
problem the feasible vector has exactly m positive coordinates, that u  an extreme point of the ensemble U. Proof. Let the feasible vector u u1 0, u m 0, 0,...,0 U has exactly m positive coordinates. We show, that u is an extreme point of the ensemble U . We suppose opposite i.e. there are the points 1 u , u 2 U , u 1 u 2 , and number , 0 1 such that
u u 1 1 u 2
(the point u U is not an extreme point).
From given presentations follows that u 1 u11 ,..., u 1m ,0,...,0 ,
u 2 u12 ,..., u m2 ,0,...,0 . Let the point u ( ) u (u 1 u 2 ),
u ( ) (u1 (u11 u12 ), u 2 (u 12 u 22 ),..., u m (u 1m u m2 ),0,...,0 . We notice, that Au ( ) Au
u u . 1
2
Consequently,
Au1 Au 2 b under any . We assume, there is negative amongst the first m coordinates of the vector u1 u 2 . Then by increasing 0 from 0 to we find the number 1 0 such that one from m first coordinates of the vector u 1 becomes equal to zero, but all rest will be nonnegative. But it isn’t possible in the nondegenerate problem. Similarly if u 1 u 2 0 that reducing from 0 till we get the contradiction. Lemma is proved. We notice, that from rangA m does not follow that problem of the linear programming (1) will be nondegenerate. Example 1. Let the ensemble
U u (u1 , u 2 , u 3 , u 4 ) E 4 u j 0, j 1,4; 3u1 u 2 u 3 u 4 3, u1 u 2 2u 3 u 4 1. In this case the matrix
3 1 1 1 a1 , a 2 , a 3 , a 4 , A 1 1 2 1 110
rangA 2.
By the extreme points of the ensemble U are u 1 1,0,0,0 ,
u 2 0,5 / 3, 4 / 3, 0, u 3 0,1,0,2 . Here u 2 , u 3 are nondegenerate extreme points, u 1 is degenerate extreme point. Since there is feasible vector u 1 , number of the positive coordinates less than rangA , the problem of the linear programming (1) is not nondegenerate. The number of the extreme points are equal to 3, that does not exceed the amount c14 c42 10 ; the vectors corresponding
to the positive coordinates of the extreme point u 2 a 2 ,a 3 ,
u a ,a 3
2
4
are linear independent.
Example 2. Let ensemble be
U u (u1 , u 2 , u 3 , u 4 ) E 4 u j 0, j 1,4; 3u1 u 2 u 3 u 4 3, u1 u 2 2u 3 u 4 1. Here rangA 2 , the extreme points u 1 1 / 2, 0, 0, 3 / 2 ,
u 2 5 / 7, 0, 6 / 7, 0,
u 3 0, 5 / 3, 0, 4 / 3, u 4 0, 1, 0, 2 .
The problem (1) is nondegenerate. We notice that in nondegenerate problem the number of the extreme points no more than C nm . Lemma 4. Any point u U can be represented as convex linear
combination points of the ensemble U, i.e. u
s
u k 1
s
k 1
k
k
k
, k 0,
1, u1 , u 2 ,..., u s  an extreme points of the ensemble U .
Proof. We prove the lemma for event, when U is convex bounded closed ensemble. In fact, for order the ensemble U * , necessary and sufficiency the ensemble U is compactly. It means that U is convex closed bounded ensemble. We prove the lemma by method of mathematical induction for events when u p U and u int U . Let u p U , u U . If n 1, that ensemble U is a segment; 111
consequently, statement of the lemma faithfully, since any point of the segment ( u p U , u int U ) can be represented in the manner of the convex combination of the extreme points (the end of the segment). Let the lemma be true for ensemble U E n 1 n 2 . We conduct supporting hyperplane to ensemble U E n through the point u pU , i.e. c, u c, u , u U . We denote by
U1 U , where u E n c, u c, u
 ensemble points
of the supporting hyperplane. We notice that ensemble U1 is
U1 E n 1 . Let u1 , u 2 ,..., u s1  extreme points of the ensemble U1 , then by
convex,
bounded
and
closed,
moreover
s1
hypothesis the point
u k u k E n 1 , k 1
s1
k 1
k
k 0, k 1, s1 ,
1 . It is remains to show, the points u1 , u 2 ,..., u s1 are extreme
points and ensembles U. Let u i w 1 v,
w, v U ,
0 1 . We show that u w v for any extreme point u i U 1 . i
In fact, since c, w c, u , c, v c, u that
and
c, u i c, u ,
c, u i c, w 1 c, v c, u c, u i . Thence
follows that c, w c, v c, u , i.e. w, v U1 . However the point u i  an extreme point U1 , consequently, u i w v . s
Therefore the points u1 , u 2 ,..., u 1 are extreme points of the ensemble U . For border point u pU . Lemma is proved. Let the point u int U . We conduct through u U a line l which crosses the borders of the ensemble U in the points a U , b U . At the point u int U is representable in the manner
of. u a 1 b, 0 1 . The points a pU ,
112
b pU
s1
a k v k , k 0,
on the strength of proved
k 1, s1 ,
k 1
s1
k 1 1
s2
k
1, b k w , k 0, k
k 1
k 1, s 2 ,
s2
k 1
k
1, where
v ,..., v s1 ; w1 ,..., w s2 are extreme points of the ensemble U . Then the point u a 1 b
k k 0, s1
s2
k 1
k
k 1
k
s1
s1
k 1
k 1
k v k 1 k wk , moreover
k 1, s1 ,
k 1 k 0,
k 1, s 2 ,
1 1 . Lemma is proved.
Lemma 5. Let U be convex bounded closed ensemble from E n , i.e. ensemble U * . Then minimum to function J (u ) on ensemble U is reached in the extreme point of the ensemble U. If minimum J (u ) on U is reached in several extreme points u 1 ,..., u k of the ensemble U, that J (u ) has same minimum value in any point k
k
u i u i , i 0,
i 1
i 1
i
1.
Proof. Let minimum J (u ) on U is reached in the point u* U * .
If u* U  an extreme point, that lemma is proved. Let u* U be certain border or internal point of the ensemble U. Then on the strength of lemma 4 we have u *
s
u , i 1
i
i
s
i 0, i 1 , where u1 ,..., u s  extreme points of the ensemble i 1
U . The value
J i J (u ) c, u , i 1, s. i
s
s
i 1
i 1
J (u* ) c, u* i c, u i i J i , where i
113
Let
J 0 min J i J (u i0 ) . Then 1i s
s
J (u* ) J 0 i J 0 i 1
J (u i0 ) . Thence we have J (u* ) J (u i0 ) , consequently, minimum J (u ) on U is reached in the extreme point u i0 . Let the value J (u* ) J (u 1 ) J (u 2 ) ... J (u k ) , where u 1 ,..., u k  extreme points of the ensemble U. We show that k k J u J i u i J u* , where i 0, i 1 . In fact, the i1 i 1
value J u
k
k
i 1
i 1
i J u i i J u* J u* . Lemma is proved.
We notice that lemmas 4, 5 are true for any problem of the linear programming in the canonical form of the type (1) with restrict closed convex ensemble U. From lemmas 1  5 follows that solution algorithm of the linear programming problem in canonical form must be based on transition from one extreme point of the ensemble U to another, moreover under such transition the value of function J (u ) in the following extreme point less, than previous. Such algorithm converges to solution of the problem (1) through finite number steps, since number of the extreme points of the ensemble U doesn’t m
exceed the number
C k 1
k n
(in the event of the nondegenerate
problem Cnm ). Under each transition from extreme point u i U to the extreme point u i 1 U necessary to make sure in that, whether the given extreme point u i U be solution of the problem (1). For this a general optimality criterion which easy checked in each extreme point must be existed. Optimality criterion for nondegenerate problem of the linear programming in canonical form is brought below. Let the problem (1) be nondegenerate and the point u* U  a solution of the problem (1). According to lemma 5 the point u* U 114
is extreme. Since the problem (1) is nondegenerate, that extreme point u* U has exactly m positive coordinates. Not derogating generalities, we consider the first m components of the vector u* U are positive, i.e.
u * u1* , u 2* ,..., u m* ,0,...,0 , u1* 0, u 2* 0,..., u m* 0 . The vector c E n and matrix А we present in the manner of c c Б , c Н , A AБ , AH , where c Б (c1 , c 2 ,..., c m ) E m ,
c H (c m 1 , c m 2 ,..., c n ) E n m ,
AБ (a 1 , a 2 ,..., a m ) ,
AH
(a m 1 , a m 2 ,..., a n ). We notice, that according to lemma 2 condition vectors a1 , a 2 ,..., a m corresponding to the positive coordinates of the extreme point u* U are linear independent, i.e. matrix AБ is nonsingular, consequently, there is the inverse matrix AБ1 . Lemma 6 (optimality criterion). In order the extreme point u* U to be a solution of the nondegenerate problem of the linear programming in canonical form (1), necessary and sufficiency to execute the inequality
c H' c Б' AБ1 AH 0.
(2)
Proof. Necessary. Let the extreme point be u* u Б , u H* U ,
where u u , u ,..., u , u 0,...,0  a solution of the nondegenerate problem (1). We show, that inequality (2) is executed. Let u uБ , uН U , where uβ = (u1, …, um), uH = * Б
* 1
* 2
* m
* H
= (um+1, …un)  an arbitrary point. We define the ensemble of the feasible directions l to the point u* U . We notice that vector l E n , l 0 is identified by feasible direction in the point u* U , if there is the number 0 0 such that vector u u* l U 115
under all 0 0 . We present the vector l E n in the manner of
l lБ , lH , where lБ l1 ,..., lm E m , lH lm1 ,..., ln E n m .
From inclusion u* l U follows that u Б* l Б 0, u H* l H
l H 0, Au* l AБ u Б* l Б AH l H b . Since u* U , that Au* AuБ* b consequently from the last equality we have AБ lБ AH lH 0 . Thence follows the vector lБ АБ1 АН lH . Since under sufficiency small 0 inequality u Б* lБ 0 is executed, the feasible directions in the point u* are defined by the correlations l H 0,
l Б AБ1 AH l H .
(3)
Finally, having chosen the arbitrary vectors lH E n m , lH 0 possible to find lБ АБ1 АН lH and construct the ensemble of the feasible directions L, which each element has the form l АБ1 АН lH , lH 0 E n , i.e. L l E n l l Б , l H , l Б
AБ1 AH l H , l H
0 E
n
. Now any point u U
can be
u u* l , l L , 0, 0 0 , 0 0 l . Consequently, u u* l , l L .
represented
in
the
manner
of
1 Since the function J (u ) c, u C U , then in the point u* U
necessary inequality J u* , u u* 0, u U (the lecture 5) is executed. Thence with provision for that J ' u* c, u u* l ,
c, l cБ , lБ сН , lH 0 . Since the number 0 , that we have cБ , lБ сН , lH 0 . Substituting the l L we get
value lБ АБ1 АН lH from formula (3), we get
cБ , АБ1 АН lН сН , lH cH cБ АБ1 АН lH 0 under all lH 0 . Thence follows the inequality (2). Necessity is proved. 116
Sufficiency. Let the inequality (2) be executed. We show, that u* U* U  an extreme point of the ensemble U . Since the
function J u C1 U is convex on convex ensemble U, that
necessary and sufficiency the execution of the inequality J (u ) J (v) J (v), u v , u, v U (the lecture 4). Hence in particular, v u* U , then we have
J (u ) J (u* ) J ' (u* ), u u* c, u u* c, l
cБ , l Б cH , l H
c
c Б' AБ1 AH l H 0, l L, u U . ' H
J (u* ) J (u ), u U . Consequently, minimum is reached in the point u* U on U . According to the lemma 5 u* U Then
an extreme point. Lemma is proved. It is easy to check the optimality criterion (2) by simplex table formed for the extreme point u* U 0 . Ba sis
с
AБ
Сб
b
а1
а1
с1
u1*
u11
…
…
…
a
i0
…
ai …
a
m
с1
ci0
u
…
…
ci
ui*
…
…
cm z
z c
* i0
u
* m
…
…
…
сj
…
с j0
…
сn
aj u1 j
…
a j0 u1 j0
…
an u1n
…
… …
…
… …
…
… …
…
z1
…
zj
0
…
zj cj …
ui0 1 …
ui1 …
um1
ui0 j uij u mj
117
… …
…
… …
…
… …
…
…
… … …
…
… …
…
… …
…
z j0
…
zn
zj0 cj0
…
zn cn
ui0 j0 uij0 umj0
ui0 n
uin umn
…
0 …
i …
The condition vectors a1 , a 2 ,..., a m corresponding to positive
a ,..., a .
coordinates of the extreme point u* u1* ,..., um* , 0,..., 0 are leaded in the first column of the table. Matrix
АБ
1
m
Coordinates of the vector c c1 ,..., cn corresponding to positive coordinates of the extreme point are given in the second column; in the third  corresponding to a i positive coordinates of the extreme point. In the following columns the decomposition coefficients of the vector a j , j 1, n on base vector ai , i 1, n are reduced. Finally, in the last column the values i which will be explained in the following lecture are shown. We consider the values specified in the last two lines more detail. Since vectors a1 , a 2 ,..., a m are linear independent (the lemma 2), that they form the base in the Euclidean space E m , i.e. any vector a j , j 1, n can be singularity decomposed by this base. Consequently, m
a j a i uij АБ u j , u j u1 j , u2 j ,..., umj E m , j 1, n . i 1
Thence m
we
z j ci u ij ,
have
u j АБ1а j , j 1, n . We denote through
j 1, n . We notice, that z j c j , j 1, m , since
i 1
u j (0,...,0,1,0,...,0), j 1, m .. Then vector
zc where
z c
Б
, z c H 0, z c H ,
z c Б z1 c1 , z 2 c2 ,..., z m cm 0,...,0, z c H
z m 1 c m 1 ,..., z n c n . 118
Since the values z j
m
c u i 1
i ij
cБ u j cБ АБ1а j , j 1, n , that
z j c j c Б А a c j , j 1, n . Consequently, vector ( z c ) 1 Б
j
(0, c Б АБ1 AH c H ) . Comparing the correlation with optimality criterion (2), we make sure in the extreme point u* U to be a solution of the problem necessary and sufficiency that values z j c j 0, j 1, n . Finally, by sign of the values in the last line of the simplextable it is possible to define whether the point u* U a solution of the nondegenerate problem (1).
119
Lecture 17 DIRECTION CHOICE. NEW SIMPLEXTABLE CONSRUCTION. THE INITIAL EXTREME POINT CONSRUCTION In the linear programming problem of the canonical form minimum to linear function is reached in the extreme point of the convex polyhedron U. A source extreme point in the simplexmethod is defined by transition from one extreme point to the following, moreover value of the linear function in the next extreme point less, than in previous. It is necessary to choose the search direction from the extreme point and find the following, in the last to check the optimality criterion etc. Necessity in determination of the initial extreme point to use the simplexmethod for problem solution is appeared. We consider the nondegenerate problem of the linear programming in canonical form
J (u ) c, u inf, u U u E n u 0, Au b 0 .
(1)
Let u U be an extreme point and in the point u U the minimum J (u ) on U isn’t reached, i.e. optimality criterion
z j c j 0, j 1, n is not executed. Then necessary to go from given extreme point to other extreme point u U , where value J u J u . It is necessary to choose the direction of the motion from given extreme point u . 120
Direction choice. Since u U is an extreme point, not derogating generalities, it is possible to consider that the first m components of the vector u are positive, i.e. u (u1 , u 2 ,..., u m ,
0,...,0), u i 0, i 1, m . According to formula (2) (the lecture 16), feasible directions in the point u are defined from correlations lH 0, lБ АБ1 АН lH . Derivative of function J (u ) on feasible direction l in the point u is equal to
J u / l c, l c Б , l Б c H , l H c H' c Б' AБ1 AH l H
z n
j m 1
j
c j l j , l j 0, j m 1, n .
Necessary to choose the feasible direction l 0 in the point u from minimum condition J u l on ensemble of the feasible directions L . The source direction l 0 L is chosen by the following in the simplexmethod: a) Index j0 I1 is defined, where I 1 j m 1 j n,
z j c j 0 from condition z j0 c j0 max ( z j c j ) . Since in the jI1
point u U
minimum J (u ) on U
is not reached, i.e. the
inequality z j c j 0 under all j , 1 j n aren’t executed, that ensemble I1 .
l H0 0 , l H0 E n m is chosen so that l H0 (0,...,0,1,0,...,0) , i.e. j0  a component of the vector lH0 is b)
Vector
equal to 1, but all rest components are equal to zero. Finally, the motion direction l 0 in the point u U is defined by the correlations:
l 0 L, l 0 l Б0 , l H0 , l H0 0,...,0, 1, 0,...0, 121
l Б0 AБ1 AH l H0 AБ1a j0 u j0
(u1 j0 , u 2 j0 ,...,u m j0 )
(2)
We notice, that derivative of the function J (u ) in the point u
in the line of l 0 is equal to J u l 0 c, l 0 z j0 c j0 0 . It is follows to note that in general event J u l
0
isn’t the
least value J u l on ensemble L . However such direction choice l 0 allows to construct a transition algorithm from one extreme point to other. We consider the points ensemble from U along chosen direction 0 These points u u l 0 U , 0. Since l .
u u1 0,..., um 0, 0,..., 0 U  an extreme point, l 0
l Б0 , l H0 u j0 ,
that
u 2 u 2 j0 ,... .., u m u
m j0
the
points
u u1 u1 j0 ,
,0,...,0, ,0,...,0 , 0 . We enter the
indexes ensemble I 2 i 1 i m, uij0 0 . We define the values
i u i u ij , i I 2 . Let 0
0 min i min ui uij0 i0 0, i0 I 2 . iI 2
iI 2
Then for values 0 vector u ( 0 ) (u1 0 u1 j0 ,..., u i0 1
0 u i0 1 j0 ,0, u i0 1 0 u i0 1 j0 ,..., u m 0 u mj0 ,0,...,0, 0 ,0,...,0). We notice that
u ( 0 ) 0,
Au ( 0 ) Au 0 Al 0 Au AБ l Б0
AH l H0 Au b, consequently, point u 0 U . On the other
hand, vector u 0 has exactly m positive coordinates. It means that
u 0 u is the extreme point of the ensemble U. We calculate value J (u ( 0 )) J (u 0 l 0 ) c, u 0l 0 c, u 0 c, l 0 122
J (u ) 0 ( z j0 c j0 ). Thence we have
J (u ( 0 )) J (u )
0 ( z j0 c j0 ) 0. Then J (u ( 0 )) J (u ) J (u ) , i.e. in the
extreme point u 0 u value J u less, than in extreme point u .
We note the following: 1) If ensemble of the indexes I 2 , i.e. uij0 0, i 1, m , then
for any 0 the point u U , moreover J (u ( )) J (u )
( z j0 c j0 ) under . In this case source problem has a no solution. However, if ensemble U is compact, that it isn’t possible. 2) It can turn out so that 0 i0 i0 , i0 I 2 , i0 I 2 . In this
case vector u 0 has m 1 positive coordinates. It means that
source problem is degenerate. In such events possible appearance "thread", i.e. through determined number steps once again render in the extreme point u 0 . There are the different methods from "recirculations". We recommend the following books on this questions: Gabasov R., Kirillova F. M. Methods of the optimization. Minsk: BGU, 1975, Karmanov V.G. Mathematical programming. M.: Science, 1975; Moiseev N.N., Ivanilov U.P., Stolyarova E.M. Methods of the optimization. M.: Science, 1978. To make sure in the extreme point u 0 u U is a solution of the problem (1), need to build the simplextable for the extreme point u with aim to check the optimality criterion (2) (lecture 16) in the point u . Construction of the new simplextable. The simplextable built for the extreme point u* U in the previous lecture, in particular for the extreme point u U is true. In the simplextable column j0 and line i0 and values i , i I 2 are indicated.
We build the simplextable for the extreme point u 0 u U
on base of simplextable for the point u* u . We notice, that 123
u U by the base vectors were condition vectors a1 ,..., a m for the point corresponding to positive coordinates of the extreme point u . By the base vectors will be
a1 , a 2 ,..., a i0 1 , a j0 , a i0 1 , a m ,
(2*)
for extreme the point u 0 u . as condition vectors corresponding to positive coordinates. Thereby, the first column of the new simplextable differs from previous, that instead of vector a i0 is written the vector a j0 . In the second column instead ci0 is written c j0 , in one third column nonnegative
u are written, since a1u1 a 2 u 2 ...
components of the vector
a i0 1ui0 1 a j0 0 a i0 1ui0 1 ... a m u m b, where u i u i 0 u ij0 , i 1, m, i i0 . In the rest columns of the new simplextable must be decompositions coefficients of the vector a j , j 1, n , on new base (2*). There were the decompositions m
a j a i uij , j 1, n . in the previous base a1 ,..., a m . Thence i 1
follows that a j
m
a u i 1 i i0
i
ij
a i0 ui0 j , j 1, n . hen under j j0 we
have m
m
i 1 i i0
i 1 i i0
a j0 a i uij0 a i0 ui0 j0 , a i0 a i
uij0 ui0 j0
Substituting value a i0 from formula (3) we get m
a j a i ui j a i0 ui0 j i 1 i i0
124
1 ui0 j0
a j0 .
(3)
m ui j ui j a i ui j 0 0 ui 0 j 0 i 1 i i0
ui0 j j a 0, ui j 0 0
j 1, n.
(3*)
The formula (3*) presents by the decompositions of the vectors a j , j 1, n on new base (2*). From expression (3*) follows that in
the new simplextable coefficients uij
u
ij нов
uij ui0 j0 uij0 ui0 j ui0 j0
нов
are determined by formula
, i i0 , but in line i0 of the new simplex
tables must be (u i0 j ) нов u i0 j u i 0 j 0 , j 1, n. Since vector a j0 in base then in column
(u i j0 ) нов 0, i i0 , (u i0 j0
j0 of the new simplextable all ) нов 1 . Finally, coefficients (u i j ) нов ,
i 1, m , j 1, n , are calculated by the known coefficients u i j , i 1, m , j 1, n of the previous simplextable. Hereinafter, the last two lines of the new simplextable are calculated by the known cБ нов с1 ,..., сi0 1 , c j0 , ci0 1 ,..., cm and (u i j ) нов , i 1, m ,
j 1, n . If it turns out ( z j c j ) нов 0, j 1, n , that u U  a problem solution. Otherwise transition to the following extreme point of the ensemble U and so on are fulfilled. Construction of the initial extreme point. As follows from lemmas 2, 3, the extreme point can be determined from system of the AБ u Б b, u Б 0 , where AÁ = algebraic equations
(a j1 , a j2 ,..., a jm ), a jk , i 1, m ,  linear independent vectors, column of the matrix A u Б (u j1 , u j2 ,..., u jm ) . However such determination way of the extreme point u (0,...,0, u j1 ,0,...,0,
u jm ,0,...,0) is complete enough, when matrix A has the great dimensionality. 1. Let the main task of the linear programming is given 125
J (u ) c, u inf, u U u E n u 0, Au b 0 ,
(4)
where c E n , b E n  the vectors; A  the matrix of the order m n . We suppose that vector b 0 . By entering of the additional variables un 1 0, i 1, m problem (4) can be represented to the canonical form
J (u ) c, u inf, [ Au ]i u n i bi , i 1, m,
u (u1 ,..., u n ) 0, u ni 0, i 1, m. Entering some denotes
, u nm E
nm
,
с (c,0) E n m ,
A A, I m A, a
n 1
,a
n2
,..., a
(5)
u u, u n 1 ,..., nm
,
a nk
(0,..., 0,1, 0,...,0), k 1, m , where I m  a unit matrix of the order m m , problem (5) is written in the manner of
J (u ) c , u inf,
u U u E n m u 0, A u b . (6)
For problem (6) initial extreme point u 0,..., 0, b1 ,..., bm Є ЄEn+m , since condition vectors a n 1 ,..., a n m (the columns of the unit matrix I m )
corresponding to positive coordinates of the
extreme point u~ , b 0 are linear independent.
2. Danzig’s method (twophase method). We consider canonical problem of the linear programming
J (u ) c, u inf, u U u E n u 0, Au b 0 , (7) where c E n , b E n ; А  a matrix of the order m n . We suppose that vector b 0 (if bi 0, i 1, m , that being multiplied by the first line, where bi 0 , always possible to provide to bi 0, i 1, m ). 126
The following canonical problem on the first stage (phase) is solved: m
u Au i u ni By
initial
i 1
ni
inf,
(8)
bi , i 1, m, u 0, u n i 0, i 1, m. extreme
point for problem (8) is vector u 0,..., 0, b1 ,..., bm 0 . Hereinafter problem (8) is solved by the
simplex method and its solution u* u1* , u2* ,..., un* , 0,..., 0
is
found. We notice, that if the source problem (7) has a solution and it is nondegenerate, that vector u* u1* , u2* ,..., un* E n has exactly m
positive coordinates and it is the initial extreme point for the problem (7), since lower bound in (8) is reached under un i 0, i 1, m . On the second stage the problem (7) with initial extreme point u* U is solved by the simplexmethod. 3. Charnes’ method (Мmethod). Charnes’ method is a generalization of Danzig’s method by associations of two stages of the problem solution. So called Мproblem of the following type instead of source problem (7) is considered: m
c, u M un i inf ,
(9)
i 1
Au i uni bi , u 0, uni 0, i 1, m , where M 0  feasible large number. For Mproblem (9) the initial extreme point u 0,..., 0, b1 , b2 ,..., bm E n m , b 0 , and it is solved by the simplexmethod. If the source problem (7) has a solution, that М problem has such solution: u~* u~1* ,..., u~n* ,0,...,0 ,
127
where components un*i 0, i 1, m . Vector u* u~1* , u~2* ,..., u~n* E n is the solution of the problem (7). We notice, that for Мproblem values z j c j j M j ,
j 1, n . So in the simplextable for lines z j , z j c j are conducted two lines: one for coefficients j , other for j . Index j0 , where
z j0 c j0 max ( z j c j ), z j c j 0 is defined by the value of j
the coefficients j , j 0 .
128
REFERENCES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
Alekseev B.M., Tychomirov B.M., F o m i n C .B. Optimal control. M.: Nauka, 1979. Boltyunsky B.G. Mathematical method of the optimal control. M.: Nauka, 1969. Brison A., Ho Yushi. Applied theory of the optimal control. M.: Mir, 1972. Vasilev F.P. Lectures on decision methods of the extreme problems. M.: MSU, 1974. Vasilev F.P. Numerical decision methods of the extreme problem. M.: Nauka, 1980. Gabasov R., Kirillova F.M. Method of the optimization. Minsk: BSU, 1980. Gelfand I.M., Fomin S.B. Various calculus. M.: PhisMath, 1961. Zubov B.I. Lectures on control theory. M.: Nauka, 1975. Karmanov B.G. Mathematical programming. M.: Nauka, 1975. Krasovsky N.N. Motion control theory. M.: Nauka, 1968. Krotov B.F., Gurman B.I. Methods and problems of the optimal control. M.: Nauka, 1973. Lie E.B., Markus L. Bases of the optimal control theory. M.: Nauka, 1972. Pontryagin L.S., Boltyansky B.G., Gamkleridze R.B., Mishenko E.F. Mathematical theory of the optimal processes. M.: Nauka, 1976. Pshenichny B.N., Danilin Yu.M. Numerical methods to extreme problems. M.: Nauka, 1975.
129
Appendix 1
TASKS FOR INDEPENDENT WORK For undertaking of the practical laboratory lessons it is reasonable to have a short base to theories required for solution of the problems and examples. To this effect in 1981 Methodical instructions on course "Methods of the optimization" prepared by S.A.Aysagaliev and T.N.Biyarov were released in alFarabi Kazakh national university. Problems and bases to the theories on sections of the course "Methods of the optimization" on base of the mentioned workbook are brought in the appendix. P.1.1. Multivariable function minimization in the absence thereof restrictions Statement of the problem. Let scalar function be determined in all space problem:
J u J u1 ,..., u n
n
E . To solve the following optimization
J u inf, u E n
u E n is identified by the point of the minimum J (u ) on E n , if J u* J u , u E n . The variable J u* is identified by least The point
E n . We notice that absolute n n (global) minimum J (u ) on E is reached in the point u* E . n The point u 0 E is identified by the point of the local minimum or minimum value of the function J (u ) on
J (u ) on E n , if J u 0 J u under all u u 0 , 0 is
sufficiently small number. Usually first define the points of the local minimum and then amongst of them find the points of the global minimum. Following theorems known from the course of the mathematical analysis: Theorem 1. If function J (u ) C the equality
1
E n , then in the point u
0
En
J u0 0 (necessary firstorder condition) is executed. 130
Theorem 2. If function J (u ) C the following conditions:
2
E n , then in the point u
0
En
J u 0 0, J u 0 0 (necessary condition
of the second order) are executed.
u0 E n be a point of the local
Theorem 3. In order to the point minimum to function J (u ) C
2
E n , necessary and sufficiency
J u0 0, J u0 0 . J(u) sup, u E n is tantamount to the n problem  J(u) inf , u E . We notice, that problem
To find points of the local and global minimum function:
1. J (u ) J (u1 ,u 2 ,u 3 ) u12 u 22 u 32 u1u 2 u1 2u 3 , u (u1 , u 2 , u 3 ) E 3 . 2. J (u1 ,u 2 ) u13u 22 (6 u1 u 2 ), u (u1 , u 2 ) E 2 . 3. J (u1 ,u 2 ) (u1 1) 2 2u 22 , u (u1 , u 2 ) E 2 . 4. J (u1 ,u 2 ) u14 u 24 2u12 4u1u 2 2u 22 , u (u1 , u 2 ) E 2 .
5. J (u1 , u 2 ) (u12 u 22 )e u1 u2 , u (u1 , u 2 ) E 2 . 1 u1 u 2 , u (u1 , u 2 ) E 2 . 6. J (u1 ,u 2 ) 2 2 1 u1 u1 2
7. J (u1 ,u 2 ,u 3 ) u1
2
u 22 u 32 2 , u1 0, u 2 0, u 3 0. 4u1 u 2 u 3
8. J (u1 ,u 2 ) u12 u1u 2 u 22 2u1 u 2 , u (u1 ,u 2 ) E 2 .
9. J (u1 ,u 2 ) sin u1 sin u 2 sin(u1 u 2 ) , 0 u1 , u 2 π. 10. J (u1 ,...,u n ) u1u 22 ...u nn (1 u1 2u 2 ... nu n ) , u i 0, i 1,n.
131
P.1.2. Convex ensembles and convex functions Let U be a certain ensemble
E n , but function J u J u1 ,..., u n
is determined on ensemble U . The ensemble U is identified by convex, if the point αu ( 1 α)ν U , u,ν U and under all , [0,1] . Function J (u ) is convex on convex ensemble U , if
J (αu ( 1 α)ν) αJ (u ) (1 α) J (ν) , u,ν U, α [0,1].
(1)
Function J (u ) is strictly convex on U , if in the equality (1) possible under only
0, 1 . Function J (u )
on U , if function
is concave (strictly concave)
J u is concave (strictly concave) on U . Function
J (u ) is strictly concave on U , if
J (αu ( 1 α)ν) αJ (u ) (1 α ) J (ν) α(1 α )κu ν2 , κ 0, u,ν U, α [0,1] Theorem 1. Let U be a convex ensemble of
(2)
E n . Then in order to the
function J u C U be a convex on U necessary and sufficiency to execute one of the following inequalities: 1
J (u ) J ( ) J v , u v , u, U ,
(3)
or
J'(u)J'(ν),uν 0,u,ν U. If int U и J(u) C necessary and sufficiency that
2
(4)
(U ) , then for convexity of J (u ) on U
J''(u) , 0, E n , Theorem 2. In order to the function
u U .
J(u) C 1(U) be strictly convex
on convex ensemble U necessary and sufficiency execution one of the following two conditions: 132
1) J (u ) J v J' v , uν κu ν2 , κ 0, u,ν U;
(5)
2) J u J v , u v u v , 2 0, u, v U . 2
(6) If int U and J(u) C (U ) , then for strictly convexity of J (u ) on U necessary and sufficiency that 2
J''(u) , 2 , 2 0, u, U . To solve the following problems on base of the determination (1), (2) and correlations (3)  (6): 1. To prove that intersection of any number convex ensembles are convex. Is this statement for unions of ensembles faithfully? To prove that closing of the convex ensemble is convex.
U E n convex, if for any points u, U the point (u ) / 2 U ? b) Is closed ensemble U convex (under 2. a) Is ensemble
performing of the previous condition)? 3.
Let
u 0 E n , but number
r E 1 , r 0 . Is ensemble
V u E n / u u0 r \ u0 (the ball without the centre) convex? 4. To show the convexity of the following function on
E1:
a) J (u ) e u ; б ) J (u ) u ; a (uc ) , a 0 , u c, в) J(u) b(uc ) , b 0 , u c;
u c, 0, г) J (u ) a(uc) , u c, a 0. 5. To prove that function J u 1 / u is strictly convex under u 0 and strictly concave under u 0 convexity (the concavity)]. 6. Let function
[using only determinations to strict
v is continuous and v 0,
133
v .
Then function
J (u ) ( u ) ( )d is convex on E 1 . To prove. u
1 ( u1 ) ( )d , 7. Let J (u1 , u 2 ) u 2 u1
u2 0, ( ) 0 ,
. Is J u1 , u 2 convex function on ensemble U u (u1 ,u 2 )/u1 E 1 , u 2 0 )? 8. To prove that J (u )
n
J (u )  a convex function on convex i
i 1
i
E n , if negative coefficients i correspond to the concave J i u , but positive i  convex functions J i u .
ensemble U of functions
9. To show, if J (u ) is convex and ensemble of values u satisfying to condition J (u ) b, b, b E is convex that J (u ) certainly is a linear function. 10. Function J (u ) is convex on U , if and only if function 1
g ( ) J (u ( u )) by one variable , 0 1 is convex under any u, U . If J (u ) is strictly convex on U , that g ( ), 0 1 is strictly convex. To prove. n 11. Function J (u ) determined on convex ensemble U from E is identified quasiconvex, if J (u (1 ) ) maxJ (u ), J ( ), u, U and under all , 0 1. Is any convex function quasiconvex, conversely? To prove that J (u ) is quasiconvex on U if and only if ensemble M ( ) u U J(u) J (ν) is convex under all U . 12. To check in each of the following exercises whether the function
J (u ) convex (concave) on given ensemble U, or indicate such points from U in the neighborhood of which J (u ) is not neither convex, nor concave:
a) J (u ) u16 u 22 u32 u 42 10u1 5u 2 3u 4 20, U E 4 ; б ) J (u ) e
2u1u2
, u E2 ; 134
в ) J (u ) u15 0,5u 32 7u1 u3 6,
U u (u1 , u 2 , u 3 ) E 3 u i 0, i 1,2,3 ; г)
J (u ) 6u12
u 23
6u32 2
13. Let J (u ) Au b matrix of the order
12u1 8u 2 7, U u E 3 / u 0 . Au b, Au b , where A is the
m n, b Е m  the vector. To prove that J (u ) is
E n . If A*A is nondegenerate, that J (u ) is strictly convex on E n . To find J u , J u .
convex on
14. Let the order
J(u) 0 ,5 Au,u b, u , where A A is the matrix of *
n n, b E n  the vector. To prove that: 1) if A A* 0 ,
J u is convex on convex ensemble U from E n ; 2) if А=А*>0, that J (u ) is strictly convex on U, moreover 1 / 2, where 1 0  the
that
least proper number of the matrix А . 15. To prove that ensemble А is convex iff, when for any numbers 0; 0 the equality A A A is executed.
16. To prove that if
A1 ,..., Am  a convex ensemble, that
n m Co Ai u E n u i u i , i 0, i 1 i 1
17. Let J (u ) moreover for any
n
i 1. i 1
be a continuous function on convex ensemble U,
u, U the inequality J u J (u ) J ( ) / 2. 2
is executed. To prove that J (u ) is convex on U. 18. To realize, under what values of the parameter
the following 4 functions are convex: a) J (u1 ,u 2 ) u u (u1 u 2 ) ; b) J (u ) u12 u 22 (u12 u 22 ) 2 . 19. To find on plane of the parameters , areas, where function J (u1 ,u 2 ) u1α u 2β , u1 0, u 2 0 is convex (strictly convex) and 2 1
concave (strictly concave). 135
2 2
20. To find on plane E2 areas, where function convex, and areas in which it is concave. 21. Let
J (u1 , u 2 ) e
u1u2
is
J i (u ), i 1,m be convex, nonnegative, monotonous m
increasing on
E 1 functions. To prove that function J (u ) J i (u ). i 1
possesses by these characteristics. 22. Let J (u ) be a function determined and convex on convex
n
ensemble U. To prove that: a) ensemble Г E / sup uU
,u
J (u ) is inempty and convex; b) function J * (u ) sup , u uU J (u ) is identified by conjugate to J (u ) . Will J (u ) convex on U? 23. Let function J (u ) be determined and convex on convex ensemble
E n . To prove that for any (including for border) points U the inequality lim J (u ) J ( ) (the semicontinuity property from U from
u
below) is executed. 24. Let function J (u ) be determined and convex on convex closed bounded ensemble U. Is it possible to confirm that: a) J (u ) is upper lower and bounded on U; b) J u reaches upper and lower border on ensemble U?
25. If J (u ) is convex (strictly convex) function in
E n and matrix
A 0 of the order n n, b E n , that J Au b is also convex (strictly convex). To prove.
136
P.1.3. Convex programming. KUNATUCKER’S theorem We consider the next problem of the convex programming:
J (u ) inf at condition
(1)
u U u E n u U 0 ; g i (u ) 0, i 1, m; g i (u ) ai , u bi 0, i m 1, p;
g i (u ) ai , u bi 0, i p 1, s ,
(2)
J (u ), g i (u ), i 1, m are convex functions determined on
where
convex ensemble
U 0 from E n ; i m 1, s, ai E n are the vectors;
bi , i m 1, s are the numbers. Theorem 1 (the sufficient existence conditions of the saddle point).
J (u ), g i (u ), i 1, m be convex functions determined on convex ensemble U 0 the ensemble U * and let the point u ri U 0 U exists Let
such that
g i (u ) 0, i 1, m. Then for each point u* U * Lagrange
coefficients
* (1* ,..., *s ) 0 E s / 1 0,..., p 0 , exist
such that pair
u , U *
*
0
0 forms saddle point to Lagrange’s function s
L (u , ) J (u ) i g i (u ), u U 0 , 0 ,
(3)
i 1
on ensemble
U 0 0 , i.e. the inequality
L(u* , ) L(u* , * ) L(u, * ), u U 0 , 0 is executed.
*
(4)
Lemma. In order to pair u* , U 0 0 be saddle point of the Lagrange’s function (3), necessary and sufficiency performing the following conditions: 137
L (u* ,* ) L (u,λ* ), u U 0 ,
(5)
*i g i (u* ) 0, i 1, s, u* U * U , * 0 , i.e. inequalities (4) is tantamount to the correlations (5). Theorem 2 (the sufficient optimality conditions). If pair
u*, *
U 0 0 is saddle point to Lagrange’s function (3), that vector
u* U * is solution of the problem (1), (2). Problem algorithm of the convex programming based on the theorems 1,2 and lemma are provided in the lectures 10. We illustrate solution rule of the convex programming problem in the following example. Example. To maximize the function
8u12 10u 22 12u1u 2 50u1 80u 2 sup
(6)
at condition
u1 u 2 1, 8u12 u 22 2, u1 0, u 2 0.
(7)
Solution. Problem (6), (7) is tantamount to the problem
J (u ) 8u12 10u 22 12u1u 2 50u1 80u 2 inf
(8)
at condition
u1 u 2 1, 8u12 u 22 2, u1 0, u 2 0.
(9)
Problem (8), (9) possible to write as
J (u ) 8u12 10u 22 12u1u 2 50u1 80u 2 inf,
(10)
u U u u1 , u 2 E 2 /u U 0 ,g1 (u ) 8u12 u 22 2 0,
g 2 (u ) u1 u 2 1 0, U 0 u u1 , u 2
E 2 / u1 0, u 2 0 .
138
(11)
U 0 E 2 is convex, function J (u ) is
We notice that ensemble convex on ensemble
U 0 , since symmetrical matrix
16  12 0, J ' ' (u )  12 20
J ' ' (u ) , 0, E 2 , u U 0 .
It is easy to check that functions
g1 (u ), g 2 (u ) are convex on ensemble
U 0 . Finally, problem (10), (11) is a problem of the convex programming. Entering ensembles
U1 u E 2 / g1 (u ) 0 , U 2 u E 2 / g 2 (u ) 0 , problem (10), (11) possible to write as
J (u ) inf, u U U 0 U1 U 2 . We notice that ensembles
U 0 , U 1 , U 2 are convex, consequently,
ensemble U is convex. Further we solve problem (10), (11) on algorithm provided in the lecture 10. 10. We are convinced of ensemble
U * u* U J (u* ) min J (u ) . In fact, ensembles
uU
U 0 , U 1 , U 2 are convex and closed, moreover
ensemble U is bounded, consequently, ensemble U is bounded and closed, i.e. U  a compact ensemble in En. As follows from correlation (10) function J (u ) is continuous (semicontinuous from below) on compact ensemble U.
U * . Now 2 problem (10), (11) possible to write as J (u ) min, u U E . Then according to the theorem 1 (lecture 2) ensemble
20. We show that Sleytera’s condition is executed. In fact, the point
u (0,1) riU 0 U , (aff U 0 U 0 , aff U 0 U 0 U 0 ) , moreover g1 (u ) 1 0 . Consequently, by theorem 1 Lagrange’s function for problem (10), (11)
139
L (u , ) 8u12 10u 22 12u1u 2 50u1 80u 2 1 (8u12 u 22 2) 2 (u1 u 2 1),
(12)
u (u1 , u 2 ) U 0 , λ ( λ1 ,λ2 ) 0 E 2 / 1 0
has saddle point. 30. Lagrange’s function has the form (12), area of its determination U 0 0, , moreover 1 0, but 2 can be as positive, as negative.
u , U to Lagrange’s function (12) on base of (5). From inequality L u , L u , , u U follows that convex function L u , , u U (  fixed vector) *
40. We define saddle point
*
0
0
*
*
*
0
*
*
*
reaches the least value on convex ensemble
0
U 0 in the point u* . Then
according to optimality criterion (lecture 5, theorem 4) necessary and sufficiently executing of the inequality
L u u* , * , u u* 0, u U 0,
(13)
where
16u * 12u 2* 50 161*u1* *2* , u (u * , u * ), L (u* , * ) 1 1 2 20u * 12u * 80 2*u * * * 2 1 1 2 2
* (1* , *2 ) Conditions
*i g i (u* ) 0, i 1,2
(the
conditions
of
the
complementing elasticity) are written:
2
2
1* 8u1* u 2* 2 0, *2 (u1* u 2* 1) 0, 1* 0. a) We suppose that
(14)
u* int U 0 . In this case from inequality (13) we have
L u (u* , ) 0. Saddle point is defined from conditions *
16u1* 12u2* 50 161*u1* *2 0, 20u*2 12u1* 80 21*u2* *2 0, 140
2
2
1* (8u1* u 2* 2) 0, u1* u 2* 1 0, *2 0, 1* 0. We notice that from the second equation (14) and
(15)
u* U follows that
u u 1 0, 0 . We solve the system of the algebraic equations * 1
* 2
* 2
(15). It is possible several events: 1)
1* 0, *2 0.
the system
2
In this case the point u* is defined from solution of
2
8u1* u 2* 2, u1* u 2* 1 . Thence we have u1* 0,47,
u 2* 0,53. Then value 1* is solution of the equation 126,2 6,461* 0 . 126,2 * Thence we have 1 0. 6,46 * It is impossible, since 1 0. * * 2) 1 0, 2 0. The point u* is defined from equations u1* u 2* 1 , 16u1* 12u 2* 50 *2 0 , 20u 2* 12u1* 80 *2 0 . * * Thence we have u 2 158 / 60, u1 98 / 60 0. The point u* U 0 . So the case is excluded. Thereby, conditions (13), (14) are not executed in the internal points of the ensemble U 0 . It remains to consider the border
U0 .
points of the ensemble
u* U , that remains to check the conditions (13), (14) in the border points (1,0), (0,1) of the ensemble U 0 . We notice that b) Since the point
border point (1,0) U , since in the point restriction 8u1 u 2 2 0. is not executed. Then the conditions (13), (14) follows to check in the singular point u* (0,1) U 0 . In the point g1 (u* ) 1 0, 2
consequently, value derivative
1* 0.
2
and equalities (14) are executed. Since
L u (u* , ) (38 *2 ,60 *2 ), that inequality (13) is *
(38 *2 )u1 (60 *2 )(u 2 1) 0 under all u1 0, u 2 0 . We choose *2 60 , then the inequality to write so: 98u1 0, u1 0, u 2 0 . Finally, the point u* (u1* 0, u 2* 1) is
written as
the saddle point to Lagrange’s function (12). 141
*
50. Problem (10), (11) has the solutions u1 To solve the following problems:
0,
u 2* 1, J (u* ) 70 .
J (u ) u12 u 22 6 sup; u1 u 2 5, u1 0. 2 2. J (u ) 2u1 2u1 4u 2 3u3 8 sup; 8u1 3u 2 3u3 40, 2u1 u 2 u3 3, u 2 0.
1.
J (u ) 5u12 u 22 4u1u 2 5u1 4u 2 3 sup, u12 u 22 2u1 2u 2 4, u1 0, u 2 0 . 4. J (u ) 3u 22 11u1 3u 2 u 3 27 sup, u1 7u 2 3u3 7, 5u 2u 2 u3 2. 5. J (u ) u1 u 2 u3 30u 4 8u5 56u6 sup, 2u1 3u 2 u 3 u 4 u 5 10u 6 20,
3.
u1 2u 2 3u 3 u 4 u 5 7u 6 11, x1 0, x 2 0, x3 0. 6. The following problem:
J (u ) c, u 0,5 Du, u sup at
conditions Au b, u 0 , where D D 0, is identified by the problem of the quadratic programming. To define the solutions of the problem. 7. To define the least distance from origin of coordinates till ensemble u1 u 2 4, 2u1 u 2 5 . 8. To formulate KunaTucker’s conditions and dual problem for the *
following problem: at condition
J (u ) 4u12 4u1u 2 2u22 3u1 e u1 2u2 inf
u1 2u 2 0, u12 u 22 10, u 2 1 / 2, u 2 0
J (u pu12 qu1u 2 sup) at condition u1 ru 22 1, u1 0, u2 0 , where  p , q , r  parameters. Under what values of the 9. To find
parameters p , q , r solution exists?
(u1 3) 2 u 22 sup 2 at condition u1 (u 2 1) 0, u1 0, u 2 0 . 10. To solve the geometric problem: J (u )
Are Kuna–Tucker’s conditions satisfied in the optimum point?
142
P.1.4. NONLINEAR PROGRAMMING We consider the next problem of the nonlinear programming:
J (u ) inf,
(1)
u U ,
(2)
U u E n / u U 0 , g i (u ) 0, i 1, m; g i (u ) 0, i m 1, s , where function J (u) C 1 (U 01 ), g i (u) C 1 (U 01 ), i 1, s, U 01 open ensemble contained the convex ensembles U 0 from E n , in particular, U 01 E n . Theorem 1 (the necessary optimality condition). If the functions
J (u ) C 1 ( E n ), g i (u ) C 1 ( E n ), i 1, s, int U 0 , U 0 is a
convex ensemble, but ensemble U * , then for each point u* U necessary exist the Lagrange’s coefficients
*
,
(*0 , 1* ,..., *s ) 0 E s 1 / 0 0, 1 0,..., m 0 such the following conditions are executed: *
0, *0 0, 1* 0, ..., *m 0,
(3)
*
L u (u* , )u u * S
*0 J (u* ) *i g i (u* ), u u* 0, u U 0 ,
(4)
i 1
*i g i (u* ) 0, i 1, s, u* U
.
(5)
Lagrange’s function for problem (1), (2) has the form S
L (u , ) 0 J (u ) i g i (u ), u U 0 , (0 , 1 ,..., s ) 0
i 1
E S 1 / 0 0, 1 0,..., m 0 . 143
At solution of the problem (1), (2) necessary to consider separately two events: 1)
*0 0
(degenerate problem); 2)
problem). In this case possible to take We suppose that the points
*0 0
(nondegenerate
1. u* , *0 0, * (1* 0,..., *m * 0
0, *m 1 ,..., *s ) from conditions (3), (4), (5) are found. The point u* U is identified by normal, if the vectors g i (u* ), i I , g m 1 (u* ),..., g s (u* ) are linear independent, where ensemble I i 1,..., m/ g i (u* ) 0. If u* U  a normal point, that problem
(1), (2) is nondegenerate, i.e.
*0 1.
In the event
U 0 E n is executed
the following theorem. Theorem 2. Let functions
J (u ), g i (u ), i 1, s, be definite,
continuous and twice continuously differentiable in the neighborhood of the normal point u* U . In order to the normal point u* U be a point of the local minimum
J (u ) on ensemble U, i.e. J (u* ) J (u ), u,
u D(u* , ) U sufficiently that quadric form 2 * 2 y L (u* , ) / u y, y 0 be positive definite on the hyperplane *
*
g (u ) g i (u* ) y 0, i I , i * y 0, i m 1, s. u u We consider the solutions of the following example on base of the theorems 1, 2 and solution algorithm of the nonlinear programming (the lectures 13). Example 1. To find the problem solution
3 6u1 2u 2 2u1 u 2 2u 22 sup
(6)
at conditions
3u1 4u 2 8, u1 4u 22 2, u1 0, u 2 0. (7) Solution. Problem (6), (7) is tantamount to the problem 144
J (u ) 2u 22 2u1 u 2 6u1 2u 2 3 inf,
(8)
u (u1 , u 2 ) U u E 2 / u U 0 , g1 (u ) 0, g 2 (u ) 0 , where U 0
(9)
u E 2 / u1 0, u2 0 ; g1 (u ) 3u1 4u2 8; g 2 (u)
) u1 4u 22 2 . Function J (u ) is not a convex function on ensemble U 0 , consequently, we have the nonlinear programming problem. 10. We show that ensemble
U * . Let U1 u E 2 / g1 (u ) 0,
U 2 u E 2 / g 2 (u ) 0 .
U U 0 U1 U 2  closed bounded ensemble, consequently, it is convex. Function J (u ) is continuous on ensemble U . Thence in effect Weierstrass theorems we have U * . Then
20. Generalized Lagrange’s function for problem (8), (9) has the form
L (u1 , u 2 , 0 , 12 ) 0 (2u 22 2u1u 2 6u1 2u 2 3) 1 (3u1 4u 2 8) 2 (u1 4u 22 2),
u (u1 , u 2 ) U 0 ,
(0 , 1 , 2 ) 0 E / 0 0, 1 0, 2 0. 3
30. According to the conditions (3)  (5) quadruple
, , * 0
* 1
* 2
U
0
u , *
*
0 is defined from correlations
 *  0, *0 0, 1* 0, *2 0,
(10)
L u u* , * , u u* 0, u U 0 ,
(11)
1* g1 (u* ) 0, *2 g 2 (u* ) 0, u* U ,
(12)
where derivative
145
*0 (2u 2* 6) 31* *2 . L u (u* , * ) * * * * * * 0 (4u 2 2u1 2) 41 82 u 2 a) We expect that u* intU 0 . Then from (11) follows that L u (u* , * ) 0 , consequently, pair (u* , * ) is defined from solution of the following algebraic equations:
*0 (2u 2* 6) 31* *2 0, *0 (4u 2* 2u1* 2) 41* 8*2 u 2* 0 . 1* (3u1* 4u 2* 8) 0, *2 (u1* 4u 2*2 2) 0 ,
(13)
where
u1* 0, u 2* 0, *0 0, 1* 0, *2 0 . We consider the event,
when
*0 0 .
Then
*2 31* , 41* (1 6u 2* ) 0 .
*2 0, consequently, * * 0 0, 1 0, *2 0 .
condition So
(10)
is
it is possible only
1* 0,
If
broken,
that since
0. * 1
Then
u 1 / 6. It is impossible, since u 0. Thence follows that 0, * 2
* 2
* 0
i.e. problem (8), (9) is nondegenerate, so it is possible to take
10 1
and
conditions (13) are written so:
2u 2* 6 31* *2 0, 4u 22 2u1* 2 41* 8*2 u 2* 0,
1* (3u1* 4u 2* 8) 0, *2 (u1* 4u 2* 2) 0
.
(14)
Now we consider the different possibilities:
1* 0, *2 0.
u* (u1* , u 2* ) is defined from * * * *2 * * system 3u1 4u2 8 0, u1 4u2 2 0, u1 0, u2 0 * * [refer to formula (14)]. Thence we have u1 1,43, u2 0,926, 1)
In this case the point
J (u1* , u2* ) 9,07 . However 1* 1,216, *2 0,5 0 , so the given point u* (1,43; 0,926) can not be solution of the problem (8), (9).
1* 0, *2 0. In this case from (14) we have 2u 2* 6 *2 0, 4u 2* 2u1* 2 8*2 u 2* 0, u1* 4u 2*2 2 0. 2)
146
*2 4,908 0 . However * * the point u* (117; 5,454) U , since the inequality 3u1 4u 2 8. Thence we have u1 117, u 2 5,454, *
*
aren’t executed. 3) 1* 0, *2 0. Equations (14) are written so: 2u 2* 6 31* 0,
4u 2* 2u1* 2 41* 0, 3u1* 4u 2* 8 0. u1* 28 / 9, u 2* 1 / 3 0, 1* 16 / 9 0 . The point u* (28 / 9, 1 / 3) U . * * 4) 1 0, 2 0. In this case from equation (14) we have Thence we have
2u 2* 6 0, 4u 2* 2u1* 2 0. u 2* 3, u1* 5. However the point u* (5; 3) U * * , since inequality u1 4u 2 2 0. are not executed. Thereby, the point u* intU 0 . Thence we have
b) We expect that point
u* ГрU 0 . Here possible the following
events: 1)
u (0, u 2 ) ГрU 0 , u 2 0; 2) u u1 ,0 ГрU 0 , u1 0. For the first type point of the restriction g1 4u 2 8 0, g 2 4u22 2 0 ,
consequently,
0 u2 1/ 2 .
Then.
u* u 0, u2 1 / 2 , J u* 3,4 . For the second type of the border point of the restriction g1 3u1 8 0, g 2 u1 2 0 . Then u* 8 / 3, 0, but value J u* 19 . * 1
Finally, problem solution (8), (9):
J u* 19.
u* u1* 8 / 3, u 2* 0 ,
Example 2. It is required from wire of the given length l to do the equilateral triangle and square which total area is maximum. Solution. Let u1 , u 2 be a sum of the lengths of the triangle sides of the
u1 u 2 l , but side of the triangle has a length u1 / 3 , side of the square  u 2 / 4 and total area is
square accordingly. Then sum
147
S u1 , u 2
3 2 1 2 u1 u 2 . 36 16
Now optimization problem can be formulated so: to minimize the function
J u1 , u 2
3 2 1 2 u1 u 2 inf 36 16
(15)
at conditions
u1 u 2 l , u1 0, u 2 0.
(16)
g1 u1 , u 2 u1 u 2 l , g 2 u1 , u 2 u1 ,
Entering indications
g 3 u1 , u 2 u 2 problem (15), (16) to write as J u1 , u 2
3 2 1 2 u1 u 2 inf 36 16
(17)
at conditions
u U u E 2 g1 (u ) u1 u 2 l 0, g 2 (u1 , u 2 ) u1 0, g 3 (u1 , u 2 ) u 2 0 ,
(18)
U 0 E n . Unlike previous example the conditions u1 0, u2 0 are enclosed in restrictions g 2 u1 , u 2 , g 3 u1 ,u 2 .
where possible
Such approach allows using the theorem 2 for study properties of the
functions J (u1 , u 2 ) in the neighborhood of the point u* u1 , u 2 U . Problem (17), (18) is the nonlinear programming problem since function *
*
J (u1 , u 2 ) is not convex on ensemble U 0 E n . 10. Ensemble U is bounded and closed, consequently, it is compact.
Function J u1 , u 2 С E , so ensemble U * 20. Generalized Lagrange’s function for problem (17), (18) to write as 2
2
148
3 2 1 2 u1 u 2 ) 36 16 1 (u1 u 2 l ) 2 u1 3 u 2 , u u1 , u 2 E 2 ,
L (u , ) L (u1 , u 2 , 0 , 1 , 2 , 3 ) 0 (
0 , 1 , 2 , 3 0 E 4 / 0 0, 2 0, 3 0. 30. Since ensemble
U 0 E 2 , that conditions (3)  (5) are written:
3 * * u1 0 1* *2 0, L (u* , ) 18 1 * * * * u 2 0 1 3 8 *
(19)
1* u1* u 2* l 0, *2 u1* 0, *3 u 2* 0 , *
0, *0 0, *2 0, *3 0 . a) We consider the event expression (19),
* 1
* 2
* 3.
*0 0 . If
(21)
In this case, as follows from
0, * 2
(20)
that
3 0 ,
consequently,
u 0, u 0 [refer to formula (20)]. It is impossible, since u u 2* l . It means, 1* *2 *3 0 . The equalities opposites to the * 1 * 2
* 2
condition (21). Then the source problem (17), (18) is nondegenerate. Consequently, value
*0 1 .
b) Since problem (17), (18) is nondegenerate, the conditions (19)  (21) are written:
3 * 1 u1 1* *2 0, u 2* 1* *3 0, u1* u 2* l 0, 18 8
1* 0, *2 u1* 0, *3u 2* 0, *2 0, *3 0. We consider the different events: 149
(22)
1.
1*
*2 0, *3 0 . Then l 3 18 8 3
u1*
9l , 94 3
u2*
4l 3 , 94 3
.
*2 0, *3 0 . In this case we have u1* 0, u2* l , 1* l / 8, *2 l / 8 . * * 3. 2 0, 3 0 . Then from systems of the equations (22) we get 2.
u1* l , u 2* 0, 1* 3 l / 18, *3 3 l / 18 .
0, 0 * 2 * 1
* 3
is excluded, since
The
event
u 0, u 0 and condition * 1
* 2
u u 2* l. is not executed. 40. Quadric form y L uu (u * , ) y *
3 2 1 2 y1 y 2 . 18 8
Hyperplane equations for events 1  3 are accordingly written: 1)
g u g1 u* y1 1 * y2 y1 y2 0 . Thence we have y1 u1 u 2
y 2 . Then y L uu (u* , * ) y 4 3 9 y12 0 , i.e. the point ( u 9l (9 4 local minimum. * 1
3 ) , u 4l 3 (9 4 3 ) ) is not the point of the * 2
g1 (u* ) g (u ) y1 1 * y 2 y1 y 2 0, u1 u 2
g 2 (u* ) g (u ) y1 2 * y 2 y1 0 . u1 u 2 Thence we have
y1 0, y 2 0; y L uu (u* , * ) y 0, y 0 . The
sufficient conditions of the local minimum are degenerated.
g 1 (u* ) g (u ) y1 1 * y 2 y1 y 2 0 , u1 u 2 g (u ) 3 * y2 y2 0 . u 2 3)
150
g 3 (u* ) y1 u1
The data of the equation have the solutions y1 0, y2 0 . Again the sufficient optimality conditions do not give the onedigit answer. Solution of the problem is found by comparison of function values J (u1 , u 2 ) in the last two events. In the second event value
J (u1* , u 2* ) l 2 / 16 , but in the
J (u1* , u 2* ) 3l 2 / 36. Then the problem solution is the * * point ( u1 0, u 2 l ), i.e. from the whole wire is made only square.
third event
To solve the following problems: 1. To find J u1 , u 2 , u3 u1u 2u3
inf at conditions:
a)
u1 u 2 u3 3 0 , b) u1 u 2 u3 8 0 ;
c)
u1u 2 u1u 3 u 2 u 3 a,
2.
To prove the equality
u i 0, i 1,3.
n
u1n u 2n u1 u 2 , n 1, u1 0, u 2 0. 2 2 3. To find the sides of the maximum rectangle area inserted in circle
u u 22 R 2 . 2 1
4. To find the most short distance from the point (1,0) till ellipse
4u 9u 22 36 . 2 1
5. To find the distance between parabola
u 2 u12 and line
u1 u2 5 . 6. To find the most short distance from ellipse line
2u12 3u 22 12 till
u1 u 2 6 .
J u u Au sup, A A* at condition u u 1 . To n show, if u* E  the problem solution, that J (u* ) is equal to the most 7. To find
characteristic root of the matrix A . 8. To find the parameters of the cylindrical tank which under given area of the surface S has the maximum volume. 9. a) J (u ) u1 u 2 u 3 inf,
151
1 1 1 1; u1 u 2 u 3
J (u ) u1u 2 u 3 inf, u1 u 2 u3 6, u1u 2 u1u3 u 2u3 12; 1 1 1 1 c) J (u ) inf, 2 1; 2 u1 u 2 u1 u 2 b)
d)
J u 2u1 3u 22 u 32 inf,
u1 u 2 u 3 8, u i 0, i 1,3; e)
J u u12 u 22 u 32 inf,
u1 u 2 u 3 12, u i 0, i 1,3 ; f)
J (u ) u1u 2 u1u 3 u 2 u 3 inf, u1 u 2 u 3 4;
g)
J (u ) u12 u 2 u 22 u1 u1u 2 u 3 inf,
u1 u2 u3 15, ui 0, i 1,3; l)
J (u ) u12 2u1u 2 u 32 inf,
u1 2u 2 u 3 1, 2u1 u 2 u 3 5, u i 0, i 1,3. 10. To find the conditional extreme to function
J (u ) (u1 2) 2
(u2 3) 2 at condition u12 u22 52 . P.1.5. LINEAR PROGRAMMING. SIMPLEXMETHOD General problem of the linear programming (in particular, basic task) is reduced to the linear programming problem in canonical form of the following type:
J (u ) c u inf,
(1)
u U u E n / u 0, Au b , where
(2)
A a1 , a 2 ,..., a m , a m1 ,..., a n  matrix of the order m n;
a i E m , i 1, n ; vectors are identified by the condition vectors, but b E m  by the restriction vector. 152
Simplexmethod is a general method of the problem solution of the linear programming in canonical form (1), (2). Since the general and the basic problems of the linear programming are reduced to type (1), (2), it is possible to consider that simplexmethod is the general method of the problem solution of the linear programming. The base of the linear programming theory is stated in the lectures 15 17. We remind briefly a rule of the nondegenerate problem solution of the linear programming in canonical form. 10. To build the initial extreme point notice, if
u 0 U ensemble U . We
U * , that lower border to linear function (1) on U is
reached in the extreme point ensemble U , moreover in the nondegenerate problem the extreme point has exactly m positive coordinates i.e.
u 0 u10 0,..., u m0 0,0,...,0 , and vectors a1 ,..., a m , corresponding to the positive coordinates of the extreme point are linear independent. The extreme point u U is defined in the general event by Мmethod (Charnes’ method), but on a number of events  on source data ensemble U directly. 0
20. Simplextable is built for extreme the point
u 0 U . Vectors
a i , i 1, m are presented in the first column, in the second  elements of the vector c with corresponding to lower indexes, in the third  positive 0 coordinates of the extreme point u U and in the rest column – i 1 m decomposition coefficients of the vectors a , i 1, n on base a ,..., a i.e. a j
m
a u i 1
i
ij
AБ u j , where AБ a1 ,..., a m
 nonsingular
u j u1 j ,..., u mj , j 1, n . Values z j ci uij , j 1, n m
matrix;
i 1
are brought in penultimate line, but in the last line – values
z j c j , j 1, n . The main purpose of the simplextable is to check the
u 0 U . If it is turn out to be that values z j c j 0, j 1, n , the extreme point u 0 U  a solution of the
optimality criterion for the point
problem (1), (2), but otherwise transition to the following extreme point
u1 U – is realized, moreover value J u1 J u 0 . 153
u1 U and corresponding to it simplextable is 0 built on base of the simplextable of the point u U . The index j0 is z j 0 c j 0 max( z j c j ) amongst defined from condition 30. The extreme point
z j c j 0 . Column j 0 of the simplextable of the point u 0 U is
a j0 to the number of base i 1 in simplextable of the point u U instead of vector a 0 . The index i0 is identified pivotal column, and vector is entered
defined from condition min
ui0 i0 amongst uij0 0 . The extreme point uij
u 1 (u10 0 u1 j0 ,..., u i00 1 0 u i0 1 j0 , 0, ui00 1 ui0 1 j0 ,..., um0 0umj0 , 0,...,0, 0 ,0,...,0), 0 i0 . The base consists of the vectors Decomposition coefficients of the vectors defined by the formula
(u ij ) нов u ij (u i0 j ) нов Hereinafter values coefficients
u ij0 u i0 j u i0 j0 u i0 j
u i0 j0
a1 ,..., a i0 1 , a j0 , a i0 1 ,..., a m . a j , j 1, n on the base are
, i i0 , j j 0 ; (3)
, j 1, n.
( z j ) нов , ( z j c j ) нов are calculated by the known
(u ij ) нов , i 1, m, j 1, n . Thereby, new simplextable is
u1 U . Further optimality criterion is checked. If it is turn 1 out to be that ( z j с j ) нов 0, j 1, n , so the extreme point u U 
built for point
solution of the problem (1), (2), but otherwise transition to the new extreme point u U realized and etc. Example 1. To solve the linear programming problem 2
J u1 , u 2 , u3 , u4 , u5 , u6 6u1 3u 2 5u3 2u 4 4u5 2u6 inf,
154
u1 2u 2 u 4 3u 5 17, 4u 2 u 3 u 5 12,
u 2 8u 4 u 5 u 6 6, u1 0, u 2 0, u 3 0, u 4 0, u 5 0, u 6 0 . Matrix A , vectors b and c are equal to 1 2 0 1 3 0 A 0 4 1 0 1 0 a1 , a 2 , a 3 , a 4 , a 5 , a 6 , 0 1 0 8 1 1
0 1 3 0 2 1 2 3 4 5 6 a 0 , a 4 , a 1 , a 0 , a 1 , a 0 , 1 8 1 0 1 0 1
17 b 12 , c (6, 3,5, 2, 4,2) . 6 Solution. The initial extreme point u 17, 0,12, 0, 0, 6 easy is defined for the example. The condition vectors corresponding to positive 0
coordinates of the extreme point 
с
Base
b
6 a1
a1 , a 3 , a 6 . 3 a2
5 a3
2 a4
4 a5
2 a6
0
u0=(17, 0, 12, 0, 0, 6)
I A1
6
17
1
2
0
1
3
0
A3
5
12
0
(4)
1
0
1
0
A6
2
6
0
1
0
8
1
1
155
17 2 12 4 6 1
i0 3
zj cj
0
37
0
24
25
0
J (u ) 174
0
z j0 c j0 37, j0 2
u1=(11, 3, 0, 0, 0, 3)
II a1 a2
6 3
11 3
1 0
0 1
1/2 1/4
1 0
5/2 1/4
0 0
11 
a6
2
3
0
0
1/4
(8)
5/4
1
3/8
0
0
 37
24
63 4
0
zj cj
4
i0 6 1
J (u ) 63
z j0 c j0 24, j0 4 u2=( 85 ,3,0,
III
3 ,0,0) 8
8
a1
6
85 8
1
0
 15
32
0
( 85 )
1 8
4
a2
3
3
0
1
1 4
0
1 4
0

a4
2
3 8
0
0
 1
32
1
 5
32
1 8

0
0
 17
0
39 2
3
zj cj
2
32
z j0 c j0
a2
2
39 2
J (u ) 54
, j0 5 u3=(0, 2, 0, 1, 4, 0)
IV a5
i0 1
4 3
4
32 85
2
 8 85
0
 15
1
25 85
85
156
0 0
1
 4
0
1 85
85
zj cj 0
j 1,6
a4
2
1 17
1
zj cj

624 85
0 0
 5
85

430 85
1 0
0 0
10 85
 177 85
3
J ( u ) 24
Example 2.
J (u ) 2u1 3u2 5u3 6u4 4u5 inf; 2u1 u 2 u3 u 4 5; u1 3u 2 u3 u 4 2u5 8; u1 4u 2 u 4 1, u j 0,
j 1,5.
Corresponding Мproblem has the form
J (u ) 2u1 3u 2 5u 3 6u 4 4u 5 Mu6 Mu7 inf; 2u1 u 2 u3 u 4 u6 5; u1 3u 2 u3 u 4 2u5 8; u1 4u 2 u 4 u 7 1, u j 0,
j 1,7.
For Mproblem matrix А, the vectors b and c are equal 2 1 1 1 0 1 0 , A 1 3 1 1 2 0 0 (a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 ) 1 4 0 1 0 0 1
5 b 8 , c (2,3,5,6,4, M , M ) . 1 Solution. The initial extreme point u0=(0, 0, 0, 0, 4, 5, 1). We present
z j c j in the manner of z j c j j M j , j 1, n , i.e.
value
instead of one line for
z j c j are entered two: in the first values j are
j . Since M – sufficiently great positive number, z j c j z k ck , if j k . But if j k , that j k .
written, in the second then
Base
с
B
2 a1
3 a2
5 a3
157
6 a4
4 a5
М а6
М а7
I J (u ) 16 6M
u 0 (0,0,0,0,4,5,1)
A6
M
5
2
1
1
1
0
1
0
A5
4
4
1/2
3/2
1/2
1/2
1
0
0
A7
M
1
1
(4)
0
1
0
0
1
j
4
9
3
4
0
0
0
j
1
5
1
2
0
0
0
zj cj
II J (u 1 ) (55 / 4) (19 / 4) M a6
M
19/4
(9/4)
0
1
3/4
0
a5
4
29/8
7/8
0
1/2
7/8
1
1 0
a2
M
1/4
1/4
1
0
1/4
0
0
j
25/4
0
3
7/4
0
j
9/4
0
1
3/4
0
zj cj
III
u 1 (0,14,0,0,29 / 8,19 / 4,0)
J (u 2 ) 5 / 9 а1 а
5
2
19/9
0
u 2 (19 / 9,7 / 9,0,0,16 / 9,0,0) 1
0
4/9
3/9
0
4
16/9
0
0
8/9
21/18
1
3
7/9
0
1
1/9
3/9
0
zj cj j
0
0
2/9
1/3
0
j
0
0
0
0
0
а2
0
As follows from the last simplextable by solution of the Мproblem is
u 2 (19 / 9,7 / 9,0,0,16 / 9,0,0) . Then solution of the source 2 2 problem is vector u (19 / 9,7 / 9,0,0,16 / 9) , J (u ) 5 / 9 . a vector
To solve the following linear programming problems: 1. J (u ) 2u1 u 2 2u3 3u 4 sup;
3u1 u3 u 4 6; u 2 u3 u 4 2; 158
u1 u 2 u3 5; u j 0, 2.
j 1,4 .
J (u ) u1 u 2 u3 3u 4 u5 u6 3u7 inf; 3u3 u5 u6 6; u 2 2u3 u 4 10; u1 u6 0; u3 u6 u7 6; u j 0,
j 1,7 .
J (u ) u1 2u 2 u3 2u4 u5 inf; u1 2u2 u3 2; 2u1 u2 u4 0; u1 3u2 u5 6; u1 0; u 2 0, u3 0, u 4 0. 4. J (u ) 2u1 3u 2 2u3 u 4 sup; 2u1 2u 2 3u3 u 4 6; u 2 u3 u4 2; u1 u2 2u3 5; u1 0; u 2 0, u3 0. 5. J (u ) u1 2u 2 u3 sup; 3u1 u2 u3 4; 2u1 3u2 6; 3.
u1 2 0 u 2 1, u j 0, j 1,3. 6.
J (u ) u1 2u2 u3 inf; u1 u2 1; u1 2u2 u3 8; u1 3u 2 3, u j 0, j 1,3.
7.
J (u ) 2u1 u 2 u3 sup; 159
u1 2u2 u3 4; u1 u2 2; u1 u3 1, u j 0, j 1,3. 8. The steel twig by length 111 cm have gone into blanking shop. It is necessary to cut them on stocking up on 19, 23 and 30 cm in amount 311, 215 and 190 pieces accordingly. To build the model on base of which it is possible to formulate the extreme problem of the variant choice performing the work under which the number cut twig are minimum. 9. There are two products which must pass processing on four machines (I, II, III, IV) in process production. The time of the processing of each product on each of these machines is specified in the following form: Machine I II III IV
А 2 4 3 1
В ј 2 1 4
The machines I, II, III and IV can be used accordingly during 45, 100, 300 and 50 hours. Price of the product А  6 tenge per unit, product В  4 tenge. In what correlation do follows to produce the product А and В to get maximum profit? To solve the problem in expectation that product А is required in amount not less 22 pieces. 10. Refinery disposes by two oil grade  А and В, moreover petrol and fuel oil are got under processing. Three possible production processes are characterized by the following scheme: a) 1 unit of the sort A + 2 unit of the sort B 2 unit of the fuel oil + 3 unit of petrol; b) 2 unit of the sort A + 1 unit of the sort B 5 unit of the fuel oil + 1 unit of petrol; c) 2 unit of the sort A + 2 unit of the sort B 2 unit of the fuel oil + 1 unit of petrol; We assume the price of the fuel oil 1 tenge per unit, but the price of the petrol 10 tenge per unit. To find the most profitable production plan if there are 10 units to oil of the sort А and 15 units oil of the sort B.
160
Appendix 2
TASKS ON MATHEMATICAL PROGRAMMING
Three term tasks for students of the 2nd, 3rd courses (5,6 terms accordingly) educating on the specialties “Applied mathematics”, “Mathematics”, “Mechanics”, “Informatics”, “Economical cybernetics” are worked out by the following parts: convex ensembles and convex function (1th task), convex and nonlinear functions (2nd task), linear programming (3rd task), Further we prefer variants of the tasks on course “Mathematical programming”. Task 1
J (u ) is convex (concave) on the ensemble U , or indicate such points from U in neighborhood of which J (u ) isn’t neither To check the function
convex or concave (1 – 89variants). 1.
J u u16 u22 u32 u42 10u1 5u2 3u4 20; U E 4 .
2. J u e
2 u1 u 2
; U E 2.
J u u13 u23 u32 10u1 u2 15u3 10; U E 3. 1 2 2 2 3 4. J u u1 u2 u3 u1u2 u3 10; U E . 2 2 2 2 5. J u u1 u 2 2u3 u1u 2 u1u3 u 2u3 3.
5u2 25; U E 3 .
1 2 u3 7u1 u3 6; U u E 3 : u 0 . 2 2 2 2 7. J u 3u1 u2 2u3 u1u2 3u1u3 u2u3 3
6. J u u2 5
3u2 6; U E 3. 161
1 2 u2 4u32 u1u2 2u1u3 2u2u3 2 u3 1; U E 3. 1 2 1 2 2 9. J u 2u1 u 2 5u 3 u1u 2 2u1u 3 u 2 u 3 2 2 3u1 2u 2 6; U E 3 . 8. J u 5u1 2
10.
J u u13 2u32 10u1 u 2 5u3 6;
U {u E 3 : u 0}. 11. 12.
J u 5u14 u26 u32 13u1 7u3 8; U E 3. J u 3u12 2u22 u32 3u1u2 u1u3
2u2u3 17; U E 3. 1 4 u3 3u1 8u2 11; 2 U {u E 3 : u 0}.
13. J u 4u13 u24
14.
J u 8u13 12u32 3u1u3 6u2 17; U {u E 3 : u 0}.
15. J u 2u12 2u 22 4u32 2u1u 2 2u1u3
2u 2u3 16; U E 3 . 16. J u 2u1 u2 2
2
1 2 u3 2u1u2 8u3 12; U E 3 . 2
1 7 1 4 u3 2u2u3 11u1 6; 2 2 U {u E 3 : u 0}. 5 3 1 18. J u u12 u 22 4u 32 u1u 2 2u1u 3 u 2 u 3 2 2 2 + 8u3 13; U E 3 .
17. J u u2
162
1 3 2 U {u E 3 : u 0}.
19. J u 3u1 u2 2u1u2 5u1u3 7u1 16; 2
3 2 1 u3 u1u2 u1u3 2 2 2u2u3 10; U E 3 . 3 5 21. J u 2u12 u32 u1u3 12u2 18; U E 3 . 2 2 20. J u 2u12 u22
22.
J u 6u12 u23 6u32 12u1 8u2 7;
U {u E 3 : u 0}. 3 2 2 2 23. J u u1 u2 2u3 u1u2 u1u3 2 2 2u2u3 8u2 ; U E 3. 5 2 2 2 3 24. J u 4u1 u2 u3 4u1u2 11u3 14; U E . 2 7 2 4 3 1 3 25. J u u1 u 2 u 3 13u1 7u 3 9; 2 2 2 3 U {u E : u 0}. 5 3 1 5 3 3 26. J u u1 u2 u3 22u2 17; 6 4 2 3 U {u E : u 0}. 5 3 3 2 2 27. J u u1 2u 2 u 3 2u1u 2 3u1u 3 u 2 u 3 ; 6 2 3 U {u E : u 0}. 28.
J u 2u12 4u22 u32 u1u2 9u1u3 u2u3 9; U E 3.
29. J u
3 2 5 2 9 2 u1 u2 u3 3u1u3 7u2u3 ; U E 3 . 2 2 2 163
7 3 5 2 5 4 1 3 u1 u 2 u 3 u 4 3u 2 ; 6 2 12 2 3 U {u E : u 0}.
30. J u
1 3 u 2 2u1u 2 5u1u 3 7u1 16; 2 U {u E 3 : u 0}.
31. J u 3u1 2
5 2 3 u1 u 22 4u 32 u1u 2 2u1u 3 2 2 1 u 2 u 3 8u 3 13; 2 1 7 1 4 33. J u u2 u3 2u2u3 11u1 6; 2 2 32. J u
U E 3.
U {u E 3 : u 0}. 1 2 u3 2u1u2 8u3 12; U E 3 . 2 2 2 2 35. J u 2u1 2u2 4u3 2u1u2 2u1u3 34. J u 2u1 u2 2
2
2u2u3 16; U E 3. 7 3 5 2 5 4 1 3 u1 u 2 u 3 u 4 3u 2 ; 6 2 12 2 4 U {u E : u 0}. 3 2 5 2 9 2 3 37. J u u1 u2 u3 3u1u3 7u2u3 ; U E . 2 2 2 2 2 2 3 38. J u 2u1 2u2 u3 u1u2 9u1u3 u2u3 9; U E . 5 3 3 2 2 39. J u u1 2u 2 u 3 2u1u 2 3u1u 3 u 2 u 3 ; 6 2 3 U {u E : u 0}. 36. J u
164
5 3 1 5 3 3 u1 u2 u3 22u2 17; 6 4 2 3 U {u E : u 0}. 7 2 4 3 1 3 41. J u u1 u 2 u 3 13u1 7u 3 9; 2 3 2 3 U {u E : u 0}. 5 2 2 2 3 42. J u 4u1 u2 u3 4u1u2 11u3 14; U E . 2 3 2 2 2 43. J u u1 u2 2u3 u1u2 u1u3 2 2u2u3 8u2 ; U E 3. 40. J u
44.
J u 6u12 u32 6u32 12u1 8u2 7;
U {u E 3 : u 0}. 3 2 5 2 3 45. J u 2u1 u3 u1u3 12u2 18; U E . 2 2 3 1 2 2 2 46. J u 2u1 u2 u3 u1u2 u1u3 2 2 2u2u3 10; U E 3. 47.
J u u16 u22 u32 u42 10u1 3u4 20; U E 4 .
48. J u e
2 u1 u 2
; U E 2.
J u u13 u23 u33 10u1 u2 15u3 10; U E 3. 1 2 2 2 3 50. J u u1 u2 u3 u1u2 u3 10; U E . 2 2 2 2 51. J u u1 u2 2u3 u1u2 u1u3 u2u3 49.
5u2 25; U E 3. 52. J u u2 5
1 2 u3 7u1 u3 6; U {u E 3 : u 0}. 2 165
53.
J u 3u12 u22 2u32 u1u2 3u1u3 u2u3 3u2 6; U E 3.
54. J u 5u1 2
1 2 u 2 4u 32 u1u 2 2u1u 3 2u 2 u 3 u 3 1; 2
U E 3. 3 3 55. J u u1 2u3 10u1 u2 5u3 6; U {u E 3 : u 0}. 56. 57.
J u 5u14 u26 u32 13u1 7u3 8; U E 3. J u 3u12 2u22 u32 3u1u2 u1u3
2u3u2 17; U E 3. 1 4 u3 3u1 8u2 11; 2 U {u E 3 : u 0}.
58. J u 4u1 u2 3
59.
4
J u 3u13 12u32 3u1u3 6u2 17; U {u E 3 : u 0}.
60.
J u 2u12 2u22 4u32 2u1u2 2u1u3 2u2u3 16; U E 3.
61.
J u u16 u22 u32 u42 10u1 3u4 20; U E 4 .
62. J u e
2 u1 u 2
; U E 2.
J u u13 u23 u33 10u1 u2 15u3 10; U E 3. 1 2 2 2 3 64. J u u1 u2 u3 u1u2 u3 10; U E . 2 2 2 2 65. J u u1 u2 2u3 u1u2 u1u3 u2u3
63.
5u2 25; U E 3. 66. J u u2 5
1 2 u3 7u1 u3 6; U {u E 3 : u 0}. 2 166
67.
J u 3u12 u22 2u32 u1u2 3u1u3 u2u3 3u2 6; U E 3.
1 2 u2 u32 u1u2 2u1u3 2u2u3 2 u3 1; U E 3. 1 2 1 2 2 69. J u 2u1 u 2 5u 3 u1u 2 2u1u 3 u 2 u 3 2 2 3u1 2u 2 6; U E 3 . 68. J u 5u1 2
70.
J u u13 u33 10u1 u2 5u3 6;
U {u E 3 : u 0}. 71. 72.
J u 5u14 u26 u32 13u1 7u3 8; U E 3.
J u 3u12 2u22 2u32 3u1u2 u1u3
2u2u3 17; U E 3. 73. J u 4u3 u24
1 4 u3 3u1 8u2 11; 2
U {u E 3 : u 0}. 74.
J u 8u13 12u32 3u1u3 6u2 17;
U {u E 3 : u 0}. J u 2u12 2u22 4u32 u1u2 2u2u3 16; U E 3. 1 2 2 2 3 76. J u 2u1 u2 u3 2u1u2 8u3 12; U E . 2 1 7 1 4 77. J u u2 u3 2u2u3 11u1 6; 2 2 3 U {u E : u 0}. 75.
167
78. J u
5 2 3 u1 u 22 4u 32 u1u 2 2u1u 3 2 2
1 u 2 u 3 8u 3 13; 2
U E 3.
1 3 u 2 2u1u 2 5u1u 3 7u1 16; 2 U {u E 3 : u 0}. 3 2 1 2 2 80. J u 2u1 u2 u3 u1u2 u1u3 2 2 2u2u3 10; U E 3. 3 2 5 2 3 81. J u u1 u3 u1u3 12u2 18; U E . 2 2 2 3 2 82. J u 6u1 u2 6u3 12u1 8u2 17;
79. J u u1 2
; U {u E 3 : u 0}. 3 2 2 2 3 83. J u u1 u2 2u2 u1u2 u1u3 2u2u3 8u2 ; U E . 2 7 2 4 3 1 3 84. J u u1 u 2 u 3 13u1 7u 3 9; 2 3 2 3 U {u E : u 0}. 5 3 1 5 3 3 85. J u u1 u2 u3 22u2 10; 6 4 2 3 U {u E : u 0}. 5 3 3 2 2 86. J u u1 2u 2 u 3 2u1u 2 3u1u 3 u 2 u 3 ; 6 2 3 U {u E : u 0}. 2 2 2 3 87. J u 2u1 4u2 u3 u1u2 9u1u3 u2u3 8; U E . 3 2 5 2 9 2 3 88. J u u1 u2 u3 3u1u3 7u2u3 ; U E . 2 2 2 168
7 3 5 2 5 4 1 3 u1 u2 u3 u4 3u2 ; 6 2 12 2 3 U {u E : u 0}.
89. J u
Task 2 To evaluate the task of the convex or nonlinear programming (1 – 87– variants):
1 2 u1 u22 u1u2 max, 2 2u1 u2 2, u1 0, u1 2u2 2, u2 0. 1 2 2 2. J u 3u1 2u2 u1 u2 u1u2 max, 2 1. J u 3u1 2u 2
u1 3, u2 6,
u1 0, u2 0.
3 2 u 2 2u1u 2 max, 2 u1 u2 3, u1 0, u1 u2 1, u2 0. 3 2 2 4. J u 4u1 8u 2 u1 u 2 2u1u 2 max, 2 u1 u2 1, u1 0, u1 4, u2 0. 3 2 2 5. J u 4u1 8u 2 u1 u 2 2u1u 2 max, 2 3u1 5u2 15, u1 u2 1, u1 0, u2 0. 1 2 2 6. J u 3u1 2u 2 u1 u 2 u1u 2 max, 2 u1 2u2 2, u1 0, 2u1 u2 2, u2 0. 3. J u 4u1 8u 2 u1 2
7. J u u1 6u 2 u1 3u 2 3u1u 2 max, 2
4u1 3u2 12, u1 0,
2
u1 u2 1, u2 0. 169
8. J u u1 6u 2 u1 3u 2 3u1u 2 max, 2
2
u1 u2 3, u1 0, 2u1 u2 2, u2 0. 2 2 9. J u u1 6u 2 u1 3u 2 3u1u 2 max, u1 u2 0, u1 0, u2 5, u2 0. 3 2 2 10. J u 6u 2 u1 u 2 2u1u 2 max, 2 3u1 4u2 12, u1 0, u1 u2 2, u2 0. 3 2 2 11. J u 6u 2 u1 u 2 2u1u 2 max, 2 u1 2u2 2, u1 0, u1 2, u2 0. 3 2 2 12. J u 6u 2 u1 u 2 2u1u 2 max, 2 3u1 4u2 12, u1 0, u1 2u2 2, u2 0. 3 2 2 13. J u 8u1 12u 2 u1 u 2 max, 2 2u1 u2 4, u1 0, 2u1 5u2 10, u2 0. 3 2 2 14. J u 8u1 12u 2 u1 u 2 max, 2 u1 2u2 2, u1 0, u1 6, u2 0. 3 2 2 15. J u 8u1 12u 2 u1 u 2 max, 2 3u1 2u2 0, u1 0, 4u1 3u2 12, u2 0. 1 2 2 16. J u 3u1 2u 2 u1 u 2 u1u 2 max, 2 2u1 u2 2, u1 0, 2u1 3u2 6, u2 0. 1 2 2 17. J u 6u1 4u 2 u1 u 2 u1u 2 max, 2 u1 2u2 2, u1 0, 2u1 u2 0, u2 0.
170
1 2 u 2 u1u 2 max, 2 2u1 u2 2, u1 0, u2 1, u2 0. 1 2 2 19. J u 6u1 4u 2 u1 u 2 u1u 2 max, 2 3u1 2u2 6, u1 0, 3u1 u2 3, u2 0.
18. J u 6u1 4u 2 u1 2
20. J u 8u1 6u2 2u1 u2 max,
u1 u2 1, u1 0, 3u1 2u2 6, u2 0. 2 2 21. J u 8u1 6u2 2u1 u2 max, u1 u2 1, u1 0, u1 3, u2 0. 2
2
22. J u 8u1 6u2 2u1 u2 max, 2
2
u1 u2 2, u1 0, 3u1 4u2 12, u2 0. 2 2 23. J u 2u1 2u 2 u1 2u 2 2u1u 2 max, 4u1 3u2 12, u1 0, u2 3, u2 0. 24. J u 2u1 2u 2 u1 2u 2 2u1u 2 max, 2
2
2u1 u2 4, u1 0, u1 u2 2, u2 0. 2 2 25. J u 2u1 2u 2 u1 2u 2 2u1u 2 max, 2u1 u2 2, u1 0, u 2 4, u 2 0. 26. J u 4u1 4u 2 3u1 u 2 2u1u 2 max, 2
2
4u1 5u2 20, u1 0, u1 4, u2 0. 2 2 27. J u 4u1 4u 2 3u1 u 2 2u1u 2 max, 3u1 6u2 18, u1 0, u1 4u2 4, u2 0. 2 2 28. J u 4u1 4u 2 3u1 u 2 2u1u 2 max, 3u1 4u2 12, u1 0, u1 2u2 2, u2 0. 2 2 29. J u 12u1 4u2 3u1 u2 max, 1 1 u1 u2 6, u1 0, u1 u2 1, u2 0. 2 2 171
11 1 2 1 u1 u 2 u12 u 22 u1u 2 max, 2 6 3 2 2u1 u2 2, u1 0, u1 2u2 2, u2 0.
30. J u
31. J u 18u1 12u 2 2u1 u 2 2u1u 2 max, 2
2
1 u1 u2 1, u2 0. 2 1 2 5 2 32. J u 6u1 16u 2 u1 u 2 2u1u 2 max, 2 2 5u1 2u2 10, u1 0, 3u1 2u 2 6, u 2 0.
u1 u2 4, u1 0,
33. J u 11u1 8u 2 2u1 u 2 u1u 2 max, 2
34. 35.
36.
37. 38.
39.
40.
2
u1 u2 0, u1 0, 3u1 4u2 12, u2 0. J u 8u 2 4u12 2u 22 4u1u 2 max, u1 4, u2 3, u1 0, u2 0. J u 18u1 20u2 u12 2u22 2u1u2 max, u1 u2 5, u1 2, u2 0. 3 1 J u 12u1 2u 2 u12 u 22 u1u 2 max, 2 2 u1 4, u2 3, u1 3u2 6, u2 0. J u 26u1 20u 2 3u12 2u 22 4u1u 2 max, 2u1 u2 4, u1 0, u2 2, u2 0. 2 J u 10u1 8u 2 u1 2u 22 2u1u 2 max, u1 u2 2, u1 0, u1 5, u2 0. 13 1 J u u1 5u 2 2u12 u 22 u1u 2 max, 2 2 u1 u2 3, u1 0, u1 u2 1, u2 0. 9 27 1 1 J u u1 u 2 u12 2u 22 u1u 2 max, 2 2 2 2 u1 u2 2, u1 0, u1 u2 4, u2 0. 172
41. J u 2u1 8u 2 u1 5u 2 4u1u 2 max, 2
42.
43.
44. 45. 46. 47. 48. 49. 50.
51.
2
u1 u2 3, u1 0, 2u1 3u2 6, u2 0. 2 J u 8u1 18u 2 u1 2u 22 2u1u 2 max, u1 u2 3, u1 0, u1 2, u2 0. 23 3 J u u1 u12 2u 22 u1u 2 max, 2 2 5u1 4u2 20, u1 0, u1 u2 2, u2 0. 2 J u 48u1 28u 2 4u1 2u 22 4u1u 2 max, 2u1 u2 6, u1 0, 2u1 u2 4, u2 0. 2 J u u1 10u 2 u1 2u 22 u1u 2 max, 3u1 5u2 15, u1 0, u1 2u2 4, u2 0. 2 J u 6u1 18u 2 2u1 2u 22 2u1u 2 max, 2u1 u2 2, u1 0, 5u1 3u2 15, u2 0. J u u1 5u 2 u12 u 22 u1u 2 max, u1 u2 3, u1 0, u1 2, u 2 0. 2 J u 14u1 10u 2 u1 u 22 u1u 2 max, 3u1 2u2 6, u1 0, u1 u2 4, u2 0. 2 2 J u u1 u2 6 max, u1 u2 5, u2 0. J u 5u12 u 22 4u1u 2 max, u1 0, u2 0. u12 u22 2u1 4u2 4,
J u u12 u1u2 u2u3 6 max,
u1 2u2 u3 3, u j 0, j 1,3.
52. J u u2u3 u1 u2 10 min,
2u1 5u2 u3 10, u 2 0, u 3 0.
53.
J u u1u2 u1 u2 5 min, 2u1 u2 3, u1 0, 5u1 u2 4, u2 0. 173
54. J u 10u1 5u 2 u1 2u 2 10 min, 2
2
2u12 u 2 4, u 2 0, u1 u2 8. 2 55. J u u3 u1u 2 6 min, 2u2 u3 3, u j 0, u1 u2 u3 2, j 1,2,3 .
56. J u 2u1 3u2 u1 6 max, 2
2
u12 u2 3, u2 0, 2u1 u2 5. 2 57. J u u1 3u1 5 min, u12 u22 2u1 8u2 16, u12 6u1 u2 7. 5 2 2 58. J u u1 u2 u1u2 7 min, 2
u12 4u1 u2 5, u12 6u1 u2 7. 59. J u u1 7 max, 5
u12 u22 4u2 0, u1u2 1 0, u1 0. 7 2 2 60. J u 4u1 u 2 u1u 2 u 2 5 min, 2 2 2 2u1 9u 2 18, u 2 0, u1 u2 1. 5 3 61. J u 2u1 3u2 11 min, u12 u22 6u1 16u2 72, u2 8 0.
62. J u 3u1 u 2 2u1u 2 5 min, 2
2
25u12 4u 22 100, u1 0, u12 1 0. 2 5 63. J u 5u1 2u 2 3u1 4u 2 18 max, 3u12 6 u2 0, u12 u 12 9, u1 0, u 2 0. 5 2 2 64. J u 3u1 u 2 3u1u 2 7 min, 2
3u1 u 2 1,
u22 2.
174
65. J u u1 2u1u2 u1 6 max, 6
3u12 15, u1 0 u1 5u 2 10, u 2 0. 2 2 66. J u 4u1 3u 2 4u1u 2 u1 6u 2 5 max, u12 u 22 3, 3u12 u2 4. 2 2 67. J u u1 2u 2 u1u 2 u1 26 max, u12 25, u1 2u2 5, u2 0. 2 2 68. J u u1 u2 max, 2u1 3u2 6, u1 5u2 10, u1 0. 2 2 69. J u 2u1 u 2 4u1u 2 u1 6 max, u1 u2 1, u1 0 , 2u1 u2 5, u2 0. 2 2 70. J u 2u1 3u 2 u1u 2 6 max, u1 u2 3. u12 u2 5, u1 0 2 2 71. J u u1 u 2 u1 5u 2 5 max, , u1 u2 5. 2 2 72. J u u1 u 2 u1u 3 min, 3u1 u 2 u 3 4, u j 0, u1 2u2 2u3 3, j 1,2,3.
73. J u 5u1 6u 2 u 3 8u1u 2 u1 max, 2
2
2
u12 u 2 u3 5, u1 5u 2 8, u1 0, u 2 0. 74.
J u u12 u22 u32 min,
u1 u2 u3 3, 2u1 u2 u3 5. 75.
J u 3u12 u22 u32 u1u2 6 min,
2
2u1 u2 u3 5, u2 3u3 8, u2 0. 76. J u u1u 2 u1u 3 2u 2 u 3 u1 5 max, u1 u2 u3 3, 2u1 u 2 5, u1 0. 1 2 3 2 2 77. J u u1 u 2 u 3 12u1 13u 2 5u 3 min, 2 2 u1 5u2 4u3 16 , 2u1 7u 2 3u 3 2, u1 0. 175
78. J u u1 2u 2 30u1 16u 3 10 min, 2
2
5u1 3u2 4u3 20, u1 6u2 3u3 0, u3 0. 1 2 2 79. J u u1 u2 5u1 u3 16 min, u 2
u1 u2 2u3 3, 2u1 u 2 3u 3 11, u j 0, j 1,2,3. 80. J u 2u1 2u1 4u 2 3u 3 8 max, 2
8u1 3u2 3u3 40, 2u1 u 2 u 3 3, 81.
J u u1u3 u1 10 min,
u 2 0.
u12 u3 3, u j 0
u22 u3 3, j 1,2,3.
82. J u 3u1 u 2 u 2 7u 3 max, 2
2
4u1 u 2 2u3 5 1, u1 0, 2u2 u3 4.
83. J u u1 u 2 u1u 2 6u1u 3 u 2 u 3 u1 25 min, 2
2
u12 u 22 u1u 2 u3 10, u1u 2 u1 2u 2 u 3 4, u j 0, j 1,2,3. 84.
J u e u1 u2 u3 max,
u12 u 22 u32 10, u 2 0 u13 5, u3 0.
85. J u 3u 2 11u1 3u 2 u 3 27 max, 2
u1 7u2 3u3 7, 5u1 2u 2 u 3 2, u 3 0. 86.
J u 4u12 8u1 u2 4u3 12 min,
3u1 u2 u3 5 , u1 2u 2 u 3 0, u j 0, j 1,2,3. 1 2 3 2 87. J u u 2 u 3 2u 2 9u 3 min, 2 2 3u1 5u2 u3 19, 2u 2 u 3 0, u j 0, j 1,2,3.
176
Task 3 To evaluate the task of the linear programming (1 – 89 variants):
J u cu min; U u E n / Au b, u 0 . 3 1 1 1 1 1. c (5,1,1,0,0), b 5,4,11, A 2 1 3 0 0 . 0 5 6 1 0 1 2 1 6 1 2. c (6,1,1,2,0), b 4,1,9, A 3 1 1 1 0 . 1 3 5 0 0 3 1 1 6 1 0 5 1 7 . 1 2 3 1 1 5 1 1 3 1 4. c (7,1,1,1,0), b 5,3,2 , A 0 2 4 1 1 . 1 3 5 0 0
3. c (0,6,1,1,0), b 6,6,6 , A 1
2 1 1 1 1 5. c (8,1,3,0,0), b 4,3,6 , A 2 0 1 3 5 . 3 0 1 6 1 2 1 2 0 0 6. c (0,1,3,1,1), b 2,8,5, A 1 1 4 1 3 . 3 1 1 0 6 2 0 1 1 1 7. c (1,2,1,1,0), b 2,7,2 , A 4 1 3 1 2 . 1 0 1 2 1
177
6 1 1 2 1 8. c (0,1,6,1,3), b 9,14,3, A 1 0 1 7 8 . 1 0 2 1 1 2 0 3 1 1 9. c (8,1,1,1,0), b 5,9,3, A 3 1 1 6 2 . 1 0 2 1 2
2 0 3 1 0 10. c (1,3,1,1,0), b 4,4,15, A 1 0 1 2 3 . 3 3 6 3 6 4 1 1 0 1 11. c (0,2,0,1,3), b 6,1,24 , A 1 3 1 0 3 . 8 4 12 4 12 8 16 8 8 24 1 . 12. c (10,5,25,5,0), b 32,1,15, A 0 2 1 1 0 3 2 1 1
4 1 1 2 1 13. c (6,0,1,1,2), b 8,2,2 , A 2 1 0 1 0 . 1 1 0 0 1 1 2 14. c (5,1,3,1,0), b 7,7,12 , A 0 3 0 4 3 4 15. c (5,3,2,1,1), b 12,16,3, A 3 2 1 3
178
3 4 1 1 4 0 . 0 8 1 1 0 0 1 1 1 . 0 0 1
1 1 1 0 0 16. c (7,0,1,1,1), b 1,12,4, A 2 2 1 1 2 . 2 1 0 0 1 1 1 1 0 0 17. c (6,1,2,1,1), b 2,11,6 , A 5 2 1 1 1 . 3 2 0 0 1 2 1 1 1 3 18. c (0,0,3,2,1), b 5,7,2 , A 3 0 2 1 6 . 1 0 1 2 1 6 3 1 1 1 19. c (1,7,2,1,1), b 20,12,6 , A 4 3 0 1 0 . 3 2 0 0 1
1 2 1 0 0 5 1 1 2 . 20. c (2,0,1,1,1), b 2,14,1, A 3 1 1 0 0 1 1 2 1 0 0 21. c (6,1,0,1,2), b 2,18,2, A 2 6 2 1 1 . 1 2 0 0 1 1 2 1 0 0 22. c (0,3,1,1,1), b 2,2,6 , A 1 1 0 1 0 . 2 1 1 1 2 2 2 1 1 1 23. c (3,0,1,2,1), b 6,2,2 , A 2 1 0 1 0 . 1 1 0 0 1
179
1 1 1 0 0 24. c (0,5,1,1,1), b 2,2,10 , A 1 2 0 1 0 . 2 1 1 1 2
3 4 1 0 0 25. c (1,5,2,1,1), b 12,1,3, A 1 1 0 1 0 . 3 2 1 1 1 1 1 1 0 0 26. c (5,0,1,1,1), b 1,3,12 , A 3 1 0 1 0 . 2 2 1 1 2 1 1 1 0 0 27. c (7,0,2,1,1), b 2,3,11, A 3 1 0 1 0 . 5 2 1 1 1 5 5 1 2 1 28. c (1,4,1,1,1), b 28,2,12 , A 1 2 0 1 0 . 3 4 0 0 1 1 29. c (0,8,2,1,1), b 2,20,6 , A 6 3 3 30. c (0,2,1,1,1), b 14,10,1, A 2 1 1 31. c (7,2,0,1,2), b 2,12,18, A 3 2
180
2 3 2 5 5 1
2 4 6
1 0 0 1 1 1 . 0 0 1 1 1 2 0 1 0 . 0 0 1 1 0 0 0 1 0 . 2 1 1
1 2 1 0 0 32. c (1,3,1,1,1), b 2,6,1, A 2 1 1 1 2 . 1 1 0 0 1 1 2 1 0 0 33. c (5,1,1,1,2), b 2,8,2 , A 4 1 1 2 1 . 1 1 0 0 1 1 1 2 2 1 34. c (1,2,1,1,1), b 11,2,3, A 1 2 0 1 0 . 1 1 0 0 1 3 1 2 1 2 35. c (10,5,2,1,1), b 17,1,3, A 1 1 0 1 0 . 1 3 0 0 1 1 0 2 1 3 36. c (2,1,3,1,1), b 6,16,7 , A 2 2 4 8 4 . 1 0 1 7 1 1 1 1 0 0 37. c (4,1,1,2,1), b 2,13,16, A 4 3 2 1 0 . 3 2 0 0 1 4 3 1 0 0 38. c ( 2,2,1,2,1), b 12,2,26 , A 1 2 0 1 0 . 6 3 1 1 1 9 1 1 1 2 39. c (5,2,1,1,1), b 26,12,6 , A 4 3 0 1 0 . 3 2 0 0 1
181
2 6 1 1 1 40. c (1,11,1,2,1), b 13,10,1, A 2 5 0 1 0 . 1 1 0 0 1 3 1 3 1 0 41. c (5,1,1,2,0), b 1,6,2 , A 2 3 1 2 1 . 3 1 2 1 0 2 1 1 1 2 42. c (0,3,1,1,1), b 6,2,1, A 1 1 0 1 0 . 1 1 0 0 1 2 1 1 1 1 43. c (8,1,3,0,0), b 4,3,6, A 2 0 1 3 5 . 3 0 1 6 1 1 1 1 0 0 44. c (2,1,1,1,1), b 2,11,3, A 1 1 2 2 1 . 1 1 0 0 1 4 3 2 45. c (5,0,1,2,1), b 13,3,6 , A 3 1 0 3 2 0 1 1 1 46. c (1,3,1,1,1), b 1,17,4 , A 5 2 2 2 1 0 3 4 47. c (9,5,2,1,1), b 12,17,3, A 2 3 1 3
182
1 1 1 0 . 0 1 0 0 1 3 . 0 1 1 0 0 1 2 1 . 0 0 1
4 3 1 0 0 48. c (1,1,1,2,1), b 12,26,12 , A 6 3 1 1 1 . 3 4 0 0 1 1 2 1 0 0 49. c (0,7,1,1,1), b 2,26,6, A 9 1 1 1 2 . 3 2 0 0 1 50.
51.
52.
53.
54.
55.
1 2 1 0 0 c (4,8,1,2,1), b 2,13,1, A 2 6 1 1 1 . 1 1 0 0 1 1 1 0 1 2 c (3,1,1,1,0), b 3,6,5, A 2 1 1 2 3 . 3 2 0 3 8 1 2 1 0 0 c (1,3,1,2,1), b 2,2,5, A 1 1 0 1 0 . 1 2 1 1 1 1 2 1 0 0 c (0,1,1,2,1), b 2,2,6 , A 2 1 0 1 0 . 2 2 1 1 1 1 1 1 0 0 c (0,5,1,1,1), b 2,2,11, A 1 2 0 1 0 . 1 1 2 2 1 3 4 1 0 0 c (9,2,1,0,1), b 12,1,17 , A 1 1 0 1 0 . 2 3 1 2 1
183
1 1 1 0 0 56. c (1,0,1,1,1), b 1,3,17 , A 3 1 0 1 0 . 5 2 2 1 3 1 1 1 0 57. c (3,2,1,2,1), b 2,3,13, A 3 1 0 1 4 3 2 1 58.
59.
60.
61.
0 0 . 1
1 2 1 0 0 c (9,0,1,1,1), b 2,12,26 , A 4 3 0 1 0 . 9 1 1 1 2 6 3 1 1 1 c (5,5,1,2,1), b 26,2,12, A 1 2 0 1 0 . 3 4 0 0 1 1 2 1 0 0 c (0,10,1,2,1), b 2,10,1, A 2 5 0 1 0 . 2 6 1 1 1 3 1 3 1 2 c (3,2,1,1,0), b 5,5,5, A 3 2 1 1 1 . 7 2 2 0 1
5 10 5 15 10 62. c (1,2,1,1,0), b 25,3,5, A 0 1 1 6 2 . 0 6 1 1 1
2 1 0 1 1 63. c (1,1,2,1,0), b 4,7,9 , A 3 2 0 1 1 . 1 1 1 2 6
184
1 2 1 1 1 64. c (1,3,1,0,0), b 4,3,6 , A 1 0 2 1 3 . 3 0 3 1 2 3 1 1 2 3 65. c (2,1,1,5,0), b 7,1,9 , A 2 0 3 2 1. 3 0 1 1 6 4 1 1 2 1 66. c (6,0,1,1,2), b 8,2,2 , A 2 1 0 1 0 . 1 1 0 0 1 1 2 3 4 1 67. c (5,1,3,1,0), b 7,7,12, A 0 3 1 4 0 . 0 4 0 8 1 3 4 1 0 0 68. c (5,3,2,1,1), b 12,16,3, A 3 2 1 1 1 . 1 3 0 0 1
1 1 1 0 0 69. c (7,0,1,1,1), b 1,12,4 , A 2 2 1 1 2 . 2 1 0 0 1 1 1 1 0 0 70. c (6,1,2,1,1), b 2,11,6 , A 5 2 1 1 1 . 3 2 0 0 1 2 1 1 1 3 71. c (0,0,3,2,1), b 5,7,2 , A 3 0 2 1 6 . 1 0 1 2 1
185
72.
73.
74.
75.
76.
77.
78.
79.
6 3 1 1 1 c (1,7,2,1,1), b 20,12,6, A 4 3 0 1 0 . 3 2 0 0 1 1 2 1 0 0 c (2,0,1,1,1), b 2,14,1, A 3 5 1 1 2 . 1 1 0 0 1 1 2 1 0 0 c (6,1,0,1,2), b 2,18,2, A 2 6 2 1 1 . 1 2 0 0 1 1 2 1 0 0 c (0,3,1,1,1), b 2,2,6, A 1 1 0 1 0 . 2 1 1 1 2 2 2 1 1 1 c (3,0,1,2,1), b 6,2,2, A 2 1 0 1 0 . 1 1 0 0 1 1 1 1 0 0 c (0,5,1,1,1), b 2,2,10, A 1 2 0 1 0 . 2 2 1 1 2 3 4 1 0 0 c (1,5,2,1,1), b 12,1,3, A 1 1 0 1 0 . 3 2 1 1 1 1 1 1 0 0 c (5,0,1,1,1), b 1,3,12, A 3 1 0 1 0 . 2 2 1 1 2
186
1 1 1 0 0 80. c (7,0,2,1,1), b 2,3,11, A 3 1 0 1 0 . 5 2 1 1 1 5 5 1 2 1 81. c (1,4,1,1,1), b 28,2,12, A 1 2 0 1 0 . 3 4 0 0 1 1 2 1 0 0 82. c (0,8,2,1,1), b 2,20,6 , A 6 3 1 1 1 . 3 2 0 0 1 1 2 1 0 0 83. c (7,2,0,1,2), b 2,12,18, A 3 4 0 1 0 . 2 6 2 1 1 1 2 1 0 0 84. c (1,3,1,1,1), b 2,6,1, A 2 1 1 1 2 . 1 1 0 0 1 1 1 2 2 1 85. c (1,2,1,1,1), b 11,2,3, A 1 2 0 1 0 . 1 1 0 0 1 3 1 2 1 2 86. c (10,5,2,1,1), b 17,1,3, A 1 1 0 1 0 . 1 3 0 0 1 1 0 2 1 3 87. c (2,1,3,1,1), b 6,16,7 , A 2 2 4 8 4 . 1 0 1 7 1
187
2 6 1 1 1 88. c (1,11,1,2,1), b 13,10,1, A 2 5 0 1 0 . 7 4 0 0 1
1 1 1 0 0 89. c ( 2,1,1,1,1), b 2,11,3, A 1 1 2 2 1 . 1 1 0 0 1
188
Appendix 3
KEYS
To the 1st task It is possible to construct the matrixes J (u ) and check its determination on U according to the theorem 3 (lecture 4). Similarly tasks 289 are executed. To the 2nd task 1.
J (u* ) 2,50 ;
17.
J (u* ) 8,00 ;
33.
J (u* ) 20,82;
2.
J (u* ) 4,75 ;
18.
J (u * ) 5,75 ;
34.
J (u* ) 15,00;
3.
J (u* ) 11,00 ;
19.
J (u * ) 8,40 ;
35.
J (u* ) 66,0;
4.
J (u* ) 11,00 ;
20.
J (u* ) 12,76;
36.
J (u* ) 25,12;
5.
J (u* ) 11,12 ;
21.
J (u* ) 17,00;
37.
J (u* ) 47,0;
6.
J (u* ) 2,50 ;
22.
J (u* ) 15,84 ;
38.
J (u* ) 23,50;
7.
J (u* ) 5,67 ;
23.
J (u* ) 4,45 ;
39.
J (u* ) 12,07;
8.
J (u* ) 5,29 ;
24.
J (u* ) 3,77 ;
40.
J (u* ) 25,13 ;
9.
J (u* ) 6,25 ;
25.
J (u* ) 4,20 ;
41.
J (u* ) 4,60 ;
10.
J (u* ) 9,59 ;
26.
J (u* ) 11,02 ;
42.
J (u* ) 40,0;
11.
J (u* ) 12,50 ;
27.
J (u* ) 10,12 ;
43.
J (u* ) 23,45;
12.
J (u* ) 9,59 ;
28.
J (u* ) 9,53 ;
44.
J (u* ) 112,0;
13.
J (u* ) 24,34 ;
29.
J (u* ) 13,00 ;
45.
J (u* ) 103,5;
14.
J (u* ) 38,91 ;
30.
J (u* ) 6,10 ;
46.
J (u* ) 41,23;
15.
J (u* ) 27,99 ;
31.
J (u* ) 41,00;
47.
J (u* ) 6,75 ;
16.
J (u* ) 4,57 ;
32.
J (u* ) 25,81 ;
48.
J (u* ) 40,0.
189
To the 3rd task 1.
J (u* ) 5 ;
17.
J (u* ) 15 ;
2.
J (u* ) 6 ;
18.
J (u* )
3.
J (u* ) 6 ;
19.
4.
J (u* )
22.
9.
14 ; 3 134 ; J (u* ) 7 137 ; J (u* ) 33 J (u* ) 0 ; 11 J (u* ) ; 3 J (u* ) 0 ;
10.
J (u* ) 7 ;
11.
J (u* )
12.
4 ; 3 J (u* ) 14 ;
33. 34. 35.
20 ; 3 22 ; J (u* ) 3 437 J (u* ) ; 13 31 ; J (u* ) 3 J (u * )
23.
43 ; 7 99 ; J (u* ) 5 17 ; J (u* ) 3 J (u* ) 7 ;
39.
89 ; 5 J (u* ) 19 ;
24.
J (u* ) 6 ;
40.
J (u* ) 21 ;
25.
U ;
41.
26.
J (u* )
27.
J (u* ) 16 ;
43.
J (u* ) 10 ;
28.
J (u* ) 26 ;
44.
J (u* ) 3 ; 17 ; J (u* ) 3 110 ; J (u* ) 7 19 ; J (u* ) 2
13.
J (u* ) 6 ;
29.
J (u* ) 12 ;
45.
14.
J (u* )
30.
J (u* )
15.
J (u* ) 26 ;
31.
16.
J (u* )
29 ; 5
32.
49.
J (u* ) 16 ;
5. 6. 7. 8.
54 ; 13
7 ; 6
20. 21.
55.
J (u* )
27 ; 5
3 ; 7 118 ; J (u* ) 5 19 ; J (u* ) 3 438 ; J (u* ) 13 190
36. 37. 38.
42.
46. 47. 48. 61.
U ; J (u* )
J (u* ) 5 ; 77 ; 5 389 ; J (u* ) 13 72 ; J (u* ) 5 J (u* )
J (u* ) 5 ;
50.
J (u* ) 23 ;
56.
J (u* ) 8 ;
57.
J (u* )
52.
23 ; 3 J (u* ) 6 ;
58.
24 ; 5 J (u* ) 22 ;
53.
J (u* ) 4 ;
59.
J (u* ) 28 ;
54.
J (u * )
60.
U ;
51.
J (u* )
26 ; 5
191
62. 63. 64. 65.
133 ; 37 21 ; J (u* ) 10 J (u* ) 1 ; 5 J (u* ) . 17 J (u* )
Appendix 4
TESTS
1) Weierstrass’ theorems define A) Sufficient conditions that ensemble
U
*
{ u U / I u min I u } * * u U
B) Necessary and sufficient conditions that ensemble
U {u U / I u min I u } * * * uU
C) Necessary conditions that ensemble U { u U / I u min I u }
*
*
*
u U
D) Necessary and sufficient conditions to ensemble convexity U {u U / I u min I u }
*
*
* uU
u* U
E) Necessary conditions that the point minimum to function
Iu on ensemble U
is a point of the
2) Sufficient conditions that the ensemble give
U {u U / I u min I u} * * * uU
A) Weierstrass’ theorems B) Farkas’s theorems C) KuhnTucker’s theorems D) Lagrange theorems E) There isn’t any correct answer amongst A)D). 3) Ensemble U E n is identified convex if
A) u U , v U , 0,1 the point
u u 1 v U
u U ,vU , 0,1 , u u 1 v C) u U , v U , 0,1, u u 1 v B)
192
u U,v U, number 0,1 for which u u 1 v U E) u U , v U , E the point u u 1 v U D)
1
4) Ensemble U E is convex iff A) it contains all convex linear combinations of any final number own points B) u U, v U possible construct the point n
which is a convex linear combination of
u u 1 v U
the points u, v C) It is limited and closed D) v U , 0 vicinity S v U E) From any sequence {u k } U it is possible to select converging subsequence
5) Function I u determined on convex ensemble U E n identified convex if: A) u U ,v U , 0,1,
is
I u 1 v I u 1 I v
B)
{u }U : lim u u, I u lim I u k k k k k u U , v U , 0,1, C) I u 1 v I u 1 I v
D)
{u } U : lim u u , I u lim I u k k k k k
E) There isn’t any correct answer amongst A)D).
6) Necessary and sufficient condition that continuously differentiated function I u is convex on convex ensemble U En : A) I u I v I v , u v , u U , v U
B)
I u I v I u 1 v , u U , v U , 0,1 193
C) D) E)
I u , u u 0, u U I u , u u 0, u U I u I v I u 1 v , u U , v U , 0,1 *
*
*
*
7) Necessary and sufficient condition that continuously differentiated function I u is convex on convex ensemble U E n
A) I u I v , u v 0, u U , v U B) I u I v , u v 0, u U , v U C) I u I v I u I v , u U , v U
D) I u 1 v I u I v , u U , v U E) I u 1 v I u I v , u U , v U
8) Function I u determined on convex ensemble U E n is identified strongly convex on U , if A) 0, I u 1 v I u 1 I v 1 u v 2,
u U ,v U , 0,1 B) C) D)
I u 1 v I u 1 I v , u U , v U , 0,1 I u 1 v I u 1 I v ,
u U , v U , 0,1
0, I u 1 v I u 1 I v 1 u v 2 ,
u U , v U , 0,1 E) There isn’t any correct answer amongst A)D).
U
9) In order to twice continuously differentiated on convex ensemble function I (u) is convex on U necessary and sufficiently
performing the condition A) I u ,
0, u U , E n B) det I u 0 u U 194
C)
I u , 0, u U , E n
I u u , v 0, u U , v U E) I u u , v 0, u U , v U D)
10) Choose correct statement A) If I u , G u are convex functions determined on convex ensemble U , 0, 0 that function
I u G u is also convex
on ensemble U B) If I u , G u are convex functions determined on convex ensemble U E n , that function I u G u is also convex on U C) If
I u , Gu
are convex functions determined on convex
ensemble U E n , that function I u G u is also convex on U D) If I u , G u are convex functions determined on convex
ensemble U E n , G u 0, u U , that I u / Gu
is convex
function on U E) There isn’t any correct answer amongst A)D). 11) Choose correct statement A) Intersection of two convex ensembles is convex ensemble B) Union of two convex ensembles is convex ensemble C) If U E n is convex ensemble, that ensemble En \ U is also convex D) If U1 En, U2 En is convex ensemble, that ensemble U1 \ U2 is also convex E) There isn’t any correct answer amongst A)D). 12) Define type of the following problem I
u U {u E n / u 0, j I , j g u ai ,u b 0,i 1, m, i i i g u a ,u b 0,i m 1, s} i i
u c, u inf
A) General problem of linear programming B) Canonical problem of linear programming 195
C) Nondegenerate problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem
I u c, u inf n u U {u E / u 0, i 1, n, Au b} j
13) Define type of the following problem
A) General problem of linear programming B) Canonical problem of linear programming C) Nondegenerate problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem 14) Function I(u) = u2 – 2u1u2 +u22 on ensemble U = En is A) convex B) concave C) neither convex, nor concave D) convex under u1 ≥ 0, u2 ≥ 0 and concave under u1 ≤ 0, u2 ≤ 0 E) convex under u1 ≤ 0, u2 ≤ 0 and concave under u1 ≥ 0, u2 ≥ 0 15) Define type of the problem I(u) = 2u12 – u22 → inf u U {u E n / 2u u 3, u 4u 5} 1 2 1 2 A) Problem of nonlinear programming B) Canonical problem of linear programming C) Convex programming problem D) General problem of linear programming E) The simplest variational problem 16) Define type of the problem I(u) = 2u1 – u2 → inf u U {u E n / u 0, u u 2 4, 2u u 2 2 1 2 1 2
A) Convex programming problem B) Canonical problem of linear programming C) General problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem 17) Define type of the problem
I (u ) 2u u inf 1 2 u U {u E n / u 0, u 4u 2, u 3u 4} 1 1 2 1 2
196
A) General problem of linear programming B) Canonical problem of linear programming C) Convex programming problem D) Problem of nonlinear programming E) The simplest variational problem 18) Sequence {uk} U is identified minimizing to function determined on ensemble U , if A) where B)
lim I (u ) I , * k k
I u
C) D)
k 1
I u
I inf I u * uU
I u , k 1,2,... k moreover
lim u u , k k
u U
k 1 I uk , k 1,2,...
Iu
E) There isn’t any correct answer amongst A)D). 19) Choose correct statement for problem
I u inf, u U E n
I (u)
and ensemble
{u } U k
for function
A) For any function minimizing sequence
U E n always exists a
I (u)
B) If function I(u) is continuously differentiable on ensemble U, that it such that reaches minimum value on U, i.e.
I u min I u * uU
u U *
C) If minimizing sequence {uk} U exists for function I (u), that such that
u U *
D) If the point
I u min I u * uU
exists such that
u U I u min I u * * uU sequence {uk} U exists for function I(u)
, that minimizing
E) There isn’t any correct answer amongst A)D). 20) Choose correct statement for problem
I u inf, u U E n , U {u U / I u min I u } * * * uU 197
A) If function I(u) is semicontinuous from below on compact ensemble U, that ensemble
U *
B) If ensemble U is convex, but function that ensemble
U *
C) If ensemble U is limited, but function U, that ensemble
I u is continuous on U,
I u is convex on ensemble
U *
D) If function
I u is semicontinuous from below on ensemble U, but
ensemble U is limited, that E) If function U, that
I u
U
*
is continuously differentiable on closed ensemble
U *
21) Simplex method is used for solution A) linear programming problem in canonical form B) linear programming problem in general form C) convex programming problem D) nonlinear programming problem E) optimal programming problem 22) KuhnTucker theorems A) define necessary and sufficient conditions that in the convex programming problem for each point
exist Lagrange’s multipliers
U {u U / I u min I u } * * * uU
0
such that pair
u , * *
forms saddle
point to Lagrange’ function B) define necessary and sufficient existence conditions of function
I (u)
minimizing sequences {uk} U C) define necessary and sufficient convexity conditions to function I u on convex ensemble U E n
D) define necessary conditions that in the convex programming problem
198
consists of single point E) define necessary and sufficient conditions that ensemble U {u U / I u min I u } * * * uU
ensemble
23)
U {u U / I u min I u } * * * uU
For
the
convex
programming
problem
of
the
I u inf, u U {u E n / u U , g u 0, i 1, m} 0 i
type
Lagrange’s function has the form A) L u , I u m
g u , u U , 0 i 1 i i m { E / 0,..., m 0} 0 1 B) Lu, m g u , 0, i i 1 i i m i 1, m, 1, u U 0 i 1 i
Lu , I (u ), u U E1 0 m D) Lu , I (u g u ), u U , E m 0 i 1 i i m E) Lu , I u g u , u U , 0 i1 i i { E m / 0,..., m 0} 0 1 C)
24) For the convex programming problem of the type I(u) →inf
u U {u E n / u U , g u 0, i 1, m, 0 i g u a i , u b 0, i m 1, s} i i
Lagrange’s function has the form
199
A) Lu, I u s g u , u U ,
0 i 1 i i {u E s / u 0, u 0, ..., u m 0 1 2 B) m Lu, g u , 0, i 1, m, i i 1 i i
0}
m 1, u U 0 i 1 i
m L u , I u g u , u U , E m 0 i 1 i i s D) Lu , I u g u , u U ,
C)
0 i 1 i i {u E s / u 0, u 0, ..., u s 0} 0 1 2
E)
Lu, I u , u U , E1
25) For the convex programming problem necessary and sufficient conditions that for any point
u U {u U / I u min I u * * * uU
exist Lagrange’s multipliers
*
such that pair
0
saddle point to Lagrange’s function are defined A) KuhnTucker’s theorems B) Lagrange’s theorems C) Weierstrass’ theorem D) Farkas’ theorems E) Bellman’s theorems 26) Let
(u , ) forms * *
I u be a convex function determined and continuously
differentiated on convex ensemble U, ensemble
min I u uU
. In order to point
u U *
U {u U / I u * * * be a point of the function
minimum I (u) on ensemble U necessary and sufficient performing the condition A)
I u , u u 0, u U * * B) I u , u u 0, u U * * 200
I u I u I u , u u , u U * * D) I u , 0, E C)
E)
* I u 0 *
27) Pair function
n
u* , * U 0 0 is identified saddle point to Lagrange’s
, if s Lu , I u g u i1 i i
L u , L u , * L u , * , u U , 0 0 * * B) * L u , L u , , u U , 0 0 * C) * L u , 0 A)
D) E)
*
L u , L u , * L u , , u U , 0 0 *
u U {u U / I u min I u }, * 0, i 1, s j * * * * uU
28) If pair
u*, * U 0 0
is a saddle point to Lagrange’s
Lu, in the convex programming problem, that A) the point u U {u U / I u min I u }
function
*
B) Lebesgue ensemble
that
*
uU is compact M u {u U / I u I u } * * C) there is a minimizing sequence for function I u , such {u } U * *
lim u k k
D) the convex programming problem is nondegenerate contains single point E) ensemble
U {u U / I u min I u } * * * uU 201
29) For the nonlinear programming problem
I u inf, u U {u E n / u U , g u 0, i 1, m, g u 0, i 1, m i 0 i
generalized Lagrange’s function has the form A) L u ,
s
I u g u , u U ,
0 0 i 1 i i { E s 1 / 0 , 0 ,..., m 0} 0 1
0
m Lu , I u g u , u U , E m 0 i1 i m C) Lu , I u g u , u U , E m 0 0 i1 i i s D) Lu , I u g u , u U , E s 1 0 0 i 1 i i
B)
E)
s Lu , I u g u , u U , 0 0 i 1 i
30) Let U E n be convex ensemble, function I u C 1 U condition is
I u , u u 0, u U * *
A) necessary condition that the point
u U {u U / I u * * * * u
min I u } uU B) necessary and sufficient condition that the point
u U {u U / I u min I u } * * * * uU
u U {u U / I u min I u } * * * * uU
C) sufficient condition that the point
D) necessary and sufficient condition that the function I(u) is convex in the point E)
U
u U *
necessary
and
sufficient
condition
{u U / I u min I u } * * * uU 202
that
ensemble
31) Indicate faithful statement A) convex programming problem is a partial case of the nonlinear programming problem B) nondegenerate problem of nonlinear programming can be reduced to the convex programming problem C) convex programming problem is a partial case of the linear programming problem D) any nonlinear programming problem can be reduced to the convex programming problem E) nonlinear programming problem is a partial case of the convex programming problem 32) For solving of the nonlinear programming problem is used A) Lagrange multipliers method B) Simplexmethod C) Method of the least squares D) Pontryagin maximum principle E) Bellman’s dynamic programming method 33) What from enumerated methods can be used for solving of the convex programming problem A) Lagrange multipliers method B) Simplexmethod C) Method of the least squares D) Pontryagin maximum principle E) Bellman’s dynamic programming method 34) What from enumerated methods can be used for solving of the linear programming problem A) Simplexmethod B) Method of the least squares C) Pontryagin maximum principle D) Bellman’s dynamic programming method E) any method from A) D) 35) In the convex programming problem the minimum to function I(u) on ensemble U can be reached A) in internal or border points ensemble U B) only in the border points ensemble U C) only in the isolated points ensemble U 203
D) only in the internal points ensemble U E) in internal, border, isolated points ensemble U 36) In the nonlinear programming problem minimum to function I(u) on ensemble U can be reached A) in internal, border, isolated points ensemble U B) only in the border points ensemble U C) only in the isolated points ensemble U D) in internal or border points ensemble U E) only in the internal points ensemble U 37) If in the linear programming problem in canonical form ensemble n
I u inf, u U {u E / u 0, j 1, n, Au b} j
U {u U / I u min I u } * * * uU
contains single point
u
*
, that the
point is A) an extreme point ensemble U B) an isolated points ensemble U C) an internal point ensemble U D) an internal or extreme point ensemble U E) extreme or isolated point ensemble U 38) The linear programming problem in canonical form has the form A) I u c, u inf,
u U {u E n / u 0, j 1, n, Au b} j I u c,u inf, B) u U {u E n / u 0, j I , g u ai ,
j
i
u b 0, i 1, m, i g u ai ,u b 0, i m 1, s} i i C) I u c , u inf, u U {u E / Au b} n
204
D) I u c,u inf,
u U {u E n / u 0, j I , g u ai , j i u b 0, i 1, m} i I u c,u inf, E) u U {u E n / 0 g u b , i 1, m} i i 39) The linear programming problem in canonical form identified nondegenerate,
I u c,u inf, is n uU {u E / u 0, j 1, n, Au b} j if:
A) any point u U has not less than rang A positive coordinates B) rangA = m, where A is a constant matrix of dimensionality
m n, m n C) ensemble
U
*
{u U / I u min I u } * * uU
D) rangA = m, where A is a constant matrix of dimensionality
m n, m n E) any point coordinates
u U
has no more than rangA = m,
positive
40) By extreme point ensemble
U {u E n / u 0, j 1, n, Au b} j
is called
u U which can not be presented in the manner u v 1 w , where 0,1, v U , w U A) Point
B) Isolated point ensemble U C) Border point ensemble U D) Point u U presented in the manner of where
0,1, v U , w U
E) Internal point ensemble U
205
of
u v 1 w ,
CONTENTS FOREWORD ..................................................................................................... 3 Introduction ........................................................................................................ 4 Lecture 1. THE MAIN DETERMINATIONS. STATEMENT OF THE PROBLEM .......................................................................................... 4 Lecture 2. WEIERSTRASS’ THEOREM ......................................................... 12 Chapter I CONVEX PROGRAMMING. ELEMENTS OF THE CONVEXANALYSIS ........................................................................................ 18 Lecture 3. CONVEX SETS ................................................................................ 18 Lecture 4. CONVEX FUNCTIONS .................................................................. 27 Lecture 5. THE ASSIGMENT WAYS OF THE CONVEX SETS. THEOREM ABOUT GLOBAL MINIMUM. OPTIMALITY CRITERIA. THE POINT PROJECTION ON SET ................................................................ 36 Lectures 6, 7. SEPARABILITY OF THE CONVEX SETS .............................. 44 Lecture 8. LAGRANGE’S FUNCTION. SADDLE POINT ............................. 52 Lectures 9, 10. KUHNTUCKER’S THEOREM ............................................. 60 Chapter II NONLINEAR PROGRAMMING .................................................. 75 Lectures 11, 12. STATEMENT OF THE PROBLEM. NECESSARYCONDITIONS OF THE OPTIMALITY .................................... 76 Lecture 13. SOLUTION ALGORITHM OF THE NONLINEAR PROGRAMMING PROBLEM ......................................................................... 88 Lecture 14. DUALITY THEORY ..................................................................... 97 Chapter III LINEAR PROGRAMMING ......................................................... 106 Lectures 15, 16. STATEMENT OF THE PROBLEM. SIMPLEXMETHOD .. 106 Lecture 17. DIRECTION CHOICE. NEW SIMPLEXTABLE CONSRUCTION. THE INITIAL EXTREME POINT CONSTRUCTION ....... 120 References .......................................................................................................... 129 Appendix I. TASKS FOR INDEPENDENT WORK ......................................... 130 Appendix II. TASKS ON MATHEMATICAL PROGRAMMING ................... 161 Appendix III. KEYS........................................................................................... 189 Appendix IV. TESTS ......................................................................................... 192
206
У ч еб но е изда ние
Айсагалиев Серикбай Абдигалиевич Жунусова Жанат Хафизовна MATHEMATICAL PROGRAMMING Учебное пособие Компьютерная верстка: Т.Е. Сапарова Дизайн обложки: Г.К. Курманова
ИБ № 5074 Подписано в печать 16.03.11. Формат 60х84 1/16. Бумага офсетная. Печать RISO. Объем 13,00 п.л. Тираж 500 экз. Заказ № 230. Издательство «Қазақ университетi» Казахского национального университета им. альФараби. 050040, г. Алматы, пр. альФараби, 71. КазНУ. Отпечатано в типографии издательства «Қазақ университетi».
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