1.1 Formulation of the problem

Operations Research

1 Linear programming

1.1 Formulation of the problem

By the general linear programming problem we mean the problem of the form:

J (u) = c₁u¹+ ::: + c_nuⁿ! min : (1)

u^k 0; k 2 I (2)

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:

a_1;1u¹+ ::: + a_1;nuⁿ b¹ :::

a_m;1u¹+ ::: + a_m;nuⁿ b^m a_m+1;1u¹+ ::: + a_m+1;nuⁿ= b^m+1

:::

a_s;1u¹+ ::: + a_s;nuⁿ = b^s

; (3)

where u = (u¹; :::; uⁿ)2 Rⁿ and cj, ai;j, bⁱ, i = 1; :::; s; j = 1; :::; n; are …xed real numbers, I f1; :::; ng is a …xed set of indexes; we do not exclude the cases: I = ;, I = f1; :::; ng, m = s, m = 0. Function J is called an objective function and inequalities and equalities in (2) - inequality and equality constraints; equalities (2) are called the nonnegativity constraints.

By hx; yi we denote the scalar product of vectors x = (x¹; :::; xⁿ), y = (y¹; :::; yⁿ), i.e.

hx; yi =Pn

i=1x_iy_i. Writing

x y

where x = (x¹; :::; xⁿ), y = (y¹; :::; yⁿ)we mean that xⁱ yⁱ; i = 1; :::; n.

The above problem can be written in the following matrix-vector form:

where

A = 2 66 64

a_1;1 ::: a_1;n ... ... a_m;1 ::: a_m;n

3 77 75; A =

2 66 64

a_m+1;1 ::: a_m+1;n ... ... a_s;1 ::: a_s;n

3 77 75;

b = 2 66 64

b¹ ... b^m

3 77 75; b=

2 66 64

b^m+1 ... b^s

3 77 75:

Each point u 2 U is called the feasible point for problem (4). A point u 2 U is called the solution of problem (4) provided that

J (u ) J (u) for any u 2 U.

By the canonical linear programming problem we mean a problem of the form 8<

:

J (u) =hc; ui ! min :

u2 U = fu = (u¹; :::; uⁿ)2 Rⁿ; u 0 , Au = bg

; (5)

where A 2 R^{m n}, b 2 R^m.

By the basic linear programming problem we mean a problem of the form 8<

:

J (u) =hc; ui ! min :

u2 U = fu = (u¹; :::; uⁿ)2 Rⁿ; u 0 , Au bg

; (6)

where A 2 R^{m n}, b 2 R^m.

1.2 Equivalence of problems

Now, we shall show that solving of the basic problem can be replaced by the solving of canonical one.

Indeed, let the problem (6) be given. Consider in the space R^n+m a problem of the form 8<

:

hd; zi ! min :

z 2 Z = fz = (u; v) 2 R^n+m; z 0, Cz = bg

; (7)

where d = (c; 0) 2 R^n+m,

C = [Aj I^{m m}] = 2 66 64

a_1;1 ::: a_1;n ... ... a_m;1 ::: a_m;n

1 ::: 0 ... ... 0 ::: 1

3 77 75

(Im m denotes the identity m m).

It is easy to see that if u 2 U is a solution of problem (6), then z = (u ; v ) where v = b Au ;

is a solution of problem (7), i.e. z 2 Z and

hd; z i hd; zi for any z 2 Z.

On the other hand, if z = (u ; v ) 2 Z is a solution of problem (7), then u is a solution of problem (6), i.e. u 2 U and

hc; u i hc; ui for any u 2 U.

Similarly, to solve the general problem (4) it is su¢ cient to solve a canonical one. Indeed, let us consider in the space R^p (p = m + I + J + J where J = f1; :::; ng I) a problem of the form 8

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he; zi =X

i2I

c_iuⁱ+X

i2J

c_iwⁱ+X

i2J

c_iwⁱ ! min : z 2 Z = fz 2 R^p; z 0;

2

4 [I_{m m} j Aⁱ; i2 I j Aⁱ; i2 J j Aⁱ; i2 J]

0j Aⁱ; i2 I j Aⁱ; i2 J j Aⁱ; i2 J 3 5 z =

2 4 b

b 3 5g

; (8)

where e = (0; ci; i 2 I; cⁱ; i 2 J; cⁱ; i 2 J) 2 R^p, z = (v; uⁱ; i 2 I; wⁱ; i 2 J; wⁱ; i 2 J) 2 R^p, Ai - i-th column of matrix A, Ai - i-th column of matrix A.

where

v = b Au ; wⁱ = maxf0; uⁱg; i 2 J;

wⁱ = maxf0; uⁱg; i 2 J;

is a solution of problem (8) (let us observe that uⁱ = wⁱ wⁱ for i 2 J).

On the other hand, if

z = (v ; uⁱ; i2 I; wⁱ; i2 J; wⁱ; i2 J);

is a solution of problem (8), then

u = (uⁱ; i2 I; wⁱ wⁱ; i2 J)

is a solution of problem (4) (with the accuracy of component order).

1.3 Geometrical method of solving of 2-dimensional linear pro- gramming problems

Let us consider problem (6) with n = 2, i.e.

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J (u) = c₁u¹+ c₂u² ! min :

u2 U = fu = (u¹; u²)2 R²; u¹ 0; u² 0;

a_i;1u¹+ a_i;2u² bⁱ; i = 1; :::; mg

: (9)

Let us denote

U_0;1 =f(u¹; u²)2 R²; u¹ 0g;

U_0;2 =f(u¹; u²)2 R²; u² 0g;

U_i =f(u¹; u²)2 R²; a_i;1u¹+ a_i;2u² bⁱg; i = 1; :::; m:

Of course

U = U_0;1\ U^0;2\ U¹ \ ::: \ U^m: We have three cases:

1⁰ the set U is empty

2⁰ the set U is non-empty convex and bounded polygon (intersection of a …nite number of closed two-dimensional half spaces)

3⁰ the set U is non-empty convex and unbounded polygon

Let us …x a number 2 R. Equality

c₁u¹+ c₂u² =

describes a level set for function J corresponding to , i.e. the set f(u¹; u²)2 R²; J (u) = g:

In case 2⁰ there exists a point which is the „…rst” common point (…rst contact point) of the moving line and polygon U (perhaps, not unique). The appropriate value is equal to

minu2UJ (u) =: J

In the other words, the optimal value J of problem (9) is the smallest value for which the corresponding line has a point in common with the constraint set.

In case 3⁰ this …rst common point exists (perhaps, not unique) or not. If such a point does not exist, the problem has not any solution; in such a case

inf

u2UJ (u) = 1:

So, we see that problem (9) may not have any solution, it may have exactly one solution or it may have in…nite many solutions. Moreover, if the set of solutions is non-empty, it contains at least one point being the extreme point of U .

Similar analysis can be given when n = 3 (with lines replaced by planes and polygons replaced by polyhedrons - by polyhedron we mean the intersection of a …nite number of closed half spaces of a dimension greater than 2).

1.4 Extreme points

A point v 2 V Rⁿ is called the extreme point of a convex closed set V , if there are no two distinct points v1 2 V , v² 2 V such that

v = v₁+ (1 )v₂; (10)

for some 2 (0; 1). In other words, an extreme point is a point that does not lie strictly within the line segment connecting two other points of the set. The notion of the extreme point plays a fundamental role in the theory of linear programming.

In the next, we shall show that if a canonical problem (with any n 2 N) possesses a solution then the set of all solutions contains an extreme point of the constraint set

U = fu 2 Rⁿ; u 0; Au = bg: (11)

We have the following characterization of extreme points of the set (11).

Theorem 1 Let U be a set of the form (11) with A 2 R^{m n} f0g. Denote r = rankA.

A point v 2 Rⁿ is the extreme point of the set U if and only if there exist indexes j1 <...<

jr 2 f1; :::; ng such that 8>

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v^j 0; j 2 fj¹; :::; j_rg v^j = 0; j =2 fj¹; :::; j_rg A_j1v^j1 + ::: + A_jrv^jr = b

columns Aj1; :::; A_jr are linearly independent in R^m

(12)

The set of columns Aj1,...,Ajr in (12) is called the basis of the extreme point v, coordinates v^j1,...,v^jr - basic coordinates of v. An extreme point, with all positive basic coordinates is called nonsingular. An extreme point possessing at least one basic coordinate which is equal to zero is called singular. Variables u^j1,...,u^jr are called basic variables and remaining variables - nonbasic variables (under …xed basis Aj1,...,Ajr).

The above theorem implies that basis of a nonsingular extreme point is uniquely deter- mined. A singular point may have many bases.

1.5 Simplex method

Simplex method is a …nite iterative method of solving canonical problems (by a "…nite"

iterative method we mean the method which allows us to solve a problem within …nite steps). It is based on checking some of the extreme points in such a way that the values of objective function at the subsequent points do not increase.

Let us consider the following problem 8<

:

J (u) =hc; ui ! min :

u2 U = fu 2 Rⁿ; u 0, Au = bg

(13)

with 0 6= A 2 R^{m n}. We assume that U 6= ? (we will return to this issue in the …nal part of the lecture).

Of course, r = rankA minfm; ng. Equality Au = b

can be written in the form of the following system of equalities Xn

j=1

ai;ju^j = bi; i = 1; :::; m:

Without loss of the generality, we may assume that r = m. It is clear that r n. If r = n, then the above system possesses exactly one solution u and u 0 (the last inequality follows from the assumption U 6= ?). In consequence, u is a solution of problem (13).

So, let us assume that r = m and r < n. Equality Au = b can be written in the following form

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a_1;1u¹+ ::: + a_1;nuⁿ = b¹ :::

a_r;1u¹+ ::: + a_r;nuⁿ = b^r

(14)

where r = rankA < n.

Now, we shall describe the simplex method. Let u 2 U and assume that the columns A₁,...,Ar are the basis of v (we will return to the question of determining of the "starting"

extreme point of U and its basis in the next). Denote

u = 2 66 64

u¹ ... u^r

3 77 75; v =

2 66 64

v¹ ... v^r

3 77 75; c =

2 66 64

c₁ ... c₁

3 77 75;

B = 2 66 64

a_1;1 ::: a_1;r ... ... a_r;1 ::: a_r;r

3 77

75= [A₁ j ::: j A^r] :

System (14) can be written in the form

Bu + A_r+1u^r+1+ ::: + A_nuⁿ = b: (15)

Linear independence of columns A1,...,Ar implies that matrix B ¹ exists. Since non basic coordinates are equal to zero, from (15) we obtain

Bv = b:

So,

v = B ¹b:

Multiplying equality (15) by B ¹, we obtain

u + Xn k=r+1

B ¹A_ku^k = B ¹b = v: (16)

Let us denote

s;k = (B ¹A_k)^s for k = r + 1; :::; n; s = 1; :::; r

where (B ¹A_k)^s is s-th coordinate of vector B ¹A_k. Now, equality (16) can be written in

the form 8

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u¹+ _1;r+1u^r+1+ ::: + _1;nuⁿ= v¹ u²+ _2;r+1u^r+1+ ::: + _2;nuⁿ= v²

:::

u^r+ _r;r+1u^r+1+ ::: + _r;nuⁿ = v^r

: (17)

Let us also denote

s;k = (B ¹A_k)^s

for k = 1; :::r; s = 1; :::; r (of course, _s:k = _s;k for k = 1; :::; r; s = 1; :::; r, where s;k is the Kroneker’s symbol).

We have shown that if we …x an extreme point and its certain basis, then constraints (14) can be written in the equivalent form (15), (16) or (17).

The value of the objective function J at a point u satisfying constraints (14), can be

written as follows

J (u) = hc; ui = Xn

i=1

c_iuⁱ =hc; ui + Xn i=r+1

c_iuⁱ

=

* c; v

Xn i=r+1

B ¹A_iuⁱ +

+ Xn i=r+1

c_iuⁱ

= hc; vi

Xn i=r+1

( c; B ¹A_i c_i)uⁱ:

Since

hc; vi = hc; vi = J(v);

therefore

J (u) = J (v)

Xn i=r+1

iuⁱ (18)

where

i = c; B ¹A_i c_i; i = r + 1; :::; n:

Let us also de…ne

i = c; B ¹A_i c_i (19)

for i = 1; :::; r. Of course,

i = c; B ¹A_i c_i =hc; eⁱi c_i = c_i c_i = 0

for i = 1; :::; r (ei denotes the i-th column of r r - dimensional identity matrix).

Let us summarize. We have shown that problem (13) can be written in the following

form 8

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J (u) = J (v) Pn i=r+1

iuⁱ ! min :

U =fu = (u¹; :::; uⁿ)2 Rⁿ; u 0 , u satis…es (17)g

(20)

We shall write the coe¢ cients _s;k, i in the form of the so-called simplex table corre- sponding to the extreme point v

Simplex table I (for point v)

u¹ ::: uⁱ ::: u^s ::: u^r u^r+1 ::: u^k ::: u^j ::: uⁿ

u¹ 1 ::: 0 ::: 0 ::: 0 _1;r+1 ::: _1;k ::: _1;j ::: _1;n v¹

... ... ... ... ... ... ... ... ... ...

uⁱ 0 ::: 1 ::: 0 ::: 0 _i;r+1 ::: _i;k ::: _i;j ::: _i;n vⁱ

... ... ... ... ... ... ... ... ... ...

u^s 0 ::: 0 ::: 1 ::: 0 _s;r+1 ::: _s;k ::: _s;j ::: _s;n v^s

... ... ... ... ... ... ... ... ...

u^r 0 ::: 0 ::: 0 ::: 1 _r;r+1 ::: _r;k ::: _r;j ::: _r;n v^r

0 ::: 0 ::: 0 ::: 0 _r+1 ::: _k ::: _j ::: _n J (v)

We can feature three cases:

1⁰ inequalities

i = c; B ¹A_i c_i 0 (21)

for i = r + 1; :::; n; are satis…ed, i.e. in the last row of simplex table all coe¢ cients i are not positive. In this case the point v is a solution of the problem. Indeed, for any u 2 U, we have

J (u) = J (v)

Xn i=r+1

iuⁱ J (v) (we used the fact that i 0, uⁱ 0).

2⁰ there exists index k 2 fr + 1; :::; ng such that

This means that in the k-th column of the simplex table the last coe¢ cient k is positive and remaining - not positive. In this case inf

u2UJ (u) = 1 (we omit the proof of this fact).

So, our problem has no solution:

3⁰ the cases 1⁰ and 2⁰ do not take place; in consequence, there exist indexes k 2 fr + 1; :::; ng and i 2 f1; :::; rg such that

k > 0; _i;k > 0: (23)

This means that in k-th column of the simplex table the last element ( k) is positive and at least one coe¢ cient _i;k is positive.

Let us assume that the case 3⁰ holds and de…ne the set I_k =fi 2 f1; :::; rg; i;k > 0g:

Let index s 2 I^k be such that

v^s

s;k

= min

i2Ik

vⁱ

i;k

(24) Coe¢ cient _s;k where k; s are given by (23) and (24), is called the solving element for simplex table I.

One can show that the system of columns

A1; :::; As 1; As+1; :::; Ar; Ak (25) is the basis of some extreme point w and

J (w) J (v):

Remark 1. From theorem characterizing extreme points it follows that the basis (25) determines the extreme point w in a unique way. So, the basic coordinates of the point w can be found with the aid of this theorem.

Now, let us consider the general case, i.e. let us assume that an extreme point v is given and

v^j1; :::; v^jr;

where 1 j1 < ::: < jr n, are its basic coordinates. The equality constraints and the objective function can be written as follows (below, Iv =fj¹; :::; j_rg)

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u^j1 = v^j1 P

k =2I^v

j1;ku^k :::

u^jr = v^jr P

k =2I^v

jr;ku^k

(26)

J (u) = J (v) X

k =2I^v

ku^k; (27)

where

ji;k = (B ¹A_k)ⁱ ; i = 1; :::; r; k = 1; :::; n (28) (in particular, _ji;k = _ji;k for i = 1; :::; r; k 2 I^v),

B = [A_j1 j ::: j A^jr] ;

v^ji = (B ¹b)ⁱ; i = 1; :::; r;

v^k = 0; k =2 I^v;

k = c; B ¹A_k c_k = Xr

i=1

c_ji(B ¹A_k)ⁱ c_k; k = 1; :::; n; (29)

where c = 2 66 64

c_j1 ... c_jr

3 77

75 (in particular, k = 0 for k 2 I^v).

In this case the simplex table for point v is the following:

Simplex table II (for point v)

u¹ ::: u^j1 ::: u^ji ::: u^k ::: u^js ::: u^j ::: u^jr ::: uⁿ

u^j1 _j1;1 ::: 1 ::: 0 ::: _j1;k ::: 0 ::: _j1;j ::: 0 ::: _j1;n v^j1

... ... ... ... ... ... ... ... ... ...

u^ji _j

i;1 ::: 0 ::: 1 ::: _j

i;k ::: 0 ::: _j

i;j ::: 0 ::: _j

i;n v^ji

... ... ... ... ... ... ... ... ... ...

u^js _j

s;1 ::: 0 ::: 0 ::: _j

s;k ::: 1 ::: _j

s;j ::: 0 ::: _j

s;n v^js

... ... ... ... ... ... ...

u^jr _jr;1 ::: 0 ::: 0 ::: _jr;k ::: 0 ::: _jr;j ::: 1 ::: _jr;n v^jr

1 ::: 0 ::: 0 ::: _k ::: 0 ::: _j ::: 0 ::: _n J (v)

Similarly as previously, we consider three cases:

1⁰ the condition

k 0; k =2 I^v (21’)

is satis…ed

2⁰ there exists k =2 I^v such that

k > 0; _j

i;k 0; i = 1; :::; r (22’)

3⁰ cases 1⁰ and 2⁰ do not take place; in consequence, there exist k =2 I^v and ji 2 I^v such that

k > 0; _ji;k > 0. (23’)

In the …rst case, v is the solution of (13), in the second one - inf

u2Uhc; ui = 1, i.e. problem (13) has no solution.

In the third case we choose _js;k basing ourselves on conditions (23’) and v^js

js;k

= min

ji2Iv;k

v^ji

ji;k

, (24’)

where Iv;k = fjⁱ 2 I^v; _j

i;k > 0g (these conditions are counterparts of conditions (23), (24)). Next, we passage to the new extreme point w. From the conditions (23’) and (24’) it follows that the columns

A_j1; :::; A_{js 1}; A_js+1; :::; A_jr; A_k,

are the basis of w (with an accuracy to their order) and inequality J (w) J (v):

will be satis…ed. The basic coordinates of the point w can be determined with aid of theorem characterizing extreme points.

Remark 2. One can show that 8>

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w^j1 = v^j1 _j1;k ^vjs .. js;k

.

w^ji = v^ji _ji;k ^vjs .. js;k

.

w^{js 1} = v^{js 1} _{js 1;k} ^vjs

js;k

w^js = v^js _js;k ^vjs

js;k = 0 w^js+1 = v^js+1 _js+1;k ^vjs

.. js;k

.

w^jr = v^jr _jr;k ^vjs

js;k

w^k = ^vjs

js;k

w^l = 0; l =2 I^v; l6= k;

Simplex table for point w takes the form

Matrix table III (for point w)

u¹ ::: u^j1 ::: u^ji ::: u^k ::: u^js ::: u^j ::: u^jr ::: uⁿ u^{j1 0}_j

1;1 ::: 1 ::: 0 ::: 0 ::: ⁰_j

1;js ::: ⁰_j

1;j ::: 0 ::: ⁰_j

1;n w^j1

... ... ... ... ... ... ... ... ... ...

u^{ji 0}_j

i;1 ::: 0 ::: 1 ::: 0 ::: ⁰_j

i;js ::: ⁰_j

i;j ::: 0 ::: ⁰_j

i;n w^ji

... ... ... ... ... ... ... ... ... ...

u^{k 0}_k;1 ::: 0 ::: 0 ::: 1 ::: ⁰_k;j

s ::: ⁰_k;j ::: 0 ::: ⁰_k;n w^k

... ... ... ... ... ... ... ... ... ...

u^{js 1 0}_{js 1;1} ::: 0 ::: 0 ::: 0 ::: ⁰_{js 1;js} ::: ⁰_{js 1;j} ::: 0 ::: ⁰_{js 1;n} w^{js 1} u^{js+1 0}_js+1;1 ::: 0 ::: 0 ::: 0 ::: ⁰_js+1;js ::: ⁰_js+1;j 0 ⁰_js+1;n w^js+1

... ... ... ... ... ... ...

u^{jr 0}_jr;1 ::: 0 ::: 0 ::: 0 ::: ⁰_jr;js ::: ⁰_jr;j ::: 1 ::: ⁰_jr;n w^jr

0

1 ::: 0 ::: 0 ::: 0 ::: ⁰_js ::: ⁰_j ::: 0 ::: ⁰_n J (w) where coe¢ cients ⁰_i;j, ⁰_j are given by formulas which are analogous to (28), (29) with

matrix B of the form

[A_j1 j ::: j A^k j ::: j A^js 1 j A^js+1 j ::: j A^jr]

(here, we assume that the rows and columns in the simplex table as well as columns in matrix B are ordered with respect to increasing indexes).

Remark 2. One can show that 8<

:

0

ji;j = _ji;j ^ji;k

js;k js;j ; i = 1; :::; r; i6= s; j = 1; :::n;

0

k;j = ^js;j

js;k; j = 1; :::n;

and

0

j = _{j k} ^js;j

js;k

for j = 1; :::; n:

So, we have described one step of the simplex method, i.e. the passage from an extreme point (v) of U to another one (w) (in the case 3⁰) in such a way that

J (w) J (v).

1.6 An anti-cycle rule

During the realization of the simplex method it can happen that 0 = min

ji2Iv;k

v^ji

ji;k

= v^js

js;k

.

From the formulas given in Remark 1 it follows that in such a case w = v

and

J (w) = J (v).

Consequently, the passage from point v to point w means only the transition form the base

A_j1; :::; A_jr of point v to the base

A_j1; :::; A_k; :::; A_{js 1}; A_js+1; :::; A_jr

of the same point. One can give an example showing that simplex method can be looped, i.e. in the subsequent iterations the extreme point will not change and only the basis of this point will change in a cycling way.

However, one can specify (in di¤erent ways) a rule of the choice of solving element, which eliminates the possibility of any loop. Such rules are called an anti-cycle rule. Now, we shall describe one of such rules.

Let us assume that we have an extreme point v. To simplify notations let us assume that the columns A1,...,Ar are the basis of it. We add to the simplex table corresponding to v the identity matrix Ir r=[di;j]. In each step of the simplex method the appended elements are transformed according to the formulas given in Remark (we do not create coe¢ cients

„ ”). We choose the solving element _s;k in the following way.

Let > 0 and I =fi 2 f1; :::; rg; > 0g 6= ?. We de…ne the set

If Ik;1 has more that one element, then we form the set I_k;2=fs 2 I^k;1; min

i2Ik;1

d_i;1

i;k

= d_s;1

s;k

g.

In general, if we have formed the set Ik;m(m 2) and it has more than one element, then we form the set

I_k;m+1 =fs 2 I^k;m; min

i2Ik;m

d_i;m

i;k

= d_s;m

s;k

g.

One can show that there exists l 2 f1; :::; r + 1g such that the set I^k;l has exactly one element s and all sets Ik;i, i = 1; :::; l 1have more than one element. The unique element of the set Ik;l is an index describing the solving element.

One proves (cf. [V]) that if we apply the above rule in each step of the simplex method, then the loop does not appear and we obtain a solution of the problem within …nite steps or ascertain that the solution of the problem does not exist.(¹).

1.7 Choice of a "starting" extreme point

Let the following problem be given

J (u) =hc; ui ! min : (30)

u2 U = fu = (u¹; :::; uⁿ)2 Rⁿ; u 0, Au = bg, (31) where ; 6= A 2 R^{m n}. Now, we shall show how to check that U 6= ? and how to …nd its extreme point (using the simplex method).

Without loss of the generality we may assume that bⁱ 0 for i = 1; :::; m (multiplying the appropriate equalities by 1, if it is necessary).

Let us consider the following auxiliary problem

J₁(z) = uⁿ⁺¹+ ::: + u^n+m! min : (32)

1So, it can not also happen that in a unordered way we shall work with bases of one extreme point and that in an "in…nite" way (cyclic or not) we shall work with extreme points (the passage to the next extreme point w 6= v means that vjs 6= 0 and, consequently, J(w) < J(v)).

z 2 Z = fz = 2 4 u

w 3

5 2 R^n+m; z 0; Cz = bg; (33)

where C = [A j I^{m m}], u = (u¹; :::; uⁿ), w = (uⁿ⁺¹; :::; u^n+m). System Cz = b

can be written as follows 8>

>>

<

>>

>:

a_1;1u¹+ ::: + a_1;nuⁿ+ uⁿ⁺¹ = b¹ :::

a_m;1u¹+ ::: + a_m;nuⁿ+ u^n+m = b^m :

It is easy to see that z0 := (0; b) 2 Z, i.e. the set Z is non-empty. Moreover, z⁰ is an extreme point of Z with basis consisting of m column e1,...,em 2 R^m of the matrix C (of course, rankC = m). So, we can apply the simplex method to problem (32)-(33) with z0

as the starting point.

Since

J₁(z) 0; z2 Z;

therefore the case

z2ZinfJ₁(z) = 1

will not happen. Thus, applying the simplex method we shall obtain, within …nite steps, an extreme point z = (v ; w ) 2 Z being the solution of problem (32)-(33).

The following two cases are possible.

1⁰ J₁(z ) > 0. In this case the set U (given by (31)) is empty. Indeed, if we assume that there exists a point u 2 U, then z = (u; 0) 2 Z and

J₁(z) = 0

which contradicts the inequality J1(z ) > 0 and optimality of the point z .

Consequently, v is an extreme point of the set U . Indeed, since z 0, v 0. Moreover, equality

Cz = b implies that

Av = b.

So,

v 2 U.

Now, let us assume that

v = u + (1 )eu;

where 2 (0; 1), u;eu 2 U. Of course, the points z = (u; 0), ez = (eu; 0) belong to Z and z = z + (1 )ez.

Since z is an extreme point of the set Z, the above equality means that z =ez.

Consequently,

u =eu.

i.e. v is an extreme point of U .

So, we have shown that applying the simplex method to auxiliary problem (32)-(33), we can check if the constraint set U given by (31) is nonempty and, if so, determine an extreme point of U .

From the above considerations the following theorem follows.

Theorem 2 If the set U given by (31) is non-empty, the it possesses at least one extreme point.

Now, using the description of the simplex method, we shall prove the following two fundamental facts of the linear programming theory.

Theorem 3 Canonical problem (30)-(31) has a solution, i.e. there exists a point u 2 U such that

hc; u i = inf

u2Uhc; ui ; if and only if

1) the set U is non-empty

2) objective function J (u) = hc; ui is bounded below on the set U.

Proof. Necessity. Necessity of conditions 1) and 2) is obvious.

Su¢ ciency. Assumption 1) and Theorem 2 imply that there exists an extreme point of the set U . So, one can apply to the problem (30)-(31) the simplex method with this point as starting one. From assumption 2) it follows that case 2⁰ (from the description of the simplex method) will not happen in any step of the method. This means that we shall obtain a solution u of the problem (30)-(31) within …nite steps . The proof is completed.

Theorem 4 If problem (30)-(31) has a solution, then it has at least one solution being an extreme point of the set U .

Proof. From Theorem 3 it follows that U 6= ? and the objective function J is bounded below on U . Theorem 2 implies that the set U has an extreme point. Applying the simplex method with this point as starting one, we shall obtain a solution of the problem (30)- (31) within …nite steps (from Theorem 2 it follows that inf

u2UJ (u) > 1 and, consequently, case 2⁰ from the description of the simplex method will not happen in any step of the method). This solution is extreme point of U as the point obtained with the aid of the simplex method. The proof is completed.