Линейное программирование. Практикум. Учебное пособие для бакалавриата

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Solution. The constraint x1 ≥1 will be consider as a non-trivial constraint, so

f = 4x1 −5x2 +8x3 −10x4 +x5 +14 → minx1 −2x2 +7x3 −x5 ≤ 37,

−4x1 −7x2 +4x4 −9x5 ≥ −28,2x1 +6x3 −4x4 +x5 = 48,

x1 ≥1,

x1 ~,x2 ≥ 0,x3 ≤ 0,x4 ≥ 0.

Thus, the primal problem has the matrix

Due to the algorithm, the dual problem has the matrix

Finally, we have the dual problem

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g = 37y1 −28y2 +48y3 +y4 +14 → max

y1 −4y2 +2y3 +y4 = 4,

−2y1 −7y2 ≤ −5, 7y1 +6y3 ≥ 8,

4y2 −4y3 ≤ −10,

−y1 −9y2 −y3 =1,

y1 ≤ 0,y2 ≥ 0,y4 ≥ 0.

6.4. The main theorems

Let we have the duel linear programming problems

Theorem 1 (the weak duality). For all feasible solutions X,Y of the pair of the dual problems it holds the inequality:

f (X)≤ g(Y).

Proof.

, AX ≤ B XT AT ≤ BT XT ATY ≤ BTY = g(Y) XT ATY ≤ g(Y).

Similarly,

ATY ≥ C XT ATY ≥ XTC = f(X) . f (X)≤ XT ATY ≤ g(Y).+

Corollary 1 (The sufficient condition for optimality). If there exist feasible solutions X*,Y* of the pair of the dual

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problems satisfying f (X* )= g(Y* ), then X*,Y* are the optimal solutions.

Proof.

, Due to the theorem 1 for any feasible solution X of the primal problem

f (X)≤ g(Y* ) f (X)≤ f (X* ) . fmax = f (X* ).

Similarly, for any feasible solution Y of the dual problem

f (X* )≤ g(Y) g(Y* )≤ g(Y) gmin = g(Y* ).+

Corollary 2. If the one of the dual linear programming

problem is unbounded ( fmax = +∞ or gmin = −∞ ) then the other problem is infeasible.

Theorem 2 (the strong duality). If there exists the optimal solution X* of the primal problem then there exists the optimal solution Y* of the dual problem and

f (X* )= g(Y* ),

i.e. fmax = gmin .

Theorem 3 (the equilibrium theorem). The optimal

solutions X = (x1 ,x2, ,xn )T and Y = (y1 ,y2, ,ym )T of the pair of the dual problems satisfy the equations

Let us write the pair of the dual problems to the canonicals form.

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Due to this correspondence the conditions above can be rewritten in the form

xj ym* +j = 0, j =1, ,n,

xi+n yi = 0, i =1, ,m.

Where

X = (x ,x , ,x ,x + , ,x + )T and

1 2 n n 1 n m

Y = (y ,y , ,y ,y + , ,y + )T

1 2 m m 1 m n

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are the optimal solutions of the pair of the dual problems. Example. Solve the linear programming problem using

the equilibrium theorem.

We can solve it by using the geometrical method.

The boundaries of the feasible region are defined by the following straight lines;

l1 : −10y1 +8y2 =22 , l2 :7y1 +2y2 = 91,

l3 : 3y1 −10y2 = −37 .

The intersection of this lines gives us the corner points:

A(9,14), B(11,7), C(1,4).

Using the gradient n = (1;22) we can easily the optimal solution of the problem.

Y = A(9;14),

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gmax = g(9;14)= 336 .

It follows from the equilibrium theorem.

Due to Y = (9;14) ( y1 ≠ 0,y2 ≠ 0 ) it follows from two last equations of the system above that

−10x1 +7x2 −1 = 0,8x1 +2x2 −22 = 0.

The point A is the intersection of l1 and l2 (A = l1 ∩l2 ) and does not belong to l3 (A l3 ). So, we have

−10y1 +8y2 −22 = 0,7y1 +2y2 −91 = 0,

3y1 −10y2 +37 ≠ 0

It follows from the equilibrium equation ‘

(3y1 −10y2 +37) x3 = 0 .

So, due to

3y1 −10y2 +37 ≠ 0

we obtain

x3 = 0 .

Thus,

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x3 = 0,

−10x1 +7x2 −1 = 0,8x1 +2x2 −22 = 0.

The solution of this gives the optimal vector

X* = (2,3,0).

f (X* )=22 2 +91 3 −37 0 +19 = 336 = g(Y* ).

Example. Solve the linear programming problem

by simplex method and find solutions of the dual problem. by the equilibrium theorem.

Solution. Let us find the matrix of the primal problem

So, the matrix of the dual problem will be

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Thus, the dual problem:

Let us choose a23 = 6 as a pivot (x5 ↔ x3 ). We get

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And get the optimal solution of the dual problem

y1* =5,5y2* = 0,5

gmin = g(Y* ) =57 5,5 +9 0,5 −18 = 300 = fmax.

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Questions for self-control

1)What are symmetric dual problems?

2)What are asymmetric dual problems?

3)How to construct dual problem in general case?

4)How are optimal solutions of dual problems related?

5)Formulate the equilibrium theorem.

Exercises for independent work

1) Construct the dual problem for the following linear programming problelems

f = x1 +2x2 −x3 → min

x1 +2x2 −5x3 ≥ 3,

b)2x1 −8x2 +6x3 ≥5,x1 −7x2 +2x3 ≥2,

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

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