introduction-lp-duality1

Introduction-lp-duality1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: we obtain m m ∑ biyi − ∑ i=1 n n ∑ ai j x j yi + ∑ i=1 j=1 m n m n j=1 i=1 j=1 ∑ ai j yix j − ∑ c j x j = ∑ biyi − ∑ c j x j = 0. j=1 i=1 144 CHAPTER 9. LINEAR PROGRAMMING Therefore, Inequalities 9.24 and 9.25 must be equalities. As the variables are positive, we further get that ￿ ￿ for all i, and for all j, n bi − ∑ ai j x j yi = 0 ￿ j=1 m ∑ ai j yi − c j i=1 ￿ x j = 0. A product is equal to zero if one of its two members is null and we obtain the desired result. Theorem 9.12. A feasible solution (x1 , . . . , xn ) of Problem 9.5 is optimal if and only if there is a feasible solution (y1 , . . . , ym ) of Problem 9.6 such that: xj > 0 ∑m ai j yi = c j i f i= yi = 0 i f ∑m 1 ai j x j < bi j= (9.26) Note that, if Problem 9.5 admits a non-degenerated solution (x1 , . . . , xn ), i.e., xi > 0 for any i ≤ n, then the system of equations in Theorem 9.12 admits a unique solution. Optimality certificates - Examples. Let see how to apply this theorem on two examples. Let us first examine the statement that ∗ ∗ ∗ ∗ ∗ ∗ x1 = 2, x2 = 4, x3 = 0, x4 = 0, x5 = 7, x6 = 0 is an optimal solution of the problem Maximize Subject to: 18x1 2x1 −3x1 8x1 4x1 5x1 − 7x2 − 6x2 − x2 − 3x2 + 2x2 + 12x3 + 2x3 + 4x3 + 5x3 + 8x3 − 3x3 + + − − + + 5x4 7x4 3x4 2x4 7x4 6x4 + 8x6 + 3x5 + 8x6 + x5 + 2x6 + 2x6 − x5 + 3x6 − 2x5 − x6 x1 , x2 , · · · , x6 ≤ ≤ ≤ ≤ ≤ ≥ 1 −2 4 1 5 0 In this case, (9.26) says: 2y∗ − 3y∗ + 8y∗ + 4y∗ + 5y∗ 1 2 3 4 5 −6y∗ − y∗ − 3y∗ + 2y∗ 1 2 3 5 3y∗ + y∗ − y∗ − 2y∗ 1 2 4 5 y∗ 2 y∗ 5 = 18 = −7 = 0 = 0 = 0 As the solution ( 1 , 0, 5 , 1, 0) is a feasible solution of the dual problem (Problem 9.6), the pro3 3 posed solution is optimal. 9.3. DUALITY OF LINEAR PROGRAMMING Secondly, is 145 ∗ ∗ ∗ ∗ ∗ x1 = 0, x2 = 2, x3 = 0, x4 = 7, x5 = 0 an optimal solution of the following problem? Maximize 8x1 Subject to: 2x1 x1 5x1 − − + + 9x2 3x2 7x2 4x2 + 12x3 + 4x3 + 3x3 − 6x3 Here (9.26) translates into: + 4x4 + 11x5 + x4 + 3x5 − 2x4 + x5 + 2x4 + 3x5 x1 , x2 , · · · , x5 ≤ ≤ ≤ ≥ 1 1 22 0 −3y∗ + 7y∗ + 4y∗ = −9 1 2 3 y∗ − 2y∗ + 2y∗ = 4 1 2 3 ∗ y2 = 0 As the unique solution of the system (3.4, 0, 0.3) is not a feasible solution of Problem 9.6, the proposed solution is not optimal. 9.3.5 Interpretation of dual variables As said in the introduction of this section, one of the major interests of the dual programme is that, in some problems, the variables of the dual problem have an interpretation. A classical example is the economical interpretation of the dual variables of the following problem. Consider the problem that consits in maximizing the benefit of a company building some products. Each variable x j of the primal problem measures the amount of product j that is built, and bi the amount of resource i (needed to build the products) that is available. Note that, for any i ≤ n, j ≤ m, ai, j represents the number of units of resource i needed per unit of product j...
View Full Document

This note was uploaded on 11/20/2013 for the course CS 101 taught by Professor Smith during the Fall '13 term at Mitchell Technical Institute.

Ask a homework question - tutors are online