introduction-lp-duality1

# 917 note that this dictionary is not feasible however

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: feasible solution if the optimal value of the auxiliary problem is null. Thus, the idea is to ﬁrst solve the auxiliary problem. Let see the details on an example. Maximise x1 − x2 + x3 Subject to : 2x1 − x2 + 2x3 ≤ 4 2x1 − 3x2 + x3 ≤ −5 −x1 + x2 − 2x3 ≤ −1 x1 , x2 , x3 ≥ 0 Maximise −x0 Subject to: 2x1 − x2 + 2x3 2x1 − 3x2 + x3 −x1 + x2 − 2x3 x1 , x2 , x3 , x0 − x0 ≤ 4 − x0 ≤ −5 − x0 ≤ −1 ≥0 We introduce the slack variables. We obtain the dictionary: x4 x5 x6 w = 4 − 2x1 + x2 − 2x3 = −5 − 2x1 + 3x2 − x3 = −1 + x1 − x2 + 2x3 = + + + − x0 x0 x0 x0 . (9.17) Note that this dictionary is not feasible. However it can be transformed into a feasible one by operating a simple pivot , x0 entering the basis as x5 exits it: x0 x4 x6 w = 5 + 2x1 = 9 = 4 +3 x1 = −5 − 2x1 − − − + 3x2 2x2 4x2 3x2 + x3 − x3 + 3x3 − x3 + + + − x5 x5 x5 x5 . 9.2. THE SIMPLEX METHOD 137 More generally, the auxiliary problem can be written as Maximise −x0 n Subject to: ∑ j=1 ai j x j − x0 ≤ bi (i = 1, 2, · · · , m) x j ≥ 0 ( j = 0, 1, 2, · · · , n) and the associated dictionary is xn+i = bi − ∑n=1 ai j x j + x0 (i = 1, 2, · · · , m) j w= − x0 This dictionary can be made feasible by pivoting x0 with the variable the ”most unfeasible”, that is the exiting variable xn+k is the one with bk ≤ bi for all i. After the pivot, the variable x0 has value −bk and each xn+i has value bi − bk . All these values are non negative. We are now able to solve the auxiliary problem using the simplex method. Let us go back to our example. After the ﬁrst iteration with x2 entering and x6 exiting, we get: x2 x0 x4 w = 1 = 2 = 7 = −2 + 0.75x1 − 0.25x1 − 1.5x1 + 0.25x1 + 0.75x3 − 1.25x3 − 2.5x3 + 1.25x3 + 0.25x5 + 0.25x5 + 0.5x5 − 0.25x5 − + + − 0.25x6 0.75x6 0.5x6 0.75x6 . After the second iteration with x3 entering and x0 exiting: x3 x2 x4 w = 1.6 − 0.2x1 + 0.2x5 + 0.6x6 = 2.2 + 0.6x1 + 0.4x5 + 0.2x6 = 3− x1 − x6 = − − + − 0.8x0 0.6x0 2x0 x0 . (9.18) The last dictionary (9.18) is optimal. As the optimal value of the auxiliary problem is null, this dictionary provides a feasible solution of the original problem: x1 = 0, x2 = 2.2, x3 = 1.6. Moreover, (9.18) can be easily transformed into a feasible dictionary of the original problem. To obtain the ﬁrst three lines of the desired dictionary, it is enough to copy the ﬁrst three lines while removing the terms with x0 : x3 = 1.6 − 0.2x1 + 0.2x5 + 0.6x6 x2 = 2.2 + 0.6x1 + 0.4x5 + 0.2x6 x4 = 3− x1 − x6 (9.19) To obtain the last line, we express the original objective function z = x1 − x2 + x3 (9.20) in function of the variables outside the basis x1 , x5 , x6 . To do so, we replace the variables of (9.20) by (9.19) and we get: z = x1 − (2.2 + 0.6x1 + 0.4x5 + 0.2x6 ) + (1.6 − 0.2x1 + 0.2x5 + 0.6x6 ) z = −0.6 + 0.2x1 − 0.2x5 + 0.4x6 (9.21) 138 CHAPTER 9. LINEAR PROGRAMMING The desired dictionary then is: x3 x2 x4 z = 1.6 = 2.2 = 3...
View Full Document

Ask a homework question - tutors are online