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Bolve the following ILPP using Gomory's cutting plane method. Minimize f(x) = 3x, + 2x, subject to + 3x. 2х, + 9 12 4 x: are integers he optimaltable for the correspondingrelaxed LPPis given below. V VI VI

Bolve the following ILPP using Gomory's cutting plane method. Minimize f(x) = 3x, + 2x, subject to + 3x. 2х, + 9 12 4 x: are integers he optimaltable for the correspondingrelaxed LPPis given below. V VI VI

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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PLZ SOLVE This question by gomory cut by integer lpp

1. Bolve the following ILPP using Gomory's cutting plane method. Minimize f(x) = 3x1 + 2x2 subject to + 3x, 2х, + X1 X2 12 4 X2 are integers The optimaltable forthe correspondingrelaxed LPPis given below. Table 1.1 Basic X2 S1 S2 S3 Solution 7 93 f(x) 5 1. X2 5 3 27 1. 5 2 14 S3 1 The optimal solution for the relaxed LPP is x1 =,x, =and f(x) =. Since the solution obtained is not integer, we use Gomory's cutting plane method. Gomory's cut corresponding to x, – row is:
Transcribed Image Text:1. Bolve the following ILPP using Gomory's cutting plane method. Minimize f(x) = 3x1 + 2x2 subject to + 3x, 2х, + X1 X2 12 4 X2 are integers The optimaltable forthe correspondingrelaxed LPPis given below. Table 1.1 Basic X2 S1 S2 S3 Solution 7 93 f(x) 5 1. X2 5 3 27 1. 5 2 14 S3 1 The optimal solution for the relaxed LPP is x1 =,x, =and f(x) =. Since the solution obtained is not integer, we use Gomory's cutting plane method. Gomory's cut corresponding to x, – row is:
Table 1.7 Basic X1 X2 S2 S3 Solution 3 f(x) 18 4 2 1 -1 X1 1 2 S3 -1 4 5 1 3 2 -2 2 3 2 Since all basic variables are nan-negative, last table is the optimai feasible table for the modified problem. The optimal feasible solution for the modified problem is x, = 6,x2 = 0 and f(x) = 18. Since both variables have integer values the optimal integer solution for the given problem is arived. HIN 2.
Transcribed Image Text:Table 1.7 Basic X1 X2 S2 S3 Solution 3 f(x) 18 4 2 1 -1 X1 1 2 S3 -1 4 5 1 3 2 -2 2 3 2 Since all basic variables are nan-negative, last table is the optimai feasible table for the modified problem. The optimal feasible solution for the modified problem is x, = 6,x2 = 0 and f(x) = 18. Since both variables have integer values the optimal integer solution for the given problem is arived. HIN 2.
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