DMOR_HW2_fall10_solutions

# DMOR_HW2_fall10_solutions - ESI 6314 Homework#2 Solutions 1...

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ESI 6314 Homework #2 Solutions 1. Problem 3, page 139 Basic Variables Basic Feasible Solution Corner Point x 1 , x 2 x 1 =150 x 2 =100 s 1 =s 2 =0 (150, 100) x 1 , s 1 x 1 =200, s 1 =150, x 2 =s 2 =0 (200, 0) x 1 , s 2 x 1 =350, s 2 =-300, x 2 =s 1 =0 Infeasible x 2 , s 1 x 2 =400, s 1 =-450, x 1 =s 2 =0 Infeasible x 2 , s 2 x 2 =175, s 2 = 225, x 1 =s 1 =0 (0, 175) s 1 , s 2 s 1 =350, s 2 =400, x 1 =x 2 =0 (0, 0) 2. Problem 5, page 139 + 332 122 26 10 144 0 14 0 will do the trick, although other correct representations are possible. (the general procedure was discussed in class on Oct 11) 3. Problem 7, page 213 z x 1 x 2 s 1 s 2 s 3 RHS Basic Var. ______________________________________ 1 -4 -1 0 0 0 0 z=0 ______________________________________ 0 2 3 1 0 0 4 s 1 =4 ______________________________________ 0 1 1 0 1 0 1 s 2 =1 ______________________________________ 0 4 1 0 0 1 2 s 3 =2 ______________________________________ z x 1 x 2 s 1 s 2 s 3 RHS Basic Var. ______________________________________ 1 0 0 0 0 1 2 z=2 ______________________________________ 0 0 5/2 1 0 -1/2 3 s 1 =3 ______________________________________ 0 0 3/4 0 1 -1/4 1/2 s 2 =1/2 ______________________________________ 0 1 1/4 0 0 1/4 1/2 x 1 =1/2 ______________________________________

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This tableau yields the optimal solution z = 2, x 1 = 1/2, x 2 = 0. Pivoting in the non-basic variable x 2 yields the alternative optimal solution z = 2, x 1 = 1/3, x 2 = 2/3. Averaging these two optimal solutions yields a third optimal solution: z = 2, x 1 = 5/12, x 2 = 1/3. Obviously, this LP has an infinite number of optimal solutions, so other correct answers for the 3
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## This note was uploaded on 11/01/2010 for the course ESI 6314 taught by Professor Vladimirlboginski during the Fall '09 term at University of Florida.

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DMOR_HW2_fall10_solutions - ESI 6314 Homework#2 Solutions 1...

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