# sol3 - CO 350 Linear Optimization (Spring 2009) [20] 1. (a)...

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CO 350 Linear Optimization (Spring 2009) Solution to Assignment 3 [20] 1. (a) B = { 1 , 4 , 6 } , x B = (2 , 0 , 0 , 5 , 0 , 4) T . (b) z = 13 (c) x 3 enters. t = min { 5 / 2 , 2 } = 2, so x 1 leaves. The new tableau, for basis { 3 , 4 , 6 } is z + 2 x 1 + x 2 + 5 x 5 = 17 2 x 1 - x 2 - 2 x 5 + x 6 = 8 - 2 x 1 + 9 x 2 + x 4 - 5 x 5 = 1 x 1 - 2 x 2 + x 3 + 2 x 5 = 2 (d) The new tableau is optimal, as all ¯ c j 0. (So the objective value cannot increase if 1 , 2 , or 5 enters the basis.) The original LP problem has an optimal solution. [0 , 0 , 2 , 1 , 0 , 8] T is an optimal solution with objective value 17. [10] 2. The LP problem is unbounded: ¯ c 4 > 0 and ¯ a i 4 0 for all i . That is, if we choose x 4 to enter, x 4 can become arbitrarily large. From the tableau we can see that we can get feasible solutions x ( t ), t 0, as follows: z ( t ) = 42 + 4 t x 1 ( t ) = 0 x 2 ( t ) = 7 + t x 3 ( t ) = 3 + 2 t x 4 ( t ) = t x 5 ( t ) = 0 x 6 ( t ) = 8 + 3 t. [10] 3. (a) The tableau corresponding to B = { 1 , 2 , 3 } , which is obtained after elementary row operations, is z - 9 2 x 4 + 5 2 x 5 - 29 x 6 = - 77 2 x 1 + x 4 + x 6 = 9 x 2 - 5 x 4 + 2 x 5 - 24 x 6 = - 31 x 3 - 1 2 x 4 + 1 2 x 5 - 2 x 6 = - 3 2 It is not feasible. The tableau corresponding to B = { 1 , 4 , 5 } is z - 2 3 x 2 - 7 3 x 3 - 25 3 x 6 = - 43 3 x 1 + 1 3 x 2 - 4 3 x 3 + 2 3 x 6 = 2 3 - 1 3 x 2 + 4 3 x 3 + x 4 + 16 3 x 6 = 25 3 - 1 3 x 2 + 10 3 x 3 + x 5 + 4 3 x 6 = 16 3 1

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This tableau is feasible. (b) Consider the feasible tableau ( T 1 ) from (a) corresponding to the basis B 1 = { 1 , 4 , 5 } . We shall use the largest coeﬃcient rule when choosing entering variables. Since
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## This note was uploaded on 04/11/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Fall '07 term at Waterloo.

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sol3 - CO 350 Linear Optimization (Spring 2009) [20] 1. (a)...

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