IOE310_AdditionalProblems_B_Solutions

IOE310_AdditionalProblems_B_Solutions - IOE 310 Additional...

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IOE 310 - Additional Problems B For additional practice only. Not collected. Problem 1 Consider the following LP: max z = 5 x 1 + 3 x 2 + x 3 s.t 2 x 1 + x 2 + x 3 6 x 1 + 2 x 2 + x 3 7 x 1 , x 2 , x 3 0 (a) Graphically solve this problem. To graph this in two-dimensions we need to find the dual: min w = 6y 1 + 7y 2 s.t. 2y 1 + y 2 5 y 1 + 2y 2 3 y 1 + y 2 1 y 1 , y 2 0
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IOE 310 – Additional Problems B Page 2 of 9 (b) Use complementary slackness to solve the max problem. Graphically we find the dual’s optimal solution to be w = 49 / 3, y 1 = 7 / 3, y 2 = 1 / 3. The fact that e 3 > 0, y 1 > 0, and y 2 > 0 shows that the optimal primal solution must have both constraints binding and have x 3 = 0. Thus the optimal values of x 1 and x 2 may be found by solving 2 x 1 + x 2 = 6 and x 1 + 2 x 2 = 7. This yields the optimal primal solution z = 49 / 3, x 1 = 5 / 3, x 2 = 8 / 3, x 3 = 0. Note that the optimal objective function values are equal. Problem 2 Consider the following LP and its optimal tableau: max z = 5 x 1 + x 2 + 2 x 3 s.t x 1 + x 2 + x 3 6 6 x 1 + + x 3 8 x 2 + x 3 2 x 1 , x 2 , x 3 0 z x 1 x 2 x 3 s 1 s 2 s 3 rhs 1 0 1/6 0 0 5/6 7/6 9 0 0 1/6 0 1 -1/6 -5/6 3 0 1 -1/6 0 0 1/6 -1/6 1 0 0 1 1 0 0 1 2 (a) Find the dual to this LP and the optimal solution to the dual. min w = 6y 1 + 8y 2 + 2y 3 s.t. y 1 + 6y 2 5 y 1 + y 3 1 y 1 + y 2 + y 3 2 y 1 , y 2 y 3 0 Optimal dual solution is w = 9, y 1 = 0, y 2 = 5 / 6, y 3 = 7 / 6. (b) Find the range of values of c 2 for which the current basis remains optimal. Since x 2 is non-basic, c 2 can increase by at most the reduced cost for the current basis to remain opti- mal. Thus current basis remains optimal as long as the reduced cost of ¯ c 2 is positive. This can be expressed as ¯ c 2 = c BV V 1 a 2 c 2 . We can also write this expression as: ¯ c 2 = [ y 1 y 2 y 3 ][ a 12 a 22 a 32 ] T c 2 . This is then ¯ c 2 = [0 5 / 6 7 / 6][1 0 1] T c 2 . In order for this expression to be 0 we see that c 2 must be 7 / 6. This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed.
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IOE 310 – Additional Problems B Page 3 of 9 Problem 3 Consider the following LP and its optimal tableau: max z = 3 x 1 + x 2 x 3 s.t 2 x 1 + x 2 + x 3 8 4 x 1 + x 2 x 3 10 x 1 , x 2 , x 3 0 z x 1 x 2 x 3 s 1 s 2 rhs 1 0 0 1 1/2 1/2 9 0 0 1 3 2 -1 6 0 1 0 -1 -1/2 1/2 1 (a) Find the dual to this LP and the optimal solution to the dual. min w = 8y 1 + 10y 2 s.t. 2y 1 + 4y 2 3 y 1 + y 2 1 y 1 - y 2 1 The optimal dual solution is w = 9, y 1 = y 2 = 1 / 2. (b) Find the range of values of b 2 for which the current basis remains optimal. If b 2 = 12, what is the new optimal solution?
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