midterm2 - NAME PAGE 2 Question 1[10 marks Consider the...

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N AME : P AGE 2 OF 8 Question 1. [10 marks] Consider the linear program: (P) max - 2 x 2 +2 x 4 subject to x 1 - 2 x 2 + x 4 = 2 - 4 x 2 + x 3 +3 x 4 = 10 x 2 - x 4 + x 5 = 1 x 1 , x 2 , x 3 , x 4 , x 5 0 . Use the simplex method to solve ( P ) starting with the feasible basis { 1 , 3 , 5 } . Do not do more than three iterations. If ( P ) has an optimal solution, give an optimal basic solution and an optimal basic dual solution. If ( P ) is unbounded, provide a parametric set of feasible solutions x ( t ) = x 0 + td , (for t 0 ) such that c T x ( t ) + as t + . Solution: The initial tableau is: z + 2 x 2 - 2 x 4 = 0 x 1 - 2 x 2 + x 4 = 2 - 4 x 2 + x 3 + 3 x 4 = 10 x 2 - x 4 + x 5 = 1 Entering variable: x 4 (since ¯ c 4 = 2 > 0 ). Leaving variable: x 1 (since min( 2 1 , 10 3 ) = 2 1 ). z + 2 x 1 - 2 x 2 = 4 x 1 - 2 x 2 + x 4 = 2 - 3 x 1 + 2 x 2 + x 3 = 4 x 1 - x 2 + x 5 = 3 Entering variable: x 2 (since ¯ c 2 = 2 > 0 ). Leaving variable: x 3 (since min( 4 2 ) = 4 2 ). z - x 1 + x 3 = 8 - 2 x 1 + x 3 + x 4 = 6 - 3 2 x 1 + x 2 + 1 2 x 3 = 2 - 1 2 x 1 + 1 2 x 3 + x 5 = 5 Entering variable: x 1 (since ¯ c 1 = 1 > 0 ). ¯ A 1 < 0 so the problem is unbounded. For each real t 0 , define x ( t ) := 0 2 0 6 5 + t 1 3 2 0 2 1 2 . Then x ( t ) is feasible for every t 0 and c T x ( t ) → ∞ as t → ∞ .
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N AME : P AGE 3 OF 8 Question 2. [10 marks] Consider the linear program ( P ) max { c T x : Ax = b, x 0 } , where A = 1 - 1 1 0 0 2 1 2 0 1 1 0 3 - 1 2 0 1 2 , b = 4 9 14 , and c = [2 , 2 , 0 , 1 , - 1 , - 2] T . Solve this linear program using the revised simplex method, starting with the feasible basic solu- tion x * = (4 , 0 , 0 , 3 , 2 , 0) T . If the problem has an optimal solution, provide one. If ( P ) has an optimal solution, give an optimal basic solution and an optimal basic dual solution. If ( P ) is unbounded, provide a parametric set of feasible solutions x ( t ) = x 0 + td , (for t 0 ) such that c T x ( t ) + as t + . Solution: Iteration 1 : Basis B = { 1 , 4 , 5 } . Solve A T B y = c B : 1 1 3 0 1 0 0 1 1 y = 2 1 - 1 . So y = (7 , 1 , - 2) T . The reduced cost for x 2 is ¯ c 2 = c 2 - y T A 2 = 5 > 0 , so x 2 enters.
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