midterm_solution

# midterm_solution - z x 1 x 2 x 3 x 4 x 5-1 2 2 1 70 1 1 2-1...

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MS&E111 Spring 2006 Introduction to Optimization May 26, 2006 Prof. Amin Saberi Solutions to Midterm Problem 1. a) If the cost of burger increases by 1%, then the total cost would be increased by \$2 × 0 . 01 × 1 . 5 = \$0 . 03. b) If the required amount of vitamin is reduced to 5.8units, then \$(6 - 5 . 8) × 2 = \$0 . 4 can be saved. c) The shadow price for vitamin is \$2, so if the cook replaces 1% of vitamin by a pill, the cook is willing to pay up to \$2 × 6 100 = \$0 . 12 for the pill. In the same way, it’s \$1 × 9 100 = \$0 . 09 for calories and \$0 for proteins. Problem 2. We first transform the linear program into a standard form: min z = - 5 x 1 - 3 x 2 - 4 x 3 subject to 2 x 1 + x 2 + x 3 + x 4 = 20 3 x 1 + x 2 + 2 x 3 + x 5 = 30 x 1 , x 2 , x 3 , x 4 , x 5 0 Then, the first basic variables are x 4 and x 5 . In a tabular form, it becomes z x 1 x 2 x 3 x 4 x 5 -1 -5 -3 -4 0 0 0 0 2 1 1 1 0 20 0 3 1 2 0 1 30 We choose x 2 as a new entering basic variable. Then, x 4 leaves out of basic variables:

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z x 1 x 2 x 3 x 4 x 5 -1 1 0 -1 3 0 60 0 2 1 1 1 0 20 0 1 0 1 -1 1 10 Since the coefficient of x 3 is still negative, we choose it as a new basic variable:
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Unformatted text preview: z x 1 x 2 x 3 x 4 x 5-1 2 2 1 70 1 1 2-1 10 1 1-1 1 10 Now, all coeﬃcients in the objective function are positive, so we’ve found an optimal solution. The optimal value is z =-70 at x 2 = 10 , x 3 = 10. Problem 3. The union of two polytopes is not a polytope. As a counterexample, { x | x ≥ 5 }∪{ x | x ≤ 1 } . Although each set is a polytope, the union is not a convex set, thus is not a polytope. As for the intersection of two polytopes, we think of two arbitrary polytopes, P 1 and P 2 . For each P i , we can ﬁnd A i , b i such that A 1 x ≥ b 1 and A 2 x ≥ b 2 . Since { x | A 1 x ≥ b 1 } ∩ { x | A 2 x ≥ b 2 } = { x | Ax ≥ b } , where A = " A 1 A 2 # , b = " b 1 b 2 # , the intersection is a polytope....
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