Ch06rp

Ch06rp - 1 SOLUTIONS TO CHAPTER 6 MP REVIEW PROBLEMS 1a min...

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SOLUTIONS TO CHAPTER 6 MP REVIEW PROBLEMS 1a. min w = 6y 1 + 3y 2 + 10y 3 s.t. y 1 + y 2 + 2y 3 4 2y 1 - y 2 + y 3 1 y 1 urs, y 2 0, y 3 0 Optimal dual solution is w = 58/3, y 1 = -2/3, y 2 = 0, y 3 = 7/3 1b. B -1 = 1 1 1 3 / 2 0 3 / 1 3 / 1 0 3 / 2 B -1 10 3 6 = 1 3 / ) 2 14 ( 3 / ) 2 ( Thus current basis remains optimal iff Δ 2, Δ -7 and Δ -1 or -1 Δ 2 or 9 b 3 12. Substituting Δ = 1 we obtain z = 58/3 + Row 3 Shadow Price = 65/3 x 2 = (2-1)/3 = 1/3, x 1 = (14 + 2)/3 = 16/3, e 2 = 1 + 1 = 2 2. When isoprofit line is flatter than x 1 + 2x 2 = 6 optimal solution shifts to A; otherwise B remains optimal. Slope of Isoprofit Line is -c 1 . Thus current basis is no longer optimal for -c 1 >-1/2 or c 1 <1/2. Current basis remains optimal for c 1 1/2. 3a. 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 0, y 2 0, y 3 0 Optimal dual solution is w = 9, y 1 = 0, y 2 = 5/6, y 3 = 7/6. 3b. c BV B -1 = [0 (5 +Δ)/6 (7-Δ)/6] _ 1

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Pricing out x 2 we obtain c 2 = [0 (5 +Δ)/6 (7-Δ)/6] 1 0 1 - 1 = (1-Δ)/6. Coefficient of s 2 in Optimal Row 0 = (5 + Δ)/6 Coefficient of s 3 in Optimal Row 0 = (7-Δ)/6 Thus current basis remains optimal iff Δ 1, Δ -5, Δ 7 or -5 Δ 1 or 0 c 1 6. Dotted line is z = 58/3. 3c. Since x 2 is non-basic, c 2 can increase by at most the reduced cost for the current basis to remain optimal. Thus current basis remains optimal as long as c 2 1 + 1/6 = 7/6. 4a. Decision variables remain the same. New z-Value = Old z-value + 10(88) = \$33,420 4b. Since the allowable increase for type 1 machines 2
is less than one, we cannot answer this question. 4c. Since we are not using all the steel that is currently available, Carco would pay \$0 for another ton of steel. 4d. Relevant Shadow Price is -\$20. Current basis remains optimal if demand is decreased by up to 3 cars, so shadow price may be used to compute new z-value. New Profit = \$32,540 + (-2) (-20) = \$32,580. 4e. The new column for jeeps is 0 0 4 2 0 2 . 1 The objective function coefficient for jeeps is \$600 and c BV B -1 = [400 350 0 0 -20 0]. Thus the coefficient of jeeps in Row 0 is given by 1.2(400)-600 = -\$120, and jeeps should be produced. 5a. min w = 6y 1 + 4y 2 + 3y 3 s.t. 3y 1 + 2y 2 + y 3 4 y 1 + y 2 + y 3 1 y 1 0 y 2 0, y 3 urs The optimal dual solution is w = 12, y 1 = y 2 = 0, y 3 = 4. 5b. c 2 1 + (reduced cost for x 2 ) = 1 + 3 = 4 5c. c BV B -1 = [4 + Δ 0 0] 2 1 0 3 0 1 1 0 0 = [ 0 0 4 + Δ] _ The current basis remains optimal iff c 2 0. But _ c 2 = [0 0 4 + Δ] 3

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1 1 1 - 1 = 3 + Δ Now 3 + Δ 0 iff Δ -3 Thus current basis remains optimal for c 1 4-3 = 1 6a. 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.
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