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Unformatted text preview: ISyE 6669 Homework 3 Solution Set 4.17 (Review Problem) (a) c ≥ 0 and b ≥ 0. (b) b ≥ 0 and c = 0. Also need a 2 > 0 and/or a 3 > 0 to ensure that when x 1 is pivoted in a feasible solution results. If only a 3 is strictly positive, then we also need b > 0. (c) c < 0, a 2 ≤ 0, a 3 ≤ 0 ensures that x 1 can be made artitrarily large. 6.2.2 BV = { x 2 ,s 1 } ,B = " 1 1 1 0 # ,B 1 = " 1 1 1 # ,c BV = [1 0] c BV B 1 = [1 0] " 1 1 1 # = [0 1] Coefficient of x 1 in row 0 = c BV B 1 a 1 c 1 = [0 1] " 2 1 # + 1 = 2 Coefficient of s 2 in row 0 = c BV B 1 " 1 # = 1 RHS of row 0 = c BV B 1 b = [0 1] " 4 2 # = 2 Column for x 1 = B 1 a 1 = " 1 1 1 #" 2 1 # = " 1 1 # Column for s 2 = B 1 " 1 # = " 1 1 # RHS of optimal tableau = B 1 b = " 1 1 1 #" 4 2 # = " 2 2 # Thus the optimal tableau is z + 2 x 1 + s 2 = 2 x 1 + x 2 + s 2 = 2 x 1 + s 1 s 2 = 2 1 6.3.6 (a) x 1 is nonbasic so changing the coefficient of x 1 in the objective function will only change the coefficient of x 1 in the optimal row 0. Let the new coefficient of x 1 in the objective function be 3 + Δ. The new coefficient of x 1 in the optimal row 0 will be c BV B 1 a 1 (3 Δ) = 3 Δ. Thus if 3 Δ ≥ 0 or Δ ≤ 3 the current basis remains optimal. Thus if profit for a Type 1 Candy Bar is ≤ 6 cents the current basis remains optimal. (b) Changing candy bar 2 profit to 7 + Δ changes c BV B 1 to [5 7 + Δ] " 3 / 2 1 / 2 1 / 2 1 / 2 # = [4 Δ / 2 1 + Δ / 2] Then coefficient of x 1 in row 0 = [4 Δ / 2 1 + Δ / 2] " 1 2 # = 3 + Δ / 2. Thus row 0 of optimal tableau is now...
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This note was uploaded on 12/30/2011 for the course ISYE 6669 taught by Professor Staff during the Fall '08 term at Georgia Tech.
 Fall '08
 Staff
 Optimization

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