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Unformatted text preview: IOE 310  Problem Set 06 Solution Suggested Reading Section 4.14 (p.184186) (skip Example 9) Section 6.1 Section 6.2 (ignore (10”)’ for artiﬁcial variables on p.272) Problem 1
Section 4.14 Problem 4 (p.189) Professor Kaufman Comments: I disagree with the solution for Problem 1, which I post below. I am interested in seeing your solutions. If you have a solution, please give it. If you agree with the solution’s manual, please give the same solution but with added justiﬁcation. The part I don’t understand is: why can a BFS not have both z’ and z” be positive? Justify this. Otherwise, if you disagree with the solutions manual, please justify why you disagree  a counterexample perhaps. You will receive full credit too if you just copy the answer below. However, I would rather you think. The solution from the textbook is: max s.t z = z + z 4x1 + x2 2x1 − x2 2x1 − 3x2 ≤4 ≤ 0. 5 = z − z All variables nonnegative First note that in any basic feasible solution, z and z cannot both be positive. Then observe that if 2x1 − 3x2 > 0, then z = 0 and the objective function will equal z + z = z = 2x1 − 3x2 = 2x1 − 3x2  while if 2x1 − 3x2 < 0, z = 0 and z = 2x1 − 3x2  so the objective function will equal z + z = z = 2x1 − 3x2 . z and z cannot both be in the basis at the same time. This problem is unbounded. z = z = c > 0 (x1 , x2 ) = (0, 0) is feasible and the objective value is 2c. IOE 310 – Problem Set 06  Solution Page 2 of 2 Problem 2
Section 6.1 Problem 5 (p.267) 1 (a) Let c1 = proﬁt from type 1 radio. Slope of isoproﬁt line is c2 . Current basis is still optimal if of iso1 proﬁt line is steeper than L1 constraint and ﬂatter than L2 constraint or 2 ≤ c2 ≤ 0.5 or 1 ≤ c1 ≤ 4 or $ 23 = 22 + 1 ≤ (Type1 price) ≤ 22 + 4 = $ 26. (b) Current basis remains optimal for 2 ≤
3 c2 ≤ 0.5 or 1.5 ≤ c2 ≤ 6 or $ 21.50 ≤ (Type2 price) ≤ 26. (c)Optimal solution still occurs where both constraints are binding or x1 + 2x2 = 30 and 2x1 + x2 = 50. This yields x1 = 70/3, x2 = 10/3, and z = 230/3. Since solution is feasible, it is optimal. (d) Optimal solution occurs where x1 + 2x2 = 40 and 2x1 + x2 = 60. This yields x1 = 80/3, x2 = 20/3, z = 280/3. (e) If 40 + ∆ laborer 1 hour are available optimal solution is where x1 + 2x2 = 40 + ∆ and 2x1 + x2 = 50. This yields x1 = 20 − ∆/3, x2 = 10 + 2∆/3, z = 80 + ∆/3. Thus laborer 1 shadow price = 1/3. If 50 + ∆ laborer 2 hours are available optimal solution is where x1 + 2x2 = 40 and 2x1 + x2 = 50 + ∆. This yields x1 = 20 + ∆/3, x2 = 10 − ∆/3, z = 80 + 4∆/3. Thus laborer 2 shadow price = 4/3. Problem 3
Section 6.2 Problem 1 (p.274) −2 11 −1 BV = {x1 , x2 } and B = and B = and cBV = [31] and cBV B −1 = [45] 2 12 Coeﬃcient of s1 in row0 = 4 Coeﬃcient of s2 in row0 = 5 2 −1 Right hand Side of row0 = cBV B b = [45] = 28 4 1 1 s1 column = B −1 = = 0 1 0 1 s2 column = B −1 = = 1 2 1 0 x1 column = and x2 column = 0 1 11 2 6 Right hand Side of Constraints = B −1 b = = 12 4 10 Thus the optimal tableau is: z + 4s1 + 5s2 = 28 x1 + s1 + s2 = 6 x2 + s1 + 2s2 = 10 2 −2 These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. ...
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 Fall '08
 Saigal

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