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Unformatted text preview: /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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