sol9 - AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Linear...

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AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Linear Programming Solutions Homework Set # 9 1). (a). Use the arcs (4,1) (4,3) (3,2) (5,4) as your starting tree solution. The edge (1,2) is non basic at its upper bound, and all other non basic edges are at the lower bound of zero. Compute the x ij and the fair prices (dual variables) w i . x 12 = 2, x 41 = 3, x 32 = 3, x 54 = 5, x 43 = 4, is the initial feasible tree solution. (Recall (1,2) is non basic at its upper bound.) To ±nd the fair prices we have: w 3 + 0 = w 2 , w 4 + 2 = w 1 , w 5 - 3 = w 4 , w 4 + 2 = w 3 , w 5 = 0. We solve: w 1 = - 1, w 2 = - 1, w 3 = - 1, w 4 = - 3, w 5 = 0. The cost of this tree is 2 · 1 + 4 · 2 + 3 · 0 + 3 · 2 + 5 · ( - 3) = 1 (b). Starting from the feasible tree soltuion of part (a), using the network simplex method, ±nd the optimal solution. We check the non basic edge (1,2) at its upper bound and ±nd that w 1 + c 12 > w 2 , so this edge enters the basis. We decrease shipment on it by Δ such that 2 - Δ 0, 3 - Δ 0, 3 + Δ 4, and 4 + Δ 6, so Δ = 1 and (3,2) becomes non basic at its upper bound. the new feasible tree solution is: x 12 = 1, x 41 = 2, x 32 = 4, x 54 = 5, x 43 = 5. Computer fair prices again: w 1 + 1 = w 2 , w 4 + 2 = w 1 , w 5 - 3 = w 4 , w 4 + 2 = w 3 , w 5 = 0. We solve: w 1 = - 1, w 2 = 0, w 3 = - 1, w 4 = - 3, w 5 = 0. The cost of this tree is 1
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This note was uploaded on 01/31/2011 for the course AMS 540 taught by Professor Arkin,e during the Fall '08 term at SUNY Stony Brook.

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sol9 - AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Linear...

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