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# 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 find 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, find the optimal solution. We check the non basic edge (1,2) at its upper bound and find 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 · 1 + 5 · 2 + 4 · 0 + 2 · 2 + 5 · ( - 3) = 0 We now check for non tree edges, and verify that this tree is optimal: (3,2) w 3 + c 32 = - 1 + 0 0 = w 2

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