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Unformatted text preview: MIT 2.098/6.255/15.093 Optimization Methods Prof. J. Vera, Fall 2007 Homework Assignment 5. Solution Problem 1: BT Ex 7.1 Solution: (a) Construct the network as follows: • For each day i , create two nodes: – Node c i for clean tableclths with supply r i , – Node d i for dirty tablecloths with demand r i (Note that this is equivalent to creating an arc ( c i ,d i ) with lower bound r i . • Create a node s for the source of new tablecloths. • Each node c i can – receive new tablecloths from purchasing: arcs ( s,c i ) with arc cost p and unlimited capacity; – receive “fast” laundered tablecloths from n days ago (or longer): arcs ( d i n j ,c i ) if i > n + j with arc cost f and unlimited capacity; – receive “slow” laundered tablecloths from m days ago (or longer): arcs ( d i m j ,c i ) if i > m + j with arc cost g and unlimited capacity. • For each node d i , create arcs ( d i ,s ) with zero cost and unlimited capacity, representing tablecloths which are not laundered and used again. We have defined a network with arc costs and capacities, with equal supply and demand. (b) It is easy to form the network from above construction. We omit this part. BT Ex 7.7 Solution: Let ¯ f ij = f ij d ij . The flow conservation constraint for each node i can be written as X j ∈ O ( i ) ( ¯ f ij + d ij ) X j ∈ I ( i ) ( ¯ f ji + d ji ) = b i , where I ( i ) (and O ( i )) is the set of start nodes (end nodes) of arcs that are incoming to (outgoing 1 from) node i . So we have X j ∈ O ( i ) ¯ f ij X j ∈ I ( i ) ¯ f ji = b i ( X j ∈ O ( i ) d ij X j ∈ I ( i ) d ji ) , where the RHS can be defined as the new supply ¯ b i of node i . BT Ex 7.8 Solution: This problem is trickier than its first look. Actually there is a counterexample that shows the optimal cost can decrease when we increase the supply at some source nodes and demand at some sink nodes even though all the costs coefficents are positive. The catch here is when the supply and demand change, the current optimal basis may become nonoptimal and the new optimal basis can give lower cost. Consider the following network with two supply nodes 1 and 2 with supply ² and 1 respectively, and two demand nodes 3 and 4 with demand ² and 1 respectively. The arcs are (1 , 3) , (1 , 4) , (2 , 3) , (2 , 4) (so it is a complete bipartite graph.) The arc costs are c 13 = c 14 = c 23 = ²,c 24 = 1, where ² is a small positive number less than 1. The current optimal solution is x 13 = 0 ,x 14 = ²,x 23 = ²,x 24 = 1 ² with cost C = 1 ² + 2 ² 2 ≈ 1. Now we increase the supply of node 1 from ² to 1, and demand of node 3 from ² to 1. The arc costs remain unchanged. The new optimal solution is x 13 = 0 ,x 14 = 1 ,x 23 = 1 ,x 24 = 0 with cost C = 2 ² . So we can see not only the optimal cost decreases from C ≈ 1 to C = 2 ² , but the ratio C/C is around 1 / 2 ² which can be arbitrarily large when ² is arbitrarily small....
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 Spring '10
 Bertsekas
 Optimization, basic feasible solution, Hessians, optimal cost

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