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HW2solution

# HW2solution - HW#3 Problem 1 – Idea of solution The...

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Unformatted text preview: HW #3 Problem 1 – Idea of solution The problem had a bit more complexity that we initially realized. To see this, note the following: Suppose we can get to some intermediate node j in 8 days for \$380 or in 9 days for \$340. Which one is better? The problem is, we do not know (yet). The additional day might be irrelevant or it might become very costly (additional \$200) if it ends up requiring to rent the car for an additional week. The above reasoning suggests that, for node j , we need to keep track of more than just the length of the cheapest (1 ,j ) path. This can be addressed by introducing a state variable standing for the number of days available for the journey. The idea of the solution follows. Notation: We denote c ij the cost (in dollars) of arc ( i,j ) and ‘ ij the length (in days) of arc ( i,j ). Optimal value function definition: Let us denote f * j ( d ) = minimum cost of getting from node 1 to node j in at most d days . Recursive formula: For any node j , we want to find optimal next-to-last node on the cheapest (1 ,j ) path. Consider node i . Then the cost of the cheapest (1 ,j ) path through node i (as path’s next-to-last node) of length at most d days is given by min d i : d i + 1 ( i> 1)+ ‘ ij ≤ d f * i ( d i ) + c ij + 200 · ( b ( d i + 1 ( i > 1) + ‘ ij ) / 7 c -b d i / 7 c ) . In the above formula, note that • f * i ( d i ) stands for the cost of the cheapest (1 ,i ) path of length at most d i days. • c ij is the cost of ( i,j ) arc. • ( b ( d i + 1 ( i > 1) + ‘ ij ) / 7 c - b d i / 7 c ) gives the number of additional weeks (for which we have to pay \$200 per week) started by adding days d i +1 ,d i +2 ,...,d i + ‘ ij + 1 ( i > 1) to the path. (Note that for i = 1, the term 1 ( i > 1) equals 0. This reflects the fact that we are not staying for an additional day in node 1.) • Since we are looking for path of at most d days, we optimize over all possible d i ’s such that d i + 1 ( i > 1) + c ij ≤ d . (Recall that we are required to stay 1 day in each intermediate node. Hence the term 1 ( i > 1).) Having evaluated minimum cost when using i as a next-to-last node, we are ready to formulate the recursive formula for node j . We have f * j ( d ) = min i : ( i,j ) ∈ E min d i : d i + 1 ( i> 1)+ ‘ ij ≤ d n f * i ( d i ) + c ij + 200 · ( b ( d i + 1 ( i > 1) + ‘ ij ) / 7 c -b d i / 7 c ) o . Solution process: Starting with the boundary condition f * 1 ( d ) = 0 for any d ≥ 0 calculate recursively the values of f * 2 ( · ) ,...,f * 10 ( · ). Then, min d f * 10 ( d ) will give the minimum cost of the cheapest (1 , 10) path. Also, arg min d f * 10 ( d ) will give the number of days this cheapest path takes. You can find out the optimal path by backtracking the optimizers in each step along the path....
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HW2solution - HW#3 Problem 1 – Idea of solution The...

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