21lagrangian_relaxation_2

# 21lagrangian_relaxation_2 - 15.082J and 6.855J Lagrangian...

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1 15.082J and 6.855J Lagrangian Relaxation 2 ± Algorithms ± Application to LPs

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2 The Constrained Shortest Path Problem (1, 10 ) (1, 1 ) (1, 7 ) (2, 3 ) (10, 3 ) (12, 3 ) (2, 2 ) (1, 2 ) (10, 1 ) (5, 7 ) 1 2 4 5 3 6 Find the shortest path from node 1 to node 6 with a transit time at most 10
3 Constrained Shortest Paths: LP Formulation Given: a network G = (N,A) c ij cost for arc (i,j) t ij traversal time for arc (i,j) (,) ij ij ij A cx Z* = Min s. t. 1 if i = s 1 if i = t 0 otherwise ij ji jj xx −= ∑∑ ij ij tx T 0 or 1 for all ( , ) ij xi j A = Complicating constraint

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4 Lagrangian Relaxation and Inequality Constraints z * = min cx subject to A x b, (P) x X. L( µ )=m i n c x + µ ( A x - b) (P( µ )) subject to x X. L* = max (L( µ ) : µ≥ 0). So we want to maximize over µ , while we are minimizing over x.
5 An alternative representation Suppo se that X = {x 1 , x 2 , x 3 , …, x K }. Possibly K is exponentially large; e.g., X is the set of paths from node s to node t. L( µ ) = min cx + µ ( A x - b) = (c + µ A) x - µ b subject to x X. L( µ ) = min {(c + µ A) x k - µ b : k = 1 to K}

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6 Solving the Lagrange Multiplier Problem L( µ ) = min {(c + µ A) x k - µ b : k = 1 to K } Determine L* = max (L( µ ) : µ≥ 0}
(1, 10 ) (1, 1 ) (1, 7 ) (2, 3 ) (10, 3 ) (12, 3 ) (2, 2 ) (1, 2 ) (10, 1 ) (5, 7 ) 1 2 4 5 3 6 Suppose we want the min cost path from 1 to 6 with transit time at most 14. We now list all K paths from 1 to 6. Pc P t P c P + µ (t P –14) 1-2-4-6 3 18 3 + 4 µ 1-2-5-6 5 15 5 + µ 1-2-4-5-6 14 14 14 etc. 7

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Paths 1-2-5-6 1-3-2-5-6 1-2-4-6 012345 1-2-4-5-6 1-3-2-4-6 1-3-2-4-5-6 1-3-4-6 1-3-4-5-6 1-3-5-6 Composite Cost Lagrange Multiplier µ 0 10 20 30 -10 L* = max (L( µ ): µ 0) Figure 16.3 The Lagrangian function for T = 14. 8
9 Solving the Lagrange Multiplier Problem L( µ ) = min {(c + µ A) x k - µ b : k = 1 to K } In the algorithm S will be a subset of {1, 2, …, K} L S ( µ ) L( µ ). So, L* S L*. Let L S ( µ ) = min {(c + µ A) x k - µ b : k S} 1. Initialize S 2. Find µ * that maximizes L* S =L S ( µ ). 3. Find x k X that minimizes (c + µ * A) x 4. If k S, quit with the optimal solution 5. Else, add k to S, and return to step 2. At end, L* S = L S ( µ *) = L( µ *) L*.

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We start with the paths 1-2-4-6, and 1-3-5-6 which are optimal for L(0) and L( ).
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## This note was uploaded on 03/15/2010 for the course IE 505 taught by Professor Yok during the Spring '10 term at Galatasaray Üniversitesi.

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21lagrangian_relaxation_2 - 15.082J and 6.855J Lagrangian...

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