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20_Lagrangian_Relaxation_2

# 20_Lagrangian_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, 1 ) (12, 3 ) (1, 2 ) (10, 1 ) 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 i j A c x j j Z* = Min s. t. 1 if i = s 1 if i = t 0 otherwise ij ji j j x x j j - = - ( , ) ij ij i j A t x T j j j 0 or 1 for all ( , ) ij x i 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 ) = min cx + 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, 1 ) (12, 3 ) (1, 2 ) (10, 1 ) 1 2 4 5 3 6 P c 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 We now list all K paths from 1 to 6. Suppose we want the min cost path from 1 to 6 with transit time at most 14.

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1-2-5-6 1-3-2-5-6 1-2-4-6 Paths 0 1 2 3 4 5 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 Figure 16.3 The Lagrangian function for T = 14. 0 10 20 30 -10 8 L* = max (L( μ ): μ 0)
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} 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. L S ( μ ) L( μ ). So, L* S L*. At end, L* S = L S ( μ *) = L( μ *) L*.

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1-2-4-6 3 + 18 μ Paths 0 1 2 3 4 5 Lagrange Multiplier μ 0 10 20 30 -10 Composite Cost 10 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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20_Lagrangian_Relaxation_2 - 15.082J and 6.855J Lagrangian...

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