15 082j and 6 855j and esd 78j
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15.082J and 6.855J and ESD.78J Lagrangian Relaxation 2 - PowerPoint PPT Presentation

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15.082J and 6.855J and ESD.78J Lagrangian Relaxation 2 Applications Algorithms Theory The Constrained Shortest Path Problem (1,1) 2 4 (1,10) (1,7) (2,3) 1 6 (1,2) (10,1) (5,7) (2,2) (10,3) 5 3 (12,3) Find the

  1. 15.082J and 6.855J and ESD.78J Lagrangian Relaxation 2 • Applications Algorithms • • Theory

  2. The Constrained Shortest Path Problem (1,1) 2 4 (1,10) (1,7) (2,3) 1 6 (1,2) (10,1) (5,7) (2,2) (10,3) 5 3 (12,3) Find the shortest path from node 1 to node 6 with a transit time at most 14. 2

  3. Constrained Shortest Paths: Path Formulation Given: a network G = (N,A) w ij cost for arc (i,j) w(P) cost of path P t ij traversal time for arc (i,j) T upper bound on transit times. t(P) traversal time for path P P set of paths from node 1 to node n Min w(P) w(P) + u (t(P) – T) L(u) = Min t(P) ≤ T s.t. P ∈ P s.t. P ∈ P Lagrangian Constrained Problem 3

  4. The Lagrangian Multiplier Problem Step 0. Formulate L(u) = Min w(P) + u t(P) - uT the Lagrangian P ∈ P s.t. Problem. L(u) = v* = Max v Step 1. Rewrite as s.t v ≤ w(P) + u t(P) – uT a maximization for all P ∈ P problem L* = max {L(u): u ≥ 0} = Step 2. Write the = Max v Lagrangian s.t v ≤ w(P) + u t(P) – u T multiplier problem for all P ∈ P u ≥ 0 4

  5. Max { v: v ≤ w(P) + u t(P) – uT ∀ P ∈ P , u ≥ 0} -uT + Min { w(P) + u t(P) : P ∈ P }  P* = Path(u) Per u fissato  Path(u) cammino minimo con pesi modificati Ogni arco ij ha peso: w ij + u t ij (w, t) (1,1) u = 0  w ij 2 4 (1,10) (1,7) u = M>>0  w ij + M t ij (2,3) u = 2.1  1 6 (10,1) (1,2) 3.1 2 4 (2,2) (10,3) (5,7) 15.7 22 8.3 5 3 (12,3) 5.2 1 6 12.1 Per ogni u una diversa istanza 16.3 6.2 19.7 del cammino minimo 5 3 18.3

  6. Discretizzazione del problema lagrangiano Ogni soluzione (cammino) definisce un iperpiano (retta) 1-3-4-6 t(P)=17 (1,1) 2 4 Paths (1,10) (1,7) (2,3) 30 1 6 (10,1) Composite Cost 20 (1,2) 1-3-4-5-6 t(P)=13 10 (2,2) (10,3) (5,7) 5 3 0 (12,3) (1,1) -10 2 4 0 1 2 3 4 5 (1,10) (1,7) (2,3) Lagrange Multiplier u w(P) + u (t(P) – T) 1 6 (10,1) (1,2) 27 + u (13 – 14) [ pendenza neg] (2,2) (10,3) (5,7) 5 16 + u (17 – 14) [ pendenza pos] 3 (12,3)

  7. Max { v: v ≤ w(P) + u t(P) – uT ∀ P ∈ P , and u ≥ 0} Paths 1-3-4-6 t(P)=17 30 1-2-4-6 Composite Cost 20 1-3-4-5-6 t(P)=13 1-2-4-5-6 1-2-5-6 10 1-3-2-4-6 0 1-3-2-4-5-6 1-3-2-5-6 L* = max (L(u): u ≥ 0) 1-3-5-6 -10 0 1 2 3 4 5 Lagrange Multiplier u Figure 16.3 The Lagrangian function for T = 14. 7

  8. The Restricted Lagrangian P P : the set of paths from node 1 to node n B ⊆ P P : a subset of paths B L * L* = v* = max u,v v B = max u,v v s.t v ≤ w(P) + u ( t(P) – T ) s.t v ≤ w(P) + u t(P) – u T for all P ∈ P for all P ∈ B u ≥ 0 u ≥ 0 Restricted Lagrangian Lagrangian Multiplier Problem Multiplier Problem If L(u) = L * L(u) ≤ L* ≤ L * B then L(u) = L*. B Optimality Conditions 8

  9. Constraint Generation for Finding L* Let Path(u) be the path that optimizes L(u). Path(u) = w(P*) + u t(P*) Let u(B) be the value of u that optimizes L B (u). M is some large number Initialize: B := {Path(0), Path(M)} Yes Quit. L(u(B)) = L* Is L(u(B)) = L* B ? No B := B ∪ Path(u(B)) 9

  10. We start with the paths 1-2-4-6, and 1-3-5-6 which are optimal for L(0) and L( ∞). Paths 30 (u(B), v(B)) 1-2-4-6 Composite Cost 20 3 + (18-14) u  Path(0) (2.1, 11.4) 10 24 + (8-14) u  Path(M) 0 1-3-5-6 -10 0 1 2 3 4 5 Lagrange Multiplier u 10

  11. Set u(B) = 2.1 and solve the shortest path problem (L(u(B) ) The optimum path is 1-3-2-5-6 Tempo di costante Costo transito 3.2 15 + 10 u(B) – 14 u(B) 2 4 15.7 22 8.3 u(B) = 2.1 15 + 21 – 29.4 = 6.6 12.1 5.2 6 1 19.7 16.3 6.2 5 3 18.3 6.6 < 11.4 Aggiungi il path a B e riottimizza 11

  12. Path(2.1) = 1-3-2-5-6. Add it to S and reoptimize. Paths 30 3 + 4 u 1-2-4-6 Composite Cost 20 10 1.5, 9 0 15 - 4 u 1-3-2-5-6 24 - 6 u 1-3-5-6 -10 0 1 2 3 4 5 Lagrange Multiplier u 12

  13. Set u(B) = 1.5 and solve the constrained shortest path problem The optimum path is 1-2-5-6. 2.5 5 + 15 u(B) – 14 u(B) 2 4 11.5 5 + 22.5 – 21 = 6.5 16 6.5 11.5 4 1 6 15.5 14.5 5 5 3 16.5 6.5 < 9 Aggiungi il path a B e riottimizza 13

  14. Add Path 1-2-5-6 and reoptimize Paths 30 3 + 4 u 1-2-4-6 Composite Cost 20 5 + u 1-2-5-6 10 2, 7 0 15 - 4 u 1-3-2-5-6 24 - 6 u 1-3-5-6 -10 0 1 2 3 4 5 Lagrange Multiplier u 14

  15. Set u(B) = 2 and solve the constrained shortest path problem The optimum paths are 1-2-5-6 and 1-3-2-5-6 3 5 + 15 u(B) – 14 u(B) 2 4 15 21 8 5 + 30 – 28 = 7 12 5 1 6 19 16 6 5 3 18 Ottimo del Duale 7 = v(B) = 7 Lagrangiano 15

  16. There are no new paths to add. u(B) = u* is optimal for the multiplier problem Paths 30 3 + 4 u 1-2-4-6 Composite Cost 20 5 + u 1-2-5-6 10 2, 7 0 15 - 4 u 1-3-2-5-6 24 - 6 u 1-3-5-6 -10 0 1 2 3 4 5 Lagrange Multiplier u 16

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