WI4131_Ex56 - Discrete (and Continuous) Optimization...

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Unformatted text preview: Discrete (and Continuous) Optimization Solutions of Exercises 5 WI4 131 Kees Roos Technische Universiteit Delft Faculteit Electrotechniek, Wiskunde en Informatica Afdeling Informatie, Systemen en Algoritmiek e-mail: C.Roos@ewi.tudelft.nl URL: http://www.isa.ewi.tudelft.nl/˜roos November – December, A.D. 2003 Course Schedule 1. Formulations (18 pages) 2. Optimality, Relaxation, and Bounds (10 pages) 3. Well-solved Problems (13 pages) 4. Matching and Assigments (10 pages) 5. Dynamic Programming (11 pages) 6. Complexity and Problem Reduction (8 pages) 7. Branch and Bound (17 pages) 8. Cutting Plane Algorithms (21 pages) 9. Strong Valid Inequalities (22 pages) 10. Lagrangian Duality (14 pages) 11. Column Generation Algorithms (16 pages) 12. Heuristic Algorithms (15 pages) 13. From Theory to Solutions (20 pages) Optimization Group 1 Exercise 5.6.7 (first part only) Let (P1) f (λ) = max (P2) h(t) = min n i=1 cj xj : n i=1 aj xj : n i=1 aj xj ≤ λ, x ∈ X ⊆ n i=1 cj xj ≥ t, x ∈ X . Rn Show that (i) f (λ) ≥ t if and only if h(t) ≤ λ. (ii) f (b) = max {t : h(t) ≤ b}. Solution: Fixing λ, let x∗ be optimal for (P1). Then x∗ ∈ X . Moreover, cT x∗ = f (λ) and aT x∗ ≤ λ. Hence f (λ) ≥ t if and only if cT x∗ ≥ t, i.e. if and only if x∗ is feasible for (P2). But by the definition of h(t) this implies h(t) ≤ aT x∗ ≤ λ. Thus we have shown that f (λ) ≥ t implies h(t) ≤ λ. The inverse implication goes in the same way. It remains to prove (ii). Given b, one has h(t) ≤ b if and only if there exists an x ∈ X such that cT x ≥ t and aT x ≤ b. Thus it follows that max {t : h(t) ≤ b} is equal to max t : cT x ≥ t, aT x ≤ b, x ∈ X = max cT x : aT x ≤ b, x ∈ X = f (b). This proves (ii). • N.B. The above result means that in the knapsack problem we can reverse the roles of the objective and constraint rows. Optimization Group 2 Exercise 5.6.9: Longest Path form of 0-1 Knapsack Problem Let D = (V, A) be the acyclic digraph with nodes (j, λ), j = 0 : n, λ = 0 : b, arcs (j − 1, λ), (j, λ′) for λ′ = λ + kaj ≤ b, k ∈ {0, 1}, with weight kcj , and a node t with zero weight arcs from (n, λ), λ = 0, 1 . . . , b to t. Then z is precisely the length of a longest path from node (0, 0) to node t. Below this digraph is shown for z = max 25x1 + 24x2 + 10x3 + 7x4 6x1 + 5x2 + 2x3 + 1x4 ≤ 7 x ∈ { 0 , 1 }4 . Solid arcs without weight are zero weight arcs. 07 06 05 04 03 02 01 00 25 17 16 15 14 13 12 11 10 27 26 25 24 24 10 37 36 35 34 33 32 7 7 7 47 46 45 44 7 t 23 22 21 20 43 42 10 31 7 41 40 3 30 Optimization Group ...
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This note was uploaded on 04/15/2010 for the course INDUSTRIAL ie513 taught by Professor Zeynephuygur during the Spring '10 term at Bilkent University.

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