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# WI4131_7 - Discrete(and Continuous Optimization WI4 131...

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Discrete (and Continuous) Optimization WI4 131 Kees Roos Technische Universiteit Delft Faculteit Electrotechniek, Wiskunde en Informatica Afdeling Informatie, Systemen en Algoritmiek e-mail: [email protected] URL: http://www.isa.ewi.tudelft.nl/ ˜ roos November – December, A.D. 2003

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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
Capter 7 Branch-and-Bound Optimization Group 2

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Divide and Conquer Consider the problem z = max n c T x : x S Z n + o . Proposition 1 Let S = S 1 . . . S K be a decomposition of S into K smaller subsets, and let z k = max n c T x : x S k o for k = 1 , . . . , K . Then z = max k z k . If the subproblems z k = max n c T x : x S k o are easy to solve, this solves the given problem. Otherwise, we can solve the hard subproblems by using the same decomposition technique: decompose S k in smaller subsets. This gives rise to an enumeration tree . For example, if S = { 0 , 1 } 3 , the tree is depicted below. S 000 S 001 S 010 S 011 S 100 S 101 S 110 S 111 S 00 S 01 S 10 S 11 S 0 S 1 S x 1 = 0 x 1 = 1 x 2 = 0 x 2 = 1 x 3 = 0 x 2 = 0 x 2 = 1 x 3 = 1 The leaves of the tree corre- spond to the singletons x = ( x 1 , x 2 , x 3 ) S , and hence the corresponding subproblems are easy indeed. However, follow- ing this approach is equivalent to complete enumeration, and hence combinatorial explosion. This has to be avoided, and this can be done by being more intelligent! Optimization Group 3
Implicit enumeration Proposition 2 Let S = S 1 . . . S K be a decomposition of S into K smaller subsets, and let z k = max n c T x : x S k o for k = 1 , . . . , K . Furthermore, for each k let z k z k ¯ z k . Then z z ¯ z , where z = max k z k and ¯ z = max k ¯ z k . Proof : One has z k z k z , whence max k z k z , and z = max k z k max k ¯ z k . By generating lower and upper bounds for the subproblems some subproblems need not be examined at all! We then say that the corresponding node in the search tree is pruned . Here are three arguments for pruning a node: ( i ) Pruning by optimality : if z k = max n c T x : x S k o has been solved, the underlying nodes need no further investigation; ( ii ) Pruning by bound : if ¯ z k z ; ( iii ) Pruning by infeasibility : if S k = . Optimization Group 4

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Pruning Pruning by optimality : S 1 20 20 S 2 15 25 S 13 27 S 1 S 2 15 25 S 20 25 The bounds for S can be improved: ¯ z = max k ¯ z k = max { 20 , 25 } = 25 z = max k z k = max { 20 , 15 } = 20 . The bounds on S 1 are equal. So z 1 = 20 : we can prune this node.
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