{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Capter 7 Branch-and-Bound Optimization Group 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
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
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern