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Branch and bound technique

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Solving traveling salesman and water jug problem using Branch and Bound Technique
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Solving traveling salesman and water jug problem using Branch and Bound Technique Prepared By Mehta Ishani
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Page 1: Branch and bound technique

Solving traveling salesman and water jug problem using Branch and

Bound Technique

Prepared By

Mehta Ishani

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IntroductionBranch and Bound method for solving optimization problems

approach developed for solving discrete and combinatorial optimization problems

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Problem domaindiscrete optimization problems

(decision variables assume discrete values)

combinatorial optimization problems(choosing the best combination)

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DifficultyNo optimality conditions

To ensure optimality perform comparison

Explicit comparison results in NP Completeness

Implicit comparison results partial enumeratoin

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SolutionBranch and Bound technique breaking up its feasible set into successively

smaller subsets calculating bounds on the objective function

value discard certain subsets

The method was first proposed by A. H. Land and A. G. Doig in 1960 for discrete programming

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Steps1)Place the start node of zero path length on the queue.2)Until the queue is empty or a goal node has been

found do the following: (i) Determine if the s first path in the queue contains a good node. (ii) If the first path contains a goal node exit with success.(iii) If the first path does not contain a good node remove the path from the queue and form new paths by extending the removed paths by on step.(iv)Compute the cost of the new paths and add them to the queue.(v) Sort the paths on the queue with lowest-cost path in front.

3)otherwise exit with failure.

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Water JugInitial state (0,0)Goal state (2,0)

Path p1:(0,0) -> (4,0) -> (1,3) -> (1,0) -> (0,1) -> (4,1) -

> (2,3) -> (2,0)

P1

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Water Jug Path 2: (0,0) -> (0,3) -> (3,3)->(3,0)->(4,0)->(0,0) Since this results infinite loop then not to put in

queue

Path 3: (0,0) -> (0,3) -> (3,3) -> (4,2) -> (0,2) -> (2,0)

P1

P3 P1

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Water Jug Now path 1 contains 8 states and path 2

contains 6 states after performing sorting the queue is

Thus our optimal solution is Path 3

P1 P3

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Traveling Salesman Problem

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Problem NP-hard Brute Force takes O(n!) to complete Algorithms to solve take one of two forms:

Exact algorithms Branch and Bound O(2n) Linear Programming

Heuristic algorithms Nearest neighbor O( log n ) Pair wise exchange Randomized improvement (genetic algorithms)

 Many real world applications:

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Solution Special properties of the Traveling Salesman

Problem that make it suitable for Branch and Bound: As you build your solution, cost increases. Partial solutions have valid costs (lower bound

costs)

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Steps Given a complete, weighted graph on n nodes,

find the least weight Hamiltonian cycle, a cycle that visits every node once. 

Though this problem is easy enough to explain, it is very difficult to solve.

Finding all the Hamiltonian cycles of a graph takes exponential time. Therefore, TSP is in the class NP.

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Solution

Path1 {A, B, C, D, E, A} - length 24

P1

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Solution

Path2 {A, B, C, E, D, A} - length 31

P2 P1

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Solution

Path 3 {A,B,D,C,E,A} - length 21

P3 P2 P1

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Solution After perform sorting queue is

Hence Optimal Solution is Path 3

P2 P1 P3

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Pros and Cons An enhancement of backtracking However, it is much slower. Indeed, it often

leads to exponential time complexities in the worst case.

On the other hand, if applied carefully, it can lead to algorithms that run reasonably fast on average.

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Summery Thus Branch and Bound is:

a general search method. minimize a function f(x), where x is restricted to

some feasible region.

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THANK YOU


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