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2nd M.E.(C.S.E)

Solving traveling salesman and water jug

problem using Branch and Bound Technique

Introduction

Branch and bound is a systematic method for solving optimization problems that applies where the greedy method and dynamic programming fail.

Branch-and-bound is an approach developed for solving discrete and combinatorial optimization problems.

Problem The discrete optimization problems are problems in which the decision variables assume discrete values from a specified set.

The combinatorial optimization problems, on the other hand, are problems of choosing the best combination out of all possible combinations. The major difficulty with these problems is that we do not have any optimality conditions to check if a given (feasible) solution is optimal or not. In order to guarantee a given feasible solution's optimality is "to compare" it with every other feasible solution.

To do this explicitly, amounts to total enumeration of all possible alternatives

which is computationally prohibitive due to the NP-Completeness. Therefore, this comparison must be done implicitly, resulting in partial enumeration of all possible alternatives.

Solution Branch and bound methods solve a discrete optimization problem by breaking up

its feasible set into successively smaller subsets, calculating bounds on the objective function value over each subset, and using them to discard certain subsets from further consideration.

The procedure ends when each subset has either produced a feasible solution, or was shown to contain no better solution than the one already in hand. The best solution found during the procedure is a global optimum.

The general idea of B&B is a BFS-like search for the optimal solution, but not all nodes get expanded (i.e., their children generated). Rather, a carefully selected criterion determines which node to expand and when, and another criterion tells the algorithm when an optimal solution has been found.

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

discrete programming

2 Mehta Ishani

2nd M.E.(C.S.E)

Steps for branch and bound algorithm 1. 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.

Water Jug Now let us see how the branch - and -bound search could be used to find the

shortest solution to the water- jug problem. Unfortunately although this algorithm is more efficient and simple. It requires exponential time. The exact amount of time it saves for particular problem depends on the order in which the paths are explored. And also in this problem, there is a possibility that the algorithm enters into an indefinite loop. For example, consider the following configuration generated in the process of generating a path. Initial state (0,0) Goal state (2,0) Path 1: (0,0) -> (4,0) -> (1,3) -> (1,0) -> (0,1) -> (4,1) -> (2,3) -> (2,0) Since goal node is reached so put in queue 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

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2nd M.E.(C.S.E)

Path 3: (0,0) -> (0,3) -> (3,3) -> (4,2) -> (0,2) -> (2,0) Goal node is reached then put in queue 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

The Traveling Salesman Problem The traveling salesman problem consists of a salesman and a set of cities. The salesman has to visit each one of the cities starting from a certain one (e.g. the hometown) and returning same city. The challenge of the problem is that the traveling salesman wants to minimize the total length of the trip.

Why the Traveling Salesman Problem?

NP-hard

Brute Force takes O(n!) to complete

Algorithms to solve take one of two forms:

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

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

Many real world applications:

Special properties of the Traveling Salesman Problem that make it suitable for Branch and Bound:

o As you build your solution, cost increases. o Partial solutions have valid costs (lower bound costs)

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Solution steps

o Given a complete, weighted graph on n nodes, find the least weight Hamiltonian cycle, a cycle that visits every node once.

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

o Finding all the Hamiltonian cycles of a graph takes exponential time. Therefore,

TSP is in the class NP.

For example the path Path1 {A, B, C, D, E, A} - length 24 Path2 {A, B, C, E, D, A} - length 31 Path 3 {A,B,D,C,E,A} - length 21 Hence our queue is

After sorting queue is Hence Optimal Solution is Path 3

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.

Summery

Thus Branch and Bound is: o a general search method. o minimize a function f(x), where x is restricted to some feasible region.

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