Assignment problem branch and bound.pptx

Assignment problem branch and bound.pptx

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  1 ASSIGNMENT PROBLEM BRANCH ANDBOUND
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  2 Branch & Bound The Branch & Bound algorithm is a method commonly used to solve optimization problems. Many problems can be solved using these two methods. Some examples of problems that the Branch & Bound algorithm can solve are Knapsack Problems, Traveling Salesman Problems, Scheduling Problems and many other optimization problems.  The Branch and Bound algorithm is an algorithm that uses a state space tree to solve a problem, in this case it is similar to the backtracking algorithm.
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  3 The way theB&B algorithm works is based on the following two principles.  1. Recursively divide the status space into smaller spaces and minimize “costs” on these spaces. This process is called branching.  2. Branching will be equivalent to brute-force enumeration. To improve performance, bound is used to limit the space in the status that is generated, eliminating candidate solutions that are proven to not contain optimal solutions. As the name implies, this algorithm has a bound or limiting function. This constraint function is useful for delimiting paths that are considered not leading to a solution node. WORKING OF B&B
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  4 Job Assignment Problem oJob Assignment Problem is one of the fundamental combinatorial optimization problems in its most common form. Examples of problems have a number of people and a number of jobs. Each person can be assigned to do any job, which has different costs depending on the job. The goal is to do as many jobs as possible by assigning one person to each job and one job per person, in such a way that the total cost is minimized. o There are many methods that can be used to solve Job Assignment Problems.
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  5 Example  The jobassignment problem testing will be carried out in one example of the following cases, namely there are 4 jobs and 4 people, each of which has a cost as in table 1.  Table 1 Job Assignment Problem matrix (4 jobs and 4 people) Job 1 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6
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  6 The assignment problem SolvingJob Assignment Problems using Branch and Bound is done by determining the lower limit by adding the minimum cost of each row Minimum Cost of each Row Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 So that Lower Bound (LB) = 4+5+3+6 = 18
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  7 The assignment problem Thereare 4 possibilities, namely A doing Job 1, A doing Job 2, A doing Job 3 and A doing Job 4. Calculate the LB of each of these possibilities. Possibility 1: Person 1 (A) does Job 1 Person 1 (A) does Job 1 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 1 is : 11+5+3+6 = 25
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  8 The assignment problem Possibility2: Person 1 (A) does Job 2 Person 1 (A) does Job 2 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 2 is : 4+5+3+6 = 18
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  9 The assignment problem Possibility3: Person 1 (A) does Job 3 Person 1 (A) does Job 3 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 3 is : 9+6+7+6 = 28
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  10 The assignment problem Possibility4: Person 1 (A) does Job 4 Person 1 (A) does Job 4 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 4 is : 10+5+3+8 = 26
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  11 The assignment problem thenode with the minimum value to be expanded is selected, namely A2 with LB = 18. Then the second person (B) is chosen to do the job or assignment.
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  12 The assignment problem Thereare 3 possibilities, namely B doing job 1, Job 3 or Job 4 (Job 2 is done by A) Possibility 1: Person 2 (A) does Job 1 Person 2 (B) does Job 1 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 1 is :4+8+3+6 = 21
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  13 The assignment problem Possibility2: Person 2 (A) does Job 3 Person 2 (B) does Job 3 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 2 is :4+5+7+6 = 22
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  14 The assignment problem Possibility3: Person 2 (A) does Job 4 Person 2 (B) does Job 4 Job 1 Job 2 Job 3 Job 4 Person 1 (A) 11 4 9 10 Person 2 (B) 8 6 5 9 Person 3 (C) 7 10 3 10 Person 4 (D) 9 8 11 6 Then LB for probability 3 is :4+9+3+9 = 25
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  15 The assignment problem Next,select the node with the minimum cost to expand, namely B→1
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  16 The assignment problem Next2 possibilities:  Possibility 1: A does Job 2, B does Job 1, C does Job 3 and D does Job 4 with the cost is 4 + 8 + 3 + 6 = 21.  Possibility 2: A does Job 2, B does Job 1, C does Job 4 and D does Job 3 with the cost is 4 + 8 + 10 + 11 = 33. Then the possibility of 1 is chosen with cost = 21, compared to all the remaining nodes, this cost is the minimum cost so that it is chosen as the solution.
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  17 The assignment problem
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  18 The assignment problem SOLUTION: A2 B1 C3 D4
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  19 The assignment problem Conclusion Theuse of branch and bound results in a shorter solution because the branch and bound algorithm performs calculations recursively, while still calculating the best value, that is what is known as branching. In addition, the best value is recorded for each calculation, so that it can improve the performance of the algorithm, what is known as bounding.