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assignment problem (1).pptx case study 1
- 1. Assignment problem SUBMITTED BY: PRADEEPAA I.MCOM 25PACOM025
- 2. What is assignmentproblem? Assignment problem refers to special class of linear programming problems that involves determining the most efficient assignment of people to project , salespeople to territories contracts to bidders and so on. It is often used to minimize the total cost or time of performing the task. One of the important characteristics of assignment problem is that only one job is assigned to one machine.
- 3. Each assignmentproblem has a matrix associated with it. The number in the table indicates the COST associated with the assignment. The most efficient linear programming algorithm to find optimum solution to the assignment problem is Hungarian method.
- 4. Assignment problem usingHungarian Method The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal- dual methods. It was developed and published in 1955 by Harold Kuhn, who gave the name “Hungarian method” because the algorithm was largely based on the earilier works of two Hungarian mathematicians: Denes Konig and Jeno Egervary.
- 5. Requirements of dummyrows and column: To arrive at the solution, assignment problem requires equal number of Row and Column. If the number of tasks that needs to be done exceeds the number of resources available, dummy rows or a column just needs to be added as the case may be. This creates a table of equal dimensions. The dummy row or column is not existent. Hence the value can be entered as zeros.
- 6. A companyhas five job to be done by five workers each worker are assigned to be one and only one job. Number of hours each worker takes to complete the job is given below: A J1 J2 J3 J4 J5 W1 28 27 24 35 38 W2 26 24 23 32 39 W3 18 20 22 30 32 W4 27 30 25 24 27 W5 29 31 28 40 36
- 7. Step 1: Find theminimum number in each row and subtract it from all the element in that particular row A J1 J2 J3 J4 J5 W1 4 3 0 11 14 W2 3 1 0 9 16 W3 0 2 4 12 14 W4 3 6 1 0 3 W5 1 3 0 12 8
- 8. Step 2: Find theminimum element in each column and subtract it from all ihe elements in that particular column. A J1 J2 J3 J4 J5 W1 4 2 0 11 11 W2 3 0 0 9 13 W3 0 1 4 12 11 W4 3 5 1 0 0 W5 1 2 0 12 5
- 9. Remark: Step 1and 2 creates at least one zero ‘0’in each row and column.
- 10. Step 3: Starting from1st row ,if there is exact one zero,make an assignment and cancel all zero’s in that column and then draw a vertical line.Similarly, starting from 1st column, if exact one zero, make an assignment and cancel an zero’s in that row, and then draw a horizontal line. Continue this step till all zero’s are an assignment or cancelled.
- 11. A J1 J2J3 J4 J5 W1 4 2 11 11 W2 3 9 13 W3 1 4 12 11 W4 3 5 1 W5 1 2 12 5 0 0 0 0 0 0 0
- 12. Step 3(Continued): If optimalassignment is not formed go to step 4 i.e. There should be 5 straight lines.
- 13. A J1 J2J3 J4 J5 W1 4 2 11 11 W2 3 9 13 W3 1 4 12 11 W4 3 5 1 W5 1 2 12 5 0 0 0 0 0 0 0
- 14. Step 4: Ensure allzero’s “0” are covered with minimum one line. Find the minimum element not covered by any line (in this sum “5” is the minimum element not covered by any line).
- 15. A J1 J2J3 J4 J5 W1 4 2 11 11 W2 3 9 13 W3 1 4 12 11 W4 3 5 1 W5 1 2 12 5 0 0 0 0 0 0 0
- 16. Step 5: Subtract theelement (i.e.5) from the elements not covered by the lines. Also add the same element (i.e.5) to the elements which are at the intersections.
- 17. A J1 J2J3 J4 J5 W1 4 2 0 6(11-5) 6(11-5) W2 3 0 0 4(9-5) 8(13-5) W3 0 1 4 7(12-5) 6(11-5) W4 8(3+5) 10(5+5) 6(1+5) 0 0 W5 1 2 0 7(12-5) 0(5-5)
- 18. Follow step 3: Coverall zero’s with straight lines again. Since five lines are needed,an optimal assignment can be made.
- 19. A J1 J2J3 J4 J5 W1 4 2 6 6 W2 3 4 8 W3 1 4 7 6 W4 8 10 6 W5 1 2 7 Assign: J1-W3=18 J2-W2=24 J3-W1=24 J4-W4=24 J5-W5=36 Minimum total time in hours =126 hours 0 0 0 0 0 0 0 0
- 20. Assign: J1-W1=18 J2-W2=24 J3-W3=24 J4-W4=24 J5-W5=36 Minimum Total Time=126Hours. Hence,the optimum solution is UNIQUE.
- 21. MAXIMISATION PROBLEM Assignmentproblems can also be used to solve casas of maximisation model. For instance, Travelling Salesman problem, milk van routings and so on. Problem can be solved by first subtracting the biggest element in the problem from all other elements (i.e. connverting cost table in to opportunity loss table. Later steps, are similar as the minimization problem.
- 22. SUMMARY Assignment problemdeals with the problem of assigning jobs to machines or men to jobs which are to be performed with varying efficiency. It can be used to solve cases of Travelling Salesman problem. It is basically a minimizing model but it can be used to solve cases of maximization by converting cost table in to opportunity loss table.
- 23. DRAWBACK OF ASSIGNMENTPROBLEM Assignment becomes a problem because each job requires different skills and the capacity or efficiency of each person with respect to these jobs can be different. This gives rise to cost difference. If each person is able to do all jobs equally efficiently then all costs will be the same and each job can be assigned to any person. When assignment is a problem it becomes a typical optimization problem it can therefore be compared to a transportation problem. The cost elements are given and is a square matrix and requirement at each destination is one and availability at each orgin is also one.
- 24. In additionwe have number of origins which equals the number of destinations hence the total demand equals total supply. There is only one assignment in each row and each column. However if we compare this to a transportation problem we find that a general transportation problem does not have the above mentioned limitations. These limitations are pecullar to assignment problem only.