1. WHAT IS ASSIGNMENT PROBLEM ? Assignment problem refers to special class of linear programming problems that involves determining the most efficient assignment of people to projects, salespeople to territories, contracts to biddersand so on. It is often used to minimize total cost or time of performing task. One important characteristic of assignment problems is that only one job (or worker) is assigned to one machine (or project).

2. WHAT IS ASSIGNMENT PROBLEM ? Each Assignment problem has a Matrix associated with it. The number in the table indicates COST associated with the assignment. The most efficient linear programming algorithm to find optimum solution to an assignment problem is Hungarian Method (It is also know as Flood’s Technique).

3. ASSIGNMENT PROBLEM USING HUNGARIAN 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 earlier works of two Hungarian mathematicians: Denes Konig and Jeno Egervary.

4. REQUIREMENTS OF DUMMY ROWS AND COLUMN To arrive at the solution, assignment problems requires equal number of Row and Column. If the number of task that needs to be done exceeds the number of resource available, dummy row 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 non existent. Hence the value can be entered as Zeros.

5. Case Study: A company has a five job to be done by 5 workers each worker are assigned to one and only one job. Number of hours each worker takes to complete a job is given with AJ1J2J3J4J5 W12827243538 W22624233239 W31820223032 W42730252427 W52931284036

6. Step 1: Find the minimum element in each row and subtract it from all the elements of that particular row. ROW SUBTRACTION AJ1J2J3J4J5 W14 (28-24)3 (27-24)0 (24- 24)11(35-24)14(38-24) W23 (26-23)1 (24-23)0 (23-23)9 (32-23)16 (39-23) W30(18-18)2(20-18)4(22-18)12 (30-18)14 (32-18) W43(27-24)6 (30-24)1 (25-24)0 (24-24)3 (27-24) W51 (29-28)3 (31-28)0 (28-28)12 (40-28)8 (36-28)

7. Step 2: Find the minimum element in each column and subtract it from all the elements of that particular column. COLUMN SUBTRACTION BRANDM1M2M3M4M5 B14 (4-0)2 (3-1)0 (0-0)11 (11-0)11 (14-3) B23 (3-0)0 (1-1)0 (0-0)9 (9-0)13 (16-3) B30 (0-0)1 (2-1)4 (4-0)12 (12-0)11 (14-3) B43 (3-0)5 (6-1)1 (1-0)0 (0-0)0 (3-3) B51(1-0)2 (3-1)0 (0-0)12 (12-0)5 (8-3)

8. Remark: Step 1 and 2 creates at least one Zero ‘0’ in each row and Column.

9. Step 3: Starting from 1 st 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 1 st column, if there is exact one zero, make an assignment and cancel all zero’s in that row, and then draw a horizontal line. Continue this step till all zero’s are an assignment or Cancelled. W111 W2 J1J2J3J4J5 42011 3009 13 W3 0 141211 W43 0 0 W51 51205120 125

10. Step 3 (Continued): If optimal assignment is not formed go to step 4 i.e. There should be 5 straight lines. J1J1J2J2J3J3J4J4J5J5 W142 0 11 W23 0 0913 W3 0 141211 W4351 0 0 W5120125

11. Step 4: Ensure all zero’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). J1J2J3J4J5 W142011 W2300913 W30141211 W435100 W5120125

12. Step 5: Subtract the element (i.e. 5) from the elements not coveredby the line. Also add the same element (i.e. 5) to the elements which are at the intersections. J1J2J3J4J5 W14206 (11-5) W23004 (9-5)8 (13-5) W30147 (12-5)6 (11-5) W48 (3+5)10 (5+5)6 (1+5)00 W51207 (12-5)0 (5-5)

13. Follow step 3. Cover all zero's with straight lines again Since five lines are needed, an optimal assignment can be made. Assign: J1 – W3.18 J2 – W2.24 J3 – W1.24 J4 – W4.24 J5 – W5.36 Minimum Total Time in hours = 126 Hours. W1 W2 W3 J1J2J3J4J5 42066 30048 01476 W4 0 0 W5 81061208106120 7 0

14. Assign: J1 – W31818 J2 – W22424 J3 – W12424 J4 – W42424 J5 – W536 Minimum Total Time = 126 Hours. Hence, the optimum solution is UNIQUE.

15. MAXIMISATION PROBLEM Assignment problems can also be used solve cases of maximization 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. converting cost table in to opportunity loss table.). Later steps, are similar as the minimization problem.

16. SUMMARYSUMMARY Assignment problem deals 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 solvecases of maximization by converting cost table in to opportunity loss table.

17. DRAWBACK OF ASSIGNMENT PROBLEM 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 differences. 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 origin is also one. In addition we 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 peculiar to assignment problem only.