ASSIGNMENT PROBLEM

1. Gurmeet Singh, Roll No: 9

2. Jyoti Singh, Roll No: 10

3. Nakul Bhardwaj, Roll No: 15

4. Prakalp Vora, Roll No: 17

5. Vicky Shah, Roll No: 31

6. Yashwant Kharat, Roll No: 34

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 bidders and 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).

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).

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.

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.

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

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

Step 1:Find the minimum element in each row and subtract it from all the elements of that particular row.

ROW SUBTRACTION

A J1 J2 J3 J4 J5

W1 4 (28-24) 3 (27-24) 0 (24- 24) 11(35-24) 14(38-24)

W2 3 (26-23) 1 (24-23) 0 (23-23) 9 (32-23) 16 (39-23)

W3 0(18-18) 2(20-18) 4(22-18) 12 (30-18) 14 (32-18)

W4 3(27-24) 6 (30-24) 1 (25-24) 0 (24-24) 3 (27-24)

W5 1 (29-28) 3 (31-28) 0 (28-28) 12 (40-28) 8 (36-28)

Step 2:Find the minimum element in each column and subtract it from all the elements of that particular column.

COLUMN SUBTRACTION

BRAND M1 M2 M3 M4 M5

B1 4 (4-0) 2 (3-1) 0 (0-0) 11 (11-0) 11 (14-3)

B2 3 (3-0) 0 (1-1) 0 (0-0) 9 (9-0) 13 (16-3)

B3 0 (0-0) 1 (2-1) 4 (4-0) 12 (12-0) 11 (14-3)

B4 3 (3-0) 5 (6-1) 1 (1-0) 0 (0-0) 0 (3-3)

B5 1(1-0) 2 (3-1) 0 (0-0) 12 (12-0) 5 (8-3)

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

Step 3:Starting from 1st 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 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.

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

Step 3 (Continued): If optimal assignment is not formed go to step 4i.e. There should be 5 straight lines.

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

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).

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

Step 5: Subtract the element (i.e. 5) from the elements not covered by the line. Also add the same element (i.e. 5) to the elements which are at the intersections.

J1 J2 J3 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)

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. 18J2 – W2. 24J3 – W1. 24J4 – W4. 24J5 – W5. 36

Minimum Total Time in hours = 126 Hours.

J1 J2 J3 J4 J5

W1 4 2 0 6 6

W2 3 0 0 4 8

W3 0 1 4 7 6

W4 8 10 6 0 0

W5 1 2 0 7 0

Assign: J1 – W3 18J2 – W2 24J3 – W1 24J4 – W4 24J5 – W5 36

Minimum Total Time = 126 Hours.Hence, the optimum solution is UNIQUE.

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.

SUMMARY

• 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 solve cases of maximization by converting cost table in to opportunity loss table.

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.

THANK YOU!