Assignment Problem Dr. Ramesh Kumar Chaturvedi

UNIT II

What an Assignment problem is?

• When we want to solve a linear programming problem with special characteristic such as a Square Matrix (i.e. No. of destinations are same as no. of sources)

• When each source (resource facility) can be associated with one and only one job/destination (One job to one machine)

• Association is to be done in such a way to maximize or minimize the total effectiveness.

Few examples of Assignment problem

• Assigning sales man to sales territory • Vehicles and drivers to different routes • Product to factories • Jobs to a machines • Contract to Bidders • Research problem to teams

Can you think of Problem that suits to Assignment

methodology?

A case Assignment problem

Mettalica Works Ltd. Is a metal working machine shop. It generally does work on contract basis. At present it has four jobs – A,B, C, C in hand. Mettalica has four machines – 1,2 ,3 and 4. The jobs can be performed on all machines. However the cost of performing a job on different machines vary because of capacity machine and characteristics of the job. The jobs are to be exclusively performed on any one machine as the setup cost is too expensive to permit setting of an other machine for same job. An estimate has been made of the cost of performing the job on each machine and estimated cost data is tabulated below:

Tables of Job –Machines Table 1

JOB/ Machine

1 2 3 4

A 20 36 31 17

B 24 32 40 12

C 22 40 38 18

D 36 39 35 16

The above matrix highlights that if job A is performed out on machine 1, it would cost Rs. 20, if performed on Machine 2 the cost would be Rs. 36, for Machine cost would be Rs. 31 and on Machine

4 cost would be RS 17.

Matellica would like to schedule the jobs on machine so that over all cost is minimum.

Back to final Assignment

The Hungarian method of solving the Assignment Problem

• Step 1: Determine the opportunity Cost table

from Table 1.

a. Subtract the lowest entry in each row with all the entries in that row.

b. Subtract the lowest entry in a column in resulting table from a above with all entries in that column.

The resultant Table is Opportunity cost table.

Row opportunity cost table

JOB/ Machine

1 2 3 4

A 3 (20-17) 19 (36-17) 14 0

B 12 20

28 0

C 4 22 20 0

D 20 23 19 0

Opportunity cost Tables

JOB/ Machine

1 2 3 4

A 0 (3 – 3) 0 (19 -19) 0 0

B 9 (12 – 3) 1

14 0

C 1 3 6 0

D 17 4 5 0

Lowest element of the column is subtracted from all other elements of column of row opportunity cost table

Step 2

• Determine weather an optimal assignment can be made( This can be done by assigning all jobs to machines where opportunity cost is 0)

Check of Optimality: Draw straight line vertically and horizontally

through the total opportunity cost table in such a manner that all Zero cells in the matrix are covered with minimum number of lines.

If number of line is equal to n (for an n X n matrix) or equal to number of rows or columns; then the optimal solutions is arrived.

Opportunity cost Tables

JOB/ Machine

1 2 3 4

A 0 0 0 0

B 9 1

14 0

C 1 3 6 0

D 17 4 5 0

There are only Two lines that could cover all 0’s in cells. But we have 4x4 matrix and for optimal solution number of lines should be equal to n (i.e. 4) Hence we can not make an optimal assignment.

Other method of drawing minimum number of straight lines

• Make assignment by selecting row/column with only one zero.

• Mark (√) the rows for which assignment has not been made.

• Mark (√) the column which has zeros in marked rows.

• Mark (√) the rows which have assignments in marked column

• Repeat sub steps above till no more marking is possible

• Draw lines through all unmarked rows and marked columns.

Step 3

Revise the total opportunity cost table as follows:

a. Select the smallest number in the table that is not covered by any straight line and subtract it from all other numbers not covered by any straight line.

b. Add this number to number lying at the intersection of any two lines. A new opportunity cost table is obtained.

C. Go to step 2 with this new opportunity cost table

New Opportunity cost Tables

JOB/ Machine

1 2 3 4

A 0 0 0 0 Now 1 (0 + 1)

B 8 (9 -10= 1 (1 – 1) = 0

13 (14-1) 0

C 0 (1-1) 2 (3 -1) 5 (6-1) 0

D 16 (17-1) 3 (4-1) 4 (5-1) 0

This is the smallest number in table not covered by any line; it is to be subtracted from all other numbers not covered by any line.

This the number at intersection of two lines, add 1 to it

Revisited Step 2

• Determine weather an optimal assignment can be made

Check of Optimality:

Draw straight line vertically and horizontally through the total opportunity cost table in such a manner that all Zero cells in the matrix are covered with minimum number of lines.

If number of line is equal to n (for an n X n matrix) or equal to number of rows or columns; then the optimal solutions is arrived.

New Opportunity cost Tables

JOB/ Machine

1 2 3 4

A 0 0 0 1

B 8 0

13 0

C 0 2 5 0

D 16 3 4 0

Now number of lines is equal to 4, equal to the order of the matrix;

hence a optimal solution is obtained.

Step 4

Making the optimal Assignment: a. Select the a row with only one ‘0’ in it. Make an Assignment

by marking‘ ’ around ‘0’ of this row. After making assignment cross ’X’ out all other ‘0’s in the corresponding column. Repeat this step for all remaining rows that fulfil ‘only one zero’ condition.

b. If there is no row (left) with zero, repeat the step a for column. Cross ’X’ all other ‘0’s in corresponding Row.

C. If all Jobs are yet not assigned and there is no Row or Column with only one zero, then chose a row or column arbitrarily and make an allocation to this row/column. Cross ’X’ all other ‘0’s in the same row as well as same column.

Making Optimal assignment JOB/ Machine

1 2 3 4

A 0 0 0 1

B 8 0

13 0

C 0 2 5 0

D 16 3 4 0

1. Row D has only one “0”, so make assignment to it (D4). And cross all other 0 in the corresponding column.

2. After Assignment of D on machine 4, now Row B and C both have only one Zero. So assignment are made to these rows ( C1 and B2) and

zeros in columns are crossed. 3. Finally only one Zero is left in Row A where Machine three has a Zero

so a job A is assigned to Machine 3 (A3).

0

0

0

0

Final assignment is as follows: JOB/ Machine

1 2 3 4

A 0 0 1

B 8

13 0

C 2 5 0

D 16 3 4

1. Cot of performing 4 jobs on 4 machines is : (See original cost Matrix) A3 = 31 B2=32 C1=22 D4=16

Total = 101

0

0

0

0

Maximization Problem

• Just Subtract all element of Cost Matrix from the largest Number in the matrix to; What we get is a loss matrix.

• Than proceed with steps for a normal minimization method.

Practice

M1 M2 M3 M4

A 5 7 11 6

B 8 5 9 6

C 4 7 10 7

D 10 4 8 3

How A,B,C, and D jobs be assigned to 4 machines M1 to M4 in order to minimize the cost?

Opportunity cost Matrix

M1 M2 M3 M4

A 0 2 2 1

B 3 0 0 1

C 0 3 2 3

D 7 1 1 0

1. 0 in red are assignments.

2nd FBS

M1 M2 M3 M4

A 0 1 1 0

B 4 0 0 1

C 0 2 1 2

D 8 1 1 0

1. 0 in red are assignments.

2nd FBS

M1 M2 M3 M4

A 0 0 0 0

B 5 0 0 2

C 0 1 0 2

D 8 0 0 0

1. 0 in red are assignments.

2. Soln. Is 23.