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ASSIGNMENT PROBLEM
2. 1 Introduction
The Assignment Problem can define as follows:
Given n facilities, n jobs and the effectiveness of each facility to each job, here the problem is to
assign each facility to one and only one job so that the measure of effectiveness if optimized.
Here the optimization means Maximized or Minimized. There are many management problems has a
assignment problem structure. For example, the head of the department may have 6 people available
for assignment and 6 jobs to fill. Here the head may like to know which job should be assigned to
which person so that all tasks can be accomplished in the shortest time possible. Another example a
container company may have an empty container in each of the location 1, 2 , 3 , 4 , 5 and requires an
empty container in each o f the locations 6, 7, 8 , 9 ,10. It would like to ascertain the assignments of
containers to various locations so as to minimize the total distance. The third example here is, a
marketing set up by making an estimate of sales performance for different salesmen as well as for
different cities one could assign a particular salesman to a particular city with a view to maximize the
overall sales.
Note that with n facilities and n jobs there are n! possible assignments. The simplest way of
finding an optimum assignment is to write all the n! possible arrangements, evaluate their total cost
and select the assignment with minimum cost. Bust this method leads to a calculation problem of
formidable size even when the value of n is moderate. For n=10 the possible number of arrangements is
2. 2 Assignment Problem Structure and Solution
The structure of the Assignment problem is similar to a transportation problem, is as follows: Jobs 1 2 … n
Workers
n
c 11 c 12 … c 1 n 1
c 21 c 21 … c 2 n 1
cn 1 cn 2 … cnn 1
Step 1:
From the given problem, find out the cost table. Note that if the number of origins is not equal to
the number of destinations then a dummy origin or destination must be added.
Step 2:
In each row of the table find out the smallest cost element, subtract this smallest cost element from
each
element in that row. So, that there will be at least one zero in each row of the new table. This new
table is known as First Reduced Cost Table.
Step 3:
In each column of the table find out the smallest cost element, subtract this smallest cost element
from each element in that column. As a result of this, each row and column has at least one zero
element. This
new table is known as Second Reduced Cost Table.
Step 4:
Now determine an assignment as follows:
- For each row or column with a single zero element cell that has not be assigned or eliminated, box that zero element as an assigned cell.
- For every zero that becomes assigned, cross out all other zeros in the same row and for column.
- If for a row and for a column there are two or more zero and one can’t be chosen by inspection, choose the assigned zero cell arbitrarily.
- The above procedures may be repeated until every zero element cell is either assigned (boxed) or crossed out.
Step 5:
An optimum assignment is found, if the number of assigned cells is equal to the number of rows (and
columns). In case we had chosen a zero cell arbitrarily, there may be an alternate optimum. If no
optimum solution is found i.e. some rows or columns without an assignment then go to Step 6.
Step 6:
Draw a set of lines equal to the number of assignments which has been made in Step 4 , covering all
the zeros in the following manner
- Mark check (√) to those rows where no assignment has been made.
- Examine the checked (√) rows. If any zero element cell occurs in those rows, check (√) the
respective columns that contains those zeros.
- Examine the checked (√) columns. If any assigned zero element occurs in those columns,
check (√) the respective rows that contain those assigned zeros.
- The process may be repeated until now more rows or column can be checked.
- Draw lines through all unchecked rows and through all checked columns.
Step 7:
Examine those elements that are not covered by a line. Choose the smallest of these elements and
subtract this smallest from all the elements that do not have a line through them.
Add this smallest element to every element that lies at the intersection of two lines. Then the resulting matrix is a new revised cost table.
Example 2.1:
Problem A work shop contains four persons available for work on the four jobs. Only one person can work on any one job. The following table shows the cost of assigning each person to each job. The objective is to assign person to jobs such that the total assignment cost is a minimum.
Jobs
1 2 3 4
A
Persons B
C
D
Solution
As per the Hungarian Method
Step 1: The cost Table
A
Persons B
C
D
Jobs 1 2 3 4
Step 2: Find the First Reduced Cost Table
A
Persons B
C
D
Jobs 1 2 3 4
Step 3 : Find the Second Reduced Cost Table
A
Persons B
C
D
Jobs 1 2 3 4
Step 4 : Determine an Assignment
By examine row A of the table in Step 3, we find that it has only one zero (cell A1) box this zero
and cross out all other zeros in the boxed column. In this way we can eliminate cell B1.
Now examine row C, we find that it has one zero (cell C2) box this zero and cross out (eliminate)
the zeros in the boxed column. This is how cell D2 gets eliminated.
There is one zero in the column 3. Therefore, cell D3 gets boxed and this enables us to eliminate cell
D4.
Therefore, we can box (assign) or cross out (eliminate) all
zeros. The resultant table is shown below:
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Step 7: Develop the new revised table.
Examine those elements that are not covered by a line in the table given in Step 6. Take the smallest
element in this case the smallest element is 1. Subtract this smallest element from the uncovered
cells and add 1 to elements (C 1 and D1) that lie at the intersection of two lines. Finally, we get the
new revised cost table, which is shown below:
Jobs
1 2 3 4
A
Persons B
C
D
Step 8:
Now, go to Step 4 and repeat the procedure until we arrive at an optimal solution (assignment).
Step 9: Determine an assignment
Examine each of the four rows in the table given in Step 7 , we may find that it is only row C which has only one zero box this cell C 2 and cross out D 2.
Note that all the remaining rows and columns have two zeros. Choose a zero arbitrarily, say A 1 and box this cell so that the cells A 3 and B 1 get eliminated.
Now row B (cell B4) and column 3 (cell D4) has one zero box these cells so that cell D4 is eliminated.
Thus, all the zeros are either boxed or eliminated. This is shown in the following table
Example 2.2:
Solve the following unbalanced assignment problem of minimizing the total time for performing all the jobs. Jobs
1 2 3 4 5
A
B
Workers C
D
E
F
Solution
In this problem the number of jobs is less than the number of workers so we have to introduce a dummy job with zero duration.
The revised assignment problem is as follows:
Jobs
1 2 3 4 5 6
A
B
Workers C
D
E
F
Now the problem becomes balanced one since the number of workers is equal to the number jobs. So that the problem can be solved using Hungarian Method.
Step 1: The cost Table
Jobs
1 2 3 4 5 6
A
Workers B
C
D
E
F
Step 4 : Determine an Assignment
By using the Hungarian Method the assignment is made as follows:
Jobs 1 2 3 4 5 6
A
B
Workers C
D
E
F
0
Step 5:
The solution obtained in Step 4 is not optimal. Because we were able to make five
assignments when six were required.
Step 6:
Cover all the zeros of the table shown in the Step 4 with five lines (since already we made
five assignments).
Check row E since it has no assignment. (^0) te that row B has a zero in column 6, therefore check
column6. Then we check row C since it has a zero in column 6. Note that no other rows and columns
are checked. Now we may draw five lines through unchecked rows (row A, B, D and F) and the
checked column (column 6 ). This is shown in the table given below:
No
Jobs 1 2 3 4 5 6
A
B^2 4 4 2
Workers C
D
E^6 1 4 6
0
F^1 4 3 3
Step 7:
0
Develop the new revised table.
Examine those elements that are not covered by a line in the table given in Step 6. Take the
smallest element in this case the smallest element is 1. Subtract this smallest element from
the uncovered cells and add 1 to elements (A6, B6, D6 and F6) that lie at the intersection of two
lines. Finally, we get the new revised cost table, which is shown below:
Jobs 1 2 3 4 5 6
0
Since the number of assignments equal to the number of rows (columns), the assignment shown in
the above tale is optimal.
Thus, the worker A is assigned to Job4, worker B is assigned to job 1, worker C is assigned
to job 6, worker D is assigned to job 5, worker E is assigned to job 2, and worker F is assigned to
job 3. Since the Job 6 is dummy so that worker C can’t be assigned.
The total minimum time is: 14 that is A 4 + B 1 + D5 + E 2 + F 3
Example 2.3:
A marketing company wants to assign three employees viz. A, B, and C to four offices located at
W, X, Y and Z respectively. The assignment cost for this purpose is given in following table.
Offices W X Y Z
A
Employees B
C
Solution
Since the problem has fewer employees than offices so that we have introduce a
dummy employee with zero cost of assignment.
The revised problem is as follows:
Offices W X Y Z
A
Employees B
C D
Now the problem becomes balanced. This can be solved by using Hungarian Method as in the case of Example 2 .2. Thus as per the Hungarian Method the assignment made as follows:
Employee A is assigned to Office X, Employee B is assigned to Office Z, Employee C is assigned to Office W and Employee D is assigned to Office Y. Note that D is empty so that no one is assigned to Office Y.
The minimum cost of assignment is: 220 + 160 + 100 = 480
2. 4 Infeasible Assignment Problem
Sometimes it is possible a particular person is incapable of performing certain job or a specific job can’t be performed on a particular machine. In this case the solution of the problem takes into account of these restrictions so that the infeasible assignment can be avoided.
The infeasible assignment can be avoided by assigning a very high cost to the cells where assignments are restricted or prohibited. This is explained in the following Example 2 .4.
Example 2.4:
A computer centre has five jobs to be done and has five computer machines to perform them. The cost of processing of each job on any machine is shown in the table below.
Jobs 1 2 3 4 5
