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Question
A departmental head has three jobs and four subordinates. The subordinates differ in their capabilities and the jobs differ in their work
contents. With the help of the performance matrix given below, find out which of the four subordinates should be assigned which jobs ?
Subordinates | Jobs | ||
I | II | III | |
A | 7 | 3 | 5 |
B | 2 | 7 | 4 |
C | 6 | 5 | 3 |
D | 3 | 4 | 7 |
Solution
As the problem is unbalanced a dummy · job IV is introduced with zero performance as, the for:
I | II | III | IV | |
A | 7 | 3 | 5 | 0 |
B | 2 | 7 | 4 | 0 |
C | 6 | 5 | 3 | 0 |
D | 3 | 4 | 7 | 0 |
As minimum element of each row is 'O' hence minimum element of each column is subtracted from all the elements of that column,
I | II | III | IV | |
A | 5 | 0 | 2 | 0 |
B | 0 | 4 | 1 | 0 |
C | 4 | 2 | 0 | 0 |
D | 1 | 1 | 4 | 0 |
All the zeros of the above matrix a re covered with minimum number of lines is as below :
As the number of lines covering zeros is equal to the number of rows and columns. hence the optional solution is obtained it is as shown below :
As D gets a dummy job N, hence the assignment is as shown below :
A → Job II
B → Job I
C → Job III
and performance is 3 + 2 + 3 = 8 units
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|||
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Solution:
Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :
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|||
I | II | III | IV | |
P | 6 | 0 | 8 | 4 |
Q | 4 | 3 | 2 | 0 |
R | 0 | 2 | 4 | 5 |
S | 4 | 2 | 0 | 6 |
Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.
Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:
Job | Machines (Processing cost in ₹) |
|||
I | II | III | IV | |
P | 6 | 0 | 8 | 4 |
Q | 4 | 3 | 2 | 0 |
R | 0 | 2 | 4 | 5 |
S | 4 | 2 | 0 | 6 |
Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.
Examine the rows one by one starting with the first row with exactly one zero is found. Mark the zero by enclosing it in (`square`), indicating assignment of the job. Cross all the zeros in the same column. This step is shown in the following table :
Job | Machines (Processing cost in ₹) |
|||
I | II | III | IV | |
P | 6 | 0 | 8 | 4 |
Q | 4 | 3 | 2 | 0 |
R | 0 | 2 | 4 | 5 |
S | 4 | 2 | 0 | 6 |
Step 5: It is observed that all the zeros are assigned and each row and each column contains exactly one assignment. Hence, the optimal (minimum) assignment schedule is :
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P | II | `square` |
Q | `square` | 21 |
R | I | `square` |
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Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`