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Question
Suggest optimum solution to the following assignment. Problem, also find the total minimum service time.
Service Time ( in hrs.)
Counters | Salesmen | |||
A | B | C | D | |
W | 41 | 72 | 39 | 52 |
X | 22 | 29 | 49 | 65 |
Y | 27 | 39 | 60 | 51 |
Z | 45 | 50 | 48 | 52 |
Solution
This problem is already 4 x 4.
Select the smallest element in each row and subtract it from every element in each row :
Select the smallest element in each column of the above matrix and subtract it form every element in that column.
Draw the minimum lines covering all zeros.
Minimum lines covering all zeros is not equal to the order of the matrix.
Minimum uncovered value 2 is subtracted from uncovered values and added to values at intersection of the lines.
Draw minimum lines covering all the zeros.
Minimum lines covering all the zeros equal to the order of the matrix.
∴ Allocation of counters can be done.
The allocation of the counters to the salesmen is
W → C, X → B, Y → A, Z → D
The minimum time = 39 + 29 + 27 +52 = 147 Hrs.
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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:
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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 :
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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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Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`