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Solve the Following Minimal Assignment Problem :

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Question

Solve the following minimal assignment problem : 

Machines A B C D E
M1 27 18 ∞ 20 21
M2 31 24 21 12 17
M3 20 17 20 ∞ 16
M4 21 28 20 16 27
Sum
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Solution

Machines A B C D E
M1 27 18 __ 20 21
M2 31 24 21 12 17
M3 20 17 20 __ 16
M4 21 28 20 16 27

Step 1 : We introduce a dummy machine M2 with time zero for each job and replace '-' by ∞

  A B C D E
M1 27 18 ∞ 20 21
M2 31 24 21 12 17
M3 20 17 20 ∞ 16
M4 21 28 20 16 27
M5 0 0 0 0 0

Step 2: Subtracting minimum element of each row from all its elememts 

As minimum number of lines = 4 ≠ order of the matrix.
. . Optimal solution is not reached.
Step 3 : We subtract minimum element (from uncovered elements) from each uncovered element and add to intersection elements.

Minimum number of lines= 5
= order of matrix.
:. Optimum solution is reached.
Step 4 : Making assignment at single zero of the row and of the column.

. . The optimal assignment of jobs to machines is

M1 → B       M2 → D     M3 → E      M4 → C     M5 → A

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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.

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(Processing cost in ₹)
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P 6 0 8 4
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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.

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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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P II `square`
Q `square` 21
R I `square`
S III 34

Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`


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