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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

Find the optimal solution for the assignment problem with the following cost matrix. Area 1 2 3 4 P 11 17 8 16 Salesman Q 9 7 12 6 R 13 16 15 12 S 14 10 12 11

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Question

Find the optimal solution for the assignment problem with the following cost matrix.

    Area
    1 2 3 4
  P 11 17 8 16
Salesman Q 9 7 12 6
  R 13 16 15 12
  S 14 10 12 11
Chart
Sum
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Solution

Here the number of rows and columns are equal.

∴ The given assignment problem is balanced.

Step 1: Select the smallest element in each row and subtract this from all the elements in its row.

    Area
    1 2 3 4
  P 3 9 0 8
Salesman Q 3 1 6 0
  R 1 4 3 0
  S 4 0 2 1

Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

    Area
    1 2 3 4
  P 2 9 0 8
Salesman Q 2 1 6 0
  R 0 4 3 0
  S 3 0 2 1

Step 3: (Assignment)
Examine the rows with exactly one zero. Mark the zero by □ Mark other zeros in its column by X

    Area
    1 2 3 4
  P 2 9 0 8
Salesman Q 2 1 6 0
  R 0 4 3 0
  S 3 0 2 1

Thus all the four assignments have been made.

The optimal assignment schedule and total cost.

Salesman Area Cost
P 3 8
Q 4 6
R 1 13
S 2 10
Total 37

The Optimum cost (minimum) = ₹ 37

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Chapter 10: Operations Research - Exercise 10.2 [Page 256]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 10 Operations Research
Exercise 10.2 | Q 7 | Page 256

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(Processing cost in ₹)
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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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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
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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 :

Job Machine Min.cost
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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