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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimates of the

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Question

A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimates of the time each man would take to perform each task is given below:

    Tasks
    1 2 3 4
Subordinates P 8 26 17 11
  Q 13 28 4 26
  R 38 19 18 15
  S 9 26 24 10

How should the tasks be allocated to subordinates so as to minimize the total manhours?

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.

    Tasks
    1 2 3 4
Subordinates P 0 18 9 3
  Q 9 24 0 22
  R 23 4 3 0
  S 0 17 15 1

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

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 9 20 0 22
  R 23 0 3 0
  S 0 13 15 1

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

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 9 20 0 22
  R 23 0 3 0
  S 0 13 15 1

Step 4: Now examine the columns with exactly one zero. Mark the zero by □. Mark other zeros in its row by X.

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 9 20 0 22
  R 23 0 3 0
  S 0 13 15 1

Step 5: Cover all the zeros of table 4 with three lines, since three assignments were made check (✓) row S since it has no assignment.

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 9 20 0 22
  R 23 0 3 0
S 0 13 15 1

Step 6: Develop the new revised tableau. Examine those elements that are not covered by a line in table 5.

Take the smallest element.

This is 1 (one) our case.

By subtracting 1 from the uncovered cells.

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 10 20 0 22
  R 24 0 3 0
  S 0 12 14 0

[Adding 1 to elements (Q, S, R) that line at the intersection of two lines]

Step 7: Go to step 3 and repeat the procedure until you arrive at an optimal assignment.

Step 8: Determine an assignment.

    Tasks
    1 2 3 4
Subordinates P 0 14 9 3
  Q 10 20 0 22
  R 24 0 3 0
  S 0 12 14 0

Thus all the four assignment have been made.

The optimal assignment schedule and total time is

Subordinates Tasks Time
P 1 8
Q 3 4
R 2 19
S 4 10
Total 41

The optimum time (minimum) = 41 Hrs.

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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 6 | Page 256

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A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table :

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Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :

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 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`
S III 34

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


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