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A job production unit has four jobs P, Q, R, and 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 - Mathematics and Statistics

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प्रश्न

A job production unit has four jobs P, Q, R, and 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:

Job Machines
(Processing cost in ₹)
I II III IV
P 31 25 33 29
Q 25 24 23 21
R 19 21 23 24
S 38 36 34 40

Find the optimal assignment to minimize the total processing cost.

संक्षेप में उत्तर

उत्तर

Step 1: Subtract smallest element of each row from every element of that row.

Jobs 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: Since each column contains element 0, column minimization will give us the some matrix.

Step 3: Drawing minimum number of horizontal and vertical lines to cover all zeros.

  Jobs I II III IV  
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
S 4 2 0 6

∵ No. of lines drawn (4) = order of the matrix (4), optimal solution has reached.

Step 4: Assignment

Jobs I II III IV
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
S 4 2 0 6

Job Assignment:

Job Machine Min.cost
P II 25
Q IV 21
R I 19
S III 34
  Total Cost ₹ 99

Minimum Processing Cost = ₹ 99.

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संबंधित प्रश्न

A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job is given in the following table:

         Jobs

 

 

                          Machines

P

Q

R

S

                Processing Cost (Rs.)

 

A

31

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33

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B

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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 :

Job Machines
(Processing cost in ₹)
I II III IV
P 31 25 33 29
Q 25 24 23 21
R 19 21 23 24
S 38 36 34 40

Complete the following activity to find the optimal assignment to minimize the total processing cost.

Solution:

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 :

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`


A plant manager has four subordinates and four tasks to perform. The subordinates differ in efficiency and task differ in their intrinsic difficulty. Estimates of the time subordinate would take to perform tasks are given in the following table:

  I II III IV
A 3 11 10 8
B 13 2 12 2
C 3 4 6 1
D 4 15 4 9

Complete the following activity to allocate tasks to subordinates to minimize total time.

Solution:

Step I: Subtract the smallest element of each row from every element of that row:

  I II III IV
A 0 8 7 5
B 11 0 10 0
C 2 3 5 0
D 0 11 0 5

Step II: Since all column minimums are zero, no need to subtract anything from columns.

Step III: Draw the minimum number of lines to cover all zeros.

  I II III IV
A 0 8 7 5
B 11 0 10 0
C 2 3 5 0
D 0 11 0 5

Since minimum number of lines = order of matrix, optimal solution has been reached

Optimal assignment is A →`square`  B →`square`

C →IV  D →`square`

Total minimum time = `square` hours.


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