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प्रश्न
A computer centre has got three expert programmers. The centre needs three application programmes to be developed. The head of the computer centre, after studying carefully the programmes to be developed, estimates the computer time in minitues required by the experts to the application programme as follows.
Programmers | ||||
P | Q | R | ||
Programmers | 1 | 120 | 100 | 80 |
2 | 80 | 90 | 110 | |
3 | 110 | 140 | 120 |
Assign the programmers to the programme in such a way that the total computer time is least.
उत्तर
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.
Programmers | ||||
P | Q | R | ||
Programmers | 1 | 40 | 20 | 0 |
2 | 0 | 10 | 30 | |
3 | 0 | 30 | 10 |
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.
Programmers | ||||
P | Q | R | ||
Programmers | 1 | 40 | 10 | 0 |
2 | 0 | 0 | 30 | |
3 | 0 | 20 | 10 |
Step 3: Examine the rows with exactly one zero, mark the zero by □. Mark other zeros in its column by X.
Programmers | ||||
P | Q | R | ||
Programmers | 1 | 40 | 10 | 0 |
2 | 0 | 0 | 30 | |
3 | 0 | 20 | 10 |
Step 4: Now examine the columns with exactly one zero mark the zero by □.
Mark other zeros in its row by X.
Programmers | ||||
P | Q | R | ||
Programmers | 1 | 40 | 10 | 0 |
2 | 0 | 0 | 30 | |
3 | 0 | 20 | 10 |
Thus all the three assignment have been made.
The optimal assignment schedule and total cost is
Programmers | Programmes | Cost |
1 | R | 80 |
2 | Q | 90 |
3 | P | 110 |
Total Cost | 280 |
The optimal assignment (minimum) cost = ₹ 280.
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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 |
25 |
33 |
29 |
B |
25 |
24 |
23 |
21 |
C |
19 |
21 |
23 |
24 |
D |
38 |
36 |
34 |
40 |
How should the jobs be assigned to the four machines so that the total processing cost is minimum?
Choose the correct alternative :
In an assignment problem if number of rows is greater than number of columns then
Fill in the blank :
When an assignment problem has more than one solution, then it is _______ optimal solution.
State whether the following statement is True or False:
In assignment problem, if number of columns is greater than number of rows, then a dummy row is added
What is the Assignment problem?
Three jobs A, B and C one to be assigned to three machines U, V and W. The processing cost for each job machine combination is shown in the matrix given below. Determine the allocation that minimizes the overall processing cost.
Machine | ||||
U | V | W | ||
Jobs | A | 17 | 25 | 31 |
B | 10 | 25 | 16 | |
C | 12 | 14 | 11 |
(cost is in ₹ per unit)
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 |
Choose the correct alternative:
If number of sources is not equal to number of destinations, the assignment problem is called ______
Choose the correct alternative:
In an assignment problem involving four workers and three jobs, total number of assignments possible are
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`