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प्रश्न
Five different machines can do any of the five required jobs, with different profits resulting from each assignment as shown below:
| Job | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 30 | 37 | 40 | 28 | 40 |
| 2 | 40 | 24 | 27 | 21 | 36 |
| 3 | 40 | 32 | 33 | 30 | 35 |
| 4 | 25 | 38 | 40 | 36 | 36 |
| 5 | 29 | 62 | 41 | 34 | 39 |
Find the optimal assignment schedule.
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उत्तर
Step 1:
Since it is a maximization problem, subtract each of the elements in the table from the largest element, i.e., 62
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 32 | 25 | 22 | 34 | 22 |
| 2 | 22 | 38 | 35 | 41 | 26 |
| 3 | 22 | 30 | 29 | 32 | 27 |
| 4 | 37 | 24 | 22 | 26 | 26 |
| 5 | 33 | 0 | 21 | 28 | 23 |
Step 2:
Row minimum Subtract the smallest element in each row from every element in its row.
The matrix obtained is given below:
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 10 | 3 | 0 | 12 | 0 |
| 2 | 0 | 16 | 13 | 19 | 4 |
| 3 | 0 | 8 | 7 | 10 | 5 |
| 4 | 15 | 2 | 0 | 4 | 4 |
| 5 | 33 | 0 | 21 | 28 | 23 |
Step 3:
Column minimum Subtract the smallest element in each column of assignment matrix obtained in step 2 from every element in its column.
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 10 | 3 | 0 | 8 | 0 |
| 2 | 0 | 16 | 13 | 15 | 4 |
| 3 | 0 | 8 | 7 | 6 | 5 |
| 4 | 15 | 2 | 0 | 0 | 4 |
| 5 | 33 | 0 | 21 | 24 | 23 |
Step 4:
Draw minimum number of vertical and horizontal lines to cover all zeros.
First cover all rows and columns which have maximum number of zeros.
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 10 | 3 | 0 | 8 | 0 |
| 2 | 0 | 16 | 13 | 15 | 4 |
| 3 | 0 | 8 | 7 | 6 | 5 |
| 4 | 15 | 2 | 0 | 0 | 4 |
| 5 | 33 | 0 | 21 | 24 | 23 |
Step 5:
From step 4, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.
∴ Select smallest element from all the uncovered elements, i.e., 4 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 14 | 3 | 0 | 8 | 0 |
| 2 | 0 | 12 | 9 | 11 | 0 |
| 3 | 0 | 4 | 3 | 2 | 1 |
| 4 | 19 | 2 | 0 | 0 | 4 |
| 5 | 37 | 0 | 21 | 24 | 23 |
Step 6:
Draw minimum number of vertical and horizontal lines to cover all zeros.
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 14 | 3 | 0 | 8 | 0 |
| 2 | 0 | 12 | 9 | 11 | 0 |
| 3 | 0 | 4 | 3 | 2 | 1 |
| 4 | 19 | 2 | 0 | 0 | 4 |
| 5 | 37 | 0 | 21 | 24 | 23 |
Step 7:
From step 6, minimum number of lines covering all the zeros are 5, which is equal to order of the matrix, i.e., 5.
∴ Select a row with exactly one zero, enclose that zero in () and cross out all zeros in its respective column.
Similarly, examine each row and column and mark the assignment ().
∴ The matrix obtained is as follows:
| Jobs | Machines (Profit in ₹) | ||||
| A | B | C | D | E | |
| 1 | 14 | 3 | 0 | 8 | 0 |
| 2 | 0 | 12 | 9 | 11 | 0 |
| 3 | 0 | 4 | 3 | 2 | 1 |
| 4 | 19 | 2 | 0 | 0 | 4 |
| 5 | 37 | 0 | 21 | 24 | 23 |
Step 8:
The matrix obtained in step 7 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
| Jobs | Machines | Profit (in ₹) |
| 1 | C | 40 |
| 2 | E | 36 |
| 3 | A | 40 |
| 4 | D | 36 |
| 5 | B | 62 |
∴ Total maximum profit = 40 + 36 + 40 + 36 + 62 = ₹ 214.
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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?
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 |
Determine `l_92 and l_93, "given that" l_91 = 97, d_91 = 38 and q_92 = 27/59`
Solve the following maximal assignment problem :
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 11 | 11 | 9 | 9 |
| Q | 13 | 16 | 11 | 10 |
| R | 12 | 17 | 13 | 8 |
| S | 16 | 14 | 16 | 12 |
The assignment problem is said to be balanced if ______.
Choose the correct alternative :
The assignment problem is said to be balanced if it is a ______.
Fill in the blank :
When an assignment problem has more than one solution, then it is _______ optimal solution.
Fill in the blank :
An _______ is a special type of linear programming problem.
Choose the correct alternative:
The assignment problem is generally defined as a problem of ______
Choose the correct alternative:
Assignment Problem is special case of ______
If the given matrix is ______ matrix, the assignment problem is called balanced problem
State whether the following statement is True or False:
The objective of an assignment problem is to assign number of jobs to equal number of persons at maximum cost
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
Give mathematical form of Assignment problem
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:
North – West Corner refers to ______
Choose the correct alternative:
The purpose of a dummy row or column in an assignment problem is to
Choose the correct alternative:
In an assignment problem involving four workers and three jobs, total number of assignments possible are
A car hire company has one car at each of five depots a, b, c, d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers(destinations) where the customers are given in the following distance matrix.
| a | b | c | d | e | |
| A | 160 | 130 | 175 | 190 | 200 |
| B | 135 | 120 | 130 | 160 | 175 |
| C | 140 | 110 | 155 | 170 | 185 |
| D | 50 | 50 | 80 | 80 | 110 |
| E | 55 | 35 | 70 | 80 | 105 |
How should the cars be assigned to the customers so as to minimize the distance travelled?
A dairy plant has five milk tankers, I, II, III, IV and V. Three milk tankers are to be used on five delivery routes A, B, C, D and E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.
| I | II | III | IV | V | |
| A | 150 | 120 | 175 | 180 | 200 |
| B | 125 | 110 | 120 | 150 | 165 |
| C | 130 | 100 | 145 | 160 | 170 |
| D | 40 | 40 | 70 | 70 | 100 |
| E | 45 | 25 | 60 | 70 | 95 |
How should the milk tankers be assigned to the chilling center so as to minimize the distance travelled?
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.
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.
