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Question
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?
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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.
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 45 | 60 | 70 | |
| B | 15 | 0 | 10 | 40 | 55 | |
| Customers | C | 30 | 0 | 45 | 60 | 75 |
| D | 0 | 0 | 30 | 30 | 60 | |
| E | 20 | 0 | 35 | 45 | 70 | |
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C | 30 | 0 | 35 | 30 | 20 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E | 20 | 0 | 25 | 15 | 15 | |
Step 3: (Assignment)
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C | 30 | 0 | 35 | 30 | 20 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E | 20 | 0 | 25 | 15 | 15 | |
Step 4: Now Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C | 30 | 0 | 35 | 30 | 20 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E | 20 | 0 | 25 | 15 | 15 | |
Step 5: Cover all the zeros of table 4 with three lives.
Since three assignments were made please note that check [✓] Row C and E which have no assignment.
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C✓ | 30 | 0 | 35 | 30 | 20 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E✓ | 20 | 0 | 25 | 15 | 15 | |
Step 6: Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5.
Take the smallest element in each row and subtract from the uncovered cells, depots
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C | 30 | 0 | 35 | 30 | 0 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E | 20 | 0 | 25 | 0 | 0 | |
Step 7: Go to step 3 and repeat the procedure until you arrive at an optimal assignments depots
Step 8: Determine an assignment
| Depots | ||||||
| a | b | c | d | e | ||
| A | 30 | 0 | 35 | 30 | 15 | |
| B | 15 | 0 | 0 | 10 | 0 | |
| Customers | C | 30 | 0 | 35 | 30 | 0 |
| D | 0 | 0 | 20 | 0 | 5 | |
| E | 20 | 0 | 25 | 0 | 0 | |
Here all the five assignments have been made.
The optimal assignment schedule and total distance is
| Customers | Depots | Total Distances |
| A | b | 130 |
| B | c | 130 |
| C | e | 185 |
| D | a | 50 |
| E | d | 80 |
| Total | 575 | |
∴ The optimum Distance (minimum) is 575 kms.
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RELATED QUESTIONS
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?
Suggest optimum solution to the following assignment. Problem, also find the total minimum service time.
Service Time ( in hrs.)
| Counters | Salesmen | |||
| A | B | C | D | |
| W | 41 | 72 | 39 | 52 |
| X | 22 | 29 | 49 | 65 |
| Y | 27 | 39 | 60 | 51 |
| Z | 45 | 50 | 48 | 52 |
Choose the correct alternative :
The assignment problem is said to be balanced if it is a ______.
In an assignment problem, if number of column is greater than number of rows, then a dummy column is added.
State whether the following statement is True or False:
In assignment problem each worker or machine is assigned only one job
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:
Number of basic allocation in any row or column in an assignment problem can be
Choose the correct alternative:
The solution for an assignment problem is optimal if
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`
