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प्रश्न
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?
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.
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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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 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 unbalance if ______
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.
Choose the correct alternative:
The assignment problem is generally defined as a problem of ______
Choose the correct alternative:
The assignment problem is said to be balanced if ______
A natural truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance(in kilometers) between the cities with a surplus and the cities with a deficit are displayed below:
To | |||||||
7 | 8 | 9 | 10 | 11 | 12 | ||
From | 1 | 31 | 62 | 29 | 42 | 15 | 41 |
2 | 12 | 19 | 39 | 55 | 71 | 40 | |
3 | 17 | 29 | 50 | 41 | 22 | 22 | |
4 | 35 | 40 | 38 | 42 | 27 | 33 | |
5 | 19 | 30 | 29 | 16 | 20 | 33 | |
6 | 72 | 30 | 30 | 50 | 41 | 20 |
How should the truck be dispersed so as to minimize the total 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`