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Question
Choose the correct alternative:
Number of basic allocation in any row or column in an assignment problem can be
Options
exactly one
at least one
at most one
none of these
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Solution
exactly one
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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 |
Solve the following minimal assignment problem and hence find minimum time where '- ' indicates that job cannot be assigned to the machine :
| Machines | Processing time in hours | ||||
| A | B | C | D | E | |
| M1 | 9 | 11 | 15 | 10 | 11 |
| M2 | 12 | 9 | - | 10 | 9 |
| M3 | - | 11 | 14 | 11 | 7 |
| M4 | 14 | 8 | 12 | 7 | 8 |
In a factory there are six jobs to be performed each of which should go through two machines A and B in the order A - B. The processing timing (in hours) for the jobs arc given here. You are required to determine the sequence for performing the jobs that would minimize the total elapsed time T. What is the value of T? Also find the idle time for machines · A and B.
| Jobs | J1 | J2 | J3 | J4 | J5 | J6 |
| Machine A | 1 | 3 | 8 | 5 | 6 | 3 |
| MAchine B | 5 | 6 | 3 | 2 | 2 | 10 |
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 for each machine is given in the following table:
| Jobs | Machines (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 31 | 25 | 33 | 29 |
| B | 25 | 24 | 23 | 21 |
| C | 19 | 21 | 23 | 24 |
| D | 38 | 36 | 34 | 40 |
Find the optimal assignment to minimize the total processing cost.
Solve the following problem :
A dairy plant has five milk tankers, I, II, III, IV and V. These 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 | 175 |
| 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?
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)
Choose the correct alternative:
North – West Corner refers to ______
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`
