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Question
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 |
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Solution
The order is A to B, hence the required sequence is
| J1 | J2 | J6 | J3 | J5 | J4 |
or
| J1 | J6 | J2 | J3 | J4 | J5 |
| Job Sequence |
Machine A | Machine B | ||
| Time in |
Time out |
Time in |
Time out |
|
| J1 | 0 | 1 | 1 | 6 |
| J2 | 1 | 4 | 6 | 12 |
| J6 | 4 | 7 | 12 | 22 |
| J3 | 7 | 15 | 22 | 25 |
| J5 | 15 | 21 | 25 | 27 |
| J4 | 21 | 26 | 27 | 29 |
Total elapsed time T = 29 hours
Idle time for A = 3 hours
Idle time for B = 1 hour.
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|
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|
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|||
|
P |
Q |
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|
|
Processing Cost (Rs.)
|
||||
|
A |
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|
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|
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|
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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`
