Question
The given problem is of n jobs and three machines. We change the problem in of n jobs and two machines.
For this either Min M1 ≤ M3 × M2 or min M3 ≤ max M2 Here min M3 = 7 = M1 × M2 Hence we write M1+ M2 = G and M2 + M3 = H the problem will be as follows
Solution
The given problem is of n jobs and three machines. We change the problem in of n jobs and two machines.
For this either Min M1 ≤ M3 × M2 or min M3 ≤ max M2
Here min M3 = 7 = M1 × M2
Hence we write M1+ M2 = G and M2 + M3 = H the problem will be as follows
Jobs | A | B | C | D |
`G=M_1+M_2` | 11 | 15 | 9 | 8 |
`H=M_2+M_3` | 13 | 15 | 12 | 14 |
The sequence as follows
Now
D | C | A | B |
Job | `M_1` | `M_2` | `M_3` | |||
Time in | Time out | Time in | Time out | Time in | Time out | |
D | 0 | 3 | 3 | 8 | 8 | 17 |
C | 3 | 10 | 10 | 12 | 17 | 27 |
A | 10 | 15 | 15 | 21 | 27 | 34 |
B | 15 | 23 | 23 | 30 | 34 | 42 |
The minimum elapsed time = 42 hrs.
Ideal time for `M_1` = 42 – 23 = 19 hrs.
Is there an error in this question or solution?
APPEARS IN
Solution The Given Problem is of N Jobs and Three Machines. We Change the Problem in of N Jobs and Two Machines. Concept: Linear Programming Problem in Management Mathematics.