Q.2 Process times for all jobs are one hour. Changeovers between families requir

Question

Q.2 Process times for all jobs are one hour. Changeovers between families require four hours. Thus, the completion time for job 1 is 5, for job 2 is 6, for job 3 is 11, and so on.

Job Number

Family Code

Due Date

1

1

5

2

1

6

3

2

12

4

2

13

5

1

15

6

1

19

7

1

20

8

2

22

9

2

25

10

1

27

a) How many possible sequences are there?

b) Compute the total tardiness of the current sequence (i.e. 1 to 10).

c) Is there any sequence with no tardiness?

d) Here, the processing times are deterministic (one hour). Suppose you are dealing with

uniformly distributed stochastic processing times (e.g. 45 < t< 75 minutes). Provide

some suggestions about handling varying operating times in scheduling activities

Job Number

Family Code

Due Date

1

1

5

2

1

6

3

2

12

4

2

13

5

1

15

6

1

19

7

1

20

8

2

22

9

2

25

10

1

27

Explanation / Answer

A. As there are 10 different tasks and no sequence priority is mentioned.

There are 10! ways possible = 3628800 ways

.

.

D.

In case of variable time, it is possible to handle with several lines parallely working for each job no and especially for each family type, this reduces the changeover time and if multiple queues are maintained for product sequences then the efficiency can be maximized and least tardiness.

Job Number Family Code Due Date Processing with Changeovers Completion Tardiness 1 1 5 1 1 0 2 1 6 1 2 0 3 2 12 5 7 0 4 2 13 1 8 0 5 1 15 5 13 0 6 1 19 1 14 0 7 1 20 1 15 0 8 2 22 5 20 0 9 2 25 1 21 0 10 1 27 5 26 0 Total 26 0 Ans B c) Is there any sequence with no tardiness? The current sequence has no tardiness

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