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Please show all written work as I am having trouble finding my way through the solution to this problem. Thank you in advance!

Question 4 (12 points: 1, 2, 2, 2, 5) Consider the three-stage process shown here (a) What is the minimum TPT? (b) Calculate hourly capacity of each stage. G1 6 min 6 min D1 4 min W2 G2 6 min min Consider the following schedule. Jobs are numbered 1, 2, 3, 4 and so on. Jobs are released in the process every four minutes (at 0, 4, 8,...). Odd numbered jobs go to W1 and G1 and even numbered jobs are scheduled on W2 and G2. Note that with this schedule you may or may not need buffers (c)Plot a Gantt chart using the diagram below (you can copy or cut and paste the diagram to your answer). Job 1 is already plotted for you. Plot jobs 2, 3, 4, 5, and 6. D1 F1 2 F2 G1 G2 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 2627 28 29 30 31 32 33 34 35 (d) Assume that the system becomes stable after four jobs. What will be the WIP contributions average TPT based on jobs 1, 2, 3 and 4? Calculate the WIP values in the table shown to the right. Then use Littles law (and results from the calculations in earlier parts) to calculate the actual cycle time. D1 F1 W1 W2 F2 G1 G2 Total (e)

2aa 33334444 ร์ 56666 D1 F1 W1 W2 F2 G1 G2 0 33 3333 5555 55 3 3 3 33 3555555 d. 1 2 3 4 7 9 |10 11 |12 13 |14 |15 |16 |17 18 |19 20 21 22 23 24 25 26 27 28-29 30 31 32 33 34 3

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Answer #1

a) The minimum TPT is 4 + 5 + 6 = 15 min

b) Stage 1: 60/4 =15 units/hr

Stage 2: (1/6+1/5)*60 = 22 units/hr

Stage 3: (1/6+1/6)*60 = 20 units/hr

Consider the following schedule. Jobs are numbered 1, 2, 3, 4 and so on. Jobs are released in the process every four minutes (at 0, 4, 8,…). Odd numbered jobs go to W1 and G1 and even numbered jobs are scheduled on W2 and G2. Note that with this schedule you may or may not need buffers.

c)

D1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6
F1
W1 1 1 1 1 1 1 3 3 3 3 3 3 5 5 5 5 5 5
W2 2 2 2 2 2 4 4 4 4 4 6 6 6 6 6
F2
G1 1 1 1 1 1 1 3 3 3 3 3 3 5 5 5 5 5 5
G2 2 2 2 2 2 2 4 4 4 4 4 4 6 6 6 6 6 6
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 22 34 35

WIP Contributions:

D1 1
F1 0
W1 6/8
W2 5/8
F2 0
G1 6/8
G2 6/8
Total 31/8

d) 1&3 = 4 + 6 + 6 = 16

2&4 = 4 + 5 + 6 = 15

Average TPT = (16+15)/2 = 15.5 min

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