Homework Answers
2.
(a)
| Station | Task | Time (Seconds) | Time left (Seconds) | Ready tasks (# followers) |
| A(6) | ||||
| 1 | A | 20 | 50 | B(2), C(3) |
| C | 40 | 10 | B(2), D(2) | |
| D | 10 | 0 | B(2), F(1) | |
| 2 | B | 40 | 30 | F(1), E(1) |
| F | 20 | 10 | E(1) | |
| 3 | E | 60 | 10 | G(0) |
| 4 | G | 50 | 20 |
(b)
| Station | Task | Time (Seconds) | Time left (Seconds) | Ready tasks (# followers) |
| A(240) | ||||
| 1 | A | 20 | 50 | B(150),C(120) |
| B | 40 | 10 | C(120),E(110) | |
| 2 | C | 40 | 30 | E(110),D(80) |
| D | 10 | 20 | E(110),F(70) | |
| F | 20 | 0 | E(110) | |
| 3 | E | 60 | 10 | G(50) |
| 4 | G | 50 | 20 |
2.
Required cycle time = 16*60*60 / 500 = 115.2 seconds.
| Cycle time (sec) | Required cycle time | Idle time | |
| Workstation #1 | 70+30+10 = 110 | 115.2 | 5.2 |
| Workstation #2 | 6+10+10 = 80 | 115.2 | 35.2 |
| Workstation #3 | 60+20 = 80 | 115.2 | 35.2 |
| Workstation #4 | 30+70 = 100 | 115.2 | 15.2 |
| Total | 90.8 |
So, total idle time per cycle = 90.8 seconds.
(b)
Percentage idle time = Idle time / (No. of stations * required cycle time) = 90.8 / (4*115.2) = 19.70%
(c)
Efficiency = 100% - 19.70% = 80.30%
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c) For one to assign the tasks to the actual workstations and
be able to use the "theoretical minimum" number, the activity
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d) Were you able to assign all the tasks to the theoretical
minimum number of workstations? Yes or No.
Stanford Rosenberg Computing wants to...
2. Show all the steps for the question to be worth any
marks.
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