(i)

(ii)
| Station | Remaining time |
Eligibles (# followers) |
Was fit | Assigned time | Time remains | Idle time | Tie-breaker |
| I | 1.3 | a(4), c(3), e(3) | a | 0.2 | 1.1 | ||
| 1.1 | c(3), e(3), b(3) | b | 0.4 | 0.7 | b | ||
| 0.7 | c(3), e(3) | c | 0.3 | 0.4 | c | ||
| 0.4 | e(3), d(2) | e | 0.1 | 0.3 | |||
| 0.3 | d(2), f(2) | - | - | 0.3 | 0.3 | ||
| II | 1.3 | d(2), f(2) | d | 1.3 | 0.0 | d | |
| 0.0 | f(2) | - | - | 0.0 | 0.0 | ||
| III | 1.3 | f(2) | f | 0.8 | 0.5 | ||
| 0.5 | g(1) | g | 0.3 | 0.2 | |||
| 0.2 | h(0) | - | - | 0.2 | 0.2 | ||
| IV | 1.3 | h(0) | h | 1.2 | 0.1 | 0.1 |
So, the assignment is as follows:
Station-I: a, b, c, e
Station-II: d
Station-III: f, g
Station-IV: h
(iii)
Percentage idle time = Total idle time per cycle / (No. of stations * cycle time) = (0.3+0+0.2+0.1) / (4*1.3) = 11.54%
(iv)
The actual cycle time is also 1.3 minutes (Station-II being the bottleneck).
So, actual output in 420 minutes = 420/1.3 = 323 units per day
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2. Show all the steps for the question to be worth any
marks.
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uestion 10 0/ Te (min) 10 For the diagram above, available time per shift is 480 minutes and 40 units are required at the end of each shift. Use the heuristic rule by assigning the task with the most followers. Break any ties with longest time first. (all time in minutes) (6 marks) 480/40-12 Hide Feedback This was not a balance line
uestion 10 0/ Te (min) 10 For the diagram above, available time per shift is 480...
) A firm is planning to set up an assembly line to assemble 40 units per hour, and 57 minutes per hour are productive. The time to perform each task and the tasks that precede each task are contained in the following table. Task Preceding Task Time to perform (min.) A -- .69 B A .55 C B .92 D B .59 E B .70 F B 1.10 G C, D, E .75 H G, F .43 I H .29...
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need help with the total idle time per cycle question, please
include steps with explanations
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