TASKS |
Predecessor |
Task Time |
|
A |
Warm tortilla |
-none |
20 sec |
B |
Rice |
-A |
10 sec |
C |
Beans |
-B |
10 sec |
D |
Meat |
-B |
10 sec |
E |
Salsa |
-C, D |
10 sec |
F |
Cheese |
-E |
5 sec |
G |
Sour Cream |
-E |
10 sec |
H |
Lettuce |
-E |
10 sec |
I |
Wrap |
-F, G, H |
15 sec |
J |
Bag |
-I |
5 sec |
K |
Checkout/pay |
-J |
25 sec |
The minimum cycle time can be equal to the maximum value of time taken by any task. Here, the task K takes 25 sec. which is the longest time. Hence the cycle time can be taken as 25.
Theoretical min. number of WS = Task time / Cycle time = 130/25=6
Assignment
WS | Tasks | Time | Idle time |
I | A | 20 | 5 |
II | B,C | 20 | 5 |
III | D,E | 20 | 5 |
IV | F,G,H | 25 | 0 |
V | I,J | 20 | 5 |
VI | K | 25 | 0 |
Idle time per cycle =20 sec.
Efficiency = task time / (cycle time x Number of WS)
= 130/25x6 = 0.8666
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