
b. If the muda or waste in the system were eliminated completely, how many containers would be needed?
a.
The Kanban formula to use
K = D*(1+SF)*KCT/C
Here
K = number of units per container = 50
D = average demand =?
SF = safety factor = 0.15
KCT = cycle time = 24/(24*60) + 2/24 = 0.1
C = number of containers = 12
We need to determine D
50 = D*(1+0.15)*0.1/12
D = 50*12/(1.15*0.1) = 5217.39
Demand is 5217
b.
If the waste is eliminated then the contingency is not needed anymore. This means we will need
C = 5217.39*0.1/50 = 10.43 or 11 containers will be enough.
b. If the muda or waste in the system were eliminated completely, how many containers would...
Answers must be to rounded to
the nearest number
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