Table $8.2$
A manufacturer using a kanban system has had problems with high inventory levels of one of the component parts it makes. Daily demand for the part is 3000 units, average waiting time during production is $0.20$ day, processing time is $0.10$ day per container, and a container holds 150 pieces.
-Use the information in Table 8.2. How many kanban containers would be required if a 5 % efficiency factor is used?
A) 3 or less
B) 4 or 5
C) 5 or 6
D) 7 or more

Question

Quizplus · Accepted Answer

The number of kanban containers is calculated using the formula: $ \text{Number of Kanbans} = \frac{(D \times (W + P) \times (1 + \text{Efficiency Factor}))}{C} $, where $ D = 3000 $ units/day, $ W = 0.20 $ day, $ P = 0.10 $ day, Efficiency Factor = 0.05, and $ C = 150 $ units/container. Plugging in the values, we get $ \frac{(3000 \times (0.20 + 0.10) \times 1.05)}{150} = 6.3 $, which rounds to 6 containers.

Table 8.28.28.2 A Manufacturer Using a Kanban System Has Had Problems with Problems

Table $8.2$ A Manufacturer Using a Kanban System Has Had Problems with Problems