Table 7.3
A company uses a kanban system. The daily demand for part A is 4000 units. The average waiting time for a container of parts is 0.2 day. The processing time for a container of part A is 0.8 day, and a container holds 300 units.
-Use the information in Table 7.3.How many kanban containers would be required if a 5 % efficiency factor is used?
A) 10 or less
B) 11 or 12
C) 13 or 14
D) 15 or more

Question

Quizplus · Accepted Answer

Efficiency factor = 5% = 0.05

Maximum inventory = (Daily demand x average waiting time) + (Daily demand x processing time)

= (4000 x 0.2) + (4000 x 0.8)

= 800 + 3200

= 4000 units

Therefore, the number of kanban containers required = (Maximum inventory / container size) x (1 + efficiency factor)

= (4000 / 300) x (1 + 0.05)

= 13.33

Since you can't have a fraction of a container, the answer is either 13 or 14.

Table 7.3 a Company Uses a Kanban System. the Daily Demand for Demand