Question

Determine the number of containers required to support a lean process which uses 200 parts per hour. The time for a container to complete a cycle (move, wait, empty, return, fill) is 0.75 hours. A single container has a capacity of 6 parts. And inefficiency factor of 0.11 is currently being used. (Round any fractional result up to the next higher whole number.)

Answer 1

Solution :

Given :

D = the rate of the planned usage of using the work center = 200 parts per hour

T = the average waiting list for the replenishment of the parts plus the average time of production of a container of parts = 0.75 hours

X = the policy variable which is set by the management - the possible inefficiency of the system = 0.11

C = the capacity of the standard container = 6 parts per container

Therefore the total number of the containers are :

$N=\frac{DT(1+X)}{C}$

$N=\frac{(200 \times 0.75)(1+0.11)}{6}$

N = 27.75  which is almost 28 containers.

The rounding up will cause a system to be a looser while rounding down will make the system tighter. Usually the rounding up is preferable.