Question

. A company is using Kanban containers. There are two adjacent work centers, a downstream (using) and an upstream (producing) one. The using work center has a production rate of 200 parts per day and each container holds 20 parts. It takes .5 days for a container to make the entire cycle from the time it leaves the upstream center until it is returned, filled with production, and leaves again. The manager wants a safety factor () of 20%. The company is interested in reducing the number of containers. a. What is the number of containers currently in use? b. If the number of parts a container holds is increased to 24 parts, how many containers are needed? c. If the company wants the number of containers (holding 20 parts each) to be 5, what must the safety factor become?

Answer 1

The formula is used to measure the number of containers.

K= d(p+w)(1+ a)/c

K= The number of kanban cards on the software.

d = total daily rate equation as determined in the launch schedule.

w = the commute time of the Bug tracking card in decimal fractions of each day.

p= total storage time, in decimal fractions per day.

C = the ability of the standard container in the proper measuring device.

a= A policy factor determined by the efficiency of the program and its netbooks and the instability of the office.

Explanation:

a) K= d(p+w)(1+ a)/c = 200 (0.5)(1+0.2) / 20 = 120/20 = 6 containers

b) K= d(p+w)(1+ a)/c = 200 (0.5)(1+0.2) / 24 = 120/24 = 5 containers

c) Let safety factor be 'X'

Therefore, K = d(p+w)(1+a)/c=200(0.5)(1+X)/20=5 containers

5(1+X) = 5 containers

1+X = 1

X = 0%

As a result, the safety factor will be 0 percent if the company requires the number of containers (with 20 pieces each) to be 5