A company is using Kanban containers. There are two adjacent work centers, a dow

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 (a) 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?

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?

Explanation / Answer

The below formula is used to calculate number of containers.

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

K= the number of Kanban cards in the operating system

d = the average daily production rate as determined from the master production schedule

w = the waiting time of Kanban cards in decimal fractions of a day(that is, the waiting time of a part)

p= the processing time per part, in decimal fractions of a day

C = capacity of a standard container in proper units of measure (parts, items, etc.)

a= a policy variable deermined by the efficiency of the process and its workstations and the uncertainty of the workplace, and therefore, a form of safety stock usually ranging from 0 to 1, however technically there is no upper limit on the value of a

a) No.of containers currently in use

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

b)If the number of parts a container holds is increased to 24 parts, how many containers are needed?

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

c)If the company wants the number of containers (holding 20 parts each) to be 5, what must the safety factor become?

Let safety factor be 'X'

Hence, 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%

Hence, the safety factor should become 0% if the company wants the number of containers (holding 20 parts each) to be 5

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.