Acompany is using a kanban system with 2 containers, each holding 50 parts, between a downstream (using) and an upstream (producing) work center. the using work center can handle 250 jobs per day. the average elapsed time for the entire cycle is currently .25 days. the company is concerned with the safety factor for the operation and the size of the containers.

a. what is the efficiency rating of the current operation?
b. if the company decides to use only one container, how many parts must it hold?
c. if the safety factor is changed to .12 and there are still 2 containers, how many parts must each container now hold?

Please find the detailed answer as follows.

Explanation:

We use the below formula to solve the questions

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)What is the efficiency rating of the current operation?

Let The efficiency rating be 'X'

Given No.of containers = 2

Hence by using formula

K= d(p+w)(1+ a)/c = 250 (0.25) (1+X) /50 = 2

=> 1.25(1+X) = 2

=> 1+X = 2/1.25 = 1.6

=> X = 1.6 - 1 = 0.6

Hence, efficiency rating of the current operation = 60%

b)If the company decides to use only one container, how many parts must it hold

Given No.of containers = 1

Let the no.of parts to e hold are X

K= d(p+w)(1+ a)/c = 250 (0.25) (1.6) /X = 1

Hence, X =   250 (0.25) (1.6) = 100 parts

Hence,If the company decides to use only one container, then it should hold 100 parts

c)If the safety factor is changed to .12 and there are still 2 containers, how many parts must each container now hold?

Given No.of containers = 2

Let the no.of parts to e hold are X

K= d(p+w)(1+ a)/c = 250 (0.25) (1.12)/X = 2

=> 2X = 250 (0.25) (1.12) = 70

=> X = 70/2 = 35 parts

Hence,If the safety factor is changed to .12 and there are still 2 containers, 35 parts are to be hold by each container.

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