Answer:
Please find the detailed answer as follows.
Step-by-step 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.