A manufacturer of automobile engines is designing a new Kanban system for engine

Question

    A manufacturer of automobile engines is designing a new Kanban system for engine #321. The demand for #321 is 75/day and they are built in groups of 8. The total process and wait time is 3 days. The manager wants an alpha (a) of 2 or 200%. How many Kanbans are required?    

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

Explanation / Answer

K= d(p+w)(1+ a)/c = 75 (3)(1 + 2) / 8 = 84.375 = 84 Kanban cards approximately required

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