A simple process has four stages – A, B, C, and D. The average amount of work ne

Question

A simple process has four stages – A, B, C, and D. The average amount of work needed to process items passing through these stages is as follows: stage A = 68 minutes, stage B = 55 minutes, stage C = 72 minutes, and stage D = 60 minutes. A spotcheck on the work-in-progress between each stage reveals the following: between stages A and B there are 82 items, between stages B and C there are 190 items, and between stages C and D there are 89 items. a. Using Little’s law (see Chapter 4), calculate the throughput time of the process. b. What is the throughput efficiency of the process?

3 In the example above, the operations manager in charge of the process re-allocates the work at each stage to improve the ‘balance’ of the process. Now each stage has an average of 64minutes of work. Also, the work-in-progress in front of stages B, C, and D is 75, 80, and 82units respectively. How has this changed the throughput efficiency of the process?

Explanation / Answer

Throughput rate of stage A = 60/68 = 0.882

Throughput rate of stage B = 60/55 = 1.091

Throughput rate of stage C = 60/72 = 0.833

Throughput rate of stage D = 60/60 = 1

Minimum throughput rate is the thrroughput rate of the process = 0.833

WIP = 82+190+89 = 361

a) Using Little's law, Throughput time (or lead time) = WIP / throughput rate = 361/0.833 = 433.2

b) Throughput efficiency = Value added time / Throughput time = (68+55+72+60)/433.2 = 0.5886 or 58.86%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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