An operation is used to machine 10 kinds of parts. The total weekly requirement

Question

An operation is used to machine 10 kinds of parts. The total weekly requirement for all 10 parts is 900. Each part requires 20 seconds machine time. Assume the machine is to be scheduled for no more than 90% of its total available time. A normal workweek is 40 hours/week, but 4 hours a week are reserved for normal machine preventive and repair maintenance. If the setup for each kind of part is 1 hour:

a. What is the maximum number of setups per week?

b. What is the smallest allowable lot size, assuming all batches are the same size?

c. To reduce the lot size in (b) by 50%, what must the setup time be reduced to?

Explanation / Answer

Gross available time = 40 hours - 4 hours = 36 hours

Net available time = 36*90% = 32.4 hours = 116640 seconds

Production time = 900 parts * 20 seconds = 18000 seconds

Total available time for setups = 116640 - 18000 = 98640 seconds

a) Maximum number of setups per week = Time available for setups / Time required for each setup = 98640 seconds / 1 hour or 3600 seconds = 27.4 setups , which means 2.74 (=27.4/10) setups per part

b) Smallest allowable lot size = Average demand / Number of setups = 900/27.4 = 32.8 ~ 32 parts

c) To reduce the lot size by 50%, number of setups should double, and therefore, setup time should be reduced by 50% , i.e. 0.5 hour

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