Question

A particular single-machine workstation has a capacity of 1,000 units per day and variability is moderate, such that V = (SCV of arrivals + SCV of effective process time)/2 = 1. Demand is currently 900 units per day. Suppose management has decided that cycle times should be no longer than 1.5 times the average process time.

A) What is the current cycle time in multiples of the process time? (i.e. if the current cycle time was 2 times longer than the process time, put 2 in the answer box)

B) If variability is not changed, what would the daily capacity have to be in order to meet the requirement that average cycle time be no longer than 1.5 times process time?

C) If capacity is not changed, what value would be needed for V in order to meet the requirement that average cycle time be no longer than 1.5 times process time?

Answer 1

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Answer:
Inter-arrival time, a = (1/900) day
Raw process time, p = (1/1000) = 0.001 day
Variability, V = 1
Utilization, u = p/a = (1/1000)/(1/900) = 0.9
Average waiting time, Tq = p*V*u/(1-u) = (1/1000)*1*0.9/(1-0.9) = 0.009 day
Cycle time, T = Tq + p = 0.009 + 1/1000 = 0.01 day
(a) Cycle time in multiples of raw process time = T/p = 0.01/0.001 = 10

(b) Cycle time should not be more than 1.5 times the raw process time. So, Capacity has to be increased such that cycle time is at most, T = 0.001*1.5 = 0.0015
Therefore, Tq should be at most = T - p = 0.0015 - 0.001 = 0.0005
Substituting this value in Tq = p*u/(1-u) = 0.0005 and solving for u, we get
2u = 1-u
or u = 1/3 = 0.33
u = p/a = p/(1/900) = 1/3 , p = 1/2700
Capacity = 1/p = 1/(1/2700) = 2700 per day
Therefore, capacity should be at least 2700 units per day
Percentage increase = (2700-1000)/1000 = 170 % increase over present capacity

(c) Keeping, p and u unchanged, p*V*u/(1-u) = 0.0005
Substituting values of p and u and solving for V, we get, V = 0.001*V*0.9/(1-0.9) = 0.0005
or, 0.009V = 0.0005
V = 0.056