Problem 4 A particular workstation has a capacity of 1,000 units per day and var
ID: 356806 • Letter: P
Question
Problem 4 A particular workstation has a capacity of 1,000 units per day and variability is moderate, such that V = 1. Demand is currently 900 units per day. Suppose management has decided that cycle times should be no longer than 1 times raw process time. (a) What is the current cycle time in multiples of the raw process time? (b) If variability is not changed, what would the capacity have to be in order to meet the cycle (? 2 time and demand requirements? What percentage increase does this represent? (c) If capacity is not changed, what value would be needed for V in order to meet the cycle time and demand requirements? What percentage decrease does this represent (compare CVs, not SCVs)?Explanation / 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
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