1. Throughput
Throughput, Utilization, Bottleneck, Effective Capacity, Theoretical Capacity, and Capacity Leverages.
November 11, 2011
Ardavan Asef-Vaziri

2. Flow Rate, Throughput, Takt Time and Capacity
Throughput is the number of flow units per unit of time.
Observe the process for a number of periods.
Measure the number of flow units that are processed per unit of time.
Compute the average number of flow units per unit of time.
Throughput = Average Flow Rate
Capacity is the maximum sustainable flow rate. In periods of heavy congestion, throughput is equal to capacity.
In a synchronized operation: Throughput = Demand < Capacity
Throughput Analysis
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3. Flow Rate, Throughput, Takt Time and Capacity
Cycle Time is defined in terms of Capacity.
Cycle time = 1/Capacity
Capacity = 15/hour
Cycle Time = 1/15 hour
Cycle Time = 4 minutes
Takt time derived from the German word “pace”.
Takt Time is defined in relation to Demand or Throughput.
Demand = 12/hour  Throughput = 12/ hour
Takt Time = 1/Demand = 1/(Throughput) = 1/R
Takt Time = 1/12 hour
Takt Time = 5 minutes
Throughput Analysis
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4. Flow Rate, Throughput, Takt Time and Capacity
R= Ra  Takt time = 1/Ra
Average inter-arrival time = Ta = 1/Ra = 1/R = Average inter- exit time
It is not uncommon for managers to state that their machines have takt time of five minutes (1/12 hour). It is wrong. Takt time is a measure of external demand, it has nothing to do with the internal measure of machine capacity.
In a synchronized system, Takt time is the time each station has to send one flow unit out to the next station.
Chapter 4 was on flow time minimization.
Chapter 5 is on throughput maximization.
Chapter 4 and 5 are both “time” minimizations. Why?
Throughput Analysis
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5. Activity Time, Unit Load, Total Unit Load
Operation B
Activity Time = 10 min
Operation A
Activity Time = 5 min
If both are done on Recourse (1) then the theoretical unit load of Resource(1) is = 15 min.
If Capacity waste factor in both stations is 25% then the total unit load 15/(1-0.25) = 20 min. We refer to this as Tp.
It is not important to memorize the theoretical unit load and the unit load. What is important is if we do not consider Capacity Waste Factor then we are talking about the theoretical capacity. If we consider, then we are talking effective capacity or simply capacity.

6. Resources, Resource Pools and Resource Pooling
Unit Load of a Resource Unit (Tp) – The amount of time the resource works to process each flow unit. This includes a share of all distractions such as maintenance, repair, setup, etc. It is a effective unit load not the theoretical unit load.
Throughput Analysis
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7. Ardavan Asef-Vaziri, June 2013
Effective Capacity
Effective Capacity of a Resource Unit Rp= 1/unit load = 1/ Tp
Effective Capacity of a Resource Pool = Rp=c/Tp
Bottleneck – The resource pool with the minimum effective capacity.
Effective capacity of a process: Effective capacity of the bottleneck. Cross train Claims Supervisor to help Mailroom clerk  Increase Capacity.
Throughput Analysis
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8. 100% Utilization is a High Risk
Imagine a a freeway where all the cars are driving exactly 65 and the distance between pairs of cars in 1 inch. As long as every one has a speed of exactly 65 that is fine. But can they do that?
What happens if one hits the break? How long does it take to clean the freeway. Do cars pass freeway easier when utilization is 1 and they are moving bumper to bumper, or when 50% of the freeway is empty, U = 0.5, or when U = How long it takes to clean up accidents in these situations?
Never make U of all the sub-processes or activities, and not even a single sub-process = 1
Throughput Analysis
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9. Ardavan Asef-Vaziri, June 2013
Capacity Utilization
R = 400 per day.
Capacity utilization of a resource pool
Up = Throughput/Effective capacity of a resource pool = R/Rp
Capacity utilization of the process
U = Throughput/Effective capacity of the bottleneck resource pool
Throughput Analysis
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10. Cycle Time, Takt Time, Flow Time
Day = 480 minutes. Rp = 480 per day Cycle time = 1/Rp CT = 1/480 day CT = 480(1/480) = 1 minute We already knew CT = 1 min.
R = 400 per day
Takt Time = 1/R
TT = 1/400 day
TT=480(1/400) = 1.2 minute
Flow Time =
FT = 16.5
Throughput Analysis
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11. Unit Load for a Product Mix
Billing: Physician claims 60%, Hospital claims 40%
Throughput is 400 units per day.
Process utilization: U =100%
Throughput Analysis
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