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 RateCapacity is the maximum sustainable flow rate. In periods of heavy congestion, throughput is equal to capacity.In a synchronized operation: Throughput = Demand < CapacityThroughput AnalysisArdavan Asef-Vaziri, June 20132
3 Flow Rate, Throughput, Takt Time and Capacity Cycle Time is defined in terms of Capacity.Cycle time = 1/CapacityCapacity = 15/hourCycle Time = 1/15 hourCycle Time = 4 minutesTakt time derived from the German word “pace”.Takt Time is defined in relation to Demand or Throughput.Demand = 12/hour Throughput = 12/ hourTakt Time = 1/Demand = 1/(Throughput) = 1/RTakt Time = 1/12 hourTakt Time = 5 minutesThroughput AnalysisArdavan Asef-Vaziri, June 20133
4 Flow Rate, Throughput, Takt Time and Capacity R= Ra Takt time = 1/RaAverage inter-arrival time = Ta = 1/Ra = 1/R = Average inter- exit timeIt 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 AnalysisArdavan Asef-Vaziri, June 20134
5 Activity Time, Unit Load, Total Unit Load Operation BActivity Time = 10 minOperation AActivity Time = 5 minIf 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 AnalysisArdavan Asef-Vaziri, June 20136
7 Ardavan Asef-Vaziri, June 2013 Effective CapacityEffective Capacity of a Resource Unit Rp= 1/unit load = 1/ TpEffective Capacity of a Resource Pool = Rp=c/TpBottleneck – 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 AnalysisArdavan Asef-Vaziri, June 20137
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 = 1Throughput AnalysisArdavan Asef-Vaziri, June 20138
9 Ardavan Asef-Vaziri, June 2013 Capacity UtilizationR = 400 per day.Capacity utilization of a resource poolUp = Throughput/Effective capacity of a resource pool = R/RpCapacity utilization of the processU = Throughput/Effective capacity of the bottleneck resource poolThroughput AnalysisArdavan Asef-Vaziri, June 20139
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 dayTakt Time = 1/RTT = 1/400 dayTT=480(1/400) = 1.2 minuteFlow Time =FT = 16.5Throughput AnalysisArdavan Asef-Vaziri, June 201310
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 AnalysisArdavan Asef-Vaziri, June 201311
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