Presentation on theme: "Based on the book: Managing Business Process Flows."— Presentation transcript:
1 Based on the book: Managing Business Process Flows. Throughput Part 2Effective Capacity&Throughput LeveragesBased on the book: Managing Business Process Flows.
2 Capacity Waste and Theoretical Capacity Effective capacity of a resource unit is 1/Tp. Unit load Tp , is an aggregation of the productive as well as the wasted time.Tp includes share of each flow unit of capacity waste and detractions such asResource breakdownMaintenanceQuality rejectsRework and repetitionsSetups between different products or batchesWe may want to turn our attention to waste elimination; and segregate the wasted capacity. Theoretical capacity is the effective capacity net of all capacity detractions.
3 ThCapacity, Capacity, ThFlowTime, FlowTime Theoretical Unit Load, Theoretical Activity TimeTheoretical Capacity is computed based on the Theoretical Unit Load (ThUL)Theoretical Flow Time is NOT computed based on Theoretical Activity TimeVery Theoretical Flow Time is computed based on theoretical Activity Time30 mins.ThUL(1+CWF) = Unit Load (Tp), Activity timeCapacity is computed based on the Unit LoadTheoretical Flow Time is computed based on Activity TimeThen What is Flow Time?10 mins.30 mins.Flow Time Ti + TpFlow time includes time in buffersCapacity does not care about buffer times3 days10 mins.30 mins.
4 Capacity Waste Factor and Theoretical Capacity An operating room (a resource unit) performs surgery every 30 min, Tp = 30 min. Tp includes all the distracts. We also refer to it as the Unit Load.Effective capacity is 1/30 per min or 60/30 =2 per hour.On average, 1/3 of the time is wasted (cleaning, restocking, changeover of nursing staff and fixing of malfunctioning equipment ).Capacity Waste Factor (CWF) = 1/3.Theoretical Unit load = Tp*(1-CWF) =30(1-1/3) = 20 min.Tp = Unit Load = ThUnit Load /(1-CWF) = 20/(1-1/3) = 30Theoretical Capacity = c/ThUnit LoadEffective Capacity = Capacity = c/Unit Load.Theoretical Capacity = 1/20 per minute or 3 per hour.Effective Capacity = Theoretical Capacity (1-CWF)
5 Problem 4. Problem 5.1 in the book A law firm processes (i) shopping centers and (ii) medical complexes contracts. The time requirements (unit loads) for preparing a standard contract of each type along with some other information is given below. In November 2012, the firm had 150 orders, 75 of each type. Assume 20 days per month, and 8 hours per day. CWF at the three resource-s are 25%, 0%, and 50%, respectively.
6 Problem 4. Problem 5.1 in the book What is the effective capacity of the process (contracts /day)?Paralegal: Theoretical Unit Load (50%Sh 50% Med): 0.5(4)+0.5(6) = 5 hrsTheoretical Capacity = 1/5 per hrCapacity Waste Factor (CWF) =Unit Load = Tp = 5/(1-0.25) = 20/3 hrsEffective Capacity = Capacity = 1/(20/3) = 3/20 per hrTax Lawyer: Theoretical Unit Load 0.5(1)+0.5(3) = 2 hrsCWF = 0Theoritical Unit Load = Tp = 2 hrsTheoretical Capacity = 1/2 per hrEffective Capacity = Capacity = 1/2 per hr
7 Problem 4. Problem 5.1 in the book Senior Partner: Theoretical Unit Load 0.5(1)+0.5(1) = 1 hrs.Theoretical Capacity = 1/1 = 1 per hrCWF = 0.5Unit Load = Tp = 1/(1-0.5) = 2 hrsEffective Capacity = Capacity = 1/2 per hrb) Compute the cycle time?4.8 units in 8 hours.Cycle time = 8/4.8 = 1.67 hours
8 Problem 4. Problem 5.1 in the book c) Compute the flow time.Vey Theoretical Flow time = = 8 !!!!!!Theoretical Flow Time = 10.67Flow Time = Waiting timesd) Compute the average inventory.Lets look at the utilization of the 3 stationsStation Capacity Throughput UtilizationStation /4.8 = 1Station /12 = 0.4Station /8 = 0.6On average 1 person with a resource in Station 1, 0.4 person with a resource in Station 2, and 0.6 person with a resource in Station 3
9 Problem 4. Problem 5.1 in the book Inventory with the processors is = 2On average there are 2 flow units with the processors; Inventoty in the processors (Ii)Now let’s look from another angle; from the Little’s Law point of viewRT=I R= 4.8 per 8 hours or 0.6 per hourT =10.67 hours I = 0.6(10.67) = 6.46.4 vs 2? Where is my mistake??1(4)+ 0.4(3)+0.6(2) = 6.4e) Can the company process all 150 cases in November?4.8 (20) = 96 < 150.150/20 = 7.5. The effective capacity of 4.8 /day is not sufficient.
10 Problem 4. Problem 5.1 in the book f) If the firm wishes to process all the 150 cases available in November, how many professionals of each type are needed?Demand per day = 150/20 = 7.5# of paralegals required = 7.5/1.2 = 6.25# of tax lawyers required = 7.5/4 = 1.875These could be rounded up to 7, 2 and 2We need 7, 2, 2. We have 4, 4,4. We may hire 3 additional paralegals.Alternatively, we may hire just 2 and have 6 paralegals.They need to work over time for 0.25 paralegal who works 8 hrs /day.That is 1.5 hours 1.5/8 = minutes over time.
11 Do you Really Need Additional Problem? A law firm handles several types of insurance policies. The theoretical unit loads for an aggregate policy (representing the combination of all policies) are as follows: 4 hours of Resource 1, 3 hours of Resource 2, and 2 hours of Resource 3. Capacity waste factor for resources 1 to 3 is 0.3, 0.1, and 0.3, respectively. Utilization of the bottleneck resource is 0.8. There are 8 working hours per day.Compute the VERY theoretical flow time for an aggregate product.4+3+2 = 9 hours2. Compute the theoretical flow time for an aggregate product.4/(1-0.3) + 3/(1-0.1) +2/(1-0.3) = 4/0.7+3/0.9+2/0.7= = 11.9
12 Do you Really Need Additional Problem? 3. Compute the daily theoretical capacity of the process. Min(1/4,1/3,1/2) = 1/4 per hour 8(1/4) = 2 per day 4. Compute the daily capacity of the process. Min(1/5.71,1/3.33,1/2.86) = 1/5.71 /hr or 8(1/5.71)= 1.4 per day Min(0.175, 0.3, 0.35) = /hr or 8(1/5.71)= 1.4 per day 5. Compute the cycle time. 8/1.4 = 5.71 hour OR 1/0.175 = 5.71 hour OR 1/(1/5.71) = 5.71 hour 6. Compute the throughput per day. Capacity = 1.4 per day U=0.8
13 Do you Really Need Additional Problem? U = Throughput/Capacity 0.8 = R/1.4 Throughput = 1.12 per day 7. Compute the takt time. 1/1.12 day OR 8/1.12 = 7.14 hours 8. Compute the utilization of the most utilized resource. The problem has already said that the utilization of the process is 0.8. That is the utilization of the most utilized resource. That is utilization of the bottleneck. 9. Compute the utilization of the least utilized resource. The three resources have capacity of 0.175, 0.3, 0.35 per hour OR OR 1.4, 2.4, 2.8 per day.
14 Do you Really Need Additional Problem? Throughput is 1.12 per day. U1 = 1.12/ 1.4 =0.8 U2 = 1.12/ 2.4 =0.47 U3 = 1.12/2.8 = On average how many flow units are with the resources (in the processors) There are three resources with U1 = 0.8, U2=0.47 , U3 = = Suppose the number of flow units in all the waiting lines are 10 units. Compute the flow time. (A day is 8 hours.) R = 1.12 per day I = = RT = I 1.12T = T = 11.67/1.12 = 10.42
15 Setup Batch and Total Unit Load Setup or Changeover: activities related to cleaning, resetting and retooling of equipment in order to process a different product.Qp : Setup batch or lot size; the number of units processed consecutively after a setup;Sp : Average time to set up a resource at resource pool p for a particular productAverage setup time per unit is then Sp/QpSp/Qp is also included in TpSetup batch (also lot size): number of units processed consecutively after a setupEXAMPLE: painting cars->how many cars before you change paint color
16 Setup Batch SizeWhat is the “right” lot size or the size of the set up batch? Lot Size or ?The higher the lot size, the lower the unit load and thus the higher the capacity.The higher the lot size, the higher the inventory and therefore the higher the flow time.Reducing the size of the setup batch is one of the most effective ways to reduce the waiting part of the flow time.Load batch: the number of units processed simultaneously. Often constrained by technological capabilities of the resource.Setup batch: the number of units processed consecutively after a setup. Setup is determined managerially.Setup batch (also lot size): number of units processed consecutively after a setupEXAMPLE: painting cars->how many cars before you change paint color
17 Capacity Analysis; Multiple Products - Single Stage Product Mix:50%-50%Set-up time30 min per productWorking hours8 hours/day10 min/unitAOperationB20 min/unit1 machine100% availableCompute the effective capacity under min cost strategy.Two set-ups each for 30 min = 60 minsAn aggregate product takes (10+20)/2 = 15Production time = 8*60-60 = 420 minsCapacity = 420/15 = 28 aggregate unitsEach aggregate unit is 0.5 A and 0.5 B (total of 14A and 14B)
18 Capacity Analysis; Multiple Products - Single Stage Compute the capacity under min flow time strategy.In a minimum inventory strategy, we produce two product at a time then switch to the other.10 min/unitAOperationB20 min/unit1 machine100% availableProduct A: 2(10)0+30 = 50Product B: 2*(20)+30 = 70An aggregate product takes (50+70)/2 = 60 = 1 hourCapacity = 8/1 = 8 aggregate units.Each aggregate unit is 0.5 A and 0.5 B (total of 4*2A and 4*2B)8A and 8B need 4*30*2 = 4 hours + 8*10+8*20 = 4 hours
20 Increasing Resource Levels ↑ c Capacity ≈ Theoretical Capacity Resources are efficiently utilized; increase the theoretical capacity.Increase the level of resources. ↑ c. buy one more ovenIncrease the size of resource units - Larger load batch - more loaves in the ovenIncrease the time of operation – ↑ Scheduled Availability, OvertimeSubcontract or OutsourceTechnology- Speed up the activities rate - Invest in faster resources or incentives for workers.
21 Reducing Resource Capacity Waste ↓ Tp Capacity << Theoretical Capacity resources are not utilized effectively; eliminate of waste; ↓ Tp or ↑ Net Availability.Eliminate non-value-adding activities.Avoid defects, rework and repetitions – These two are exactly the same as what was stated for flow time reduction. For flow time we focus on activities along the critical path. For flow rate we focus on activities performed by bottleneck resources.Increase Net Availability – Reduce breakdown and work stoppage by improved maintenance policies and effective problem-solving, to reduce the frequency and duration of breakdowns and maintenance outside of working hours. Reduce absenteeism.
22 Reducing Resource Capacity Waste Reduce setup time – setup time per unit is Sp/Qp. Decrease the frequency of changeovers, reduce the time required for each setup and manage the product mix to decrease changeover time from one product to the next.Move some of the work to non-bottleneck resources –This may require greater flexibility on the part of non-bottleneck resources as well as financial investments in tooling and cross-training.Reduce interference waste – Eliminate starvation and blockage among work-stations.Methods improvement.Training.Management.
23 Total Unit Load for Product mix STOPCompute the unit load and the total unit load for each Load batch of Regular tile, Jumbo tile and a product mix of 75% Regular and 25% Jumbo . Load Batches are 4 and 9 for regular and jumbo, respectively. Set-up Batches are 300 and 100 for regular and jumbo, respectively.
24 Effective Capacity of a Resource Unit STOPTheoretical Capacity of a resource unit =(1/Unit Load) × Load Batch × Scheduled AvailabilityScheduled Availability – the scheduled time period during which a resource unit is available for processing flow units.Availability factor = Net Availability/Scheduled AvailabilityEffective Capacity of a resource unit =(1/Total Unit Load) × Load Batch × Scheduled AvailabilityEffective Capacity of a pool =(c/Tp) × Load Batch ×Scheduled Availability