Throughput and Capacity Management Examples

Throughput Key Problems Effective Capacity & Utilization Based on the book: Managing Business Process Flows.

Please read the material up to slide 16. The solution to the problem on slides 9-15 is on slide 8. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 2

Problem 1 Example: The average arrival rate to a store is 6 customers per hour. The average service time is 5 min per customer. R = 6 customers per hour Rp =1/5 customer per minute, or 60(1/5) = 12/hour U= R/Rp = 6/12 = 0.5 a) How long does a customer stay in the processor (with the server)? Tp = 5 minutes b) On average how many customers are there with the server? RTp = Ip = 6(5/60) = 0.5 Alternatively; Ip = cU =1(0.5) = 0.5 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 3

Problem 2 What if the arrival rate is 11 per hour? Processing rate is still Rp=12 U= R/Rp U=11/12 a) How long does a customer stay in the processor (with the server)? Tp = 5 minutes b) On average how many customers are there with the server? RTp = Ip = (11/60)(5) = 11/12 Alternatively Ip = cU = 1(11/12) = 11/12 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 4

Problem 3 A local GAP store on average has 10 customers per hour for the checkout line. The store has two cashiers. The average service time for checkout is 5 minutes . Arrival rate: R = 10 per hour Average inter-arrival time: Ta = 1/R = 1/10 hr = 6 min Average service time: Tp = 5 min Number of servers: c =2 Rp = c/Tp = 2/5 per min or 24 per hour U= R/Rp = 10/24 = 0.42 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 5

Problem 3 a) How long does a customer stay in the processors (with the servers)? Average service time: Tp = 5 min b) On average how many customers are therewith the servers? Ip =? RTp = Ip = (1/12)(10) = 0.84 Alternatively Ip = cU = 2(0.42) = 0.84 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 6

Problem 4 A call center has 11 operators. The average arrival rate of calls is 200/hr. Each operator on average can serve 20 customers per hr. a) Compute the capacity 11*20 = 220 Alternatively, Tp = 1/Capacity of one server =1/20 hrs. = 3 mins. Rp=c/Tp = 11/3 per min. or 60*(11/3) = 220 per hr. b) Compute the utilization. U=200/220 = 0.91 c) How long does a customer stay in the processors? Tp= 3 min. d) On average how many customers are with the servers (Ip)? RTp = Ip = (200/60)(3) = 10 Ip = cU = 11(200/220) = 10 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 7

Lecture on the Next Problem https://youtu.be/iIT9VWpdqY4 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 8

Problem 4. Problem 5.1 in the book A law firm processes shopping centers and medical complexes contracts. There are four Paralegals, three Tax lawyers, and two Senior partners. The unit loads of the resources to handle one standard contract is given below. Assume 8 hours per day, and 20 days per month. It takes a Paralegal 20 hours to complete 3 contract. That is 20/3 = 6.666666666666666667 hours to complete a contract. It takes a Tax lawyer 2 hours to complete a contract. It takes a Senior partner 2 hours to complete a contract. a) What is the Theoretical Flow Time of a contract? 6.667+2+2=10.667. Flow Time = Theoretical Flow Time + Waiting times Flow Time = 10.67 + Waiting times Compute the Capacity of each of the three Resource Pools Throughput-Part 2 Ardavan Asef-Vaziri April-2015 9

Paralegals Operation 3 Senior Partners Operation 1 Paralegals Operation 2 Tax Lawyers Tp=2 hr., c = 2 Tp=20/3 hr., c= 4 Tp=2 hr., c=3 A Paralegal can complete 1 contract in 20/3 = 6.667 hour How many contracts in one hour? 1/6.667 , or 1/(20/3) = 0.15 How many contracts all the Paralegals can complete in one hour. There are 4 Paralegals: c = 4 Four Paralegals 4(0.15) = 0.6 contracts per hours We could have also said Tp = 20/3 = 6.6667. Capacity of one resource unit is 1/Tp. Capacity of one resource unit is 1/6.667 = 0.15. Capacity of the resource units: Rp=c/Tp ; c=4 and Tp = 20/3 =6.667 Rp = 4/(20/3) = 4/6.667 = 0.6 per hour Capacity of the resource pool is 0.6 contracts per hour. It is 8(0.6) = 4.8 contracts per day Throughput-Part 2 Ardavan Asef-Vaziri April-2015 10

Tax Lawyers A Tax Lawyer can complete 1 contract in 2 hour How many contracts in one hour? 1/2 = 0.5 How many contracts all the Tax Lawyers can complete in one hour. There are 3 Tax Lawyers: c = 3 There Tax Lawyers 3(0.5) = 1.5 contracts per hours We could have also said Tp = 2. Capacity of one resource unit is 1/Tp. Capacity of one resource unit is 1/2 = 0.5. Capacity of all resource units: Rp=c/Tp where c=3 and Tp = 2 Rp = 3/2 = 1.5 per hour Capacity of the resource pool is 1.5 contracts per hour. It is 8(1.5) = 12 contracts per day Throughput-Part 2 Ardavan Asef-Vaziri April-2015 11

Senior Partners A Senior Partners can complete 1 contract in 2 hours. How many contracts in one hour? 1/2 = 0.5 How many contracts all the Senior Partners can complete in one hour. There are 2 Senior Partners: c = 2 There Senior Partners 2(0.5) = 1 contracts per hours We could have also said Tp = 2. Capacity of one resource unit is 1/Tp. Capacity of one resource unit is 1/2 = 0.5. Capacity of all resource units: Rp=c/Tp where c=2 and Tp = 2 Rp = 2/2 = 1 per hour Capacity of the resource pool is 1 contracts per hour. It is 8(1) = 8 contracts per day Throughput-Part 2 Ardavan Asef-Vaziri April-2015 12

Problem 4. Problem 5.1 in the book c) Compute the capacity of the process. Unit Load 50%SH 50%MD Capacity of a Resource Unit /hr Capacity of the Resource Pool/hr Cap of the R- Pool / day # Of Resourse Units Paralegal 0.6 4.8 6.667 0.15 4 Tax lawyer Senior partner 1.5 12 2 0.5 3 1 8 2 0.5 2 d) Compute the cycle time? Capacity is 4.8 per day. Cycle Time = 1/Capacity Cycle Time = 1/4.8 days. That is 8(1/4.8) = 8/4.8 = 1.67 hrs. Capacity is 0.6 per hr. Cycle Time = 1/Capacity Cycle Time = 1/0.6 hrs. Cycle Time = 1.67 hrs. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 13

Problem 4. Problem 5.1 in the book d) Compute the average inventory (assume U=1). Lets look at the utilization of the 3 stations Station Capacity Throughput Utilization Station 1 4.8 Station 2 12 Station 3 8 On 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. 4.8 4.8 4.8 4.8/4.8 = 1 4.8/12 = 0.4 4.8/8 = 0.6 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 14

Problem 4. Problem 5.1 in the book Inventory with the processors is 1+ 0.4+0.6 = 2 On average there are 2 flow units with the processors; Inventory in the processors (Ii) Now let s look from another angle; from the Little s Law point of view RT=I R= 4.8 per 8 hours or 0.6 per hour T =10.67 hours I = 0.6(10.67) = 6.4 6.4 vs 2? Where is my mistake?? 1(4)+ 0.4(3)+0.6(2) = 6.4 e) There are 150 cases in November can the company process all 150 cases? 150/20 = 7.5 per day 4.8 (Capacity) < 7.5 (Demand). Throughput-Part 2 Ardavan Asef-Vaziri April-2015 15

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? # of paralegals required = 7.5/1.2 = 6.25 # of tax lawyers required = 7.5/4 = 1.875 # of tax lawyers required = 7.5/4 = 1.875 These could be rounded up to 7, 2 and 2 We need 7, 2, and 2. We have 4, 3, and 2. 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. 0.25(8) = 2 hours total over time. There will be 6 paralegals; over time pf each = 2/6 = 1/3 hour Or 20 minute per paralegal. PLUS some safety Capacity. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 16

A Little More Practice Now suppose throughput is 3.6 contracts per day. Compute the average inventory. Station Capacity Throughput Utilization Station 1 4.8 Station 2 12 Station 3 8 Managerial Observation: Note that the utilization of bottleneck resource is not necessarily 100%. On average 0.75 person with a resource in Station 1 0.3 person with a resource in Station 2 0.45 person with a resource in Station 3 3.6 3.6 3.6 3.6/4.8 = 0.75 3.6/12 = 0.3 3.6/8 = 0.45 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 17

A Little More Practice Station Station1 Station2 Station3 I = 3+0.9+0.9= 4.8 Lets check it through the Little s Law Theoretical Flow Time = 6.66667+2+2= 10.66667 RT=I 3.6(10.66667/8) = I I = 4.8 (3.6/8)(10.66667) = 4.8 Now suppose there are 16.8 contracts are waiting in different waiting lines? What is the Flow Time ? RT=I 3.6*T= 16.8+4.8 = 21.6 3.6T = 21.6 T=6 days Rp 4.8 12 8 R 3.6 3.6 3.6 U 0.75 0.3 .45 c 4 3 2 Ip 4* 0.75 = 3 3* 0.3 = 0.9 2*0.45 = 0.9 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 18

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 as Resource breakdown Maintenance Quality rejects Rework and repetitions Setups between different products or batches We 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. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 19

Activity Time and Unitload Throughput-Part 2 Ardavan Asef-Vaziri April-2015 20

ThCapacity, Capacity, VThFlowTime, ThFlowTime, FlowTime Activity time Capacity is computed based on the Unit Load Theoretical Flow Time is computed based on Activity Time Then What is Flow Time? 10 mins. 30 mins. Flow Time Ti + Tp Flow time includes time in buffers Capacity does not care about time in buffers 3 days 10 mins. 30 mins. Theoretical Unit Load, Theoretical Activity Time Theoretical Capacity is computed based on the Theoretical Unit Load (ThUL) Theoretical Flow Time is NOT computed based on Theoretical Activity Time Very Theoretical Flow Time is computed based on theoretical Activity Time ThUL(1+CWF) = Unit Load (Tp), 30 mins. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 21

ThCapacity, Capacity, VThFlowTime, ThFlowTime, FlowTime Unit Load Activity time Capacity Theoretical Flow Time 10 mins. 30 mins. Flow Time Capacity does not care about buffer times 3 days 10 mins. 30 mins. Theoretical Unit Load Theoretical Activity Time Theoretical Capacity Very Theoretical Flow Time 30 mins. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 22

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) = 30 Theoretical Capacity = c/ThUnit Load Effective Capacity = Capacity = c/Unit Load. Theoretical Capacity = 1/20 per minute or 3 per hour. Effective Capacity = Theoretical Capacity (1-CWF) Throughput-Part 2 Ardavan Asef-Vaziri April-2015 23

Problem 4. Problem 5.1 in the book=CWF 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. ThUL Shopping (hrs /contract) 4 ThUL Load Medical (hrs No. Of Professionals Paralegal 6 4 Tax lawyer 1 3 3 Senior partner 1 1 2 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 24

Problem 4. Problem 5.1 in the book-CWF a) 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 hrs. Theoretical Capacity = 1/5 per hr. Capacity Waste Factor (CWF) = 0.25 Unit Load = Tp = 5/(1-0.25) = 20/3 hrs. Effective Capacity = Capacity = 1/(20/3) = 3/20 per hr. Tax Lawyer: Theoretical Unit Load 0.5(1)+0.5(3) = 2 hrs. CWF = 0 Theoretical Unit Load = Tp = 2 hrs. Theoretical Capacity = 1/2 per hr. Effective Capacity = Capacity = 1/2 per hr. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 25

Problem 4. Problem 5.1 in the book-CWF Senior Partner: Theoretical Unit Load 0.5(1)+0.5(1) = 1 hrs. Theoretical Capacity = 1/1 = 1 per hr. CWF = 0.5 Unit Load = Tp = 1/(1-0.5) = 2 hrs. Effective Capacity = Capacity = 1/2 per hr. ThUnit Load SH (hrs) ThUnit Load MD (hrs) Unit Load 50%SH 50%MD Teoretical Capacity of a Resource Unit /hr Unit Load 50%SH 50%MD Capacity of a Resource Unit /hr Th Capacity of the Resource Pool/hr Capacity of the Resource Pool/hr Cap of the R- Pool / day # Of Resourse Units Th Cap of R- Pool / day CWF Paralegal 4 6 5 0.2 0.25 6.667 0.15 4 0.8 0.6 6.4 4.8 Tax lawyer Senior partner 1 3 2 0.5 0 2 0.5 3 1.5 1.5 12 12 1 1 1 1 0.5 2 0.5 2 2 1 16 8 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 26

A Second Example on Throughput Part 2a The theoretical unit loads for an aggregate policy (combination of all policies) in a law firm are: 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. 1. Compute the VERY theoretical flow time for an aggregate product. 4+3+2 = 9 hours 2. 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 = 5.71+3.33+ 2.86 = 11.9 What is the average CWF UL(1-CWF) = ThUL 11.9(1-CWF) = 9 11.9-9 = 11.9CWF CWF = 0.24 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 27

A Second Example on Throughput Part 2a 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) = Min(0.175, 0.3, 0.35) = 0.175 /hr. Min[1/(4/0.7),1/(3/0.9), 1/(2/0.7)]=Min(0.7/4,0.9/3,0.7/2) = 0.175 /hr. or 8(0.175)= 1.4 per day 5. Compute the cycle time. 1/1.4 = 0.7143 day or 8(0.7143) = 5.71 hour OR 1/0.175 = 5.71 hour OR 1/(1/5.71) = 5.71 hour 6. Compute the throughput per day if process utilization is 80%. Capacity = 1.4 per day U=0.8 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 28

A Second Example on Throughput Part 2a U = Throughput/Capacity 0.8 = R/1.4 Throughput = 1.12 per day 7. Compute the takt time. 1/1.12 = 0.893 day OR 8(0.893) = 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 1.4, 2.4, 2.8 per day. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 29

A Second Example on Throughput Part 2a 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 = 0.4 10. 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 = 0.4 0.8+0.47+0.4 = 1.67 11. 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 = 1.67+10 = 11.67 RT = I 1.12T = 11.67 T = 11.67/1.12 = 10.42 Throughput-Part 2 Ardavan Asef-Vaziri April-2015 30

Multiple Choice Which of the following 2 statements is true? I. A process can have more than 1 bottleneck resource. II. Having flexible equipment can increase utilization. A) Only I B) Only II C) Both I and II D) Neither I nor II E) Cannot be determined Which of the following statement is false? A) Throughput rate is always smaller than or equal to the capacity B) Customers may wait even if the utilization rate of the service process is smaller than 100% C) Bottleneck resource(s) always has 100% utilization rate D) Increasing WIP may increase utilization rate E) None of the above Throughput-Part 2 Ardavan Asef-Vaziri April-2015 31

Multiple Choice To improve the utilization rate, we can I. Cross-train the workers II. Adopt flexibility equipment III. Shift from MTS systems to MTO system Choose the most appropriate. A) I B) II C) III D) I and II E) I, II, and III Throughput-Part 2 Ardavan Asef-Vaziri April-2015 32

Throughput Improvement Mapping Throughput Effective Capacity Theoretical Capacity Throughput << Capacity External Bottleneck External blockage (demand) - sales efforts, advertising budget, . Make them an offer they can t refuse. prices, quality, time variety. External starvation (supply) - identifying additional suppliers, more reliable suppliers. Throughput = Capacity Internal Bottleneck Increase financial capacity - modifying the product mix. Increase physical capacity - c Tp if Capacity Theoretical Capacity If Capacity << Theoretical Capacity. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 33

Reducing Resource Capacity Waste Tp Capacity << Theoretical Capacity 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. resources are not utilized Throughput-Part 2 Ardavan Asef-Vaziri April-2015 34

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. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 35

Increasing Resource Levels c Capacity Theoretical Capacity Resources are efficiently utilized; increase the theoretical capacity. Increase the level of resources. c. buy one more oven Increase the size of resource units - Larger load batch - more loaves in the oven Increase the time of operation Scheduled Availability, Overtime Subcontract or Outsource Technology- Speed up the activities rate - Invest in faster resources or incentives for workers. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 36

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 product Average setup time per unit is then Sp/Qp Sp/Qp is also included in Tp Throughput-Part 2 Ardavan Asef-Vaziri April-2015 37

Setup Batch Size: Throughput or Flow Time What 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. Throughput-Part 2 Ardavan Asef-Vaziri April-2015 38

Throughput or Flow Time Product Mix: 50%-50% Set-up time 30 min per product Working hours 8 hours/day Compute the effective capacity under min cost strategy. Two set-ups each for 30 min = 60 mins An aggregate product takes (10+20)/2 = 15 Production time = 8*60-60 = 420 mins Capacity = 420/15 = 28 aggregate units Each aggregate unit is 0.5 A and 0.5 B (total of 14A and 14B) 10 min/unit A Operation B 20 min/unit 1 machine 100% available Throughput-Part 2 Ardavan Asef-Vaziri April-2015 39

Throughput & Cost or Flow Time 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. Product A: 2(10)0+30 = 50 Product B: 2*(20)+30 = 70 An aggregate product takes (50+70)/2/2 = 30 = 0.5 hour Capacity = 8/0.5 = 16 aggregate units per day. Each aggregate unit is 0.5 A and 0.5 B (total of 8A and 8B) 8A and 8B need 4*30*2 = 4 hrs. Setup + 8*10+8*20 = 4 hrs. Production, 10 min/unit A Operation B 20 min/unit 1 machine 100% available Throughput-Part 2 Ardavan Asef-Vaziri April-2015 40