1. 1 3. Process Flow Measures 1. Coffee Shop You enter a Starbucks coffee shop. You have experienced that in this time of day the entrance door opens two times per minute; once for a customer to come in, once for a customer to leave. What is throughput of this system. Every minutes one customer comes in and one customer leaves. R = 1 per minute. There are 5 customers in the line, how long does it take you to get your coffee and leave. I = 5, R=1, RT=I 1T=5  T=5

2. 2 3. Process Flow Measures 1. Coffee Shop Now suppose there are two waiting lines. Suppose R is still 1 per minutes and still on average there are 5 customers in the first line to pay for their order and get their non-exotic orders. In addition, suppose there are 4 people in the exotic order (latte, cappuccino, etc.) waiting line. 40% of the customers place exotic orders. What is the flow time of a person who orders latte, cappuccino, etc. Such a customer spends 5 minutes in the first line. Throughput of the second line is R= 0.4(1) = 0.4 customer per minute. Inventory of the second line is 4. RT=I  0.4T=4  T=10 Simple order T =5, Exotic order T= 5+10 = 15

3. 3 3. Process Flow Measures 1. Coffee Shop What is the flow time of a customer. S/he is neither a customer who puts a simple order nor one with exotic order, but s/he is both. Procedure 1- Not good. 60% simple order: T = 5 40% exotic order: T=5+10= 15 A customer: T= 0.6(5) + 0.4(15) = 9 minutes Procedure 2- Not bad. Every one goes through the first process and spends 5 minutes. 60% spend no additional time, 40% spend 10 minutes additional. 0.6(0) + 0.4(10) = 4  4+5 =9.

4. 4 3. Process Flow Measures 1. Coffee Shop Procedure 3- Good. Throughput of the system is 1 per minute. There are 9 people in the system (5 at the register and 4 in the second line). RT= I 1T=9 T=9 Throughput in this system was 1 per minute or 60 per hour or 720 per day (assuming 12 hours per day). But inventory in the system is always 9.

5. 5 3. Process Flow Measures b) 50% of all the claims that TTIS receives are car insurance claims, 10% motorcycle, 10% boat, and the remaining are house insurance claims. On average, there are, 300 car, 114 motorcycle, and 90 boat claims in process. How long, on average, does it take to process a car insurance claim? 4. Insurance Company 240 car motorcycle boat house 300 114 90 ? 120 24 72 0.5 0.1 0.3 I = 300 cars R = 0.5(240) =120 claims/wk TR = I  T(120) = 300 T = 300/120= 2.5 weeks

6. 6 3. Process Flow Measures c) How long, on average, does it take to process a house insurance claim? Average # of claims in process = 720 720–300 car–114 motorcycle–90 boat = 216 Average # of claims for house: I = 216 House claims are 1- 0.5-0.1-0.1 = 0.3 of all claims R = 0.3(240) = 72 TR = I  T = I/R  T = 216/72= 3 weeks 240 car motorcycle boat house 300 114 90 ? 120 24 72 0.5 0.1 0.3 4. Insurance Company

7. 7 3. Process Flow Measures 5. Fresh Juice and Fruit 5. Consider a roadside stand that sells fresh oranges, and fresh orange juice. Every hour, 40 customers arrive to the stand, and 60% purchase orange juice, while the remaining purchase oranges. Customers first purchase their items. Customers that purchased oranges leave immediately after purchasing their oranges. Any customer that ordered orange juice must wait while the juice is squeezed. There are 3 customers on average waiting to purchase either oranges or orange juice, and 1 customer on average waiting for orange juice to be squeezed. 40 31 0.6(40) = 24 How long on average must customers that purchase fresh orange juice wait?

8. 8 3. Process Flow Measures 5. Fresh Juice and Fruit How long on average must customers that purchase fresh orange juice wait? In the ordering process RT =I  40T = 3  T = 3/40 hours  T = 60(3/40) = 4.5 minutes In the juice squeezing process RT =I  24T = 1  T = 1/24 hours  T = 60(1/24) = 2.5 minutes Average waiting time = T = 4.5 + 2.5 = 7 minutes 40 31 24

9. 9 3. Process Flow Measures 6. Cold Beverage 6. A recent CSUN graduate has opened up a cold beverage stand “CSUN-Stop” in Venice Beach. She takes life easy and does a lot of surfing. It sounds crazy, but she only opens her store for 4 hours a day. She observes that on average there are 120 customers visiting the stand every day. She also observes that on average a customer stays about 6 minutes at the stand. a) How many customers on average are waiting at “CSUN- Stop”? R = 120 in 4 hours  R = 120/4 = 30 per hour R = 30/60 = 0.5 per minute T = 6 minutes RT = I  I = 0.5(6) = 3 customers are waiting

10. 10 3. Process Flow Measures 6. Cold Beverage She is thinking about running a marketing campaign to boost the number of customers per day. She expects that the number of customers will increase to 240 per day after the campaign. She wants to keep the line short at the stand and hopes to have only 2 people waiting on the average. Thus, she decides to hire an assistant. b) What is the average time a customer will wait in the system after all these changes? R = 240/(4hrs*60min) = 1 person/minute I = 2 Average time a customer will wait: T = I/R = 2/1 = 2 minutes

11. 11 3. Process Flow Measures 6. Cold Beverage The business got a lot better after the marketing campaign and she ended up having about 360 customers visiting the stand every day. So, she decided to change the processes. She is now taking the orders and her assistant is filling the orders. They observe that there are about 2 people at the ordering station of the stand and 1 person at the filling station. c) How long does a customer stay at the stand? R = 360/4 hours = 1.5 people/minute I = 2 (ordering station) and 1 (filling station) = 3 RT = I  1.5T = 3  T = 2 minutes 360/4hrs 21

12. 12 3. Process Flow Measures 6. Cold Beverage A recent UCLA graduate has opened up a competing cold beverage stand “UCLA-Slurps”. The UCLA grad is not as efficient as the CSUN grad, so customers must stay an average of 15 minutes at “UCLA-Slurp”, as opposed to 6 minutes at the “CSUN-Stop”. Suppose there is an average of 3 customers at “UCLA-Slurps”. The total number of customers remains at 120, as it was before the marketing campaign. But now the 120 is divided between the “CSUN-Stop” and “UCLA-Slurps”. d) By how much has business at the “CSUN-Stop” decreased? First find how many customers “CSUN-Stop” is losing to “UCLA-Slurps”

13. 13 3. Process Flow Measures 6. Cold Beverage At “UCLA-Slurps” we have I = 3 people and T = 15 minutes TR = I  15R = 3  R= 1/5 per minute R = 60(1/5) = 12 customers per hour or = 48 customers/day Business at “CSUN-Stop” has decreased by 48 customers/day The new (lower) arrival rate to the “CSUN-Stop”? 120-48 = 72 customers/day e) What is now the average number of customers waiting at the “CSUN-Stop” if the flow time at CSUN-Stop remains 6 mins? R = 72/day = 72/(60min×4hrs) = 0.3/minute T = 6 minutes RT = I  I = 0.3(6) = 1.8 customers

14. 14 3. Process Flow Measures 7. Business School The following example may also work in differentiating R and I. On average 1600 students per year enter a business school and there are the same number of graduates. They are 25% accounting major, 20% marketing, 20% management, 15% finance, and the rest are other majors. The business school has 9000 students. 3000 accounting, 1500 Management, 2000 Marketing, 1000 finance, and 1500 other majors. a) On average how many accounting students get graduated each year. Racc = 0.25(1600) = 400 per year b) On average how long does it take a finance student to get graduated. Rfin = 0.15(1600) = 240 /year Ifin = 1000  TFIN = 1000/240 = 4.17 year

15. 15 3. Process Flow Measures 7. Business School a) Students of what majors spent the least and the most amount of time at this school. ​

16. 16 3. Process Flow Measures 8. Call Center 3. A call center employs 1000 agents. Every month 50 employees leave the company and 50 new employees are hired. a) How long on average does an agent work for this call center? Now suppose the cost of hiring and training a new agent is $1000. The manager of this call center believes that increasing agents’ salary would keep them working longer term in the company. The manager wants to increase the average time that an agent works for the call center to 24 months, or 2 years. b) If an agent works for the call center for 24 months on average, how much can the company save on hiring and training costs over a year? Hint: first determine the current annual cost for hiring and training, then determine the new annual cost for hiring and training. c) How much the monthly salary of each agent can be increased?

17. 17 3. Process Flow Measures 8. Call Center 1000 Agents 50/month a) How long on average does an agent work for this call center? R = 50 people/month I = 1000 people Average working time = Throughput Time = I/R = 1000/50 = 20 months or 20/12 = 1.67 years Suppose the cost of hiring and training a new agent is $1000. The manager of this call center believes that increasing agents’ salary would keep them working longer term in the company. The manager wants to increase the average time that an agent works for the call center to 24 months, or 2 years.

18. 18 3. Process Flow Measures 8. Call Center b) If an agent works for the call center for 24 months on average, how much can the company save on hiring and training costs over a year? Hint: first determine the current annual cost for hiring and training, then determine the new annual cost for hiring and training. 1000 Agents ?/month 2 years 1000 Agents 50/month 20 months Current annual cost for hiring and training: Throughput Rate = 50 people/month = 600 people/year Annual hiring and training cost is 600 (1000) = $600,000

19. 19 3. Process Flow Measures 8. Call Center New annual cost for hiring and training: Average working time = Throughput Time = 24 months = 2 years Throughput Rate  R= I/T = 1000 people / 2 years = 500 people/year  41.7 per month Annual hiring and training cost is 500 (1000) = $500,000 Annual saving on hiring and training cost is $600,000-$500,000 = $100,000 c) How much the monthly salary of each agent can be increased? Average # of employees = 1000 Annual saving on hiring and training cost = $100,000 Monthly saving = $8,333.33 8333.33/1,000 = $8.33 per employee/month

20. 20 3. Process Flow Measures Auto-Moto Financial Services- The Old Process Auto-Moto receives 1,000 applications per month. In the old process, each application is handled in a single activity, with 20% of applications being approved. 500 were in the process at any time. Average flow time T = ? The firm recently implemented a new loan application process. In the new process, applicants go through an initial review and are divided into three categories. Discussion: How operational power destroys the walls of poverty. RT = I T = I/R = 500/1,000 months = 0.5 month or 15 days. Process I p =500 1000/month 200/month 800/month

21. 21 3. Process Flow Measures New Process: The Same R, But smaller I Initial Review Subprocess A Review Subprocess B Review Accepted Rejected 1000/month 25% 50% 25% 70% 30% 90% 10% 800/month 200/month R = 1000 I = I IR + I A + I B = 200 + 25 + 150 = 375 Inventory reduced to 375 from 500 in the old process. Since R is constant, therefore T has reduced. T = I/R = 375/1000 = 0.375 month or 0.375(30) = 11.25 days The new process has decreased the processing time from 15 days to 11.25 days. I B = 150 I A = 25 I R = 200

22. 22 3. Process Flow Measures Compute average flow time. Compute average flow time at Initial Review Process. Compute average flow time at Subprocess A. Compute average flow time at Subprocess B. Compute average flow time of an Accepted application. Compute average flow time of a Rejected application.

23. 23 3. Process Flow Measures Flow Time at Each Sub-process (or activity) Average Flow Time for sub-process IR. Throughput R IR = 1,000 applications/month Average Inventory I IR = 200 applications T IR = 200/1,000 = 0.2 months = 6 days in the IR sub-process Average Flow Time for sub-process A. Throughput R A = 250 applications/month Average Inventory I A = 25 applications T A = 25/250 months = 0.1 months = 3 days in sub-process A. Average Flow Time for sub-process B. Throughput R B = 250 applications/month Average Inventory I B = 150 applications T B = 150/250 months = 0.6 months = 18 days in sub-process B

24. 24 3. Process Flow Measures Routing, Flow Time, and Percentage of Each Flow units One flow unit at very macro level: Application 1000 flow units/month at very micro level: Each specific application Two flow units: Accepted and rejected Accepted-A: IR, A Accepted-B: IR, B Five flow units: Accepted-A, Accepted-B Rejected-IR, Rejected-A, Rejected-B T IR = 6 days T A = 3 days T B = 18 days We also need percentages of each of the five flow units Rejected-IR: IR Rejected-A: IR, A Rejected-B: IR, B

25. 25 3. Process Flow Measures New Process: Intermediate Probabilities Initial Review T = 6 Subprocess A Review T = 3 Subprocess B Review T = 18 Accepted Rejected 100% 25% 50% 25% 70% 30% 90% 10% 80% 20%

26. 26 3. Process Flow Measures New Process: Intermediate Probabilities Initial Review T = 6 Subprocess A Review T = 3 Subprocess B Review T = 18 Accepted Rejected 100% 25% 50% 25% 17.5% 7.5% 22.5% 2.5% 80% 20% 50%

27. 27 3. Process Flow Measures Flow Time of the Accepted Applications Under the Original Process – the average time spent by an application in the process is 15 days (approved or rejected). In the new process: 15 days reduced to 11.25 days. On average, how long does it take to approve an applicant? On average, how long does it take to reject an applicant? Accepted-A: IR, A  Accepted-A(T) = 6 + 3 = 9  Accepted-A = 17.5 % Accepted-B: IR, B  Accepted-B(T) = 6 + 18 = 24  Accepted-B = 2.5 % Average Flow time of an accepted application = [0.175(9)+0.025(24)] /(0.175+.025) = 10.875 The average flow time has reduced from 15 to 11.25. In addition, the flow time of accepted applications has reduced to 10.875. That is what the firm really cares about, the flow time of the accepted applications.

28. 28 3. Process Flow Measures Flow Time of Rejected Applications Rejected-IR: IR  Rejected-IR(T) = 6  Rejected-IR(%) = 50% Rejected-A: IR, A  Rejected-A(T) = 6+3 = 9  Rejected-A(%) = 7.5% Rejected-B: IR, B  Rejected-B(T) = 6+18 = 24  Rejected-B(%) = 22.5% Average Flow time of a rejected application = = 11.343 Check our computations: Average flow time of an application 0.8(11.343)+0.2(10.875) = 11.25

29. 29 3. Process Flow Measures A hospital emergency room (ER) is currently organized so that all patients register through an initial check-in process. Each patient is seen by a doctor and then exits the process, either with a prescription or with admission to the hospital. 55 patients per hour arrive at the ER, 10% are admitted to the hospital and the rest will leave with a simple prescription. On average, 7 people are waiting to be registered and 34 are registered and waiting to see a doctor. The registration process takes, on average, 2 minutes per patient. Among patients who receive prescriptions, average time spent with a doctor is 5 minutes. Among those admitted to the hospital, average time is 30 minutes. Assume the process to be stable; that is, average inflow rate equals average outflow rate. ER1; Problem 3.4 MBPF

30. 30 3. Process Flow Measures Directions o) Draw the flow process chart b) On average how long a patient spend in ER? c) On average how many patients are in ER? Hints: Compute flow time in buffer 1 Compute average activity time of Doctor Compute the average flow time in this process Compute average flow time for a simple prescription patient Compute average flow time for a potential admission patient Compute number of patients in Doctor activity Compute the average number of patients in the process.

31. 31 3. Process Flow Measures Re-Check 7.6 7.5 37.1 54.2 T D = 0.9(5)+0.1(30) = 7.5 T SP = 7.6+2+37.1+5 = 51.7 T P = 0.1(76.7)+0.9(51.7) = 54.2 Flow Time T PA = 7.6+2+37.1+30 = 76.7

32. 32 3. Process Flow Measures 6.9 Inventory 1.8

33. 33 3. Process Flow Measures 49.7 I = 49.7, R = 55. Is our T= 54.2 correct? RT= I R = 55/60 = 0.916667 minutes RT = I  T=I/R = 49.7/0.916667 = 54.2 Recheck The Little’s Law

34. 34 3. Process Flow Measures A triage system has been proposed for the ER described in Exercise 3.4. As mentioned earlier, 55 patients per hour arrive at the ER. Patients will be registered as before and it takes an average of 2 minutes per patient. They will then be quickly examined by a nurse practitioner who will classify them as Simple Prescriptions or Potential Admits. Planners anticipate that the initial examination by triage nurses will take 3 minutes. They expect that, on average, 20 patients will be waiting to register and 5 will be waiting to be seen by the triage nurse. Planners expect eh Simple Prescriptions area to have, on average, 15 patients waiting to be seen. As before, once a patient’s turn come, each will take 5 minutes of a doctor’s time. The hospital anticipates that, on average, the emergency area will have only 1 patient waiting to be seen. As before, once that patient’s turn comes, he or she will take 30 minutes of a doctor’s time. Assume that, as before, 90% of all patients are Simple Prescriptions, assume, too, that the triage nurse is 100% accurate in making classifications. ER2; Problem 3.5 MBPF

35. 35 3. Process Flow Measures Directions o) Draw the flow process chart a) On average how many patients are in ER? b) On average, how long a patient spend in ER? c) Compute average flow rate in buffer 3 and buffer 4. d) Compute average flow time in all buffers. e) Compute average number of patients in all activities. f) On average, how long does a simple prescription patient spend in the ER? d) On average, how long does a Potential Admit patient spend in the ER?

36. 36 3. Process Flow Measures Process Flow and TR=I Table

37. 37 3. Process Flow Measures I ER = 52.5 a) On average, how many patients are in ER? Method 1: Macro Method, a single flow unit R ER = 55/60 flow units / min, I ER = 52.5 b) On average, how long will a patient spend in ER? T ER = 52.5/(55/60) = 57.2 minutes Inventory and Flow Time; Macro Method

38. 38 3. Process Flow Measures Method 2: Micro Method, two flow units, Potential Admission and Simple Prescription T PA = 21.8+2+5.5+3+10.9 +30 = 73.2 T SP = 21.8+2+5.5+3+18.2 +5 = 55.5 T ER =.1(73.2)+.9(55.5) = 57.3 c) On average, how long will a potential admission patient spend in ER? 73.2 Flow Time; Micro Method

39. 39 3. Process Flow Measures Refer again to Exercise 3.5. Once the triage system is put in place, it performs quite close to expectations. All data conform to planners’ expectations except for one set-the classifications made by the nurse practitioner. Assume that the triage nurse has been sending 91% of all patients to the Simple Prescription area when in fact only 90% should have been so classified. The remaining 1% is discovered when transferred to the emergency area by a doctor. Assume all other information from Exercise 3.5 to be valid. ER3; Problem 3.6 MBPF

40. 40 3. Process Flow Measures Directions o) Draw the flow process chart a) On average how many patients are in ER? b) On average, how long a patient spend in ER? c) Compute average flow rate in buffer 3 and buffer 4. d) Compute average flow time in all buffers. e) Compute average number of patients in all activities. f) On average, how long does a simple prescription patient spend in the ER? d) On average, how long does a Potential Admit patient spend in the ER?

41. 41 3. Process Flow Measures Problem 3.6 5.5 Process Flow and Throughputs

42. 42 3. Process Flow Measures Macro: Average number of patients in the system = 20+1.8+5+2.8+1+2.8+15+4.2 = 52.6 Average flow time = I/R = 52.6/(55/60) = 57.3 Inventory and Flow Time; Macro Method

43. 43 3. Process Flow Measures Micro Method Compute SP and PA first Common = 21.8 +2+5.5+3 = 32.3 T SP = 32.3+18+5 = 55.3 Flow Time; Simple Prescription

44. 44 3. Process Flow Measures Flow Time; Potential Admission and Overall 0.9 0.01 0.09

45. 45 3. Process Flow Measures Recheck T= 55.3 (0.9) + 73.2 (0.09) + 96.2 (0.01) T= 57.32 T PA1 = 32.3 +10.9+30 = 73.2 ……(4.95 PA patients out of 5.5 PA patient: 90%) T PA2 = 73.2+18+5 = 96.2 (0.55 PA patients out of 5.5 PA patient: 10%) T PA = 73.2(.9) + 96.2(.1) =75.5 Flow Time; Potential Admission and Overall Recheck Again T SP = 55.3 is 90% of flow units, and T PA =75.5 is for 10% of flow units T = 55.3 (.9) + 75.5(.1) = 57.32