1. 1 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Made-to-stock (MTS) operations  Product is manufactured and stocked in advance of demand  Inventory permits economies of scale and protects against stockouts due to variability of inflows and outflows  Make-to-order (MTO) process  Each order is specific, cannot be stored in advance  Process manger needs to maintain sufficient capacity  Variability in both arrival and processing time  Role of capacity rather than inventory  Safety inventory vs. Safety Capacity  Example: Service operations Make to stock vs. Make to Order

2. 2 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Banks (tellers, ATMs, drive-ins)  Fast food restaurants (counters, drive-ins)  Retail (checkout counters)  Airline (reservation, check-in, takeoff, landing, baggage claim)  Hospitals (ER, OR, HMO)  Service facilities (repair, job shop, ships/trucks load/unload)  Some production systems- to some extend (Dell computer)  Call centers (telemarketing, help desks, 911 emergency) Examples

3. 3 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 The DesiTalk Call Center Sales Reps Processing Calls (Service Process) Incoming Calls (Customer Arrivals) Calls on Hold (Service Inventory) Answered Calls (Customer Departures) Blocked Calls (Due to busy signal) Abandoned Calls (Due to long waits) Calls In Process (Due to long waits) The Call Center Process

4. 4 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Tp : processing time. Tp units of time. Example, on average it takes 5 minutes to serve a customer. Rp : processing rate. Rp flow units are handled per unit of time. If Tp is 5 minutes. Compute Rp. Rp= 1/5 per minute. Or 60/5 = 12 per hour. Service Process Attributes

5. 5 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Tp : processing time. Rp : processing rate. What is the relationship between Rp and Tp? If we have one resource  Rp = 1/Tp Service Process Attributes What is the relationship between Rp and Tp Rp = c/Tp Each customer always stay for Tp unites of time with the server In general when we have c recourses

6. 6 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 T p = 5 minutes. Processing time is 5 minute. Each customer on average is with the server for 5 minutes. c = 3, we have three servers. Processing rate of each server is 1/5 customers per minute. Or 12 customer per hour. Rp is the processing rate of all three servers. Rp = c/Tp Rp = 3/5 customers/minute. Or 36 customers/hour. Service Process Attributes

7. 7 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Service Process Attributes T a : customer inter-arrival time. On average each 10 minutes one customer arrives. R i : customer arrival (inflow) rate. What is the relationship between Ta and Ri Ta = every ten minutes one customer arrives How many customers in a minute? 1/10; Ri = 1/Ta= 1/10 Ri = 1/10 customers per min; 6 customers per hour Ri = 1/Ta

8. 8 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Service Process Attributes Ri MUST ALWAYS <= Rp We will show later that even Ri=Rp is not possible Incoming rate must be less than processing rate Throughput = Flow Rate R = Min (Ri, Rp Stable Process = Ri < Rp,, so that R = Ri Safety Capacity Rs = Rp – Ri

9. 9 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Waiting time in the servers (processors)? Throughput? Operational Performance Measures Flow time T = T i + T p Inventory I = I i + I p T i : waiting time in the inflow buffer I i : number of customers waiting in the inflow buffer

10. 10 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 = Utilization = inflow rate / processing rate = throughout / process capacity = R/ Rp < 1 Safety Capacity = Rp – R For example, R = 6 per hour, processing time for a single server is 6 min  Rp= 12 per hour, = R/ Rp = 6/12 = 0.5 Safety Capacity = Rp – R = 12-6 = 6 Service Process Attributes

11. 11 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Given a single server. And a utilization of = 0.5 How many flow units are in the server ? Operational Performance Measures Given 2 servers. And a utilization of r = 0.5 How many flow units are in the servers ?

12. 12 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Operational Performance Measures Flow time T T i + T p Inventory I = I i + I p R I = R T R = I/T = Ii/Ti = Ip/Tp Ii = R Ti Ip = R  Tp Tp  if 1 server  Rp = 1/Tp In general, if c servers  Rp = c/Tp

13. 13 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Operational Performance Measures  = R/ R p = (Ip/Tp)/(c/Tp) = Ip/c And it is obvious that  = Ip/c Because on average there are Ip people in the processors and the capacity of the servers is to serve c costomer at a time. = R/R p = Ip/c From the Little’s Law  Rp = c/Tp R = Ip/Tp From definition of Rp 

14. 14 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Sales Throughput Rate  Cost Capacity utilization Number in queue / in system  Customer service Waiting Time in queue /in system Financial Performance Measures

15. 15 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 What is the queue size? What is the capacity utilization? Arrival Rate at an Airport Security Check Point Customer NumberArrival Time Departure Time Time in Process 1055 24106 38157 412208 516259 6203010 7243511 8284012 9324513 10365014

16. 16 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 What is the queue size? What is the capacity utilization? Flow Times with Arrival Every 6 Secs Customer NumberArrival Time Departure Time Time in Process 1055 26115 312175 418235 524295 630355 736415 842475 948535 1054595

17. 17 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 What is the queue size? What is the capacity utilization? Flow Times with Arrival Every 6 Secs Customer NumberArrival TimeProcessing Time Time in Process 1-A077 2-B1011 3-C2077 4-D2227 5-E3288 6-F33714 7-G36415 8-H43816 9-I52512 10-J54111

18. 18 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 What is the queue size? What is the capacity utilization? Flow Times with Arrival Every 6 Secs Customer NumberArrival TimeProcessing Time Time in Process 1-E088 2-H1088 3-D2022 4-A2277 5-B3211 6-J3311 7-C3677 8-F4377 9-G5244 10-I5457

19. 19 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  If inter-arrival and processing times are constant, queues will build up if and only if the arrival rate is greater than the processing rate  If there is (unsynchronized) variability in inter-arrival and/or processing times, queues will build up even if the average arrival rate is less than the average processing rate  If variability in interarrival and processing times can be synchronized (correlated), queues and waiting times will be reduced Conclusion

20. 20 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Two key drivers of process performance are (i) Capacity utilization, and (ii) interarrival time and processing time variability  High capacity utilization ρ= R/ Rp or low safety capacity Rs =R – Rp, due to High inflow rate R Low processing rate Rp=c / Tp, which may be due to small-scale c and/or slow speed 1 /Tp  High, unsynchronized variability in Interarrival times Processing times Causes of Delays and Queues

21. 21 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Variability in the interarrival and processing times can be measured using standard deviation. Higher standard deviation means greater variability.  Coefficient of Variation: the ratio of the standard deviation of interarrival time (or processing time) to the mean.  Ci = coefficient of variation for interarrival times  Cp = coefficient of variation for processing times Drivers of Process Performance

22. 22 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Operational Performance Measures Flow time T = T i + T p Inventory I = I i + I p T i : waiting time in the inflow buffer = ? I i : number of customers waiting in the inflow buffer =?

23. 23 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  R i / R p, where R p = c / T p  C i and C p are the Coefficients of Variation  (Standard Deviation/Mean) of the inter-arrival and processing times (assumed independent) The Queue Length Formula Utilization effect Variability effect

24. 24 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 This part factor captures the capacity utilization effect, which shows that queue length increases rapidly as the capacity utilization p increases to 1 Factors affecting Queue Length The second factor captures the variability effect, which shows that the queue length increases as the variability in interarrival and processing times increases. Whenever there is variability in arrival or in processing queues will build up and customers will have to wait, even if the processing capacity is not fully utilized.

25. 25 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Throughput- Delay Curve Variability Increases Average Flow Time T Utilization (ρ)  100% TpTp

26. 26 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Coefficient of Variations

27. 27 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 A sample of 10 observations on Interarrival times in seconds 10,10,2,10,1,3,7,9, 2, 6 =AVERAGE ()  Avg. interarrival time = 6 R i = 1/6 arrivals / sec. =STDEV()  Std. Deviation = 3.94 C i = 3.94/6 = 0.66 C 2 i = (0.66) 2 = 0.4312 Example 8.4

28. 28 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 A sample of 10 observations on processing times in seconds 7,1,7 2,8,7,4,8,5, 1 T p = 5 seconds R p = 1/5 processes/sec. Std. Deviation = 2.83 C p = 2.83/5 = 0.57 C 2 p = (0.57) 2 = 0.3204 Example 8.4

29. 29 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 R i =1/6 < R P =1/5  R = R i  = R/ R P = (1/6)/(1/5) = 0.83 With c = 1, the average number of passengers in queue is as follows: Example 8.4 On average 1.56 passengers waiting in line, even though safety capacity is Rs= R P - Ri = 1/5 - 1/6 = 1/30 passenger per second, or 2 per minutes I i = [(0.83 2 )/(1-0.83)] ×[(0.66 2 +0.57 2 )/2] = 1.56

30. 30 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Other performance measures: T i =I i /R = (1.56)(6) = 9.4 seconds Compute T? T = Ti+Tp Since T P = 5  T = T i + T P = 14.4 seconds Total number of passengers in the process is: I = RT = (1/6) (14.4) = 2.4 Alternatively, 1.56 in the buffer. How many in the process? I = 1.56 + 0.83 = 2. 39 Example 8.4

31. 31 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Compute R, Rp and : Ta= 6 min, Tp = 5 min R = Ri= 1/6 per minute Processing rate for one processor 1/5 for 2 processors Rp = 2/5  = R/Rp = (1/6)/(2/5) = 5/12 = 0.417 Now suppose we have two servers. On average 0.076 passengers waiting in line. safety capacity is Rs= R P - Ri = 2/5 - 1/6 = 7/30 passenger per second, or 14 passengers per minutes

32. 32 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Other performance measures: T i =I i /R = (0.076)(6) = 0.46 seconds Compute T? T = Ti+Tp Since T P = 5  T = T i + T P = 0.46+5 = 5.46 seconds Total number of passengers in the process is: I= 0.08 in the buffer and 0.417 in the process. I = 0.076 + 2(0.417) = 0.91 Example 8.4 cρRsRs IiIi TiTi TI 10.830.031.569.3814.382.4 20.4170.230.0770.465.460.91

33. 33 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processor (with the server)? On average how many customers are with the server? On average how many customers are in the system? On average how long a customer stay in the system? Assignment- Problem 1: M/M/1 Performance Evaluation

34. 34 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Assignment- Problem 2: M/M/1 Performance Evaluation What if the arrival rate is 11 per hour?

35. 35 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processors (with the servers)? On average how many customers are with the servers? On average how many customers are in the system ? On average how long a customer stay in the system ? Assignment- Problem 3: M/G/c

36. 36 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)? Assignment- Problem 4: M/M/c Example

37. 37 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2  Suppose the service time is a constant  What is the answer of the previous question? Assignment- Problem 5: M/D/c Example