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Variability and Its Impact on Process Performance: Waiting Time Problems Chapter 7 A Somewhat Odd Service Process (Chapters 1-6) Patient Arrival Service Time Time 1 2 3 4 5 6 7 8 9 10 11 12 0 5 10 15 20 25 30 35 40 45 50 55 4 4 4 4 4 4 4 4 4 4 4 4 7:00 7:10 7:20 7:30 7:40 7:50 8:00 A More Realistic Service Process Patient 1 Patient 3 Patient 5 Patient 7 1 2 3 4 5 6 7 8 9 10 11 12 0 7 9 12 18 22 25 30 36 45 51 55 5 6 7 6 5 2 4 3 4 2 2 3 Patient 2 Patient 4 Patient 6 Patient 8 Patient 11 Patient 10 Patient 12 Time 7:00 7:10 7:20 7:30 7:40 7:50 3 Number of cases Arrival Service Patient Time Time Patient 9 2 1 0 2 min. 3 min. 4 min. 5 min. 6 min. 7 min. Service times 8:00 Variability Leads to Waiting Time Pt 1 2 3 4 5 6 7 8 9 10 11 12 Arrival Time 0 7 9 12 18 22 25 30 36 45 51 55 p 5 6 7 6 5 2 4 3 4 2 2 3 Service time Wait time 7:00 7:10 7:20 7:30 7:40 7:50 8:00 7:00 7:10 7:20 7:30 7:40 7:50 8:00 5 4 3 Inventory (Patients at lab) 2 1 0 Sources of Variability Buffer Processing The “Memoryless” Exponential Function • Interarrival times follow an exponential distribution P{IA t} 1 e t a • Exponential interarrival times = Poisson arrival process • Coefficient of Variation (CV ) CV Quite possibly the worst run of three slides you will see this semester Waiting Time Flow rate Inventory waiting Iq Inventory In process Ip Inflow Outflow Entry to system Average flow time T Increasing Variability Begin Service Departure Time in queue Tq Service Time p Total Flow Time T=Tq+p Theoretical Flow Time Utilization 100% A Waiting Time Formula CVa2 CVp2 utilization Time in queue Activity Time 2 1 utilization Service time factor Utilization factor Variability factor Waiting Time for Multiple, Parallel Resources Inventory in service Ip Inflow Inventory waiting Iq Entry to system Outflow Begin Service Time in queue Tq Flow rate Departure Service Time p Total Flow Time T=Tq+p The Waiting Time Formula for Multiple (m) Servers 2 ( m 1) 1 CVa2 CV p2 Activity tim e utilization Tim ein queue m 2 1 utilization Summary of Queuing Analysis Utilization Flow unit u Server Inventory in service Ip 1 a p am m 1 p Time related measures 2 ( m 1) 1 CVa2 CV p2 Activity tim e utilization Tim ein queue m 1 utilizatio n 2 Outflow Inflow T Tq p Inventory waiting Iq Inventory related measures Entry to system Begin Service Waiting Time Tq Departure Service Time p Flow Time T=Tq+p Iq Tq a I p um I I p Iq 7.2 E-mails arrive to My-Law.com from 8 a.m. to 6 p.m. at a rate of 10 emails per hour (cv=1). At each moment in time, there is exactly one lawyer “on call” waiting these e-mails. It takes on average five minutes to respond with a standard deviation of four minutes. a. What is the average time a customer must wait for a response? b. How many e-mails will a lawyer receive at the end of the day? c. When not responding to e-mails, the lawyer is encouraged to actively pursue cases that potentially could lead to large settlements. How much time during a ten hour shift can be devoted to this pursuit? To increase responsiveness of the system, a template will be used for most e-mail responses. The standard deviation for writing the response now drops to 0.5 minutes but the average writing time is unchanged. d. e. Now how much time can a lawyer spend pursuing cases? How long does a customer have to wait for a response to an inquiry? Service Levels in Waiting Systems Fraction of 1 customers who have to wait x 0.8 seconds or less 90% of calls had to wait 25 seconds or less Waiting times for those customers who do not get served immediately 0.6 0.4 Fraction of customers who get served without waiting at all 0.2 0 0 50 100 150 Waiting time [seconds] • Target Wait Time (TWT) • Service Level = Probability{Waiting TimeTWT} • Example: Deutsche Bundesbahn Call Center - now (2003): 30% of calls answered within 20 seconds - target: 80% of calls answered within 20 seconds 200 Data in Practical Call Center Setting Number of customers Per 15 minutes Distribution Function 1 160 140 120 100 80 60 40 20 0 0.8 Exponential distribution 0.6 0:01:09 0:01:00 0:00:52 0:00:43 0:00:35 0:00:26 Time 0:00:17 0:15 2:00 3:45 5:30 7:15 9:00 10:45 12:30 14:15 16:00 17:45 19:30 21:15 23:00 0 0:00:00 0.2 0:00:09 Empirical distribution (individual points) 0.4 The Power of Pooling Independent Resources 2x(m=1) Waiting Time Tq 70.00 60.00 m=1 50.00 40.00 - OR Pooled Resources (m=2) 30.00 20.00 10.00 m=2 m=5 m=10 0.00 60% 65% 70%75%80%85% 90% 95% Utilization u Priority Rules in Waiting Time Systems Service times: A: 9 minutes B: 10 minutes C: 4 minutes D: 8 minutes C A 9 min. 19 min. 4 min. B 13 min. C 23 min. D Total wait time: 9+19+23=51min •SPT •FCFS •EDD •Priority •WNIFOARB D 21 min. A B Total wait time: 4+13+21=38 min 7.3The airport branch of a car rental company maintains a fleet of 50 SUVs. The interarrival time between requests for an SUV is 2.4 hours with a standard deviation of 2.4 hours. Assume that, if all SUVs are rented, customers are willing to wait until an SUV is available. An SUV is rented, on average for 3 days, with a standard deviation of 1 day. What is the average number of SUVs in the company lot? What is the average time a customer has to wait to rent an SUV? The company discovers that if it reduces its daily $80 rental price by $25, the average demand would increase to 12 rentals per day and the average rental duration will become 4 days. Should they go for it? How would the waiting time change if the company decides to limit all SUV rentals to exactly 4 days? Assume that if such a restriction is imposed, the average interarrival time will increase to 3 hours as will the std deviation.