1. D-1 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Operations Management Waiting-Line Models Module D

2. D-2 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 You manage a call center which can answer an average of 20 calls per hour. Your call center gets 17.5 calls in an average hour. On average, what is the time a customer spends on hold waiting for service? a) on average, customers should not have to wait on hold since capacity is greater than demand. b) average customer wait will be less than 10 minutes. c) average customer wait will be between 10 and 20 minutes. d) average customer wait will be greater than 20 minutes. e) who knows? There’s no way to tell. Queues / Lines © 1995 Corel Corp.

3. D-3 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Two ways to address waiting lines  Queuing theory  Certain types of lines can be described mathematically  Requires that assumptions are valid for your situation  Systems with multiple lines that feed each other are too complex for queuing theory  Simulation  Building mathematical models that attempt to act like real operating systems  Real-world situations can be studied without imposing on the actual system

4. D-4 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Why do we have to wait? © 1995 Corel Corp. Why do services (and most non-MTS manufacturers) have queues?  Processing time and/or arrival time variance  Costs of capacity – can we afford to always have more people/servers than customers?  Efficiency – e.g. scheduling at the Doctor’s office

5. D-5 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Waiting Costs and Service Costs Total expected cost Cost of waiting time Cost Low level of service Optimal service level High level of service Minimum total cost Cost of providing service

6. D-6 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Costs of Queues Too Little Queue Too Much Queue Cost of capacity Wasted capacity Annoyed customers Lost customers Space Possible opportunity: e.g. wait in the bar for a restaurant table

7. D-7 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 BankCustomersTellerDeposit, etc. Doctor’sPatientDoctorTreatment office Traffic CarsLightControlled intersection passage Assembly linePartsWorkersAssembly 1–800 softwareUser call-insTech supportTechnical support support personnel Situation Arrivals ServersService Process Waiting Line Examples

8. D-8 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Service Facility Population  Pattern of arrivals  Scheduled  Random – estimated by Poisson distribution Arrival Characteristics Characteristics of a Waiting Line System  Size of the source population  Limited  Unlimited  Behavior of arrivals  Join the queue, wait until served  Balk – refuse to join the line  Renege – leave the line Waiting Line = average arrival rate

9. D-9 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Poisson Distributions for Arrival Rates =2 =4 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0123456789101112 x Probability 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0123456789101112 x Probability = average arrival rate

10. D-10 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Service Facility Waiting Line Population Waiting Line Characteristics  Length of the queue  Limited  Unlimited  Queue discipline  FIFO  Other Characteristics of a Waiting Line System

11. D-11 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Service Facility Waiting Line Population Service Characteristics  Number of channels  Single  Multiple  Number of phases in service system  Single  Multiple  Service time distribution  Constant  Random – estimated by negative exponential distribution Characteristics of a Waiting Line System  = average service rate

12. D-12 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Negative Exponential Distribution Average Service Rate (  ) = 3 customers per hour Average Service Time = 20 minutes per customer Average Service Rate (  ) = 1 customer per hour Probability that Service Time is greater than t=e -  t, for t > 0 Time t in Hours Probability that Service Time  t 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.000.250.500.751.001.251.501.752.002.252.502.753.00 Average Service Rate (  ) = 3 customers per hour Average Service Time = 20 minutes per customer Average Service Rate (  ) = 1 customer per hour Probability that Service Time is greater than t = e for t > 0 –t–t

13. D-13 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Basic Queuing System Designs Arrivals Served units Service facility Queue Single-Channel, Single-Phase Service facility Arrivals Served units Service facility Queue Service facility Multi-Channel, Multi-Phase Arrivals Served units Service facility Queue Service facility Multi-Channel, Single-Phase e.g. U.S. Post Office e.g. drive-through bank e.g. Suds & Suds Laundromat Arrivals Served units Service facility Queue Service facility Single-Channel, Multi-Phase e.g. McDonald’s drive-through

14. D-14 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 You manage a call center which can answer an average of 20 calls per hour. Your call center gets 17.5 calls in an average hour. On average, what is the time a customer spends on hold waiting for service? Call Center – Solution (1) = average arrival rate = 17.5 calls/hr  = average service rate = 20 calls/hr ρ = average utilization of system = /  = 87.5%

15. D-15 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 You manage a call center which can answer an average of 20 calls per hour. Your call center gets 17.5 calls in an average hour. On average, what is the time a customer spends on hold waiting for service? Call Center – Solution (2) = average arrival rate = 17.5 calls/hr  = average service rate = 20 calls/hr ρ = average utilization of system = /  = 87.5% L = average number of customers in service system (line and being served) = / (  - ) = 7 calls L q = average number waiting in line = ρL = 6.125 calls

16. D-16 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 You manage a call center which can answer an average of 20 calls per hour. Your call center gets 17.5 calls in an average hour. On average, what is the time a customer spends on hold waiting for service? Call Center – Solution (3) = average arrival rate = 17.5 calls/hr  = average service rate = 20 calls/hr ρ = average utilization of system = /  = 87.5% W = average time in system (wait and service) = 1 / (  - ) =.4 hr = 24 min W q = average time waiting = ρW = 21 min

17. D-17 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Single server equations = average arrival rate  = average service rate ρ = average utilization of system = /  P n = probability that n customers are in the system = (1- ρ) ρ n L = average number of customers in service system (line and being served) = / (  - ) L q = average number waiting in line = ρL W = average time in system (wait and service) = 1/ (  - ) W q = average time waiting = ρW

18. D-18 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458  You are opening an ice cream stand that has a single employee (you). You expect to see about 25 customers an hour. It takes you an average of 2 minutes to serve each customer. Customers are served in a FCFS manner.  Your research suggests that if there is a line of more than 4 people that some customers will leave without buying anything. In addition, if customers have to wait more than 6 minutes to get their order filled they are not likely to come back.  How well will this system do at satisfying customers?  What assumptions are you making to answer this question? Izzy’s Ice Cream Stand

19. D-19 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Izzy’s Ice Cream Stand (2) = 25 customers/hr  = 30 customers/hr ρ = /  =.833 P n = probability that n customers are in the system = (1- ρ) ρ n P 0 = (1 –.833) x.833 =.167.167 P 1 = (1 –.833) x.833 =.139.306 P 2 = (1 –.833) x.833 =.116.422 P 3 = (1 –.833) x.833 =.097.519 P 4 = (1 –.833) x.833 =.080.599 P more than 4 = 1 –.599 =.401 1 2 3 4 0 cumulative L = / (  - ) = 5 customers Customers in the system

20. D-20 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Izzy’s Ice Cream Stand (3) = 25 customers/hr  = 30 customers/hr ρ = /  =.833 L = / (  - ) = 25 / (30 – 25) = 5 customers L q = ρL =.833 x 5 = 4.17 customers W = 1 / (  - ) = 1 / (30 – 25) =.2 hr = 12 min W q = ρW =.833 x 12 min = 10 min Why not L – 1 ?

21. D-21 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 Why do we have to wait? © 1995 Corel Corp. Why do services (and most non-MTS manufacturers) have queues?  Processing time and/or arrival time variance  Costs of capacity – can we afford to always have more people/servers than customers?  Efficiency – e.g. scheduling at the Doctor’s office

22. D-22 © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458 x Time t Probability that Service Time  t Probability ARRIVALS (Poisson) SERVICE (Exponential) CUSTOMERS IN THE SYSTEM Probability n