1. D - 1© 2011 Pearson Education, Inc. publishing as Prentice Hall D D Waiting-Line Models PowerPoint presentation to accompany Heizer and Render Operations Management, 10e Principles of Operations Management, 8e PowerPoint slides by Jeff Heyl

2. D - 2 © 2011 Pearson Education, Inc. publishing as Prentice Hall Outline  Queuing Theory  Characteristics of a Waiting-Line System  Arrival Characteristics  Waiting-Line Characteristics  Service Characteristics  Measuring a Queue’s Performance  Queuing Costs

3. D - 3 © 2011 Pearson Education, Inc. publishing as Prentice Hall Outline – Continued  The Variety of Queuing Models  Model A(M/M/1): Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times  Model B(M/M/S): Multiple-Channel Queuing Model  Model C(M/D/1): Constant-Service- Time Model  Little’s Law  Model D: Limited-Population Model

4. D - 4 © 2011 Pearson Education, Inc. publishing as Prentice Hall Outline – Continued  Other Queuing Approaches

5. D - 5 © 2011 Pearson Education, Inc. publishing as Prentice Hall Learning Objectives When you complete this module you should be able to: 1.Describe the characteristics of arrivals, waiting lines, and service systems 2.Apply the single-channel queuing model equations 3.Conduct a cost analysis for a waiting line

6. D - 6 © 2011 Pearson Education, Inc. publishing as Prentice Hall Learning Objectives When you complete this module you should be able to: 4.Apply the multiple-channel queuing model formulas 5.Apply the constant-service-time model equations 6.Perform a limited-population model analysis

7. D - 7 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Theory  The study of waiting lines  Waiting lines are common situations  Useful in both manufacturing and service areas

8. D - 8 © 2011 Pearson Education, Inc. publishing as Prentice Hall Common Queuing Situations SituationArrivals in QueueService Process SupermarketGrocery shoppersCheckout clerks at cash register Highway toll boothAutomobilesCollection of tolls at booth Doctor’s officePatientsTreatment by doctors and nurses Computer systemPrograms to be runComputer processes jobs Telephone companyCallersSwitching equipment to forward calls BankCustomerTransactions handled by teller Machine maintenance Broken machinesRepair people fix machines HarborShips and bargesDock workers load and unload Table D.1

9. D - 9 © 2011 Pearson Education, Inc. publishing as Prentice Hall Characteristics of Waiting- Line Systems 1.Arrivals or inputs to the system  Population size, behavior, statistical distribution 2.Queue discipline, or the waiting line itself  Limited or unlimited in length, discipline of people or items in it 3.The service facility  Design, statistical distribution of service times

10. D - 10 © 2011 Pearson Education, Inc. publishing as Prentice Hall Parts of a Waiting Line Figure D.1 Dave’s Car Wash EnterExit Population of dirty cars Arrivals from the general population … Queue (waiting line) Service facility Exit the system Arrivals to the system Exit the system In the system Arrival Characteristics  Size of the population  Behavior of arrivals  Statistical distribution of arrivals Waiting Line Characteristics  Limited vs. unlimited  Queue discipline Service Characteristics  Service design  Statistical distribution of service

11. D - 11 © 2011 Pearson Education, Inc. publishing as Prentice Hall Arrival Characteristics 1.Size of the population  Unlimited (infinite) or limited (finite) 2.Pattern of arrivals  Scheduled or random, often a Poisson distribution 3.Behavior of arrivals  Wait in the queue and do not switch lines  No balking or reneging

12. D - 12 © 2011 Pearson Education, Inc. publishing as Prentice Hall Poisson Distribution P(x) = for x = 0, 1, 2, 3, 4, … e - x x! whereP(x)=probability of x arrivals x=number of arrivals per unit of time =average arrival rate e=2.7183 (which is the base of the natural logarithms)

13. D - 13 © 2011 Pearson Education, Inc. publishing as Prentice Hall Poisson Distribution Probability = P(x) = e - x x! 0.25 – 0.02 – 0.15 – 0.10 – 0.05 – – Probability 0123456789 Distribution for = 2 x 0.25 – 0.02 – 0.15 – 0.10 – 0.05 – – Probability 0123456789 Distribution for = 4 x 1011 Figure D.2

14. D - 14 © 2011 Pearson Education, Inc. publishing as Prentice Hall Waiting-Line Characteristics  Limited or unlimited queue length  Queue discipline - first-in, first-out (FIFO) is most common  Other priority rules may be used in special circumstances

15. D - 15 © 2011 Pearson Education, Inc. publishing as Prentice Hall Service Characteristics  Queuing system designs  Single-channel system, multiple- channel system  Single-phase system, multiphase system  Service time distribution  Constant service time  Random service times, usually a negative exponential distribution

16. D - 16 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Departures after service Single-channel, single-phase system Queue Arrivals Single-channel, multiphase system Arrivals Departures after service Phase 1 service facility Phase 2 service facility Service facility Queue A family dentist’s office A McDonald’s dual window drive-through

17. D - 17 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Multi-channel, single-phase system Arrivals Queue Most bank and post office service windows Departures after service Service facility Channel 1 Service facility Channel 2 Service facility Channel 3

18. D - 18 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Multi-channel, multiphase system Arrivals Queue Some college registrations Departures after service Phase 2 service facility Channel 1 Phase 2 service facility Channel 2 Phase 1 service facility Channel 1 Phase 1 service facility Channel 2

19. D - 19 © 2011 Pearson Education, Inc. publishing as Prentice Hall Negative Exponential Distribution Figure D.4 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – Probability that service time ≥ 1 ||||||||||||| 0.000.250.500.751.001.251.501.752.002.252.502.753.00 Time t (hours) Probability that service time is greater than t = e -µt for t ≥ 1 µ = Average service rate e = 2.7183 Average service rate (µ) = 1 customer per hour Average service rate (µ) = 3 customers per hour  Average service time = 20 minutes per customer

20. D - 20 © 2011 Pearson Education, Inc. publishing as Prentice Hall Measuring Queue Performance 1.Average time that each customer or object spends in the queue 2.Average queue length 3.Average time each customer spends in the system 4.Average number of customers in the system 5.Probability that the service facility will be idle 6.Utilization factor for the system 7.Probability of a specific number of customers in the system

21. D - 21 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Costs Figure D.5 Total expected cost Cost of providing service Cost Low level of service High level of service Cost of waiting time Minimum Total cost Optimal service level

22. D - 22 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models The four queuing models here all assume:  Poisson distribution arrivals  FIFO discipline  A single-service phase

23. D - 23 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 ModelNameExample ASingle-channel Information counter system at department store (M/M/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialUnlimitedFIFO

24. D - 24 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 ModelNameExample BMultichannel Airline ticket (M/M/S) counter NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline Multi-SinglePoissonExponentialUnlimitedFIFO channel

25. D - 25 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 ModelNameExample CConstant- Automated car service wash (M/D/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonConstantUnlimitedFIFO

26. D - 26 © 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 ModelNameExample DLimited Shop with only a population dozen machines (finite population) that might break NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialLimitedFIFO

27. D - 27 © 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel 1.Arrivals are served on a FIFO basis and every arrival waits to be served regardless of the length of the queue 2.Arrivals are independent of preceding arrivals but the average number of arrivals does not change over time 3.Arrivals are described by a Poisson probability distribution and come from an infinite population

28. D - 28 © 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel 4.Service times vary from one customer to the next and are independent of one another, but their average rate is known 5.Service times occur according to the negative exponential distribution 6.The service rate is faster than the arrival rate

29. D - 29 © 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3 =Mean number of arrivals per time period µ=Mean number of units served per time period L s =Average number of units (customers) in the system (waiting and being served) = W s =Average time a unit spends in the system (waiting time plus service time) = µ – 1 µ –

30. D - 30 © 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3 L q =Average number of units waiting in the queue = W q =Average time a unit spends waiting in the queue =  =Utilization factor for the system = 2 µ(µ – ) µ(µ – ) µ

31. D - 31 © 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3 P 0 =Probability of 0 units in the system (that is, the service unit is idle) =1 – P n > k =Probability of more than k units in the system, where n is the number of units in the system = µ µ k + 1

32. D - 32 © 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example =2 cars arriving/hourµ= 3 cars serviced/hour L s = = = 2 cars in the system on average W s = = = 1 hour average waiting time in the system L q = = = 1.33 cars waiting in line 2 µ(µ – ) µ – 1 µ – 2 3 - 2 1 3 - 2 2 3(3 - 2)

33. D - 33 © 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example W q = = = 2/3 hour = 40 minute average waiting time  = /µ = 2/3 = 66.6% of time mechanic is busy µ(µ – ) 2 3(3 - 2) µ P 0 = 1 - =.33 probability there are 0 cars in the system =2 cars arriving/hourµ= 3 cars serviced/hour

34. D - 34 © 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example Probability of more than k Cars in the System kP n > k = (2/3) k + 1.667 0.667  Note that this is equal to 1 - P 0 = 1 -.33 1.444 2.296.198 3.198  Implies that there is a 19.8% chance that more than 3 cars are in the system 4.132 5.088 6.058 7.039

35. D - 35 © 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Economics Customer dissatisfaction and lost goodwill= $10 per hour W q = 2/3 hour Total arrivals= 16 per day Mechanic’s salary= $56 per day Total hours customers spend waiting per day = (16) = 10 hours 2323 2323 Customer waiting-time cost = $10 10 = $106.67 2323 Total expected costs = $106.67 + $56 = $162.67

36. D - 36 © 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Model Table D.4 M=number of channels open =average arrival rate µ=average service rate at each channel P 0 = for Mµ > 1 1M!1M! 1n!1n! Mµ Mµ - M – 1 n = 0 µ n µ M + ∑ L s = P 0 + µ( /µ) M (M - 1)!(Mµ - ) 2 µ

37. D - 37 © 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Model Table D.4 W s = P 0 + = µ( /µ) M (M - 1)!(Mµ - ) 2 1µ1µ L s L q = L s – µ W q = W s – = 1µ1µ L q

38. D - 38 © 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Example = 2 µ = 3 M = 2 P 0 = = 1 12!12! 1n!1n! 2(3) 2(3) - 2 1 n = 0 2323 n 2323 2 + ∑ 1212 L s = + = (2)(3(2/3) 2 2323 1! 2(3) - 2 2 1212 3434 W q = =.0415.083 2 W s = = 3/4 2 3838 L q = – = 2323 3434 1 12

39. D - 39 © 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Example Single ChannelTwo Channels P0P0.33.5 LsLs 2 cars.75 cars WsWs 60 minutes22.5 minutes LqLq 1.33 cars.083 cars WqWq 40 minutes2.5 minutes

40. D - 40 © 2011 Pearson Education, Inc. publishing as Prentice Hall Waiting Line Tables Table D.5 Poisson Arrivals, Exponential Service Times Number of Service Channels, M ρ12345.10.0111.25.0833.0039.50.5000.0333.0030.752.2500.1227.0147.903.1000.2285.0300.0041 1.0.3333.0454.0067 1.62.8444.3128.0604.0121 2.0.8888.1739.0398 2.64.9322.6581.1609 3.01.5282.3541 4.02.2164

41. D - 41 © 2011 Pearson Education, Inc. publishing as Prentice Hall Waiting Line Table Example Bank tellers and customers = 18, µ = 20 From Table D.5 Utilization factor  = /µ =.90 W q = L q Number of service windowsM Number in queueTime in queue 1 window18.1.45 hrs, 27 minutes 2 windows2.2285.0127 hrs, ¾ minute 3 windows3.03.0017 hrs, 6 seconds 4 windows4.0041.0003 hrs, 1 second

42. D - 42 © 2011 Pearson Education, Inc. publishing as Prentice Hall Constant-Service Model Table D.6 L q = 2 2µ(µ – ) Average length of queue W q = 2µ(µ – ) Average waiting time in queue µ L s = L q + Average number of customers in system W s = W q + 1µ1µ Average time in the system

43. D - 43 © 2011 Pearson Education, Inc. publishing as Prentice Hall Net savings= $ 7 /trip Constant-Service Example Trucks currently wait 15 minutes on average Truck and driver cost $60 per hour Automated compactor service rate (µ) = 12 trucks per hour Arrival rate ( ) = 8 per hour Compactor costs $3 per truck Current waiting cost per trip = (1/4 hr)($60) = $15 /trip W q = = hour 8 2(12)(12 – 8) 1 12 Waiting cost/trip with compactor = (1/12 hr wait)($60/hr cost)= $ 5 /trip Savings with new equipment = $15 (current) – $5(new)= $10 /trip Cost of new equipment amortized= $ 3 /trip

44. D - 44 © 2011 Pearson Education, Inc. publishing as Prentice Hall Little’s Law  A queuing system in steady state L = W (which is the same as W = L/ L q = W q (which is the same as W q = L q /  Once one of these parameters is known, the other can be easily found  It makes no assumptions about the probability distribution of arrival and service times  Applies to all queuing models except the limited population model

45. D - 45 © 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Model Table D.7 Service factor: X = Average number running: J = NF(1 - X) Average number waiting: L = N(1 - F) Average number being serviced: H = FNX Average waiting time: W = Number of population: N = J + L + H T T + U T(1 - F) XF

46. D - 46 © 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Model D =Probability that a unit will have to wait in queue N =Number of potential customers F =Efficiency factorT =Average service time H =Average number of units being served U =Average time between unit service requirements J =Average number of units not in queue or in service bay W =Average time a unit waits in line L =Average number of units waiting for service X =Service factor M =Number of service channels

47. D - 47 © 2011 Pearson Education, Inc. publishing as Prentice Hall Finite Queuing Table Table D.8 XMDF.0121.048.999.0251.100.997.0501.198.989.0602.020.999 1.237.983.0702.027.999 1.275.977.0802.035.998 1.313.969.0902.044.998 1.350.960.1002.054.997 1.386.950

48. D - 48 © 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Example Service factor: X = =.091 (close to.090) For M = 1, D =.350 and F =.960 For M = 2, D =.044 and F =.998 Average number of printers working: For M = 1, J = (5)(.960)(1 -.091) = 4.36 For M = 2, J = (5)(.998)(1 -.091) = 4.54 2 2 + 20 Each of 5 printers requires repair after 20 hours (U) of use One technician can service a printer in 2 hours (T) Printer downtime costs $120/hour Technician costs $25/hour

49. D - 49 © 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Example Service factor: X = =.091 (close to.090) For M = 1, D =.350 and F =.960 For M = 2, D =.044 and F =.998 Average number of printers working: For M = 1, J = (5)(.960)(1 -.091) = 4.36 For M = 2, J = (5)(.998)(1 -.091) = 4.54 2 2 + 20 Each of 5 printers require repair after 20 hours (U) of use One technician can service a printer in 2 hours (T) Printer downtime costs $120/hour Technician costs $25/hour Number of Technicians Average Number Printers Down (N - J) Average Cost/Hr for Downtime (N - J)$120 Cost/Hr for Technicians ($25/hr) Total Cost/Hr 1.64$76.80$25.00$101.80 2.46$55.20$50.00$105.20

50. D - 50 © 2011 Pearson Education, Inc. publishing as Prentice Hall Other Queuing Approaches  The single-phase models cover many queuing situations  Variations of the four single-phase systems are possible  Multiphase models exist for more complex situations

51. D - 51 © 2011 Pearson Education, Inc. publishing as Prentice Hall All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.