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# Heizer om10 mod_d-waiting-line models

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### Heizer om10 mod_d-waiting-line models

1. 1. 10/16/2010 Outline D Waiting-Line Models Queuing Theory Characteristics of a Waiting-Line System PowerPoint presentation to accompany p Heizer and Render p y Arrival Characteristics Operations Management, 10e Principles of Operations Management, 8e Waiting-Line Characteristics PowerPoint slides by Jeff Heyl Service Characteristics Measuring a Queue’s Performance Queuing Costs© 2011 Pearson Education, Inc. publishing as Prentice Hall D-1 © 2011 Pearson Education, Inc. publishing as Prentice Hall D-2 Outline – Continued Outline – Continued The Variety of Queuing Models Model A(M/M/1): Single-Channel Other Queuing Approaches Queuing Model with Poisson Arrivals and Exponential Service Times Model B(M/M/S) M lti l Ch M d l B(M/M/S): Multiple-Channel l Queuing Model Model C(M/D/1): Constant-Service- Time Model Little’s Law Model D: Limited-Population Model© 2011 Pearson Education, Inc. publishing as Prentice Hall D-3 © 2011 Pearson Education, Inc. publishing as Prentice Hall D-4 Learning Objectives Learning Objectives When you complete this module you When you complete this module you should be able to: should be able to: 1. Describe the characteristics of 4. Apply the multiple-channel arrivals, arrivals waiting lines and service lines, queuing model formulas systems 5. Apply the constant-service-time 2. Apply the single-channel queuing model equations model equations 6. Perform a limited-population 3. Conduct a cost analysis for a model analysis waiting line© 2011 Pearson Education, Inc. publishing as Prentice Hall D-5 © 2011 Pearson Education, Inc. publishing as Prentice Hall D-6 1
2. 2. 10/16/2010 Common Queuing Queuing Theory Situations Situation Arrivals in Queue Service Process The study of waiting lines Supermarket Grocery shoppers Checkout clerks at cash register Waiting lines are common Highway toll booth Automobiles Collection of tolls at booth situations Doctor’s office Patients Treatment by doctors and nurses Useful in both Computer system Programs to be run Computer processes jobs manufacturing Telephone company Callers Switching equipment to forward calls and service Bank Customer Transactions handled by teller areas Machine Broken machines Repair people fix machines maintenance Harbor Ships and barges Dock workers load and unload© 2011 Pearson Education, Inc. publishing as Prentice Hall D-7 © 2011 Pearson Education, Inc. publishing as Prentice Hall Table D.1 D-8 Characteristics of Waiting- Waiting- Parts of a Waiting Line Line Systems Population of Arrivals Queue Service Exit the system dirty cars from the (waiting line) facility 1. Arrivals or inputs to the system general population … Dave’s Population size, behavior, statistical Car Wash distribution 2. 2 Queue discipline or the waiting line discipline, Enter Exit itself Limited or unlimited in length, discipline Arrivals to the system In the system Exit the system of people or items in it Arrival Characteristics Waiting Line Service Characteristics 3. The service facility Size of the population Characteristics Service design Behavior of arrivals Limited vs. Statistical distribution Design, statistical distribution of service Statistical distribution unlimited of service of arrivals Queue discipline times Figure D.1© 2011 Pearson Education, Inc. publishing as Prentice Hall D-9 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 10 Arrival Characteristics Poisson Distribution 1. Size of the population Unlimited (infinite) or limited (finite) e-λλx P(x) = for x = 0, 1, 2, 3, 4, … 2. Pattern of arrivals x! Scheduled or random, often a Poisson distribution where P(x) = probability of x arrivals 3. Behavior of arrivals x = number of arrivals per unit of time Wait in the queue and do not switch lines λ = average arrival rate e = 2.7183 (which is the base of the No balking or reneging natural logarithms)© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 11 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 12 2
3. 3. 10/16/2010 Poisson Distribution Waiting- Waiting-Line Characteristics e-λλx Probability = P(x) = x! Limited or unlimited queue length 0.25 – 0.25 – 0.02 – 0.02 – Queue discipline - first-in, first-out (FIFO) is most common lity lity Probabil Probabil 0.15 – 0.15 – Other priority rules may be used in 0.10 – 0.10 – special circumstances 0.05 – 0.05 – – – 0 1 2 3 4 5 6 7 8 9 x 0 1 2 3 4 5 6 7 8 9 10 11 x Distribution for λ = 2 Distribution for λ = 4Figure D.2© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 13 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 14 Service Characteristics Queuing System Designs A family dentist’s office Queuing system designs Queue Single-channel system, multiple- Service Departures Arrivals channel system facility after service Single phase Single-phase system, multiphase Single-channel, single phase Single channel single-phase system system A McDonald’s dual window drive-through Service time distribution Queue Phase 1 Phase 2 Departures Constant service time Arrivals service facility service facility after service Random service times, usually a Single-channel, multiphase system negative exponential distribution Figure D.3© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 15 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 16 Queuing System Designs Queuing System DesignsMost bank and post office service windows Some college registrations Service facility Phase 1 Phase 2 Channel 1 service service Queue Queue facility facility Channel 1 Channel 1 Service Departures Departures Arrivals facility Arrivals after service Channel 2 after service Phase 1 Phase 2 service service facility facility Service Channel 2 Channel 2 facility Channel 3 Multi-channel, single-phase system Multi-channel, multiphase system Figure D.3 Figure D.3© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 17 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 18 3
4. 4. 10/16/2010 Negative Exponential Measuring Queue Distribution Performance Probability that service time is greater than t = e-µt for t ≥ 1 1. Average time that each customer or object µ = Average service rate 1.0 – e = 2.7183 spends in the queue Probability that servic time ≥ 1 0.9 – Average service rate (µ) = 3 customers per hour 2. Average queue length 0.8 – ⇒ Average service time = 20 minutes per customer 3. Average time each customer spends in the ce 0.7 07 – 0.6 – system 0.5 – 4. Average number of customers in the system 0.4 – Average service rate (µ) = 0.3 – 1 customer per hour 5. Probability that the service facility will be idle 0.2 – 6. Utilization factor for the system 0.1 – 0.0 |– | | | | | | | | | | | | 7. Probability of a specific number of customers 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 in the system Time t (hours)Figure D.4© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 19 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 20 Queuing Costs Queuing Models Cost The four queuing models here all assume: Poisson distribution arrivals Minimum Total Total expected cost FIFO discipline cost Cost of providing service A single-service phase Cost of waiting time Low level Optimal High level of service service level of service Figure D.5© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 21 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 22 Queuing Models Queuing Models Model Name Example Model Name Example A Single-channel Information counter B Multichannel Airline ticket system at department store (M/M/S) counter (M/M/1) Number Number Arrival Service Number Number Arrival Service of of Rate Time Population Queue of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Channels Phases Pattern Pattern Size Discipline Single Single Poisson Exponential Unlimited FIFO Multi- Single Poisson Exponential Unlimited FIFO channel Table D.2 Table D.2© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 23 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 24 4
5. 5. 10/16/2010 Queuing Models Queuing Models Model Name Example Model Name Example C Constant- Automated car D Limited Shop with only a service wash population dozen machines (M/D/1) (finite population) that might break Number Number Arrival Service Number Number Arrival Service of of Rate Time Population Queue of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Channels Phases Pattern Pattern Size Discipline Single Single Poisson Constant Unlimited FIFO Single Single Poisson Exponential Limited FIFO Table D.2 Table D.2© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 25 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 26 Model A – Single-Channel Single- Model A – Single-Channel Single- 1. Arrivals are served on a FIFO basis and 4. Service times vary from one customer every arrival waits to be served to the next and are independent of one regardless of the length of the queue another, but their average rate is 2. 2 Arrivals are independent of preceding known arrivals but the average number of 5. Service times occur according to the arrivals does not change over time negative exponential distribution 3. Arrivals are described by a Poisson 6. The service rate is faster than the probability distribution and come from arrival rate an infinite population© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 27 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 28 Model A – Single-Channel Single- Model A – Single-Channel Single- λ = Mean number of arrivals per time period Lq = Average number of units waiting in the µ = Mean number of units served per time period queue Ls = Average number of units (customers) in the = λ2 system (waiting and being served) µ(µ – λ) = λ Wq = Average time a unit spends waiting in the µ–λ queue Ws = Average time a unit spends in the system λ = (waiting time plus service time) µ(µ – λ) = 1 ρ = Utilization factor for the system µ–λ λ = Table D.3 µ Table D.3© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 29 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 30 5
6. 6. 10/16/2010 Model A – Single-Channel Single- Single- Single-Channel Example P0 = Probability of 0 units in the system (that is, λ = 2 cars arriving/hour µ = 3 cars serviced/hour the service unit is idle) 2 λ λ Ls = = = 2 cars in the system on average = 1– µ–λ 3-2 µ Pn > k = Probability of more than k units in the 1 1 system, where n is the number of units in Ws = = = 1 hour average waiting time in µ–λ 3-2 the system the system k+1 λ λ2 22 = Lq = = = 1.33 cars waiting in line µ µ(µ – λ) 3(3 - 2) Table D.3© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 31 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 32 Single- Single-Channel Example Single- Single-Channel Example Probability of more than k Cars in the System λ = 2 cars arriving/hour µ = 3 cars serviced/hour k Pn > k = (2/3)k + 1 λ 2 0 .667 ← Note that this is equal to 1 - P0 = 1 - .33 Wq = = = 2/3 hour = 40 minute µ(µ – λ) 3(3 - 2) 1 .444 average waiting time 2 .296 ρ = λ/µ = 2/3 = 66.6% of time mechanic is busy 3 .198 ← Implies that there is a 19.8% chance that more than 3 cars are in the system λ 4 .132 P0 = 1 - = .33 probability there are 0 cars in the µ 5 .088 system 6 .058 7 .039© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 33 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 34 Single- Single-Channel Economics Multi- Multi-Channel Model Customer dissatisfaction M = number of channels open and lost goodwill = \$10 per hour λ = average arrival rate Wq = 2/3 hour µ = average service rate at each channel Total arrivals = 16 per day Mechanic’s salary = \$56 per day 1 P0 = for Mµ > λ Total hours M–1 1 λ n 1 λ M Mµ customers spend waiting per day = 2 3 (16) = 10 2 3 hours ∑ n! µ + M! µ Mµ - λ n=0 2 Customer waiting-time cost = \$10 10 = \$106.67 M 3 λµ(λ/µ) λ Ls = P0 + (M - 1)!(Mµ - λ) 2 µ Total expected costs = \$106.67 + \$56 = \$162.67 Table D.4© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 35 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 36 6
7. 7. 10/16/2010 Multi- Multi-Channel Model Multi- Multi-Channel Example λ = 2 µ = 3 M = 2 M Ls λµ(λ/µ) 1 Ws = P0 + = 1 1 (M - 1)!(Mµ - λ) 2 µ λ P0 = = 1 n 2 2 1 2(3) ∑ n! 2 3 + 1 2! 2 3 2(3) - 2 n=0 λ Lq = Ls – µ (2)(3(2/3)2 1 2 3 Ls = + = 2 3 4 1! 2(3) - 2 2 1 Lq Wq = Ws – = µ λ 3/4 3 3 2 1 .083 Ws = = Lq = – = Wq = = .0415 2 8 4 3 12 2 Table D.4© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 37 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 38 Multi- Multi-Channel Example Waiting Line Tables Poisson Arrivals, Exponential Service Times Number of Service Channels, M Single Channel Two Channels ρ 1 2 3 4 5 .10 .0111 P0 .33 .5 .25 .0833 .0039 .50 .5000 .0333 .0030 Ls 2 cars .75 cars 75 .75 75 2.2500 2 2500 .1227 1227 .0147 0147 .90 3.1000 .2285 .0300 .0041 Ws 60 minutes 22.5 minutes 1.0 .3333 .0454 .0067 1.6 2.8444 .3128 .0604 .0121 Lq 1.33 cars .083 cars 2.0 .8888 .1739 .0398 Wq 40 minutes 2.5 minutes 2.6 4.9322 .6581 .1609 3.0 1.5282 .3541 4.0 2.2164 Table D.5© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 39 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 40 Waiting Line Table Example Constant- Constant-Service Model Bank tellers and customers Average length λ2 λ = 18, µ = 20 Lq = of queue 2µ(µ – λ) Lq Utilization factor ρ = λ/µ = .90 Wq = λ Average waiting time λ in queue Wq = From Table D.5 D5 2µ(µ – λ) Number of Number service windows M in queue Time in queue Average number of λ Ls = Lq + 1 window 1 8.1 .45 hrs, 27 minutes customers in system µ 2 windows 2 .2285 .0127 hrs, ¾ minute Average time 1 3 windows 3 .03 .0017 hrs, 6 seconds Ws = Wq + in the system µ 4 windows 4 .0041 .0003 hrs, 1 second Table D.6© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 41 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 42 7
8. 8. 10/16/2010 Constant- Constant-Service Example Little’s Law Trucks currently wait 15 minutes on average Truck and driver cost \$60 per hour A queuing system in steady state Automated compactor service rate (µ) = 12 trucks per hour Arrival rate (λ) = 8 per hour Compactor costs \$3 per truck L = λW (which is the same as W = L/λ Current waiting cost per trip = (1/4 hr)(\$60) = \$15 /trip g p p ( )( ) p Lq = λWq (which is the same as Wq = Lq/λ 8 1 Wq = = hour Once one of these parameters is known, the 2(12)(12 – 8) 12 other can be easily found Waiting cost/trip It makes no assumptions about the probability with compactor = (1/12 hr wait)(\$60/hr cost) = \$ 5 /trip Savings with distribution of arrival and service times = \$15 (current) – \$5(new) = \$10 /trip new equipment Applies to all queuing models except the limited Cost of new equipment amortized = \$ 3 /trip population model Net savings = \$ 7 /trip© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 43 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 44 Limited- Limited-Population Model Limited- Limited-Population Model D = Probability that a unit N = Number of potential T will have to wait in customers Service factor: X = T+U queue Average number running: J = NF(1 - X) F = Efficiency factor T = Average service time H = Average number of units U = Average time between Average number waiting: L = N(1 - F) being b i servedd unit service it i requirements Average number being serviced: H = FNX J = Average number of units W = Average time a unit T(1 - F) not in queue or in waits in line Average waiting time: W = XF service bay L = Average number of units X = Service factor Number of population: N = J + L + H waiting for service M = Number of service channels Table D.7© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 45 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 46 Finite Queuing Table Limited- Limited-Population Example X M D F .012 1 .048 .999 Each of 5 printers requires repair after 20 hours (U) of use .025 1 .100 .997 One technician can service a printer in 2 hours (T) .050 1 .198 .989 Printer downtime costs \$120/hour Technician costs \$25/hour .060 2 .020 .999 1 .237 .983 2 Service factor: X = = .091 (close to .090) 091 090) .070 2 .027 .999 2 + 20 1 .275 .977 For M = 1, D = .350 and F = .960 .080 2 .035 .998 For M = 2, D = .044 and F = .998 1 .313 .969 .090 2 .044 .998 Average number of printers working: 1 .350 .960 For M = 1, J = (5)(.960)(1 - .091) = 4.36 .100 2 .054 .997 For M = 2, J = (5)(.998)(1 - .091) = 4.54 Table D.8 1 .386 .950© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 47 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 48 8
9. 9. 10/16/2010 Limited- Limited-Population Example Other Queuing Approaches Average Average Number Each of 5 printers require Cost/Hrafter 20 Cost/Hr(U) of use repair for hours for Number of Printers Downtime Technicians One technician can service a printer in 2 hours (T) Total The single-phase models cover many Technicians Down (N - J) (N - J)\$120 Printer downtime costs \$120/hour (\$25/hr) Cost/Hr queuing situations Technician costs \$25/hour \$76.80 1 .64 \$25.00 \$101.80 2 Variations of the four single-phase Service factor: X = 2 .46 46 \$55.20 091 \$50 00 090) \$55 = .091 (close to .090) 20 2 + 20 20 \$50.00 \$105.20 \$105 systems are possible For M = 1, D = .350 and F = .960 Multiphase models For M = 2, D = .044 and F = .998 exist for more Average number of printers working: complex situations For M = 1, J = (5)(.960)(1 - .091) = 4.36 For M = 2, J = (5)(.998)(1 - .091) = 4.54© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 49 © 2011 Pearson Education, Inc. publishing as Prentice Hall D - 50 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.© 2011 Pearson Education, Inc. publishing as Prentice Hall D - 51 9