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© 2008 Prentice Hall, Inc.D – 1 Operations Management Module D – Waiting-Line Models PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 7e Operations Management, 9e
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© 2008 Prentice Hall, Inc.D – 2 Outline Characteristics of a Waiting-Line System Arrival Characteristics Waiting-Line Characteristics Service Characteristics Measuring a Queue’s Performance Queuing Costs
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© 2008 Prentice Hall, Inc.D – 3 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 Model D: Limited-Population Model
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© 2008 Prentice Hall, Inc.D – 4 Outline – Continued Other Queuing Approaches
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© 2008 Prentice Hall, Inc.D – 5 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
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© 2008 Prentice Hall, Inc.D – 6 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
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© 2008 Prentice Hall, Inc.D – 7 Waiting Lines Often called queuing theory Waiting lines are common situations Useful in both manufacturing and service areas
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© 2008 Prentice Hall, Inc.D – 8 Common Queuing Situations Situation Arrivals in Queue Service Process Supermarket Grocery shoppers Checkout clerks at cash register Highway toll booth Automobiles Collection of tolls at booth Doctor’s office Patients Treatment by doctors and nurses Computer system Programs to be run Computer processes jobs Telephone company Callers Switching equipment to forward calls BankCustomer Transactions handled by teller Machine maintenance Broken machines Repair people fix machines Harbor Ships and barges Dock workers load and unload Table D.1
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© 2008 Prentice Hall, Inc.D – 9 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
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© 2008 Prentice Hall, Inc.D – 10 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
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© 2008 Prentice Hall, Inc.D – 11 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) Servicefacility 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
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© 2008 Prentice Hall, Inc.D – 12 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 =average arrival rate e=2.7183 (which is the base of the natural logarithms)
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© 2008 Prentice Hall, Inc.D – 13 Poisson Distribution Probability = P(x) = e - x x! 0.25 0.25 – 0.02 0.02 – 0.15 0.15 – 0.10 0.10 – 0.05 0.05 – – Probability 0123456789 Distribution for = 2 x 0.25 0.25 – 0.02 0.02 – 0.15 0.15 – 0.10 0.10 – 0.05 0.05 – – Probability 0123456789 Distribution for = 4 x 1011 Figure D.2
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© 2008 Prentice Hall, Inc.D – 14 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
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© 2008 Prentice Hall, Inc.D – 15 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
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© 2008 Prentice Hall, Inc.D – 16 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
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© 2008 Prentice Hall, Inc.D – 17 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
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© 2008 Prentice Hall, Inc.D – 18 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
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© 2008 Prentice Hall, Inc.D – 19 Negative Exponential Distribution Figure D.4 1.0 1.0 – 0.9 0.9 – 0.8 0.8 – 0.7 0.7 – 0.6 0.6 – 0.5 0.5 – 0.4 0.4 – 0.3 0.3 – 0.2 0.2 – 0.1 0.1 – 0.0 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
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© 2008 Prentice Hall, Inc.D – 20 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
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© 2008 Prentice Hall, Inc.D – 21 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 MinimumTotalcost Optimal service level
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© 2008 Prentice Hall, Inc.D – 22 Queuing Models The four queuing models here all assume: Poisson distribution arrivals FIFO discipline A single-service phase
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© 2008 Prentice Hall, Inc.D – 23 Queuing Models Table D.2 ModelNameExample ASingle-channel Information counter system at department store system at department store(M/M/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialUnlimitedFIFO
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© 2008 Prentice Hall, Inc.D – 24 Queuing Models Table D.2 ModelNameExample BMultichannel Airline ticket (M/M/S) counter (M/M/S) counter NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline Multi-SinglePoissonExponentialUnlimitedFIFO channel channel
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© 2008 Prentice Hall, Inc.D – 25 Queuing Models Table D.2 ModelNameExample CConstant- Automated car service wash service wash(M/D/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonConstantUnlimitedFIFO
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© 2008 Prentice Hall, Inc.D – 26 Queuing Models Table D.2 ModelNameExample DLimited Shop with only a population dozen machines population dozen machines (finite population) that might break NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialLimitedFIFO
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© 2008 Prentice Hall, Inc.D – 27 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
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© 2008 Prentice Hall, Inc.D – 28 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
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© 2008 Prentice Hall, Inc.D – 29 Model A – Single-Channel =Mean number of arrivals per time period =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 – µ – 1 Table D.3
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© 2008 Prentice Hall, Inc.D – 30 Model A – Single-Channel L q =Average number of units waiting in the queue = W q =Average time a unit spends waiting in the queue = p=Utilization factor for the system = 2 µ(µ – ) µ Table D.3
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© 2008 Prentice Hall, Inc.D – 31 Model A – Single-Channel 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 Table D.3
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© 2008 Prentice Hall, Inc.D – 32 Single-Channel Example =2 cars arriving/hourµ= 3 cars serviced/hour =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 – µ – 1 2 3 - 2 1 2 2 3(3 - 2)
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© 2008 Prentice Hall, Inc.D – 33 Single-Channel Example W q = = = 2/3 hour = 40 minute average waiting time p= /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 =2 cars arriving/hourµ= 3 cars serviced/hour
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© 2008 Prentice Hall, Inc.D – 34 Single-Channel Example Probability of more than k Cars in the System kP n > k = (2/3) k + 1 0.667 Note that this is equal to 1 - P 0 = 1 -.33 1.444 2.296 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
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© 2008 Prentice Hall, Inc.D – 35 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 Customer waiting-time cost = $10 10 = $106.67 23 Total expected costs = $106.67 + $56 = $162.67
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© 2008 Prentice Hall, Inc.D – 36 Multi-Channel Model M=number of channels open =average arrival rate =average arrival rate µ=average service rate at each channel P 0 = for Mµ > P 0 = for Mµ > 11 11M!M!11M!M!1 11n!n!11n!n! Mµ Mµ - Mµ - M – 1 n = 0 µ n µ M + ∑ L s = P 0 + µ( /µ) µ( /µ)M (M - 1)!(Mµ - ) 2 µ Table D.4
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© 2008 Prentice Hall, Inc.D – 37 Multi-Channel Model Table D.4 W s = P 0 + = µ( /µ) µ( /µ) M (M - 1)!(Mµ - ) 2 1µ L s L q = L s – µ W q = W s – = 1µ L q
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© 2008 Prentice Hall, Inc.D – 38 Multi-Channel Example = 2 µ = 3 M = 2 = 2 µ = 3 M = 2 P 0 = = 11 112!2!112!2! 1 11n!n!11n!n! 2(3) 2(3) - 2 1 n = 0 23 n 23 2 + ∑ 12 L s = + = (2)(3(2/3) 2 23 1! 2(3) - 2 2 1234 W q = =.0415.0832 W s = = 3/4238 L q = – = 23 34112
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© 2008 Prentice Hall, Inc.D – 39 Multi-Channel Example Single Channel Two Channels P0P0P0P0.33.5 LsLsLsLs 2 cars.75 cars WsWsWsWs 60 minutes 22.5 minutes LqLqLqLq 1.33 cars.083 cars WqWqWqWq 40 minutes 2.5 minutes
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© 2008 Prentice Hall, Inc.D – 40 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 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
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© 2008 Prentice Hall, Inc.D – 41 Waiting Line Table Example Bank tellers and customers = 18, µ = 20 = 18, µ = 20 From Table D.5 Utilization factor ρ = /µ =.90 Wq =Wq =Wq =Wq = L q Number of service windows M Number in queue Time in queue 1 window 18.1.45 hrs, 27 minutes 2 windows 2.2285.0127 hrs, ¾ minute 3 windows 3.03.0017 hrs, 6 seconds 4 windows 4.0041.0003 hrs, 1 second
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© 2008 Prentice Hall, Inc.D – 42 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 + 1µ Average time in the system
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© 2008 Prentice Hall, Inc.D – 43 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) 112 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
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© 2008 Prentice Hall, Inc.D – 44 Limited-Population Model 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 Table D.7
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© 2008 Prentice Hall, Inc.D – 45 Limited-Population Model 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 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
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© 2008 Prentice Hall, Inc.D – 46 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
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© 2008 Prentice Hall, Inc.D – 47 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
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© 2008 Prentice Hall, Inc.D – 48 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
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© 2008 Prentice Hall, Inc.D – 49 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
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