1. D – 1 Operations Management Module D – Waiting-Line Models © 2006 Prentice Hall, Inc. PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 6e Operations Management, 8e

2. D – 2 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 maintenanceBroken machinesRepair people fix machines HarborShips and bargesDock workers load and unload Table D.1

3. D – 3 Characteristics of Waiting-Line Systems 1. 1.Arrivals or inputs to the system   Population size, behavior, statistical distribution 2. 2.Queue discipline, or the waiting line itself   Limited or unlimited in length, discipline of people or items in it 3. 3.The service facility   Design, statistical distribution of service times

4. D – 4 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

5. D – 5 Arrival Characteristics 1. 1.Size of the population   Unlimited (infinite) or limited (finite) 2. 2.Pattern of arrivals (statistical distribution)   Scheduled or random, often a Poisson distribution 3. 3.Behavior of arrivals   Wait in the queue and do not switch lines   Balking or reneging

6. D – 6 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)

7. D – 7 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

8. D – 8 Waiting-Line Characteristics   Limited or unlimited queue length   Queue discipline: first-in, first-out is most common   Other priority rules may be used in special circumstances

9. D – 9 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

10. D – 10 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 Your family dentist’s office McDonald’s dual window drive-through

11. D – 11 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

12. D – 12 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

13. D – 13 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 in 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

14. D – 14 Measuring Queue Performance 1. 1.Average time that each customer or object spends in the queue 2. 2.Average queue length 3. 3.Average time in the system 4. 4.Average number of customers in the system 5. 5.Probability the service facility will be idle 6. 6.Utilization factor for the system 7. 7.Probability of a specified number of customers in the system

15. D – 15 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

16. D – 16 Queuing Models Table D.2 ModelNameExample ASingle channel Information counter system at department store (M/M/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialUnlimitedFIFO

17. D – 17 Queuing Models Table D.2 ModelNameExample BMultichannel Airline ticket (M/M/S) counter (M/M/S) counter NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline Multi-SinglePoissonExponentialUnlimitedFIFO channel channel

18. D – 18 Queuing Models Table D.2 ModelNameExample CConstant Automated car service wash service wash(M/D/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonConstantUnlimitedFIFO

19. D – 19 Queuing Models Table D.2 ModelNameExample DLimited Shop with only a population dozen machines population dozen machines (finite) that might break NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialLimitedFIFO

20. D – 20 Model A - Single Channel 1. 1.Arrivals are FIFO and every arrival waits to be served regardless of the length of the queue 2. 2.Arrivals are independent of preceding arrivals but the average number of arrivals does not change over time 3. 3.Arrivals are described by a Poisson probability distribution and come from an infinite population

21. D – 21 Model A - Single Channel 4. 4.Service times vary from one customer to the next and are independent of one another, but their average rate is known 5. 5.Service times occur according to the negative exponential distribution 6. 6.The service rate is faster than the arrival rate

22. D – 22 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

23. D – 23 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

24. D – 24 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

25. D – 25 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)

26. D – 26 Single Channel Example W q = = = 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

27. D – 27 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

28. D – 28 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

29. D – 29 Measuring Queue Performance 1. 1.Average time that each customer or object spends in the queue 2. 2.Average queue length 3. 3.Average time in the system 4. 4.Average number of customers in the system 5. 5.Probability the service facility will be idle 6. 6.Utilization factor for the system 7. 7.Probability of a specified number of customers in the system

30. D – 30 Queuing Models Table D.2 ModelNameExample BMultichannel Airline ticket (M/M/S) counter (M/M/S) counter NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline Multi-SinglePoissonExponentialUnlimitedFIFO channel channel

31. D – 31 P0= Probability of 0 units in the system (that is, the service unit is idle) Ls = Average number of units (customers) in the system (waiting and being served) Lq = Average number of units waiting in the queue Ws = Average time a unit spends in the system (waiting time plus service time) Wq = Average time a unit spends waiting in the queue Multi-Channel Model

32. D – 32 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

33. D – 33 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

34. D – 34 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

35. D – 35 Queuing Models Table D.2 ModelNameExample CConstant Automated car service wash service wash(M/D/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonConstantUnlimitedFIFO

36. D – 36 Constant Service Model Table D.5 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 waiting time in system

37. D – 37 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