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

D – 2 Negative Exponential Distribution Figure D – – – – – – – – – – – Probability that service time ≥ 1 ||||||||||||| Time t in hours Probability that service time is greater than t = e -µt for t ≥ 1 µ = Average service rate e = Average service rate (µ) = 1 customer per hour Average service rate (µ) = 3 customers per hour  Average service time = 20 minutes per customer

D – 3 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

D – 4 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

D – 5 Queuing Models Table D.2 ModelNameExample ASingle channel Information counter system at department store (M/M/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonExponentialUnlimitedFIFO

D – 6 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

D – 7 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

D – 8 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

D – 9 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 = ρ=Utilization factor for the system = 2 µ(µ – ) µ Table D.3

D – 10 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

D – 11 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 – (3 - 2)

D – 12 Single Channel Example W q = = = 40 minute average waiting time ρ= /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

D – 13 Single Channel Example Probability of More Than k Cars in the System kP n > k = (2/3) k  Note that this is equal to 1 - P 0 =  Implies that there is a 19.8% chance that more than 3 cars are in the system

D – 14 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 = $ Total expected costs = $ $56 = $162.67

D – 15 Queuing Models Table D.2 ModelNameExample BMultichannel Airline ticket (M/M/S) counter (M/M/S) counter NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline Multi-SinglePoissonExponentialUnlimitedFIFO channel channel

D – 16 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

D – 17 Multi-Channel Example = 2 µ = 3 M = 2 = 2 µ = 3 M = 2 P 0 = = !2!112!2! 1 11n!n!11n!n! 2(3) 2(3) n = 0 23 n ∑ 12 L s = + = (2)(3(2/3) ! 2(3) W q = = W s = = 3/4238 L q = – =

D – 18 Multi-Channel Example Single ChannelTwo Channels P0P LsLs 2 cars.75 cars WsWs 60 minutes22.5 minutes LqLq 1.33 cars.083 cars WqWq 40 minutes2.5 minutes

D – 19 Queuing Models Table D.2 ModelNameExample CConstant Automated car service wash service wash(M/D/1) NumberNumberArrivalService ofofRateTimePopulationQueue ChannelsPhasesPatternPatternSizeDiscipline SingleSinglePoissonConstantUnlimitedFIFO

D – 20 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

D – 21 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

D – 22 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

D – 23 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 – 24 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

D – 25 A paper distributor employs 1 worker who loads pallets of paper on outgoing trucks An average of 24 trucks arrive during the day The worker can load 4 trucks per hour The manager is thinking about adding another worker (effectively increasing the number of channels to 2) Analyze the effect on the queue for this type of change Another Example

D – 26 Another Example, cont’d. M/M/1M/M/S Average server utilization (ρ) Average number of customers in the queue (L q ) Average number of customers in the system (L) Average waiting time in the queue (W q ) Average time in the system (W) Probability (% of time) system is empty (P 0 )

D – 27 Truck drivers make $10/hour and warehouse workers make $6/hour Truck drivers earn their hourly salary even if they are sitting idle waiting for a truck to get loaded Does it pay to hire the second warehouse worker to help load trucks? Another Example – Economic Analysis

D – 28 Another Example – Economic Analysis, cont’d. Truck Driver Hourly Cost $ Warehouse Worker Hourly Cost $ 6.00 M/M/1M/M/S Truck Driver Idle Time Costs per Hour $ $ 8.73 Loading Costs per Hour $ 6.00 $ Total Expected Cost per Hour $ $ 20.73