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QueuingSession 6 & 7M/M/1 and M/M/SEveryday QueuesFast Food McDonalds, Burger King,Retail supermarkets, stores, banksMedical doctors office, access to diagnostic procedures, specialist referralsAirports check-in, baggage collection, runway delays, waiting to landTraffic - congestionQueues in ManufacturingOrder backlogsWork in process inventoriesDistribution inventoriesOften queue (inventory) size is limitedNumber of Kanbans limits work in process Fixed number of AGVs circulate in a loopQueuing Theory - Basics

What determines queue lengths and waiting times?

arrivals Queuedepartures (Customers, Jobs)

Arrival rate= = 1/mean time between arrivalsService rate = = 1/mean time to serve one customerSERVERKey IdeasQueue length, waiting time determined by(1) Utilization of server = = /= Job arrival rate mean time to serve each job(2) Variability of arrivals (lowest with equally spaced arrivals)(3) Variability of service time (lowest is when all jobs require exactly the same time)Sources of Service Time VariabilityServer not consistent (some variability inherent in all repetitive tasks: depends on task (cognitive) complexity)Different people performing same task have different mean times (best 2worst in manual assembly tasks, ratio increases with task complexity)Mixture of different customer types, each type takes a different timeFinite QueuesCall centers, hospitalsNumber of waiting spaces is finite (and fixed)If waiting spaces occupied arrivals are lostQuality of Service measures:Fraction of calls accepted Number of attempts to obtain serviceFraction of customers served within 5 minutes (How does a manager behave to meet this type of service target?)Closed QueuesFinite

Finite number of customers circulate in systemPallets in a manufacturing cellKanban cards

Notations = average number of customers that arrive per unit of time (also called as arrival rate of the system) = average number of customers that can be serviced by a single channel per unit of time ( also called as service rate of a channel)M =number of serving channels

Behaviour of simple systemsM, and are the three inputs that are required to predict the behaviour of the queues in the systemFor the system with M channels, M is the service rate of the systemIn a queuing system, the average number of customers that can served per hour must be greater than the average number that arrive per hour (that is, it must be that M > ExampleSuppose that a hamburger stand has one server who can serve 70 customers per hour on average. (the service rate is 70/60 =0.86 minutes per customer)Customer arrive at an average rate of 49 per hour in the busy periodThe given data are:M=1 = average arrival rate = 49 = average service rate = 70

ExampleThe average number of people on the queue/ = 49/70 = 0.7No. of servers = 1Using Exhibit 3,

We can estimate that average queue length(Lq) is 1.6333personsMeasures of system performanceWq = average (expected waiting time)= Lq/L = average number of people in the system (waiting or being served) = Lq + /W = average time in system (waiting or being served) = Wq + 1/ = Lq/ + 1/

Performance measuresExpected waiting time is 1.6333/49 = 0.033 hours or 2 minutesAverage number of people in the system = 1.633 +0.70 = 2.333 peopleThe average time in the system = 0.033 + 1/70 hours = 2.86 minutes (2 minutes of waiting time and 0.86 minutes of serving time)Summary of results(per hour)(per hour)M/Lq(people)Wq (minutes)L (people)W (minutes)497010.700.701.63322.3332.86What if the number of servers is raised to 2, but still have a single queue?

By using the Exhibit 3 we obtain the value of Lq as 0.0976(per hour)(per hour)M/Lq(people)Wq (minutes)L (people)W (minutes)497010.700.701.63322.3332.86497020.700.350.0980.120.800.98Inferences Average queue length has reduced by 94.5%But the average time spent in the system (W) has been reduced by only 66%Serving time still remain at 0.86, though the expected waiting time has dropped to 0.12 minutesWhat-if analysisWhat if the average service time is reduced to half rather increasing the channels?Means, M=1, but we have a new value of By cutting the average service time in half, we double the number of customers served per hour. Therefore is 140 and / = =0.35What-if analysisFor M=1 and =0.35, average queue length is 0.1884 people, and other values are

The second channel reduces average queue length (Lq) more,Reduction in service time reduces the total time in the system(per hour)(per hour)M/Lq(people)Wq (minutes)L (people)W (minutes)497010.700.701.63322.3332.86497020.700.350.0980.120.800.984914010.350.350.1880.230.540.66What if another hamburger stand is opened some blocks away and half the customers turn to the new stand?Specification of Queueing SystemsCustomer arrival and service stochastic modelsStructural Parameters Number of serversStorage capacityOperating policiesCustomer class differentiation (are all customers treated the same or do some have priority over others?)Scheduling/Queueing policies (which customer is served next)Admission policies (which/when customers are admitted)Queueing System Notation A/B/m/K/NArrival ProcessM: Markovian D: DeterministicEr: ErlangG: GeneralService ProcessM: Markovian D: DeterministicEr: ErlangG: GeneralNumber of servers m=1,2,Storage Capacity K= 1,2, (if then it is omitted)Number of customers N= 1,2, (for closed networks otherwise it is omitted)Single channel waiting lineDistribution of arrivalPoisson DistributionProvides the probability function of x arrivals in a specific time period. Probability function is

Where x = number of arrivals in the time period = the mean number of arrivals per time period

Example problemSuppose that Burger Dome analysed data on customer and concluded that the arrival rate is 45 customers per hour.What is ?What is the probability of 0,1, and 2 customer arrivals during a one minute period?Distribution of service timeExponential probability distributionP(service time t) = 1-e-tThe mean number of units that can be served per time period,, is called the service rate Suppose that Burger dome estimated that it can process 60 customer orders per hour.What is the probability that an order can be processed in minute or less1 minute or less2 minute or lessPerformance Measures of Interest We are interested in steady state behaviorEven though it is possible to pursue transient results, it is a significantly more difficult task. E[S] average system time (average time spent in the system)E[W] average waiting time (average time spent waiting in queue(s))E[X] average queue lengthE[U] average utilization (fraction of time that the resources are being used)E[R] average throughput (rate that customers leave the system)E[L] average customer loss (rate that customers are lost or probability that a customer is lost)M/M/1 Performance Measures = the mean number of arrivals per time period (the arrival rate) = the mean number of services per time period (the service rate)The probability that no units are in the system

The average number of units in the waiting line

The average number of units in the system

M/M/1 Performance MeasuresThe average time a unit spends in the waiting line

The average time a unit spends in the system

The probability that an arriving unit has to wait for service

The probability of n units in the system

M/M/1 Performance Measures