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Ch-10. Queuing models Learning objectives: After completing this chapter, you should be able to: Why waiting lines can occur in service system, and explain that. 2. Identify typical goals for design of service system with respect to waiting. Summary

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Ch-10Queuing models Learning objectives:After completing this chapter, you should be able to:Why waiting lines can occur in service system, and explain that. 2. Identify typical goals for design of service system with respect to waiting.SummaryWaiting line problems represent an important class of management science models. Waiting lines are commonly found in a wide range of production and service system that encounter variable rates and service time.Management science interest in these problems centers on predicting system performance for the purpose of capacity design.GlossaryCalling population: pool of potential customers.Queuing system: system in which waiting lines tend to developed.System: A queuing system consists of a waiting line and a Service facility.System utilization : the proportion of time that servers are busy .Ch-10Queuing modelsIn our daily lives, we commonly encounter waiting lines at :Stop signs Banks Another places : factories buses post officesWaiting lines tend to from in these system due to overloads created by variability in either service or the arrival rates. Goals of queuing system designAvery common goals in queuing design is to attempt to balance the cost of providing service capacity with the cost of customers waiting cost for service.These two costs are in direct conflict :A decrease in customers waiting cost, increasing in the cost of service ?So , how to manage these two costs effectively. Elements and characteristics of queuing systems look at this figure : Prosing orderThe two major elements of queuing system are :1.Arrivals rate2.Service rateCallingpopulation Service exit Arrivals Waiting line Measures of system performanceA number of different performance measures can be computed summarize queuing behavior given the customer arrival rate the number of servers the service rate Among the most commonly used measures are the followingLq the average number waiting for serviceL the average number in the system, waiting or being served.Po the probability of zero units in the system.P the system utilization % of time servers are busy serving customers.Wq the average time customers must wait for serviceW the average time customers spend in the system , waiting for service and service timeM the expected maximum number waiting for service for a given level of confidence Basic Relationships the average number being served r = where = customer arrival rate M= service rate2. The average number in the systemL = Lq + rWhere L average number in the system Lq average number in the line3. The average time in line Wq = 4. The average time in the system including service Ws = Wq + 5. System utilization ( proportion of time servers are busy) P = Where S number of channels or servers.Note :- System utilization must be less than 1.00Go to the example Example : the owner of a car wash franchise intends to construct another car wash in a suburban location.Based on experience , the owner estimates that the arrival rate for the proposed facility will be 20 cars per hour and service rate will be 25 cars per hour.Service time will be variable because all cars are washed by hand rather than by machine. Cars will be processed one at a time, this is single-channel.Determine the following : the average number of cars being washed. the average number of cars in the system , either being washed or waiting to be washed , for a case where the average number waiting in line is 3.2 . The average time in line , average time cars wait to get washed. The average time cars spend in the system waiting in line and being washed. The system utilization. SolutionArrival rate ,, = 20 cars per hr.Service rate ,M, = 25 cars per hr .Number of servers ,S, = 1.Lq= 3.2 r= = = .80 cars being served . L= Lq + r = 3.2 + .8 = 4.0 cars Wq = = = .16 hrs.wich 9.6minutes (.16 x 60 minutes)Ws = Wq + = .16 + = .20 hr/12 minute P = = = .80 busy Queuing models!! First we are to solve for basic – single channel .?? Condition : one server. first _come , first served . calling population is infinite. No limit on queue length.Performance measures system utilizationAverage number is lineAverage number in systemAverage time in lineAverage time in systemProbability of zero units in the systemProbability of n units in the systemFormulaP= Lq= L = Lq + Wq = W = Wq + Po = 1 – ( )Pn = Po( ) n Formulas for basic single server model8. Probability the waiting line won't exceed K units9. Average waiting time for an arrival not served immediatelyP K = 1-( )K+1Wa = Example : -The mean arrival rate of customers at a ticket counter with one server is 3 per minute , and the mean service rate is 4 customers per minute , calculate each of the performance measures. Suppose that n=2 and k=4 .Solution : - P= = .75 or 75 percent Lq = = 2.25 customers L = 2.25 + = 3.00 customers Wq = = .75 minute W = .75 + = 1.00 minute Po = 1 - = .25 this means that the probability is 25 percent that an arriving unit will not have to wait for service.Hence , the probability that an arrival will have to wait service is 75 percent .7. Pn = 2 = .25 ( ) 2 = .1406Pn 5 = .1 – ( ) 5+1 = .8’22 Wa = = 1.0 minute . For the multiple channels or more than one server use the computer solutions?? Determining maximum line lengthsits important to design the amount of space that will be needed to accommodate waiting customers.An approximate line length that will satisfy any stated probability can be determined by solving the following equation for n :Pn = K for n Where K = Go to example Example :-Determine the maximum number of customers in line for probabilities of both 95 percent and 99 percent for this situation.S = 1 , = 4 per hour , M 5 per hour .Solution :-P = = = .80Lq = = = 3.2For 95 percent , K = = .078 n = = 11.43 ; n max 12For 99 percent , K = = 0.0156 n = = 18.64 , n max 19 Cost ConsiderationThe design of service system often reflects the desire of management to balance the cost of capacity with the expected cost of customers waiting in the system .Minimize= + Go to exampleCustomersWaitingcostCapacityCostTotalCost

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