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## Ch-10

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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**Waiting 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.**Glossary**Calling 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-10**Queuing models In our daily lives, we commonly encounter waiting lines at : • Stop signs • Banks • Another places : factories buses post offices Waiting lines tend to from in these system due to overloads created by variability in either service or the arrival rates.** Goals of queuing system design**Avery 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 order The two major elements of queuing system are : 1.Arrivals rate 2.Service rate Calling population Service exit Arrivals Waiting line** Measures of system performance**A 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 following Lq the average number waiting for service L 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 service**W the average time customers spend in the system , waiting for service and service time M 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 rate**2. The average number in the system**L = Lq + r Where L average number in the system Lq average number in the line 3. 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 1.00 Go 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.** Solution**Arrival 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 utilization Average number is line Average number in system Average time in line Average time in system Probability of zero units in the system Probability of n units in the system Formula P= Lq= L = Lq + Wq = W = Wq + Po = 1 – ( ) Pn = Po( ) n Formulas for basic single server model**8. Probability the waiting line won't exceed K units**9. Average waiting time for an arrival not served immediately P K = 1-( )K+1 Wa =** 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 = .1406 • Pn 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 lengths**its 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 = = = .80 Lq = = = 3.2 For 95 percent , K = = .078 n = = 11.43 ; n max 12**For 99 percent , K = = 0.0156**n = = 18.64 , n max 19** Cost Consideration**The 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 example Customers Waiting cost Capacity Cost Total Cost