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1. Queuing Theory Chapter 12

3. Analyzing queuing systems requires a clear understanding of the appropriate service measurement. Possible service measurements Average time a customer spends in line. Average length of the waiting line. The probability that an arriving customer must wait for service.

4. The Arrival Process There are two possible types of arrival processes Deterministic arrival process. Random arrival process. The random process is more common in businesses. Under three conditions a Poisson distribution can describe the random arrival process.

5. The Poisson Arrival Distribution

6. The Waiting Line Line configuration A Single service Queue. Multiple service queue with single waiting line. Multiple service queue with multiple waiting lines. Tandem queue (multistage service system). Jockeying Jockeying occurs if customers switch lines when they perceived that another line is moving faster. Balking Balking occurs if customers avoid joining the line when they perceive the line to be too long.

7. Priority rules Priority rules define the line discipline. These rules select the next customer for service. There are several commonly used rules: First come first served (FCFS). Last come first served (LCFS). Estimated service time. Random selection of customers for service. Homogeneity An homogeneous customer population is one in which customers require essentially the same type of service. A Nonhomogeneous customer population is one in which customers can be categorized according to: Different arrival patterns Different service treatments.

8. The Service Process Some service systems require a fixed service time. In most business situations, however, service time varies widely among customers. When service time varies, it is treated as a random variable. The exponential probability distribution is used sometimes to model customer service time.

9. The Exponential Service Time Distribution

11. 12.3 Measures of Queuing System Performance Performance can be measured by focusing on: Customers in queue. Customers in the system. Transient and steady state periods complicate the service time analysis

12. The transient period occurs at the initial time of operation. Initial transient behavior is not indicative of long run performance. The steady state period follows the transient period. In steady state, long run probabilities of having �n� customers in the system do not change as time goes on. In order to achieve steady state, the effective arrival rate must be less than the sum of effective service rates . l< m l< m1 +m2+�+mk l< km For one server For k servers For k servers each with service rate m

13. Steady State Performance Measures P0 = Probability that there are no customers in the system. Pn = Probability that there are �n� customers in the system. L = Average number of customers in the system. Lq = Average number of customers in the queue. W = Average time a customer spends in the system. Wq = Average time a customer spends in the queue. Pw = Probability that an arriving customer must wait for service. r = Utilization rate for each server (the percentage of time that each server is busy).

14. Little�s Formulas Little�s Formulas represent important relationships between L, Lq, W, and Wq. These formulas apply to systems that meet the following conditions: Single queue systems, Customers arrive at a finite arrival rate l, and The system operates under steady state condition. L = l W Lq = l Wq L = Lq + l / m For the infinite population case

15. Classification of Queues Queuing system can be classified by: Arrival process. Service process. Number of servers. System size (infinite/finite waiting line). Population size. Notation M (Markovian) = Poisson arrivals or exponential service time. D (Deterministic) = Constant arrival rate or service time. G (General) = General probability for arrivals or service time.

16. 12.4 M / M / 1 Queuing System Characteristics Poisson arrival process. Exponential service time distribution. A single server. Potentially infinite queue. An infinite population.

17. Performance Measures for the M / M /1 Queue P0 = 1- (l / m) Pn = [1 - (l / m)] (l/ m)n L = l / (m - l) Lq = l 2 / [m(m - l)] W = 1 / (m - l) Wq = l / [m(m - l)] Pw = l / m r = l / m

18. MARY�s SHOES Customers arrive at Mary�s Shoes every 12 minutes on the average, according to a Poisson process. Service time is exponentially distributed with an average of 8 minutes per customer. Management is interested in determining the performance measures for this service system.

19. SOLUTION Input l = 1/ 12 customers per minute = 60/ 12 = 5 per hour. m = 1/ 8 customers per minute = 60/ 8 = 7.5 per hour. Performance Calculations

20. 12.5 M / M / k Queuing Systems Characteristics Customers arrive according to a Poisson process at a mean rate l. Service time follow an exponential distribution. There are k servers, each of which works at a rate of m customers. Infinite population, and possibly infinite line.

21. Performance measure

23. LITTLE TOWN POST OFFICE Little Town post office is opened on Saturdays between 9:00 a.m. and 1:00 p.m. Data On the average 100 customers per hour visit the office during that period. Three clerks are on duty. Each service takes 1.5 minutes on the average. Poisson and Exponential distribution describe the arrival and the service processes respectively.

24. SOLUTION This is an M / M / 3 queuing system. Input l = 100 customers per hour. m = 40 customers per hour (60 / 1.5). Does steady state exist (l < km )? l = 100 < km = 3(40) = 120.

25. 12.6 M / G / 1 Queuing System Assumptions Customers arrive according to a Poisson process with a mean rate l. Service time has a general distribution with mean rate m. One server. Infinite population, and possibly infinite line.

26. Pollaczek - Khintchine Formula for L

27. TED�S TV REPAIR SHOP Ted�s repairs television sets and VCRs. Data It takes an average of 2.25 hours to repair a set. Standard deviation of of the repair time is 45 minutes. Customers arrive at the shop once every 2.5 hours on the average, according to a Poisson process. Ted works 9 hours a day, and has no help. He considers purchasing a new piece of equipment. New average repair time is expected to be 2 hours. New standard deviation is expected to be 40 minutes.

29. SOLUTION This is an M / G / 1 system (service time is not exponential (because s 1/m). Input The current system (without the new equipment) l = 1/ 2.5 = 0.4 customers per hour. m = 1/ 2.25 = 0.4444 customers per hour. s = 45/ 60 = 0.75 hours. The new system (with the new equipment) m = 1/2 = 0.5 customers per hour. s = 40/ 60 = 0.6667 hours.

30. 12.7 M / M / k / F Queuing System Many times queuing systems have designs that limit their size. When the potential queue is large, an infinite queue model gives accurate results, even though the queue might be limited. When the potential queue is small, the limited line must be accounted for in the model.

31. Characteristics of the M / M / k / F system Poisson arrival process at mean rate l. k servers, each having an exponential service time with mean rate m. Maximum number of customers that can be present in the system at any one time is �F�. Customers are blocked (and never return) if the system is full.

32. The Effective Arrival Rate A customer is blocked if the system is full. The probability that the system is full is PF. The effective arrival rate = the rate of arrivals that make it through into the system (le).

33. RYAN ROOFING COMPANY Ryan gets most of its business from customers who call and order service. Data One appointment secretary takes phone calls from 3 telephone lines. Each phone call takes three minutes on the average. Ten customers per hour call the company on the average.

34. When a telephone line is available but the secretary is busy serving a customer, a new calling customer is willing to wait until the secretary becomes available. When all the lines are busy, a new calling customer gets a busy signal and calls a competitor. Arrival process is Poisson, and service process is Exponential.

36. SOLUTION This is an M / M / 1 / 3 system Input l = 10 per hour. m = 20 per hour (1/ 3 per minute). WINQSB gives: P0 = 0.5333, P1 = 0.2667 , P2 = 0. 1333 , P3 = 0.0667 6.7% of the customers get a busy signal. This is above the goal of 2%.

37. 12.8 M / M / 1 / / m Queuing Systems In this system the number of potential customers is finite and relatively small. As a result, the number of customers already in the system effects the rate of arrivals of the remaining customers. Characteristics A single server. Exponential service and interarrivall time, Poisson arrival process. A population size of m customers (m is finite).

38. PACESETTER HOMES Pacesetter Homes runs four different development projects. Data A stoppage occurs once every 20 working days on the average in each site. It takes 2 days on the average to solve a problem. Each problem is handled by the V.P. for construction. How long on the average a site does not operate? With 2 days to solve a problem (current situation) With 1.875 days to solve a problem (new situation)

39. SOLUTION This is an M / M / 1 // 4 system. The four sites are the four customers. The V.P. for construction can be considered a server. Input l = 0.05 (1/ 20) m = 0.5 (1/ 2 using the current car). m = 0.533 (1/1.875 using a new car).

41. 12.9 ECONOMIC ANALYSIS OF QUEUING SYSTEMS The performance measures previously developed are used next to determine a minimal cost queuing system. The procedure requires estimated costs such as: Hourly cost per server . Customer goodwill cost while waiting in line. Customer goodwill cost while being served.

42. WILSON FOODS TALKING TURKEY HOT LINE Wilson Foods has an 800 number to answer customers� questions. Data On the average 225 calls per hour are received. An average phone call takes 1.5 minutes. A customer will stay on the line waiting at most 3 minutes. A customer service representative is paid \$16 per hour. Wilson pays the telephone company \$0.18 per minute when the customer is on hold or when being served. Customer goodwill cost is \$20 per minute while on hold. Customer goodwill cost while in service is \$0.05.

43. SOLUTION The total cost model

44. Input Cw= \$16 Ct = \$10.80 per hour [0.18(60)] gw= \$12 per hour [0.20(60)] gs = \$0.05 per hour [0.05(60)] The Total Average Hourly Cost is TC(K) = 16K + (10.8+3)L + (12 - 3)Lq = 16K + 13.8L + 9Lq

45. Assuming a Poisson arrival process and an Exponential service time, we have an M / M / K system. l = 225 calls per hour. m = 40 per hour (60/ 1.5). The minimal possible value for K is 6 to ensure that steady state exists (l<Km). WINQSB was used to generate results for L, Lq, and Wq.

46. Summary of results of the runs for k=6,7,8,9,10

47. Example: M/M/1 vs M/M/2 Which is better? Select 1 machine with certain speed and certain cost or 2 machines with half speed and half cost? M/M/1 Input l = 20 m = 30 Cw = \$5 gw = 2 gs = 1 M/M/2 Input l = 20 m = 15 Cw = \$2.5 gw = 2 gs = 1

48. Example: M/M/1 vs M/M/2 (cont)

49. Example: M/M/1 vs M/M/2 (cont)

50. 12.10 Tandem Queuing Systems In a Tandem Queuing System a customer must visit several different servers before service is completed. For cases in which customers arrive according to a Poisson process and service time in each station is Exponential,

51. BIG BOYS SOUND, INC. Big Boys sells audio merchandise. The sale process is as follows: A customer places an order with a sales person. The customer goes to the cashier station to pay for the order. After paying, the customer is sent to the pickup desk to obtain the good.

52. Data for a regular Saturday Personnel. 8 sales persons are on the job. 3 cashiers. 2 workers in the merchandise pickup area. Average service times. Average time a sales person wait on a customer is 10 minutes. Average time required for the payment process is 3 minutes. Average time in the pickup area is 2 minutes. Distributions. Exponential service time in all the service stations. Poisson arrival with a rate of 40 customers an hour.

53. SOLUTION This is a Three Station Tandem Queuing System