Introduction

1. Introduction • Queuing is the study of waiting lines, or queues. • The objective of queuing analysis is to design systems that enable organizations to perform optimally according to some criterion. • Possible Criteria • Maximum Profits. • Desired Service Level.

2. Introduction • 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.

3. Elements of the Queuing Process • A queuing system consists of three basic components: • Arrivals: Customers arrive according to some arrival pattern. • Waiting in a queue: Arriving customers may have to wait in one or more queues for service. • Service: Customers receive service and leave the system.

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.

5. The Arrival Process • Under three conditions the arrivals can be modeled as a Poisson process • Orderliness : one customer, at most, will arrive during any time interval. • Stationarity: for a given time frame, the probability of arrivals within a certain time interval is the same for all time intervals of equal length. • Independence : the arrival of one customer has no influence on the arrival of another.

6. The Poisson Arrival Process (lt)ke- lt k! P(X = k) = Where l = mean arrival rate per time unit. t = the length of the interval. e = 2.7182818 (the base of the natural logarithm). k! = k (k -1) (k -2) (k -3) … (3) (2) (1).

7. HANK’s HARDWARE – Arrival Process • Customers arrive at Hank’s Hardware according to a Poisson distribution. • Between 8:00 and 9:00 A.M. an average of 6 customers arrive at the store. • What is the probability that k customers will arrive between 8:00 and 8:30 in the morning (k = 0, 1, 2,…)?

8. 0 HANK’s HARDWARE – An illustration of the Poisson distribution. • Input to the Poisson distribution l = 6 customers per hour.t = 0.5 hour.lt = (6)(0.5) = 3. 1 2 3 4 5 6 7 8 k 1 0 2 3 (lt) e- lt k ! 1 0 = P(X = k )= 2 0.224042 0.224042 0.149361 0.049787 3 1! 0! 2! 3!

9. HANK’s HARDWARE – Using Excel for the Poisson probabilities • Solution • We can use the POISSON function in Excel to determine Poisson probabilities. • Point probability: P(X = k) = ? • Use Poisson(k, lt, FALSE) • Example: P(X = 0; lt = 3) = POISSON(0, 1.5, FALSE) • Cumulative probability: P(X£k) = ? • Example: P(X£3; lt = 3) = Poisson(3, 1.5, TRUE)

10. HANK’s HARDWARE – Excel Poisson

11. The Waiting Line Characteristics • Factors that influence the modeling of queues • Line configuration • Jockeying • Balking • Priority • Tandem Queues • Homogeneity

12. 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).

13. Jockeying and Balking • Jockeying occurs when customers switch lines once they perceived that another line is moving faster. • Balking occurs if customers avoid joining the line when they perceive the line to be too long.

14. Priority Rules • 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.

15. Tandem Queues • These are multi-server systems. • A customer needs to visit several service stations (usually in a distinct order) to complete the service process. • Examples • Patients in an emergency room. • Passengers prepare for the next flight.

16. Homogeneity • A homogeneous customer population is one in which customers require essentially the same type of service. • A non-homogeneous customer population is one in which customers can be categorized according to: • Different arrival patterns • Different service treatments.

17. The Service Process • In most business situations, 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.

18. f(t) = me-mt The probability that the service time X is less than some “t.” P(X £ t) = 1 - e-mt The Exponential Service Time Distribution m = the average number of customers who can be served per time period. Therefore, 1/m = the mean service time.

19. Schematic illustration of the exponential distribution The probability that service is completed within t time units P(X £ t) = 1 - e-mt X = t

20. HANK’s HARDWARE – Service time • Hank’s estimates the average service time to be 1/m = 4 minutes per customer. • Service time follows an exponential distribution. • What is the probability that it will take less than 3 minutes to serve the next customer?

21. Using Excel for the Exponential Probabilities • We can use the EXPDIST function in Excel to determine exponential probabilities. • Probability density: f(t) = ? • Use EXPONDIST(t, m, FALSE) • Cumulative probability: P(X£k) = ? • Use EXPONDIST(t, m, TRUE)

22. 3 minutes = .05 hours HANK’s HARDWARE – Using Excel for the Exponential Probabilities • The mean number of customers served per minute is ¼ = ¼(60) = 15 customers per hour. • P(X < .05 hours) = 1 – e-(15)(.05) = ? • From Excel we have: • EXPONDIST(.05,15,TRUE) = .5276

23. =EXPONDIST(B4,B3,TRUE) HANK’s HARDWARE – Using Excel for the Exponential Probabilities =EXPONDIST(A10,$B$3,FALSE) Drag to B11:B26

24. The Exponential Distribution -Characteristics • The memoryless property. • No additional information about the time left for the completion of a service, is gained by recording the time elapsed since the service started. • For Hank’s, the probability of completing a service within the next 3 minutes is (0.52763) independent of how long the customer has been served already. • The Exponential and the Poisson distributions are related to one another. • If customer arrivals follow a Poisson distribution with mean rate l, their interarrival times are exponentially distributed with mean time 1/l.

25. 9.3 Performance Measures of Queuing System • Performance can be measured by focusing on: • Customers in queue. • Customers in the system. • Performance is measured for a system in steady state.

26. 9.3 Performance Measures of Queuing System • The transient period occurs at the initial time of operation. • Initial transient behavior is not indicative of long run performance. n Roughly, this is a transient period… Time

27. 9.3 Performance Measures of Queuing System • The steady state period follows the transient period. n This is a steady state period……….. Roughly, this is a transient period… • Meaningful long run performance measures can be calculated for the system when in steady state. Time

28. k servers 9.3 Performance Measures of Queuing System In order to achieve steady state, theeffective arrival rate must be less than the sum of the effective service rates . l< m For one server l< m1 +m2+…+mk For k servers with service rates mi • l< km • Each with • service rate of m

29. 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).

30. 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 a steady state condition. L = l W Lq = l Wq L = Lq + l/m For the case of an infinite population

31. 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. Example: M / M / 6 / 10 / 20

32. M/M/1 Queuing System - Assumptions • Poisson arrival process. • Exponential service time distribution. • A single server. • Potentially infinite queue. • An infinite population.

33. M / M /1 Queue - Performance Measures P0 = 1 – (l/m) Pn = [1 – (l/m)](l/m)n L = l /(m – l) Lq = l2 /[m(m – l)] W = 1 /(m – l) Wq = l /[m(m – l)] Pw = l / m r = l / m The probability that a customer waits in the system more than “t” is P(X>t) = e-(m - l)t

34. 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.

35. m –l = 7.5 – 5 = 2.5 per hr. P(X<10min) = 1 – e-2.5(10/60) = .565 Pw = l/m =0.6667 r = l/m =0.6667 MARY’s SHOES - 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 • P0 = 1 - (l/m) = 1 - (5/7.5) = 0.3333 • Pn = [1 - (l/m)](l/m)n = (0.3333)(0.6667)n • L = l/(m - l) = 2 • Lq = l2/[m(m - l)] = 1.3333 • W = 1/(m - l) = 0.4 hours = 24 minutes • Wq = l/[m(m - l)] = 0.26667 hours = 16 minutes

36. =A11-B4/B5 =1-B4/B5 =B4/B5 =1-E11 =B4/(B5-B4) =C11-1/B5 =H11*($B$4/$B$5) Drag to Cell AL11 =A11/B4 MARY’s SHOES - Spreadsheet solution

37. 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.

38. Meats Beverage Tandem Queuing Systems • In a Tandem Queuing System a customer must visit several different servers before service is completed. • Examples • All-You-Can-Eat restaurant

39. Tandem Queuing Systems • In a Tandem Queuing System a customer must visit several different servers before service is completed. Meats Beverage • Examples • All-You-Can-Eat restaurant

40. Tandem Queuing Systems • In a Tandem Queuing System a customer must visit several different servers before service is completed. Meats Beverage • Examples • All-You-Can-Eat restaurant • A drive-in restaurant, where first you place your order, then pay and receive it in the next window. • A multiple stage assembly line.

41. Tandem Queuing Systems • For cases in which customers arrive according to a Poisson process and service time in each station is exponential, …. Total Average Time in the System = Sum of all average times at the individual stations

42. 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.

43. BIG BOYS SOUND, INC. • 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 waits 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 at all the service stations. • Poisson arrival with a rate of 40 customers an hour.

44. BIG BOYS SOUND, INC. What is the average amount of time, a customer who makes a purchase spends in the store? Only 75% of the arriving customers make a purchase!

45. BIG BOYS SOUND, INC. – Solution • This is a Three Station Tandem Queuing System Pickup desk M / M / 2 (.75)(40)=30 l = 30 Cashiers M / M / 3 Sales Clerks M / M / 8 l = 30 W3 = 2.67 minutes l = 40 W2 = 3.47 minutes Total = 20.14 minutes. W1 = 14 minutes