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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 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 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 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 5 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. The Arrival Process

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

7 7 HANKs HARDWARE – Arrival Process Customers arrive at Hanks 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 8 k Input to the Poisson distribution = 6 customers per hour. t = 0.5 hour. t = (6)(0.5) = 3. t e - t k ! ! ! ! P(X = k )= HANKs HARDWARE – An illustration of the Poisson distribution.

9 9 HANKs 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, t, FALSE) Example: P(X = 0; t = 3) = POISSON(0, 1.5, FALSE ) –Cumulative probability: P(X k) = ? Example: P(X 3; t = 3) = Poisson(3, 1.5, TRUE )

10 10 HANKs HARDWARE – Excel Poisson

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

12 12 A single service queue. Multiple service queue with single waiting line. Multiple service queue with multiple waiting lines. Tandem queue (multistage service system). Line Configuration

13 13 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. Jockeying and Balking

14 14 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. Priority Rules

15 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 16 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. Homogeneity

17 17 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. The Service Process

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

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

20 20 HANKs HARDWARE – Service time Hanks estimates the average service time to be 1/ = 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 21 We can use the EXPDIST function in Excel to determine exponential probabilities. Probability density: f(t) = ? –Use EXPONDIST(t,, FALSE ) Cumulative probability: P(X k) = ? –Use EXPONDIST(t,, TRUE ) Using Excel for the Exponential Probabilities

22 22 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 HANKs HARDWARE – Using Excel for the Exponential Probabilities 3 minutes =.05 hours

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

24 24 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 Hanks, the probability of completing a service within the next 3 minutes is ( ) 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, their interarrival times are exponentially distributed with mean time 1 / The Exponential Distribution - Characteristics

25 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 26 Roughly, this is a transient period… n Time 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.

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

28 28 k Each with service rate of k Each with service rate of … For k servers with service rates … For k servers with service rates For one server For one server In order to achieve steady state, the effective arrival rate must be less than the sum of the effective service rates. 9.3 Performance Measures of Queuing System k servers

29 29 P 0 = Probability that there are no customers in the system. P n = Probability that there are n customers in the system. L = Average number of customers in the system. L q = Average number of customers in the queue. W = Average time a customer spends in the system. W q = Average time a customer spends in the queue. P w = Probability that an arriving customer must wait for service. P w = Probability that an arriving customer must wait for service. = Utilization rate for each server (the percentage of time that each server is busy). = Utilization rate for each server (the percentage of time that each server is busy). Steady State Performance Measures

30 30 Littles Formulas represent important relationships between L, L q, W, and W q. These formulas apply to systems that meet the following conditions: – Single queue systems, – Customers arrive at a finite arrival rate and – The system operates under a steady state condition. L = W L q = W q L = L q + Littles Formulas For the case of an infinite population

31 31 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 Example: M / M / 6 / 10 / 20 Classification of Queues

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

33 33 The probability that a customer waits in the system more than t is P(X>t) = e - ( - )t The probability that a customer waits in the system more than t is P(X>t) = e - ( - )t P 0 = 1 – ( ) P n = [1 – ( )]( ) n L = ( – ) L q = 2 [ ( – )] W = 1 ( – ) W q = [ ( – )] P w = = M / M /1 Queue - Performance Measures

34 34 MARYs SHOES Customers arrive at Marys 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 35 MARYs SHOES - Solution –Input = 1 / 12 customers per minute = 60 / 12 = 5 per hour. = 1 / 8 customers per minute = 60 / 8 = 7.5 per hour. –Performance Calculations P 0 = 1 - ( ) = 1 - (5 7.5) = P n = [1 - ( )]( ) n = (0.3333)(0.6667) n L = ( - ) = 2 L q = 2 [ ( - )] = W = 1 ( - ) = 0.4 hours = 24 minutes W q = ( - )] = hours = 16 minutes P 0 = 1 - ( ) = 1 - (5 7.5) = P n = [1 - ( )]( ) n = (0.3333)(0.6667) n L = ( - ) = 2 L q = 2 [ ( - )] = W = 1 ( - ) = 0.4 hours = 24 minutes W q = ( - )] = hours = 16 minutes P w = = = = P(X<10min) = 1 – e -2.5(10/60) =.565 – = 7.5 – 5 = 2.5 per hr.

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

37 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 38 Tandem Queuing Systems In a Tandem Queuing System a customer must visit several different servers before service is completed. Beverage Meats Examples –All-You-Can-Eat restaurant

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

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

41 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 Total Average Time in the System = Sum of all average times at the individual stations

42 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 43 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. BIG BOYS SOUND, INC.

44 44 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! BIG BOYS SOUND, INC.

45 45 BIG BOYS SOUND, INC. – Solution This is a Three Station Tandem Queuing System Sales Clerks M / M / 8 Cashiers M / M / 3 Pickup desk M / M / 2 = 40 = 30 W 1 = 14 minutes W 2 = 3.47 minutes W 3 = 2.67 minutes Total = minutes. (.75)(40)=30


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