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1 Queuing Models

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2 Introduction (1/2) 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.

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3 Introduction (2/2) 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.

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4 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.

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5 The Arrival Process (1/2) There are two possible types of arrival processes –Deterministic arrival process. –Random arrival process. The random process is more common in businesses.

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6 Under three conditions the arrivals can be modeled as a Poisson process – Orderliness : one customer, at most, will arrive during a predefined 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 (2/2)

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7 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

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8 George’s HARDWARE – Arrival Process Customers arrive at George’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,…)?

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9 kk Input to the Poisson distribution = 6 customers per hour. t = 0.5 hour. t = (6)(0.5) = 3. t e - t k ! 0 ! ! ! P(X = k )= 8 George’s HARDWARE – An illustration of the Poisson distribution.

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

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11 A single service queue. Multiple service queue with single waiting line. Multiple service queue with multiple waiting lines. Multistage service system. Line Configuration

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12 Balking occurs if customers avoid joining the line when they perceive the line to be too long Reneging occurs when customers abandon the waiting line before getting served Jockeying (switching) occurs when customers switch lines once they perceived that another line is moving faster Balking, Reneging, Jockeying

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13 These rules select the next customer for service. There are several commonly used rules: –First come first served (FCFS - FIFO). –Last come first served (LCFS - LIFO). –Estimated service time. –Random selection of customers for service. Priority Rules

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14 Multistage Service 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.

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15 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

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16 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

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17 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

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

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19 George’s HARDWARE – Service time George 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?

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20 The mean number of customers served per minute is ¼ = ¼(60) = 15 customers per hour. P(X <.05 hours) = 1 – e -(15)(.05) = GEORGE’s HARDWARE 3 minutes =.05 hours

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

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22 The memoryless property (Markov) –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 George’s, the probability of needing more than 3 minutes is ( = ) independent of how long the customer has been served already. –P(T≥s+t / T≥s) = P(T≥t) 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

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23 Performance Measures (1/4) Performance can be measured by focusing on: –Customers in queue. –Customers in the system. Performance is measured for a system in steady state.

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24 Roughly, this is a transient period… n Time Performance Measures (2/4) The transient period occurs at the initial time of operation. Initial transient behavior is not indicative of long run performance.

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25 This is a steady state period……….. n Time Performance Measures (3/4) 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…

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26 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. Performance Measures (4/4) k servers

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27 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

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28 Little’s 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 + Little’s Formulas For the case of an infinite population

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29 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

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

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31 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

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32 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.

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33 MARY’s 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 = = = =

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34 M M k Queuing Systems Characteristics –Customers arrive according to a Poisson process at a mean rate –Service times follow an exponential distribution. –There are k servers, each of who works at a rate of customers (with k > . –Infinite population, and possibly infinite line.

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35 M / M /k Queue - Performance Measures

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36 The performance measurements L, L q, W q,, can be obtained from Little’s formulas. M / M /k Queue - Performance Measures

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37 LITTLE TOWN POST OFFICE Little Town post office is open 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 distributions describe the arrival and the service processes respectively.

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38 LITTLE TOWN POST OFFICE The Postmaster needs to know the relevant service measures in order to: –Evaluate the current service level. –Study the effects of reducing the staff by one clerk.

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39 This is an M / M / 3 queuing system. –Input 100 customers per hour. 40 customers per hour (60 1.5). Does steady state exist ( < k 100 < k (40) = 120. LITTLE TOWN POST OFFICE - Solution

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40 LITTLE TOWN POST OFFICE – solution continued First P 0 is found by P 0 is used to determine all the other performance measures.

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41 LITTLE TOWN POST OFFICE

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42 LITTLE TOWN POST OFFICE

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

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44 Note: It is not necessary to know the particular service time distribution. Only the mean and standard deviation of the distribution are needed. M G 1 Queuing System – Pollaczek - Khintchine Formula for L

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45 Vassilis repairs television sets and VCRs. Data –It takes an average of 2.25 hours to repair a set. –Standard deviation 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. –Vassilis 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. VASSILIS’S TV REPAIR SHOP (1/2)

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46 Vassilis wants to know the effects of using the new equipment on – 1. The average number of sets waiting for repair; 2. The average time a customer has to wait to get his repaired set. VASSILIS’S TV REPAIR SHOP (2/2)

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47 This is an M G 1 system (service time is not exponential (note that 1 ). Input –The current system (without the new equipment) = 1 / 2.5 = 0.4 customers per hour. = 1 / 2.25 = costumers per hour. = 45 / 60 = 0.75 hours. –The new system (with the new equipment) = 1 / 2 = 0.5 customers per hour. = 40 / 60 = hours. VASSILIS’S TV REPAIR SHOP - Solution

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48 VASSILIS’S TV REPAIR SHOP - Results

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49 M / M / k / F Queuing System Many times queuing systems have designs that limit their line 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.

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50 Poisson arrival process at mean rate k servers, each having an exponential service time with mean rate 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. Characteristics of M M k F Queuing System

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51 A customer is blocked if the system is full. The probability that the system is full is P F (100P F % of the arriving customers do not enter the system). The effective arrival rate = the rate of arrivals that make it through into the system ( e ). e = (1 - P F ) M M k F Queuing System – Effective Arrival Rate

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52 ANTONIS’S ROOFING COMPANY (1/3) Antonis gets most of its business from customers who call and order service. –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.

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53 Data –Arrival process is Poisson, and service process is Exponential. –Each phone call takes 3 minutes on the average. –10 customers per hour call the company on the average. –One appointment secretary takes phone calls from 3 telephone lines. Antonis’s ROOFING COMPANY (2/3)

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54 Management would like to design the following system: – The fewest lines necessary. – At most 2% of all callers get a busy signal. Management is interested in the following information: – The percentage of time the secretary is busy. – The average number of customers kept on hold. – The average time a customer is kept on hold. – The actual percentage of callers who encounter a busy signal. ANTONIS’S ROOFING COMPANY (3/3)

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55 This is an M M 1 3 system Input = 10 per hour. = 20 per hour (1 3 per minute). Excel spreadsheet gives: P 0 = 0.533, P 1 = 0.133, P 3 = % of the customers get a busy signal. This is above the goal of 2%. P 0 = 0.516, P 1 = 0.258, P 2 = 0.129, P 3 = 0.065, P 4 = % of the customers get the busy signal Still above the goal of 2% ANTONIS’S ROOFING COMPANY - Solution M M 1 4 system M M 1 5 system P 0 = 0.508, P 1 = 0.254, P 2 = 0.127, P 3 = 0.063, P 4 = P 5 = % of the customers get the busy signal The goal of 2% has been achieved. See spreadsheet next

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56 ANTONIS’S ROOFING COMPANY - Spreadsheet Solution

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57 Probability Analysis, F=1, 2, 3, 4, 5

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58 Sensitivity Analysis for F

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59 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 affects the rate of arrivals of the remaining customers. Characteristics –A single server. –Exponential service time, Poisson arrival process. –A population size of a (finite) m customers.

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60 Progress Homes Progress Homes runs four different development projects. Data –At each site running a project is interrupted once every 20 working days on the average. –The V.P. for construction handles each stoppage. How long on the average a site is non-operational? –If it takes 2 days on the average to restart a project’s progress (the V.P. is using the current car). –If it takes days on the average to restart a project’s progress (the V.P. is using a new car)

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61 Progress Homes – Solution This is an M M 1 4 system, where: –The four sites are the four customers. –The V.P. for construction is the server. Input = 0.05 (1 20) = 0.5 = (1 2 days, using the current car) (1 / days, using a new car).

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62 Progress Homes – Computer Output

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63 Progress Homes – Summary of Results

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64 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.

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65 WILSON FOODS (1/2) Wilson Foods has an 800 number to answer customers’ questions. If all the customer representatives are busy when a new customer calls, he/she is asked to stay on the line. A customer stays on the line if the waiting time is not longer than 3 minutes.

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66 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 $0.20 per minute while on hold. –Customer goodwill cost while in service is $0.05. How many customer service representatives should be used to minimize the hourly cost of operation? How many customer service representatives should be used to minimize the hourly cost of operation? WILSON FOODS (2/2)

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67 TC(K) = C w k + (C t + g s )L q + (C t + g s )(L – L q ) WILSON FOODS – Solution The total hourly cost model TC(K) = C w k + C t L + g w L q + g s (L - L q ) Total hourly wages Total average hourly Telephone charge Average hourly goodwill cost for customers on hold Average hourly goodwill cost for customers in service

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68 Input C w = $16 C t = $10.80 per hour [0.18(60)] g w = $12 per hour [0.20(60)] g s = $3 per hour [0.05(60)] –The Total Average Hourly Cost = TC(K) = 16K + (10.8+3)L + (12 - 3)L q = 16K L + 9L q WILSON FOODS – Solution continued

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69 Assuming a Poisson arrival process and an Exponential service time, we have an M M K system. = 225 calls per hour. = 40 per hour (60 1.5). –The minimal possible value for K is 6 to ensure that steady state exists (

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70 Summary of results of the runs for k=6,7,8,9,10 Conclusion: employ 8 customer service representatives. WILSON FOODS – Solution re-re-continued

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71 WILSON FOODS – Results for k=6

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72 WILSON FOODS – Sensitivity Analysis

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