Chapter 9 Queuing Models 1 9 1 Introduction

Chapter 9 Queuing Models 1

9. 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

9. 1 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

9. 2 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 ke- lt (lt) P(X = k) = 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

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. (lt) e P(X =0123 k )= k! 1! 0! 2! 3! 10 k 23 lt 0 1 2 3 4 5 6 7 8 = 0. 224042 0. 149361 0. 049787 0. 224042 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 – Priority – Line configuration – Tandem Queues – Jockeying – Homogeneity – Balking 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

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

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) = ? =. 05 hours 3 minutes • From Excel we have: – EXPONDIST(. 05, 15, TRUE) =. 5276 23

HANK’s HARDWARE – Using Excel for the Exponential Probabilities =EXPONDIST(B 4, B 3, T RUE) =EXPONDIST(A 10, $B$3, F ALSE) Drag to B 11: B 26 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. • Meaningful long run performance measures can be calculated for the system when in steady state. n Roughly, this is a transient period… This is a steady state period…… …. . Time 28

9. 3 Performance Measures of Queuing System In order to achieve steady state, the effective arrival rate must be less than k servers the sum of the effective service l< m 1 +m 2+…+mk l< km rates l< m. For k servers For one server with service rates mi Each with service rate of m 29

Steady State Performance Measures P 0 = 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 L q = l Wq l/mcase of an infinite population For the L = Lq + 31

Classification of Queues • Queuing system can be classified by: – – – Arrival process. Example: Service process. Number of servers. M / 6 / 10 / 20 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. 32

9. 4 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 P 0 = 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 The probability that a customer waits in the system more tha “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 35

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 m –l = 7. 5 – 5 = 2. 5 per hr. per hour. P 0 = 1 - (l/m) = 1 - (5/7. 5) = 0. 3333 –PPerformance Calculations P(X<10 min) = 1 – e-2. 5(10/60 n n n = [1 - (l/m)](l/m) = (0. 3333)(0. 6667) =. 565 L = l/(m - l) = 2 Pw = l/m = 0. 6667 2 Lq = l /[m(m - l)] = 1. 3333 r = l/m = 0. 6667 W = 1/(m - l) = 0. 4 hours = 24 minutes Wq = l/[m(m - l)] = 0. 26667 hours = 16 minutes 36

MARY’s SHOES Spreadsheet solution =A 11 B 4/B 5 =B 4/(B 5 B 4) =A 11/ B 4 =B 4/B 5 =C 111/B 5 =1 B 4/B 5 =1 E 11 =H 11*($B$4/$ B$5) Drag to Cell AL 11 37

12. 5 M/M/k Queuing Systems • Characteristics – Customers arrive according to a Poisson process at a mean rate l. – Service times follow an exponential distribution. – There are k servers, each of who works at a rate of m customers (with km> l). 38 – Infinite population, and possibly infinite line.

M / M /k Queue - Performance Measures 39

M / M /k Queue - Performance Measures The performance measurements L, Lq, Wq, , can be obtaine from Little’s formulas. 40

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 41 the

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. 42

LITTLE TOWN POST OFFICE Solution • This is an 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. 43

LITTLE TOWN POST OFFICE – solution continued • First P 0 is found by • P 0 is used now to determine all the other performance measures. 44

LITTLE TOWN POST OFFICE – Spreadsheet Solution 45

9. 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. 46

9. 6 M/G/1 Queuing System – Pollaczek - Khintchine Formula for L Note: It is not necessary to know the particular service time dis Only the mean and standard deviation of the distribution are ne 47

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 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. 48

TED’S TV REPAIR SHOP Ted wants to know the effects of using the ne equipment on – 1. The average number of sets waiting for re 2. The average time a customer has to wait to get his repaired set. 49

TED’S TV REPAIR SHOP Solution • This is an M/G/1 system (service time is not exponential (note that s ¹ 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 costumers 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. 50

9. 7 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 51

Characteristics of M/M/k/F Queuing 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”. 52

M/M/k/F Queuing System – Effective Arrival Rate • • • A customer is blocked if the system is full. The probability that the system is full is PF (100 PF% of the arriving customers do not enter the system). The effective arrival rate = the rate of arrivals that make it through into the system (le). le = l(1 - PF) 53

RYAN ROOFING COMPANY • Ryan 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. 54

RYAN ROOFING COMPANY • 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. 55

RYAN ROOFING COMPANY • 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. 56

RYAN ROOFING COMPANY Solution M/M/1/4 system • This is. M/M/1/5 an M/M/1/system 3 system • Input l = 10 per hour. See spreadsheet next = 20 per (1/3 per P 0 m = 0. 516, P 1 hour = 0. 258, P 2 minute). = 0. 129, P 3 = 0. 065, • P 4 Excel spreadsheet gives: = 0. 032 P 0 = 0. 508, P 1 = 0. 254, P 2 = 0. 127, P 3 = 0. 063, P 4 = 0 P 0 = 0. 533, P 1 = 0. 133, P 3 = 0. 06 P 5 = 0. 016 3. 2% the customers getsignal. the busy signal 6. 7% of 1. 6% theofcustomers get a busy of the customers get busy signal Still above the goal of achieved. 2% The goal of 2% has been This is above the goal of 2%. 57

RYAN ROOFING COMPANY Spreadsheet Solution 58

12. 8 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. 59

PACESETTER HOMES • Pacesetter 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 nonoperational? – 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 1. 875 days on the average to restart a 60 project’s progress (the V. P. is using a new car)

PACESETTER 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 l = 0. 05 m = 0. 5 (1/2 days, using the current car) (1/20) m = 0. 533 (1/1. 875 days, using a new car). 61

PACESETTER HOMES – Solution continued Summary of results 62

PACESETTER HOMES – Spreadsheet Solution 63

9. 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. 64

WILSON FOODS TALKING TURKEY HOT LINE • 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. 65

WILSON FOODS TALKING TURKEY HOT LINE • Data – On the average 225 calls per hour are received. – An average phone call takes 1. 5 minutes. How many customer – A customer will stay on the line waiting at most 3 minutes. service representatives should used is paid $16 per – A customer service be representative hour. to minimize the – Wilson pays the telephone $0. 18 per hourly cost ofcompany operation? minute when the customer is on hold or when being served. – Customer goodwill cost is $20 per minute while on 66

WILSON FOODS – Solution • The total hourly cost model Total hourly wages Average hourly goodwill cost for customers on hold TC(K) = Cwk + Ct. L + gw. Lq + gs(L - Lq) Total average Average hourly goodwill hourly Telephone charge cost for customers in serv TC(K) = Cwk + (Ct + gs)Lq + (Ct + gs)(L – Lq 67

WILSON FOODS – Solution continued • Input Cw= $16 Ct = $10. 80 per hour [0. 18(60)] gw= $12 per hour [0. 20(60)] gs = $3 per hour [0. 05(60)] – The Total Average Hourly Cost = TC(K) = 16 K + (10. 8+3)L + (12 - 3)Lq = 16 K + 13. 8 L + 9 Lq 68

WILSON FOODS – Solution continued • 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). – Excel MMk worksheet was used to generate results for L, Lq, and Wq. 69

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

WILSON FOODS – Spreadsheet Solution 71

HARGROVE HOSPITAL MATERNITY WARD • Hargrove Hospital is experiencing cutbacks, and is trying to reorganize operations to reduce operating costs. • There is a trade off between – the costs of operating more birthing stations and – the costs of rescheduling surgeries in the surgery room when women give birth there, if all the birthing stations are occupied. • The hospital wants to determine the optimal number of birthing stations that will minimize operating costs. 72

HARGROVE HOSPITAL MATERNITY WARD • Data – Cutting one birthing station saves $25, 000 per year. – Building a birthing station costs $30, 000. – Maternity in the surgery room costs $400 per hour. – Six women on the average need a birthing station a day. The arrival process is Poisson. – Every birthing process occupies a birthing 73

HARGROVE HOSPITAL • Solution • Analysis of the current situation – Currently there are two birthing stations – The current problem can be modeled as a M/G/2/2 queuing system. – Using the MGkk Excel worksheet with l = 6 and m = 12/day we have: • • • r =. 23077 W =. 0833 days Pw =. 076923 L =. 46154 P 0 =. 6154 7. 7% of the arriving women are sent to the surgery room to give birth. 74

HARGROVE HOSPITAL • Solution – continued • The birthing stations problem can be modeled as a M/G/k/k queuing model. • Input – l = 6 women per day; m = 12 women per day (24/2); – k = the number of birthing stations used • The total cost for the hospital is TC(k) = Cost of using the surgery room for maternity + Additional cost of operating 75

HARGROVE HOSPITAL • Solution – continued • Average daily cost of using the surgery room for maternity: Pk(l)(average time in the surgery room)(hourly cost) • Average additional daily cost of operating k stations 25, 000/365 = $68. 49 per day. • Average daily total cost 76

HARGROVE HOSPITAL - Solution • From repeated runs of the MGkk worksheet to determine Pk we got the following results: k Current Optimal 1 2 3 4 Pk Addition Total net al $4800 P average k Cost of Daily cost stations . 33333 $1, 600 3 369. 23. 07692 60. 76 3 7. 58. 01265 $– 68. 49 0 82. 19 $1, 531. 51 364. 23 142. 95 171. 96 77

HARGROVE HOSPITAL – Spreadsheet Solution 78

9. 10 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 79

9. 10 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 80

9. 10 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. 81

9. 10 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 82

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. 83

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. 84

BIG BOYS SOUND, INC. Only 75% of the arriving customers make a purchase! What is the average amount of time, a customer who makes a purchase spends in the store? 85

BIG BOYS SOUND, INC. – Solution • This is a Three Station Tandem Queuing System Pickup desk (. 75)(40)=30 l = 30 Cashiers M/M/3 Sales Clerks M/M/8 M/M/2 l = l W 3 = 2. 67 minutes 40 30 = W 1 = 14 minutes W 2 = 3. 47 minutes Total = 20. 14 minutes 86

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