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Queuing Models.

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1 Queuing Models

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.

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.

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.

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.

6 The Arrival Process (2/2)
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.

7 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 = (the base of the natural logarithm). k! = k (k -1) (k -2) (k -3) … (3) (2) (1).

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,…)?

9 George’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 2 3 (lt) e- lt k ! P(X = k )= 1 2 3 = 0! 1! 2! 3!

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

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

12 Balking, Reneging, Jockeying
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

13 Priority Rules 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.

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.

15 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.

16 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.

17 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 “t.” P(X £ t) = 1 - e-mt

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

19 George’s HARDWARE – Service time
George 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?

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

21 GEORGE’s HARDWARE Using Excel for the Exponential Probabilities

22 The Exponential Distribution - Characteristics
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 l, their interarrival times are exponentially distributed with mean time 1/l.

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.

24 Performance Measures (2/4)
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

25 Performance Measures (3/4)
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

26 Performance Measures (4/4)
In order to achieve steady state, the effective arrival rate must be less than the sum of the effective service rates . k servers l< m For one server l< m1 +m2+…+mk For k servers with service rates mi l< km Each with service rate of m

27 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).

28 For the case of an infinite population
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

29 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

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

31 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

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.

33 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) = Pn = [1 - (l/m)](l/m)n = (0.3333)(0.6667)n L = l/(m - l) = 2 Lq = l2/[m(m - l)] = W = 1/(m - l) = 0.4 hours = 24 minutes Wq = l/[m(m - l)] = hours = 16 minutes Pw = l/m = r = l/m =

34 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). Infinite population, and possibly infinite line.

35 M / M /k Queue - Performance Measures

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

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.

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.

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.

40 LITTLE TOWN POST OFFICE – solution continued
First P0 is found by P0 is used to determine all the other performance measures.



43 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.

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

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 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.

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 = 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 = hours.


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.

50 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”. Customers are blocked (and never return) if the system is full.

51 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 (100PF% 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)

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.

53 Antonis’s ROOFING COMPANY (2/3)
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.

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.

This is an M/M/1/3 system Input l = 10 per hour. m = 20 per hour (1/3 per minute). Excel spreadsheet gives: P0 = 0.533, P1 = 0.133, P3 = 0.06 6.7% of the customers get a busy signal. This is above the goal of 2%. M/M/1/5 system M/M/1/4 system See spreadsheet next P0 = 0.516, P1 = 0.258, P2 = 0.129, P3 = 0.065, P4 = 0.032 3.2% of the customers get the busy signal Still above the goal of 2% P0 = 0.508, P1 = 0.254, P2 = 0.127, P3 = 0.063, P4 = 0.032 P5 = 0.016 1.6% of the customers get the busy signal The goal of 2% has been achieved.

56 ANTONIS’S ROOFING COMPANY - Spreadsheet Solution

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

58 Sensitivity Analysis for F

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.

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)

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 l = (1/20) m = m = 0.533 (1/2 days, using the current car) (1/1.875 days, using a new car).

62 Progress Homes – Computer Output

63 Progress Homes – Summary of Results

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.

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.

66 WILSON FOODS (2/2) 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?

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

68 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) = 16K + (10.8+3)L + (12 - 3)Lq = 16K L + 9Lq

69 WILSON FOODS – Solution re-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).

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

71 WILSON FOODS – Results for k=6

72 WILSON FOODS – Sensitivity Analysis

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