Title: Queuing Models
1Queuing Models
2Introduction (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.
3Introduction (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.
4Elements 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.
5The 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.
6The 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.
7The 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).
8Georges HARDWARE Arrival Process
- Customers arrive at Georges Hardware according
to a Poisson distribution. - Between 800 and 900 A.M. an average of 6
customers arrive at the store. - What is the probability that k customers will
arrive between 800 and 830 in the morning (k
0, 1, 2,)?
9Georges 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
0.049787
0.149361
0.224042
0.224042
3
0!
1!
2!
3!
10The Waiting Line Characteristics
- Factors that influence the modeling of queues
- Line configuration
- Balking
- Reneging
- Jockeying
- Priority
- Tandem Queues
- Homogeneity
11Line Configuration
- A single service queue.
- Multiple service queue with single waiting line.
- Multiple service queue with multiple waiting
lines. - Multistage service system.
12Balking, 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
13Priority 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.
14Multistage 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.
15Homogeneity
- 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.
16The 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.
17The Exponential Service Time Distribution
m the average number of customers who
can be served per time period. Therefore, 1/m
the mean service time.
18Schematic illustration of the exponential
distribution
The probability that service is completed within
t time units P(X t) 1 - e-mt
X t
19Georges 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?
20GEORGEs HARDWARE
- The mean number of customers served per minute is
¼ ¼(60) 15 customers per hour. - P(X lt .05 hours) 1 e-(15)(.05) 0.5276
21GEORGEs HARDWARE Using Excel for the
Exponential Probabilities
EXPONDIST(A10,B3,FALSE) Drag to B11B26
22The 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 Georges, the probability of needing more
than 3 minutes is (1-0.527630.47237) independent
of how long the customer has been served already. - P(Tst / Ts) P(Tt)
- 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.
23Performance 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.
24Performance 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
25Performance 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
26Performance Measures (4/4)
- In order to achieve steady state, theeffective
arrival rate must be less than the sum of the
effective service rates .
llt km Each with service rate of m
llt m1 m2mk For k servers with service rates mi
llt m For one server
27Steady 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).
28Littles Formulas
- Littles 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
29Classification 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
30M/M/1 Queuing System - Assumptions
- Poisson arrival process.
- Exponential service time distribution.
- A single server.
- Potentially infinite queue.
- An infinite population.
31M / 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(Xgtt) e-(m - l)t
32MARYs 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.
33MARYs 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
34M/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 kmgt l). - Infinite population, and possibly infinite line.
35M / M /k Queue - Performance Measures
36M / M /k Queue - Performance Measures
The performance measurements L, Lq, Wq,, can be
obtained from Littles formulas.
37LITTLE TOWN POST OFFICE
- Little Town post office is open on Saturdays
between 900 a.m. and 100 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.
38LITTLE 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.
39LITTLE TOWN POST OFFICE - Solution
- 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 lt km )?
-
- l 100 lt km 3(40) 120.
40LITTLE TOWN POST OFFICE solution continued
- P0 is used to determine all the other performance
measures.
41LITTLE TOWN POST OFFICE
42LITTLE TOWN POST OFFICE
43M/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.
44M/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.
45VASSILISS TV REPAIR SHOP (1/2)
- 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.
46VASSILISS TV REPAIR SHOP (2/2)
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.
47VASSILISS 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.
48VASSILISS TV REPAIR SHOP - Results
49M / 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.
50Characteristics 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.
51M/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).
52ANTONISS 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.
53Antoniss 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.
54ANTONISS ROOFING COMPANY (3/3)
- 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.
55ANTONISS ROOFING COMPANY - Solution
- 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/1/4 system
M/M/1/5 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
56ANTONISS ROOFING COMPANY - Spreadsheet
Solution
57Probability Analysis, F1, 2, 3, 4, 5
58Sensitivity Analysis for F
59M / 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.
60Progress 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
projects progress (the V.P. is using the current
car). - If it takes 1.875 days on the average to restart
a projects progress (the V.P. is using a new car)
61Progress 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 (1/20)
- m 0.5 m 0.533
(1/2 days, using the current car)
(1/1.875 days, using a new car).
62Progress Homes Computer Output
63Progress Homes Summary of Results
64Economic 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.
65WILSON 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.
66WILSON 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?
67WILSON FOODS Solution
- The total hourly cost model
Average hourly goodwill cost for customers on
hold
Total hourly wages
Total average hourly Telephone charge
Average hourly goodwill cost for customers in
service
TC(K) Cwk (Ct gs)Lq (Ct gs)(L Lq)
68WILSON 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 (10.83)L (12 - 3)Lq 16K
13.8L 9Lq
69WILSON 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 (lltKm).
70WILSON FOODS Solution re-re-continued
- Summary of results of the runs for k6,7,8,9,10
Conclusion employ 8 customer service
representatives.
71WILSON FOODS Results for k6
72WILSON FOODS Sensitivity Analysis