Transcript B.7 Waiting Line Predictions

Readings
Readings
Chapter 11, Sections 1, 2, 3, 5
Waiting Line Models
BA 452 Lesson B.7 Waiting Line Predictions
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Overview
Overview
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Overview
Exponential Probability is a continuous probability distribution (the probability of
hitting any single value is zero, like a dart hitting a single point). It models the
time between independent events.
Poisson Probability is a discrete probability distribution of the number of events
occurring in a fixed period of time if those events occur independently, with a
known average rate.
Structure of a Waiting Line System has four characteristics: customer arrivals,
service time, the order of service (like first-come, first-served), the number and
configuration of servers in the system.
Analytical Formulas for operating characteristics (average units in system,
average time in the system, …) have been derived, under the queue discipline
first-come, first-served, for several queuing models.
M/M/1 Queuing System designates M = Markov (memoryless) arrival
distribution (Poisson), M = Markov service-time distribution (exponential), and 1
service channel, under first-come, first-served.
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Exponential Probability
Exponential Probability
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Exponential Probability
Overview
Exponential Probability is a continuous probability
distribution (the probability of hitting any single value is zero,
like a dart hitting a mathematical point). In general, it
models the time between independent events that can
happen at a constant average rate.
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Exponential Probability
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How many servers should be available at Malibu Yogurt?
What do you need to know about Malibu Yogurt to
answer? Three things:
First, how long does it take to serve customers? That is a
random variable. The customer may be indecisive; the
yogurt machine may be slow; … . We often use an
exponential distribution for service times.
Second, how many customer arrivals do you expect?
That is another random variable. Even regular customers
may be stuck in traffic; … We often use a Poisson
distribution for arrival times.
Third, how valuable is your customers’ time? (Malibu
attracts wealthy customers, so you want to have enough
servers so customers do not have to wait long.)
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Exponential Probability
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The exponential is a continuous probability distribution
(the probability of hitting any single value is zero, like a
dart hitting a mathematical point). In general, it models
the time between independent events that can happen at
a constant average rate.
The probability density function (pdf) of an exponential
distribution is f(t) = µe-µt for variables t in the interval [0,∞).
Often, t is an instant in time. f(t) can thus be the
instantaneous (infinitesimal) probability that a customer
takes exactly t units of time to serve.
The cumulative distribution function is P(T < t ) = 1 - e-µt
(that is, f(t) integrated from 0 to t) for variables t in [0,∞).
P(T < t ) can thus model the probability that a customer
can be served in t units of time or less.
P(T < t ) = P(T < t ).
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Exponential Probability
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µ > 0 is the distribution’s only parameter, often called the
rate parameter. A higher µ means events occur more
quickly. Thus consider the probability density functions
for rate parameters µ = 0.5, µ = 1.5, µ = 1.5:
µ = 0.5
µ = 1.0
µ = 1.5
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The expected value of the exponential distribution (the
integral of tf(t)) is 1/µ.
The variance (the integral of (t- 1/µ) 2f(t)) is 1/µ2.
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Exponential Probability
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We will assume that customers’ service times have an
exponential probability distribution. Why? That is, why
assume P(T < t ) = 1 - e-µt is the probability that a
customer can be served in t units of time or less?
One good reason is the exponential probability function
has only a single parameter, and that parameter, µ, is the
rate of customer service.
One bad reason is the exponential attaches positive
probability to even unreasonably large service times (like
2 billion years) or to small times (like 1 second).
One other property, which may be a good or a bad
reason to assume exponential probability (depending on
the type of service) is the exponential probability
distribution is memoryless.
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Exponential Probability
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Exponential probability is the only continuous probability
distribution that is memoryless.
For example, its conditional probability obeys
P(T > 40 | T > 30) = P(T > 10)
That is, the probability that you have to wait more than
another 10 seconds before your service is completed
(your Malibu yogurt is ready), given that you already had
to wait 30 seconds, is no different from the initial
probability that you have to wait more than 10 seconds
after you first arrive for service. Is that realistic?
When you hang up on a 30-minute phone transaction
and decide to save time by calling back later, you believe
P(T > 40 | T > 30) is greater than P(T > 10)
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Exponential Probability
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In the simplest waiting-line models, customers’ service
time is assumed to have an exponential probability
distribution. The time between the arrival of each new
customer is also assumed to be an exponential variable.
In particular, P(T > 40 | T > 30) = P(T > 10) means the
probability that you have to wait more than another 10
minutes before a new customer comes in for service,
given that you already had to wait 30 minutes, is no
different from the initial probability that you have to wait
more than 10 minutes before a new customer comes in
for service. Is that realistic?
When you close your store early because you have not
had any customers for 30 minutes, and decide to start
fresh tomorrow, you believe
P(T > 40 | T > 30) is greater than P(T > 10)
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Poisson Probability
Poisson Probability
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Poisson Probability
Overview
Poisson Probability is a discrete probability distribution that
expresses the probability of a number of events occurring
in a fixed period of time if those events occur with a known
average rate, and are independent of the time since the
last event.
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Poisson Probability
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An alternative and equivalent way to measure the time
between the arrival of each new customer is to measure
the probability f(k) that a number k of customers arrives
within a fixed period of time. When the time between
arrivals has an exponential distribution, the probabilities
f(k) have a Poisson distribution.
In general, the Poisson distribution is a discrete
probability distribution that expresses the probability of a
number of events occurring in a fixed period of time if
those events occur with a known average rate, and are
independent of the time since the last event.
The probability mass function of a Poisson distribution
has the form f(k) = k e-/k! for non-negative integers k =
0, 1, … , where k is the number of events that occur, and
k! is the factorial of k (for example, 3! = 3x2x1, … .).
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Poisson Probability
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 > 0 is the Poisson distribution’s parameter, often called
its arrival rate.  equals the distribution’s expected
value; it equals the expected number of arrivals that
occur during the given interval. A higher  means
arrivals can occur more quickly. Thus consider the
probability mass functions for parameters  = 1,  = 4, 
= 10:
=1
=4
=
10
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Structure of a Waiting Line System
Structure of a Waiting Line System
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Structure of a Waiting Line System
Overview
The Structure of a Waiting Line System has four
characteristics:
 the manner in which customers arrive (the simplest case
is a random variable with a Poisson distribution)
 the time required for service (the simplest case is a
random variable with an exponential distribution)
 the priority determining the order of service (the most
common is first come, first served; it has less waiting
variance than first come, last served)
 the number and configuration of servers in the system.
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Structure of a Waiting Line System
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The order of service
• Most common queue discipline is first come, first
served (FCFS).
• An elevator is an example of last come, first
served (LCFS) queue discipline (because the last
one on is the first one off).
• Other disciplines assign priorities to the waiting
units and then serve the unit with the highest
priority first.
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Structure of a Waiting Line System
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Single service channel
Customer
arrives
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Waiting line
Multiple service channels
System
S1
Customer
leaves
System
S1
Customer
arrives
Waiting line
S2
Customer
leaves
S3
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Structure of a Waiting Line System
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A three part code A/B/k describes various queuing
systems.
A identifies the arrival distribution, B the service
(departure) distribution, and k the number of
channels for the system.
Symbols used for the arrival and service processes
are: M = Markov (memoryless) distributions
(Poisson/exponential), D = Deterministic
(constant), and G = General distribution (with a
known mean and variance).
For example, M/M/k refers to a system in which
arrivals occur according to a Poisson distribution,
service times follow an exponential distribution,
and there are k servers working at identical
service rates.
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Structure of a Waiting Line System
So, how well is Malibu Yogurt serving customers?
Determine
1)
2)
3)
4)
5)
6)
7)
The probability of no units in the system: P0
The average number of units in the waiting line: Lq
The average number of units in the system: L
The average time a unit spends in the waiting line: Wq
The average time a unit spends in the system: W
The probability that an arriving unit has to wait: Pw
Probability of n units in the system: Pn
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Analytical Formulas
Analytical Formulas
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Analytical Formulas
Overview
Analytical Formulas for operating characteristics (average
units in system, average time in the system, …) have been
derived, under the queue discipline FCFS, for several
different queuing models including the following:
• M/M/1
• M/M/k
 Analytical formulas are not available for all possible
queuing systems. With no formula, insights may be
gained through a simulation of the system (starting with
our Lesson III.1).
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Analytical Formulas
The following analytical formulas apply to an M/M/1 system
in a steady state (after the initial customers have passed
through)
1) The probability of no units in the system: P0 = 1-/m
2) The average number of units in the waiting line: Lq =
2/(m(m-))
3) The average number of units in the system: L = Lq + /m
4) The average time a unit spends in the waiting line: Wq =
Lq/
5) The average time a unit spends in the system: W =
1/(m-)
6) The probability that an arriving unit has to wait: Pw =
/m
7) Probability of n units in the system: Pn = (/m)nP0
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Analytical Formulas
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Poisson arrival-rate distribution
Exponential service-time distribution
Single channel
Unlimited maximum queue length
Infinite calling population of potential customers
Examples (which are approximations when
“unlimited” and “infinite” are really just “large”):
• Single-window theatre ticket sales booth
• Single-scanner airport security station
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M/M/1 Queuing System
M/M/1 Queuing System
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M/M/1 Queuing System
Overview
M/M/1 Queuing System designates M = Markov
(memoryless) arrival distribution (exponential), M = Markov
service-time distribution (Poisson), and 1 service channel,
under first-come, first-served.
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M/M/1 Queuing System
Question: Joe Ferris is a stock trader on the floor of
the New York Stock Exchange for the firm of Smith,
Jones, Johnson, and Thomas, Inc. Stock
transactions arrive at a mean rate of 20 per hour.
Each order received by Joe requires an average of
two minutes to process. Assume Poisson arrival
times and exponential service times. Answer the
following questions using the Analytical Formulas for
a M/M/1 queuing system: M/M/1:
P0 = 1-/m
Lq = 2/(m(m-))
L = Lq + /m
Wq = Lq/
W = 1/(m-)
Pw = /m
Pn = (/m)nP0
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M/M/1 Queuing System
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Arrival rate distribution questions: P (x=k) = (ke -)/k!
What is the probability that no orders are received
within a 15-minute period?
Answer: Orders arrive at a mean rate of 20 per hour or one
order every 3 minutes. Therefore, in a 15-minute interval
the average number of orders arriving will be  = 15/3 = 5.
Hence, P (x = 0) = (50e -5)/0! = e -5 = .0067
What is the probability that exactly 3 orders are received
within a 15-minute period?
Answer: P (x = 3) = (53e -5)/3! = 125(.0067)/6 = .1396
What is the probability that more than 6 orders arrive within
a 15-minute period?
Answer: P(x>6) = 1 - P(x=0) - P(x=1) - P(x=2) - P(x=3)
- P(x=4) - P(x=5) - P(x=6) = 1-.762 = .238
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M/M/1 Queuing System
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Service rate distribution questions: P (T < t ) = 1 - e-µt
What is the mean service rate per hour?
Answer: Since Joe Ferris can process an order in an
average time of 2 minutes (= 2/60 hr.), the mean service
rate, µ, is µ = 1/(mean service time), or 60/2. m = 30/hr.
What percentage of the orders will take less than one
minute to process?
Answer: Since the units are expressed in hours,
P (T < 1 minute) = P (T < 1/60 hour).
Use the exponential distribution, P (T < t ) = 1 - e-µt.
Hence, P (T < 1/60) = 1 - e-30(1/60)
= 1 - .6065 = .3935 = 39.35%
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M/M/1 Queuing System
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Service rate distribution questions: P (T < t ) = 1 - e-µt
What percentage of the orders will be processed in
exactly 3 minutes?
Answer: Since the exponential distribution is a
continuous distribution, the probability a service time
exactly equals any specific value is 0.
What percentage of the orders will require more than 3
minutes to process?
Answer: The percentage of orders requiring more than 3
minutes to process is:
P (T > 3/60) = e-30(3/60) = e -1.5 = .2231 = 22.31%
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M/M/1 Queuing System
M/M/1:
P0 = 1-/m
Lq = 2/(m(m-))
L = Lq + /m
Wq = Lq/
W = 1/(m-)
Pw = /m
Pn = (/m)nP0
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What is the average time an order must wait
from the time Joe receives the order until it is
finished being processed (i.e. its turnaround
time)?
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Answer: This is an M/M/1 queue with  = 20 per hour
and m = 30 per hour. Analytical Formula #5 says the
average time an order waits in the system is:
W = 1/(µ -  )
= 1/(30 - 20)
= 1/10 hour or 6 minutes
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M/M/1 Queuing System
M/M/1:
P0 = 1-/m
Lq = 2/(m(m-))
L = Lq + /m
Wq = Lq/
W = 1/(m-)
Pw = /m
Pnper
= (/m)nP0
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What is the average number of orders Joe has
waiting to be processed?
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Answer: This is an M/M/1 queue with  = 20
hour and m = 30 per hour. Analytical Formula #2
says the average number of orders waiting in the
queue is:
Lq = 2/[µ(µ - )]
= (20)2/[(30)(30-20)]
= 400/300
=
4/3
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M/M/1 Queuing System
M/M/1:
P0 = 1-/m
Lq = 2/(m(m-))
L = Lq + /m
Wq = Lq/
W = 1/(m-)
Pw = /m
Pn = (/m)nP0
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What percentage of the time is Joe processing
orders?
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Answer: The percentage of time Joe is
processing orders is equivalent to the utilization
factor, /m (the probability of waiting Pw). Thus, the
percentage of time he is processing orders is:
/m = 20/30
= 2/3 or 66.67%
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M/M/1 Queuing System
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M/M/1 Queuing System
M/M/1:
P0 = 1-/m
Lq = 2/(m(m-))
L = Lq + /m
Wq = Lq/
W = 1/(m-)
Pw = /m
Pn = (/m)nP0
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BA 452
Quantitative Analysis
End of Lesson B.7
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