Transcript Managerial Decision Modeling with Spreadsheets

Queuing Models
Stay in Queue: Short Video
• http://www.youtube.com/watch?v=IPxBKxU8GIQ
&feature=related
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© 2002 Prentice-Hall, Inc
Ch 9-2
Introduction
•Queues are everyday occurrence: affect people –Shopping for groceries.
–Buying gasoline.
–Making bank deposit.
–Waiting on telephone.
•Discuss how analytical models of queues help managers
evaluate cost effectiveness of service systems.
•Look at queuing system costs.
•Describe characteristics of queues and underlying
mathematical assumptions used to develop queuing decision
models.
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Queuing System Costs
• Most queuing problems are involved with finding ideal level
of services firm should provide.
– Supermarkets decide how many cash register checkout
positions should be opened.
– Gasoline stations decide how many pumps should be
available.
– Manufacturing plants determine optimal number of
mechanics on duty each shift to repair machines that
break down.
• Most managers recognize trade-off:
• cost of providing good service vs. cost of customer
waiting time.
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Queuing Costs and Service Levels
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Three Rivers Shipping Company Example
•Three Rivers runs huge docking facility.
•Approximately five ships arrive to unload cargoes every
12-hour work shift.
•Each hour ship sits idle in line costs firm $1,000.
•If one team on duty to handle unloading work, each ship
waits average 7 hours to be unloaded.
–If two teams are working, average waiting time drops
to 4 hours.
–For three teams, it is 3 hours.
–For four teams, it is only 2 hours.
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Waiting Line Cost Analysis
Three Rivers Shipping
•Team salaries are computed as: number of people in team
(assumed to be 50), times number of hours each person
works per day (12 hours), times hourly salary of $10 per
hour.
•If two teams are employed, rate is doubled.
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Waiting Line Cost Analysis
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Characteristics Of Queuing System
• Three primary parts of a queuing system.
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(1) Arrivals or inputs to system referred to as
calling population.
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(2) Queue or waiting line itself.
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(3) Service facility.
• These three components have certain characteristics that
need to be examined prior to development of mathematical
queuing models.
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Arrival Characteristics
• Major three characteristics of input source that
generates arrivals or customers for service system.
•
(1) Size of arrival population.
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(2) Pattern of arrivals or arrival distribution at
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queuing system.
•
(3) Behavior of arrivals.
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Size of the Arrival Population
• Population sizes are unlimited (essentially infinite) or
(limited) finite.
• When number of arrivals at any given moment is small
portion of potential arrivals, arrival population is
considered unlimited.
• Examples of unlimited populations include – Cars arriving at highway tollbooth.
– Shoppers arriving at supermarket.
– Students arriving to register for classes.
• Most queuing models assume infinite arrival
population.
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Arrival Distribution
•Customers either arrive at service facility according to some
known schedule (i.e., one patient every 15 minutes or one
student for advising every half hour) or they arrive randomly.
•For any given arrival rate, such as two customers per hour, or
four trucks per minute, a discrete Poisson distribution can be
established:
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•
•
•
•
•
P(X) = (e -l l X) / X !
where:
for X = 0, 1, 2, …
X = number of arrivals per unit of time.
P(X) = probability of exactly X arrivals.
l = average arrival rate (average number
of arrivals per unit of time).
e = 2.7183 (exponential constant).
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Examples Poisson Distribution
for Arrival Times
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Behavior of Arrivals
•Most queuing models assume arriving customer is
patient customer.
•Patient customers are people or machines wait in queue
until served and do not switch between lines.
•Life and decision models are complicated by fact
people balk or renege.
•Balking refers to customers who refuse to join queue
because it is too long to suit their needs.
•Reneging customers enter queue but become
impatient and leave without completing transaction.
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Queue Characteristics
•Length of queue either limited (finite) or unlimited
(infinite).
•Queue limited when it cannot increase to infinite length
due to physical or other restrictions.
•Queue defined as unlimited when size is unrestricted.
•Assume queue lengths are unlimited.
•Queue discipline. Rule by which customers in line
receive service.
•Most systems use queue discipline known as First-In,
First-Out rule (FIFO).
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Service Facility Characteristics
•Important to examine two basic properties:
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(1) Configuration of service facility.
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(2) Pattern of service times.
•Number of service channels in queuing system is
number of servers.
•Single-phase means customer receives service at only
one station before leaving system.
•Multiphase implies two or more stops before leaving
system.
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Queuing System Configurations
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Queuing System Configurations
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Service Time Distribution
•Assume random service times are described by
exponential probability distribution.
•For any given service rate, exponential distribution can be
established using formula:
•P(t) = e -mt for t  0
• where:
•
•
•
•
•
•
t
= service time.
P(t) = probability service time will be
greater than t.
m
= average service rate (average number of
customers served per unit of time).
e
= 2.7183 (exponential constant).
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Examples Exponential Distribution
for Service Times
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Measuring Queue Performance
•Queuing models help manager obtain performance measures
(operating characteristics) of waiting line system.
•Some measures commonly used in practice.
= Utilization factor of system
Lq = Average length of queue.
•L = Average number of customers in system.
•Wq= Average time each customer spends in queue.
•W= Average time each customer spends in system.
•P0 = Probability no customers in system.
•Pn = Probability exactly n customers in system.
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Kendall’s Notation for Queuing Systems
•Queuing theory commonly uses three-symbol notation known as
Kendall’s notation:
• A/B/S
• where:
• A denotes arrival probability distribution choices are:
•
M (Markovian) for a Poisson distribution,
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D for constant or deterministic distribution, or
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G for general distribution with known mean and variance.
• B denotes service time probability distribution choices are:
•
M for exponential service times,
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D for constant or deterministic service times, or
•
G for general service times with known mean and variance.
• S denotes the number of servers (or channels).
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Queuing Models Reviewed
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Single-server Queuing System (M/M/1 Model)
•Assumptions M/M/1 Queuing Model
•Poisson arrivals served on FIFO (first-in, first-out) basis.
•Every arrival waits to be served regardless of line length.
•Arrivals independent of preceding arrivals, but average number of
arrivals (the arrival rate) does not change over time.
•Arrivals are described by Poisson probability distribution and
come from infinite or very large population.
•Exponential service times vary from one customer to next and
are independent of one another, but average rate is known.
•Service times occur according to the exponential probability
distribution.
•Average service rate is greater than average arrival rate.
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Operating Characteristics Equations M/M/1
Queuing System
• Operating characteristic equations for the M/M/1 queuing
system are as follows:
• l = average number of arrivals per time period.
• m = average number of people or items served
per time period.
1. Average server utilization.
 = l/m
2. Average number of customers waiting in line.
Lq = l2 / [ m (m - l) ]
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Operating Characteristics Equations M/M/1
Queuing System
3.
Average number of customers or units in system.
L = Lq + l / m
4. Average time customer or unit spends waiting in
line for service.
Wq = Lq / l = Lq = l / [ m (m - l) ]
5.
Average time a customer or unit spends in system.
W = Wq + 1 / m
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Operating Characteristics Equations M/M/1
Queuing System
6. Probability zero customers or units in system.
P0 = 1 - l / m
7.
Probability n customers or units in system.
Pn = ( l / m ) n Po
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Example-1
• Our local food court uses an average of 10,000
pounds of potatoes per week.
• The average number of pounds of potatoes on
hand is 5000 pounds.
• On the average, how long do potatoes stay in the
restaurant before being used?
© 2002 Prentice-Hall, Inc
Ch 9-28
Solution
• We are given that L=5000 pounds
and λ = 10,000 pounds/week.
Therefore W = 5000 pounds/(10,000
pounds/week)=0.5 weeks.
© 2002 Prentice-Hall, Inc
Ch 9-29
Example-2
•
Suppose that all car owners fill up when their tanks are
exactly half full.
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At the present time, an average of 7.5 customers per hour
arrive at a single-pump gas station.
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It takes an average of 4 minutes to service a car.
Assume that inter arrival and service times are both
exponential.
For the present situation, compute L and W.
© 2002 Prentice-Hall, Inc
Ch 9-30
Suppose that a gas shortage occurs and panic
buying takes place.
– To model the phenomenon, suppose that all
car owners now purchase gas when their
tank are exactly three-fourths full.
– Since each car owner is now putting less gas
into the tank during each visit to the station,
we assume that the average service time has
been reduced to 3.333 minutes.
– How has panic buying affected L and W?
© 2002 Prentice-Hall, Inc
Ch 9-31
We have an M/M/1/GD/∞/∞ system with
λ = 7.5 cars per hour
µ = 15 cars per hour.
 = 7.5/15 =0 .50.
L = 0.50/1-0.50 = 1
W = L/λ = 1/7.5 = 0.13 hour.
Hence, in this situation, everything is under control, and long
lines appear to be unlikely.
© 2002 Prentice-Hall, Inc
Ch 9-32
We now have an M/M/1/GD/∞/∞ system
with λ = 2(7.5) = 15 cars per hour.
Now µ = 60/3.333 = 18 cars per hour,
and  = 15/18 = 5/6. Then
Thus, panic buying has cause long lines.
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L 5 1
6
L
 5 cars and W    hours  20minutes
5
l 15 3
1
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© 2002 Prentice-Hall, Inc
Ch 9-33
Example-3
• An average of 10 machinists per hour arrive seeking tools. At
present, the tool center is staffed by a clerk who is paid $6 per
hour and who takes an average of 5 minutes to handle each
request for tools. Since each machinist produces $10 worth of
goods per hour, each hour that a machinists spends at the tool
center costs the company $10.
• The company is deciding whether or not it is worthwhile to hire
(at $4 per hour) a helper for the clerk. If the helper is hired the
clerk will take an average of only 4 minutes to process
requirements for tools. Assume that service and arrival times are
exponential. Should the helper be hired?
© 2002 Prentice-Hall, Inc
Ch 9-34
Solution
• Goal: Minimize the sum of the hourly service cost and expected
hourly cost due to the idle times of machinists
• Delay cost is the component of cost due to customers waiting in
line
• Goal: Minimize Expected cost/hour = service cost/hour +
expected delay cost/hour
• Expected delay cost/hour = (expected delay cost/customer)
(expected customers/hour)
• Expected delay cost/customer = ($10/machinist-hour)(average
hours machinist spends in the system) = 10W
• Expected delay cost/hour = 10Wl
• Now compute expected cost/hour if the helper is not hired and
also compute the same if the helper is hired
© 2002 Prentice-Hall, Inc
Ch 9-35
• If the helper is not hired l = 10 machinists per hour and m = 12
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machinists per hour
W = 1/(m-l) for M/M/1/GD//. Therefore, W = 1/(12-10)
= ½ = 0.5 hour
Service cost /hour = $6/hour and expected delay cost/hour =
10(0.5)(10) = $50
Without the helper, the expected hourly cost is $6 + $50 = $56
With the helper, m = 15 customers/hour. Then W = 1/(m-l)=
1/(15-10) = 0.2 hour and the expected delay cost/hour =
10(0.2)(10) = $20
Service cost/hour = $6 + $4 = $10/hour
With the helper, the expected hourly cost is $10 + $20 = $30
© 2002 Prentice-Hall, Inc
Ch 9-36
Introducing Costs Into the Model
•Cost of customer waiting time is $10 per hour.
•Cost of providing service is mechanic’s salary of $12 per hour.
•Total cost is sum of waiting cost and service cost.
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Total cost = Cw x L + Cs x s
• where: Cw = customer waiting cost per unit time.
•
L = average number of customers in system.
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Cs = cost of providing service per server per unit
•
•
time period.
s = number of servers in queuing system.
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Introducing Costs Into Model
• Total cost = Cw x L + Cs x s
• where: Cw = customer waiting cost per unit time.
•
L = average number of customers in system.
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Cs = cost of providing service per server per
•
unit time.
•
s = number of servers in queuing system.
Cw == $12
per hour,
• With
Total cost
($10)(2)
+ ($12)(1) = $32 per hour.
L = 2
Cs = $12 per hour,
s = 1
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Multiple-server Queuing System (M/M/S
Model)
•Multiple-server system assumes Poisson arrivals follow Poisson
probability distribution and exponential service times follow
exponential probability distributions.
•Service is first come, first served and all servers are assumed to
perform at same rate.
•Other assumptions listed earlier for single-channel model apply.
• Operating Characteristics Equations for M/M/s Queuing
System
• l = average number of arrivals per time.
• m = average number of customers served per time per server.
• s = number of servers.
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Operating Characteristic Equations M/M/s
Queuing System
• Operating characteristic equations for M/M/s queuing
system are as follows:
• 1. Average server utilization in the system.
•
 = l/(sm)
• 2. Probability zero customers or units in system.
•
•
•
•
•
•
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Operating Characteristic Equations M/M/s
Queuing System
• 3.
•
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Average number of customers or units waiting
in line for service.
•
•
4. Average number of customers or units in system.
L = Lq + l / m
5. Average time a customer or unit spends waiting in
line for service.
Wq = Lq / l
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Operating Characteristic Equations M/M/s
Queuing System
•
• 6. Average time customer or unit spends in system.
•
W = Wq + 1 / m
•
• 7. Probability n customers or units in system.
•
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Example-4
• Bank Staffing Example: The manager of a bank must
determine how many tellers should work on Fridays. For every
minute a customer stands in line, the manager believes that a
delay cost of 5 cents is incurred. An average of 2 customers per
minute arrive at the bank. On the average it takes, a teller 2
minutes to complete a customer’s transaction. It costs the bank
$9 per hour to hire a teller. Inter-arrival times and service times
are exponential. To minimize the sum of service costs and delay
costs, how many tellers should the bank have working on
Fridays?
• l = 2 customer per minute and m = 0.5 customer per minute,
l/sm requires that 4/s < 1. Thus, there must be at least 5 tellers,
or the number of customers present will “blow up.”
• Now compute for s = 5, 6…. Expected service cost/minute +
expected delay cost/minute
© 2002 Prentice-Hall, Inc
Ch 9-43
Solution
• Each teller is paid $9/60 = 15 cents per minute. Expected service
cost/minute = 0.15s
• Expected delay cost/minute = (expected customers/minute)
(expected delay cost/customer)
• Expected delay cost/customer = 0.05Wq
• Expected delay cost/minute = 2(0.05) Wq = 0.10 Wq
• For s = 5,  =l/sm = 2/.5(5) = 0.8
© 2002 Prentice-Hall, Inc
Ch 9-44
•
•
•
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P(j  5)=0.55
Wq = .55/(5(.5)-2) = 1.1 minutes
For s = 5, expected delay cost/minute = 0.10(1.1) = 11 cents
For s = 5, total expected cost/minute = 0.15(5) + 0.11 = 86
cents
• Since s = 6 has a service cost per minute of 6(0.15) = 90 cents, 6
tellers cannot have a lower total cost than 5 tellers. Hence, having
5 tellers serve is optimal
© 2002 Prentice-Hall, Inc
Ch 9-45
Multiple-server Queuing System (M/M/S/
Finite /N Model)
1. There are s servers with identical service time distributions.
2. Finite population size of units seeking service is size N.
3. Poisson arrival distribution of each customer in
population follows Poisson distribution with mean rate of
l.
4. Exponential service times are exponentially distributed
with mean rate of m.
5. Both l and m are specified for same time period.
6. Customers are served on first-come, first-served basis.
•
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Operating Characteristics Equations Finite
Population Queuing System
• l = average number of arrivals per time period.
• m = average number of people or items served per
•
time.
• s = number of servers.
• N = size of the population.
• Operating characteristic equations follow.
• 1. Probability zero customers or units in system.
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Operating Characteristics Equations Finite
Population Queuing System
2. Probability n customers in system.
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Operating Characteristics Equations Finite
Population Queuing System
• 3. Average number of customers or units in line
•
waiting for service.
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•
•
• 4.
Average number of customers or units in system.
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Operating Characteristics Equations Finite
Population Queuing System
• 5. Average time a customer or unit spends in the
queue waiting for service.
•
•
•
• 6. Average time a customer or unit spends in the
system.
•
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