C.2 Waiting Line Simulation

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Transcript C.2 Waiting Line Simulation

Readings
Readings
Chapter 12
Simulation
BA 452 Lesson C.2 Waiting Line Simulation
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Overview
Overview
BA 452 Lesson C.2 Waiting Line Simulation
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Overview
Bidding Simulations estimate the probability that various bids will win, then
performs Risk Analysis. The simplest analysis determines the bid that
maximizes expected gains.
Product Failure Simulations estimate the probability that a product fails after
various amounts of use, then performs risk analysis, trading off consumer
confidence of warranties with expected costs.
Waiting Line Simulations model queuing when there are no analytical formula,
such as non-Markov arrival distributions or service-time distributions.
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Overview
Tool Summary
 Use the Excel 2003 workbook general-purpose template
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Template.xls
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Simulate Bidding with rival buyers or sellers. If buyers, the
high bid wins; if sellers, the low bid. In either case, the
objective is uncertain, so maximize expected value.
Use Summary Statistics to simulate the probability of
product failure.
Simulate Single Channel Waiting Lines with the
specialized Excel 2003 workbook template
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWait.xls
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Bidding
Bidding
BA 452 Lesson C.2 Waiting Line Simulation
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Bidding
Overview
Bidding Simulations estimate the probability that various
bids will win, then perform Risk Analysis. The simplest
analysis determines the bid that maximizes expected gains.
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Bidding
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Strassel Investors buys real estate, develops it, and
resells it for a profit. A new property is available, and
Bud Strassel believes [is certain] it can be sold for
$160,000.
The current property owner asked for bids and stated
that the property will be sold for the highest bid in excess
of $100,000.
Two competitors will be submitting bids for the property.
Strassel assumes, for planning purposes, that the bid by
each competitor is [independent and] uniformly
distributed between $100,000 and $150,000.
Simulate the probability that Strassel can win with a bid
of $130,000. Use a 500 trial simulation.
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Bidding
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The bid by each
competitor is
[independent and]
uniformly
distributed between
$100,000 and
$150,000.
Strassel must beat
the maximum of
the two bids.
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Bidding
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Count the number
of times out of 500
that the $130,000
bid is better than
the maximum bid
by competitors.
Then, compute the
frequency
(probability) of
outbidding
competitors.
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Bidding
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Compare that
probability of
0.3400 for a
$130,000 bid with
probabilities for
$140,000 and
$150,000 bids.
It is better to
compute both
probabilities on a
single simulation,
rather than running
two separate
simulations (as on
the right).
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Bidding
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Compute expected profit for each bid.
Profit ($130,000 bid) = 0.3400 x ($160,000-$130,000) =
$10,200.
Profit ($140,000 bid) = 0.6260 x ($160,000-$140,000) =
$12,520.
Profit ($150,000 bid) = 1.0000 x ($160,000-$150,000) =
$10,000.
So, bid $140,000, and loose some of the time.
It is more accurate to compute all of those probabilities
from a single simulation, rather than from 3 separate
simulations.
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Product Failure
Product Failure
BA 452 Lesson C.2 Waiting Line Simulation
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Product Failure
Overview
Product Failure Simulations estimate the probability that a
product fails after various amounts of use, then performs
Risk Analysis, trading off the gains from consumer
confidence of offering warranties with the expected cost.
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Product Failure
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Goodyear Tire Company has produced a new tire with
an estimated mean lifetime mileage of 36,500 miles.
Management also believes that the standard deviation is
5000 miles, and mileage is normally distributed. [Of
course, negative miles make no sense, but that is
unlikely since it negative values of more than 7 standard
deviations from the mean.]
Simulate the probability that a tire lasts more than
40,000 miles; more than 32,000 miles; more than 30,000
miles; and more than 28,000 miles.
What tire mileage guarantee do you recommend if
management wants no more than 10% of tires to fail the
guaranteed amount?
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Product Failure
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… a new tire with
an estimated mean
lifetime mileage of
36,500 miles …
standard deviation
is 5000 miles, and
mileage is normally
distributed.
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Product Failure
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Simulate the
probability that a
tire lasts more than
40,000 miles; more
than 32,000 miles;
more than 30,000
miles; and more
than 28,000 miles.
What tire mileage
guarantee do you
recommend if
management wants
no more than 10%
of tires to fail the
guaranteed
amount? 30,000
miles has 9%
failure.
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Waiting Line
Waiting Line
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Waiting Line
Overview
Waiting Line Simulations model queuing when there are no
analytical formula, such as non-Markov arrival distributions
(non-Poisson) or non-Markov service-time distributions
(non-exponential).
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Waiting Line
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For the next example, download Excel 2003 workbook
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWait.xls
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Waiting Line
Skandinaviska Enskilda Banken AB (SEB) will open several
new branch banks during the coming year. Each new
branch is designed to have one automated teller machine
(ATM). A concern is that during busy periods several
customers may have to wait to use the ATM. This concern
prompted the bank to study the ATM waiting line system.
The bank established the guideline that the average
customer waiting time should be 1 minute or less.
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Waiting Line
One probabilistic input to the ATM simulation model is the
arrival times of customers who use the ATM. In waiting line
simulations, arrival times are determined by randomly
generating the time between two successive arrivals. For
the SEB ATM system, the interarrival times are
continuously uniformly distributed between 0 and 5
minutes.
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Waiting Line
The other probabilistic input is the service time, which is the
time a customer spends using the ATM machine. Past data
indicates a normal probability distribution with a mean of 2
minutes and a standard deviation of 0.5 minutes.
(Although negative service times will be generated by the
normal distribution, they are more than 4 standard
deviations from the mean, and so have an insignificant
effect on the simulation.)
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Waiting Line
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All formulae for Customers 3 through 1000 are copied from
the second customer.
Results for Customers 6 through 995 are hidden below.
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Waiting Line
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Summary Statistics are computed for the steady state of the
system.
Results are only used for Customers 101 through 1000
because Customers 1 through 100 face unusually short lines
for the ATM.
The steady state is relevant for an ATM open 24 hours per
day. (Of course, using the same interarrival distribution for
4am as for 4pm is then inaccurate.)
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Waiting Line
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Statistics for
Customers 100, … use:
Waiting Time (E116, …,
E1015)
Service Time (F116, …,
F1015)
Completion Time
(G116, …, G1015)
Number Waiting counts
the number of
customers with positive
Waiting Times.
Utilization of ATM sums Service Times and divides that sum by the
time between when Customer 100 could have been served (G115) to
when Customer 1000 has completed service (G1015).
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Waiting Line
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The central limit theorem predicts the Average Waiting Time from
many repeated simulations is most likely close to the expected value
(mean) of waiting times for the true probability distribution of waiting
times.
The expected value of waiting times is greater than 1 minute.
Therefore: 1 ATM is insufficient for the bank to meet satisfy the
guideline that the average customer waiting time should be 1 minute
or less.
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Review Questions
Review Questions
 You should try to answer some of the following
questions before the next class.
 You will not turn in your answers, but students may
request to discuss their answers to begin the next class.
 Your upcoming Final Exam will contain some similar
questions, so you should eventually consider every
review question before taking your exams.
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Review 1: Waiting Lines
Review 1: Waiting Lines
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Review 1: Waiting Lines

For the next example, download Excel 2003 workbook
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWaitInNOut.xls
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Review 1: Waiting Lines
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A fast-food franchise has received several complaints
about long waits for service at its Newbury Park location.
The franchise is especially concerned about waits of
over 5 minutes.
Hence, the franchise collects data on interarrival times of
customers, and service times.
Show how to use that data to estimate the probability of
a customer waiting over 5 minutes for service.
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Review 1: Waiting Lines
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Assume the interarrival times between customers at its
Newbury Park location are estimated to be continuously
uniformly distributed between 0 and 7minutes.
Assume customer service times from workers of average
abilities follow a finite distribution:
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With probability .20, the customer is chatty and pays in exact
change, and so takes 4 minutes to serve.
With probability .30, the customer is chatty and pays with a
credit card, and so takes 3 minutes to serve.
With probability .15, the customer is quiet and pays in exact
change, and so takes 2.5 minutes to serve.
With probability .35, the customer is quiet and pays with a credit
card, and so takes 2 minutes to serve.
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Review 1: Waiting Lines
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The Excel 2003 workbook
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWaitInNOut.xls
constructs and explains the single-channel waiting line
simulation.
 Most estimates for the probability of a customer waiting over
5 minutes for service are between 10% and 18%.
 Save that workbook to use as a template for other singlechannel waiting line simulation that use other distributions
for interarrival times or service times.
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BA 452
Quantitative Analysis
End of Lesson C.2
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