Session 2a

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Session 2a
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Overview

• Operations Simulation Models
• Reliability Analysis
• RANK, VLOOKUP, MIN
• Inventory Order Quantities
• Single Product (MAX)
• Multiple Products with Correlated Demand
• Project management
• PERT Analysis

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Example 1 Reliability
Consider a device that requires two batteries to
function. If either of these batteries dies, the
device will not work. Currently there are two
brand new batteries in the device, and there are
three extra brand new batteries. Each battery,
once it is placed in the device, lasts a random
amount of time that is lognormally distributed
with mean 20 hours and standard deviation 5
hours. When any of the batteries in the device
dies, it is immediately replaced by an extra (if
an extra is still available). Simulate the time
the device can last with the batteries currently
available. What is the estimated probability that
the device will last longer than 50 hours?
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B3B4 contain the parameters for the random
lifetimes of batteries. D1I6 keep track of all
five batteries, including when they start (and in
what order) and when they fail (and in what
order). Note that the random variables are G2G6,
which will be simulated by Crystal Ball. These
are the lifetimes of the batteries.
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We calculate the fail time of a battery by using
the formula Fail time Start time Lifetime
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For example, in this realization, Batteries A and
B both start at time 0. Battery B fails at time
26.41, and is replaced by Battery C (see row 10).
Battery A fails at time 28.23 and is replaced
by Battery D (row 11). Battery D fails at time
50.75 and is replaced by E (D started at time
28.23 and lasted 22.52 see row 12). Finally,
Battery C fails at time 52.08 (it started at time
26.41 and lasted 25.67 see row 13). At this
point the device fails, having lasted 52.08
hours.
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Using the frequency charts certainty feature,
we can drag the lower grabber to a point where
the left window reads 50 (or Crystal Ball will
allow you simply to type 50 into the left window).

The certainty window indicates the estimated area
under the curve in the blue shaded region. In
this case, we estimate a 13.55 probability that
the device will last longer than 50 hours.
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Example 2 Blockbuster Publishers
At present, 2000 copies of the book are in stock,
and Blockbuster must determine how many copies of
the book to print for the next year. Demand has
a triangular distribution with parameters 4000,
6000, and 9000. Each copy sold during brings
the publisher a revenue of 35. Any copies left
at the end of the year can be sold for 5. The
cost of printing the book is 50,000 plus 15 per
book printed.
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Somewhat arbitrarily, we have decided to consider
possible order quantities of 0, 1000, 2000, 3000,
4000, 5000, 6000, and 7000 (B1I1). For each
possible order quantity, we have a fairly simple
income statement-like analysis in rows 2 through
9. All columns get their demand number from B3.
(Note that the demand level in B3 will be a
random variable later.) The lower part of the
spreadsheet (rows 11 through 16) contains the
basic parameters of the problem.
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We run for 1000 trials, and then click on the
extract data button.
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Does your answer change if 4000 copies are
currently in stock?
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Example 3 Walton Bookstore
Consider two competing products sold by a
company. Sales of either product tend to take
away sales from the other product. That is,
demands for the two products are negatively
correlated. The company first places an order
for each product. Then during a period of time,
there is demand D1 for product 1 and demand D2
for product 2. These demands are normally
distributed with means 1000 and 1200 and standard
deviations 250 and 350. The correlation between
D1 and D2 is ?, where ? is a negative number
between 1 and 0.
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The unit cost of each product is 7.50, the unit
price for each product is 10, and the unit
refund for any unit of either product not sold is
2.50. The company must decide how many units
of each product to order. Use simulation to help
the company by experimenting with different order
quantities. Try this for ? -0.3, ? -0.5, and
? -0.7. What recommendations can you give about
the best order quantities as the demands become
more highly correlated (in a negative
direction)?
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We have laid out (somewhat arbitrarily) 16
different combinations of order quantities for
the two products (B2Q3). Each of the columns
from B to Q represents a model of one order
quantity strategy.
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Beta Distribution
The Beta distribution is a continuous probability
distribution defined by four parameters

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The Beta distribution is popular among simulation
modelers because it can take on a wide variety of
shapes, as shown in the graphs above. The Beta
can look similar to almost any of the important
continuous distributions, including Triangular,
Uniform, Exponential, Normal, Lognormal, and
Gamma. For this reason, the Beta distribution
is used extensively in PERT, CPM and other
project planning/control systems to describe the
time to completion of a task.
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PERT Approximations
The project management community has evolved
approximations for the Beta distribution which
allow it to be handled with three parameters,
rather than four. The three parameters are the
minimum, mode, and maximum activity times
(usually referred to as the optimistic,
most-likely, and pessimistic activity
times). This doesnt give exactly the same
results as the mathematically-correct version,
but has important practical advantages. Most
real-life managers are not comfortable talking
about things like probability functions and
Greek-letter parameters, but they are comfortable
talking in terms of optimistic, most-likely, and
pessimistic.
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3-step Procedure
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Beta Distributions in Crystal Ball
The Crystal Ball distribution gallery includes
the Beta distribution, but in a form slightly
different from the description above.
Specifically, Crystal Ball assumes the minimum
is zero. Instead of maximum or pessimistic,
it asks for a Scale parameter.
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Example
Assume we are given optimistic, most-likely, and
pessimistic times of 1, 2, and 3 time units,
respectively. We first use these parameters to
calculate the mean (formula (iii)), standard
deviation (formula (iv)), alpha (formula (v)),
beta (formula (vi)), and the difference between
the maximum and minimum, as shown here
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Next, we create a Crystal Ball assumption cell in
A2, using the parameters shown
We make a cell next to the assumption cell,
adding the random number to the minimum.
Cell B3 will now be a Beta-distributed random
variable with the optimistic, most-likely, and
pessimistic activity times we specified.
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Operations Example Project Management (PERT)
Sharon Katz is project manager in charge of
laying the foundation for the new Brook Museum of
Art in New Haven, Connecticut. Liya Brook, the
benefactor and namesake of the museum, wants to
have the work done within 41 weeks, but Sharon
wants to quote a completion time that she is 90
confident of achieving. The contract specifies
a penalty of 10,000 per week for each week the
completion of the project extends beyond week 43.
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Heres an activity-on-arc diagram of the problem
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We start a spreadsheet model like this,
calculating the mean and standard deviation using
the PERT formulas
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Now we calculate shape and scale parameters
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Model Overview
A section for simulating the times of the
activities
A section to keep track of each path through the
network, to identify the critical path in each
simulated project completion
A section to keep track of each node and when it
occurs
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Example Activity C
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Its important to be careful with the nodes that
have multiple activities leading into them (in
this model, Nodes 3 and 8). The times for those
nodes must be the maximum ending time for the set
of activities leading in. Nodes with only one
preceding activity are easier (see Nodes 4 and 8
below).
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Now we set up an area in the spreadsheet to track
each of the paths through the network, to see
which one is critical. This network happens to
have six paths, so we set up a cell to add up all
of the activity times for each of these paths
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Now, for each path, and for each activity, we can
set up an IF statement to say whether the path
(or activity) was critical for any particular
realization of the model
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Heres a cell to tell whether the project was
completed by week 43
Heres a cell to keep track of the penalty (if
any) Sharon will have to pay. Note that we have
assumed that the penalty applies continuously to
any part of a week.
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Crystal Ball
For each of the random activities, we create an
assumption cell, as shown here for Activity A
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Heres the model after doing this for every
random activity time (Activities D, F, I, L, M,
and the Dummy activity have no variability)
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Now we create forecast cells to track the
completion time of the whole project (B30) as
well as the criticalities of the various paths
(H19H24) and activities (N2N15). We also make
forecast cells to track whether the project took
longer than 43 weeks, and what the penalty was.
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Question 7 Compare the PERT results to those you
would have found using (a) basic CPM using the
most-likely times, (b) the by-hand PERT method
from the textbook, and (c) HOM. CPM analysis
gives a completion time of 42 weeks. The critical
path is A-B-D-E-F-G-J-K-L-M
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Textbook Method
The textbook method involves (a) finding the
means and standard deviations for each path, (b)
determining which path has the longest expected
total time, and (c) summing the variances of the
activities on that path to get the variance of
the path. In our case, the longest path would be
A-B-D-E-F-G-J-K-L-M, with a mean of 42.83 weeks
and a variance of 2.92 weeks.
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HOM Methods
The HOM program allows for two possibilities,
depending on whether the Run Simulation box is
checked in the HOM parameters.
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When the Run Simulation box is checked, HOM
yields the following output
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90 Completion Time

• All of these results are consistent with each
  other the estimates are all within a narrow
  range.
• The HOM output provides somewhat arbitrary
  intervals, which makes it difficult to specify
  the completion time associated with a particular
  probability.
• The Textbook method is based on the assumption
  that the probability distribution of the total
  project time is normal.

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Probability of Completion by Week 43

• Again, the estimates are all consistent with each
  other.
• In this case the HOM intervals are perhaps too
  wide to be useful.
• The Textbook method, as above, is based on a
  normal distribution for the total project time.

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Expected Penalty

• Crystal Ball has a distinct advantage in
  answering this question not only does it provide
  a precise estimate of the expected penalty, but
  it also provides a standard error for this
  estimate, which would be necessary if we were
  interested in constructing a confidence interval
  around the estimate.
• The HOM simulation method can be used to come up
  with an estimate, but it is complicated. For the
  estimate shown above, we took the midpoints of
  the bins in the completion time frequency
  histogram, calculated the penalties associated
  with the midpoints, and calculated the weighted
  average penalty using HOMs estimated
  probabilities as weights. (Strangely, the
  frequencies added up to 999, not 1000.) A similar
  approach was used to create the HOM
  non-simulation estimate.
• The Textbook method cannot be used to answer
  this question without employing some difficult
  calculus on the normal distribution.

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Criticality Paths
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Criticality Activities
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• The non-simulation methods assume that there is
  only one path that could be critical (the one
  with the longest expect total time). With these
  models, any discussion of criticality is not very
  interesting. To the extent that several paths
  have the potential to be critical, these methods
  may underestimate the total time of the project
  and/or the variability of the project time.
• The HOM and Crystal Ball simulation models come
  up with very similar estimated probabilities,
  although HOM is not set up to look at the paths.
• Note that we dont have enough information to
  completely dissect the frequencies that each path
  was critical in the HOM output.

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General Observations

• HOM is the easiest of these methods to set up and
  run.
• There are a number of questions here where
  Crystal Ball provides easier/better answers, but
  it requires more work to set up the model to
  begin with.
• This HOM module was created specifically to deal
  with project management problems, whereas Crystal
  Ball is a more general simulation program.
• The choice of which to use would depend on ones
  need for flexibility (Crystal Ball having more
  inherent flexibility), along with other factors.
• With respect to the non-simulation methods, they
  are clearly inferior in terms of answering some
  of the probabilistic questions in this case
  (highlighting the advantages of learning how to
  do simulations), but they do provide reasonable
  estimates for the other questions.

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Summary

• Operations Simulation Models
• Reliability Analysis
• RANK, VLOOKUP, MIN
• Inventory Order Quantities
• Single Product (MAX)
• Multiple Products with Correlated Demand
• Project management
• PERT Analysis