### C.1 Risk Simulation

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Review
We will spend up to 30 minutes reviewing Exam 1
cumulative, and may contain similar questions.
BA 452 Lesson C.1 Risk Simulation
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Chapter 12
Simulation
BA 452 Lesson C.1 Risk Simulation
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Overview
Overview
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Overview
The Law of Large Numbers proves the average of many independent random
variables converges to its expected value with certainty.
Simulation models equations of random variables that are too complex to be
solved with analytical formulas. (Like queuing models where arrivals are do not
have a Poisson distribution.)
Repeated Risk differs from one-time risk because of the central limit theorem.
Many independent risks average to a certainty, so each should be accepted if it
has positive expected value.
Risk Analysis predicts the outcome of an uncertain decision. The simplest case
concerns best-case and worst-case outcomes. Complicated cases concern
expected value and confidence intervals.
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The Law of Large Numbers
The Law of Large Numbers
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The Law of Large Numbers
Overview
The Law of Large Numbers proves the average of many
independent random variables converges to its expected
value with certainty.
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The Law of Large Numbers
The Law of Large Numbers proves the average of many
independent and identically-distributed random variables,
with mean m and finite variance is a random variable with
almost all of its probability weight on values near m.
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The more variables, the greater the concentration of
probability weight near m.
For example, the average (mean) height for all males in
the U.S. is 5’10”. If you measured 2 randomly-selected
males in the U.S., their average height may be a lot
different from 5’10”. But if you randomly measured
2,000 males in the U.S., their average height will most
likely be almost equal to 5’10”.
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The Law of Large Numbers
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The expected value (mean) of the sum of any two
random variables equals the sum of the expected values
of each individual variable:
E(X+Y) = E(X)+E(Y).
The expected value (mean) of the product of two
independent random variables equals the product of the
expected values of each individual variable:
E(XY) = E(X)E(Y).
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Simulation
Simulation
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Simulation
Overview
Simulation models equations of random variables that
are too complex to be solved with analytical formulas.
(Like queuing models where arrivals are do not have
a Poisson distribution.) One disadvantage of
simulation is that there is no guarantee that you
preformed enough trials so your conclusions are
accurate.
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Simulation
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Simulation is a common management science
technique.
It is typically used to model random processes that
are too complex to be solved with analytical
formulas. (Like queuing models where arrivals are
do not have a Poisson distribution.)
One disadvantage of simulation is that there is no
guarantee that you preformed enough trials so
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Simulation
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One begins a simulation by developing a
mathematical statement of the problem.
The model should be realistic yet solvable within the
speed and storage constraints of the computer
system being used.
Input values for the model as well as probability
estimates for the random variables must then be
determined.
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Simulation
The simplest random variable used in simulation has a
finite distribution is the simplest random input into a
simulation. For example, a change in a stock price is
random but it can take on only one of a finite number of
possible values.
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Simulation
Another random variable used in simulation has the
continuous uniform distribution
 The continuous uniform distribution has all intervals of
the same length in the distribution's support are equally
probable. The support is defined by the two parameters,
a and b, which are its minimum and maximum values.
 The standard uniform distribution has support [0,1]. The
probability of each interval subset of [0,1] equals the
length of that interval, Prob[a,b] = b-a.
 The graph of the probability density of any uniform
distribution is flat:
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Simulation
The most common random variable used in simulation
has the normal distribution. That is a continuous
probability distribution that is convenient to manipulate
when used in an equation: A linear function of a normal
random variable is another normal random variable.
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Simulation
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The normal distribution is a family of distributions of the
same general form, differing in their location and scale
parameters: the mean ("average") m and standard
deviation ("variability") s.
f(x) =
The probability density function is
The standard normal distribution is the normal
distribution with mean m = 0 and standard deviation s = 1.
The graph of its probability density resembles a bell (see
the green curve).
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Simulation
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About 68% of values drawn from a normal distribution
are within 1 standard deviation away from the mean;
about 95% of the values are within two standard
deviations and about 99.7% lie within 3 standard
deviations.
There is positive probability weight on all values, both
positive and negative, but the weight becomes negligible
as values become very positive or very negative.
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Simulation
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One convenient feature of the normal distribution is that
a linear function of a normal random variable is another
normal random variable.
For any number m and any positive number s > 0, if the
random variable X has the standard normal distribution,
then the linear combination m + sX is a normal random
variable with mean m and standard deviation s.
Here are graphs of normal probability densities for
various means m and variances s2.
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Repeated Risk
Repeated Risk
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Repeated Risk
Overview
Repeated Risk differs from one-time risk because of the
central limit theorem. Many independent risks average
to a certainty, so each should be accepted if it has
positive expected value.
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Repeated Risk
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Risk management depends on whether an investor
will face a risk just once, like a person one year from
living off retirement savings, or repeatedly, like a
young person 40 years from living off retirement
savings.
Consider an investment (gamble) where you receive
\$11 if a coin comes up heads, and you loose \$10 if
the coin comes up tails.
Accepting that gamble just once involves risk. There
is a 50% chance you will loose.
But that risk of loss becomes negligible (it is
eventually much less than the risk of dying in a
accident today) as the gamble is repeated.
Simulate the results of 10, 100, and 1000 coin flips
for someone with an initial net worth of \$500,000.
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Repeated Risk
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http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Example 1.xls
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Repeated Risk
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Simulate each coin flip by a
standard continuous uniform
random variable, with values
between 0 and 1.
simulated by the variable having
a value between 0 and 0.5.
“Tales” (net gain -\$10.00) by a
value between 0.5 and 1.
Since the random variable is
continuous, the probability that
its value is exactly 0.5 is zero, so
the intervals can overlap at the
value 0.5.
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Repeated Risk
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The net gain from each flip is
computed by the function
VLOOKUP.
RAND() refers to the standard
continuous uniform variable.
\$A\$8:\$C\$9 refers to the
dimensions of the probability
table with A8 the upper left cell,
and C9 the lower right cell.
The 3 specifies that the third
column of the table (Net Gain) is
returned as the result of the coin
flip.
Repeat that formula for each
row.
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Repeated Risk
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Add the net gain from the initial
coin flip to the initial net worth.
Repeat for flips 2, …, 10.
Hit F9 to recalculate a simulation.
Do 20 simulations, and compute
the percentage of times the 10
coin flips increases net worth.
According to the law of large
numbers, you will most likely get
an increase in net worth more
than 50% of the time. (I
computed an increase 70% of
the time.)
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Repeated Risk
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Do similar simulations for 100 flips and 1000 flips.
Use the Excel hide command to view your results.
1000 flips has the highest likelihood of increasing worth.
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Risk Analysis
Risk Analysis
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Risk Analysis
Overview
Risk Analysis predicts the outcome of an uncertain decision.
The simplest case concerns finding best-case and worst-case
outcomes. Complicated cases concern expected value and
confidence intervals.
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Risk Analysis
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Hewlett Packard makes computer equipment. Hewlett
Packard’s product design group developed a prototype
for a new portable printer. Market and financial research
has accurately predicted the following parameters
(constants):
 Selling price = \$249 per unit.
However, there is uncertainty about three other terms
affecting profit, which are considered to be random
variables.
 The mean (expected value) of direct labor cost on
each unit = \$45.
 The mean of the parts cost on each unit = \$90.
 The mean of the first year demand = 15,000.
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Risk Analysis
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What-if analysis generates values for the probabilistic
inputs then computes the resulting profit.
 Let c1 = direct labor cost on each unit.
 Let c2 = parts cost on each unit.
 Let x = first-year demand.
 Hence, profit = P = (249 - c1 – c2)x – 1,000,000.
If all probabilistic inputs equaled their means, then profit
= p = (249-Ec1-Ec2)Ex – 1,000,000
= (249-45-90)15,000 – 1,000,000 = \$710,000
Assume random variables c1, c2, and x are independent.
Hence, Ep = E{(249 - c1 – c2)x – 1,000,000}
= E{(249 - c1 – c2)x} – 1,000,000
= E(249 - c1 – c2)Ex – 1,000,000
= (249 - Ec1 – Ec2)Ex – 1,000,000 =
\$710,000.
Profit is at its mean when variables are at their means.
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Risk Analysis
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If Hewlett Packard were neutral toward risk, then all they
would care about is expected profit. (In the same way,
gamblers seek wagers with positive expected profit. In
particular, gamblers do not fixate on the probability of a
loss.)
 In that case, no further analysis is required, Ep =
\$710,000.
Risk analysis is required when, rather than being neutral,
Hewlett Packard is adverse toward risk.
Simulations are needed when problems are too complex
to be computed exactly.
 Some problems are so complex, the expected value
cannot be computed exactly.
 The Hewlett Packard problem allows the expected
value to be computed exactly, but not other statistics,
such as the probability of a loss.
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Risk Analysis
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Hewlett Packard believes possible values of c1 (direct
labor cost for each unit) range from \$43 to \$47 per unit.
Hewlett Packard believes possible values of c2 (parts cost
for each unit) range from \$80 to \$100 per unit.
Hewlett Packard believes possible values of x (first-year
demand) range from 1,500 to 28,000.
In the best-case scenario, cost is low, demand is high,
and profit = p = (249-43–80)28000 – 1,000,000 =
\$2,591,000.
In the worst-case scenario, cost is high, demand is low,
and profit = p = (249-47–100)1500 – 1,000,000
= -\$847,000.
 (Negative profit is a loss.)
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Risk Analysis
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On the one hand, if Hewlett Packard were perfectly
pessimistic, then all they would care about is the worstcase scenario.
On the other hand, if Hewlett Packard were perfectly
optimistic, then all they would care about is the best-case
scenario.
For all cases in between, Hewlett Packard cares about
the probabilities of alternative scenarios. For that, they
first need to assess probabilities for all random variables.
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Template.xls
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That template facilitates simulations with a mixture of
inputs of random variables with one of three types of
distributions: a general finite distribution, a continuous
uniform distribution, and a general normal distribution.
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Risk Analysis
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Risk Analysis
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Hewlett Packard believes possible values of c1 (direct
labor cost for each unit) have the following finite
probability distribution. (Hence, adapt the general
distribution part of the template.)
Direct labor cost
c1 = \$43
c1 = \$44
c1 = \$45
c1 = \$46
c1 = \$47
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Probability
0.1
0.2
0.4
0.2
0.1
Prob( c1 = \$47 ) = 0.1 = 1.0 – 0.9
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Risk Analysis
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Hewlett Packard believes possible values of c2 (parts cost
for each unit) depend on the general economy, the
overall demand for parts, and the pricing policy of Hewlett
Packard’s parts suppliers. Specifically, they believe c2
has a continuous uniform distribution that ranges from
\$80 to \$100 per unit. (Hence, adapt the general
distribution part of the template.)
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Risk Analysis
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Hewlett Packard reassesses its beliefs about possible
values of x (first-year demand).
Rather than ranging from 1,500 to 28,000, they now
believe that demand can possibly have any value.
Specifically, Hewlett Packard believes x has a normal
distribution with a mean m = 15,000 and standard
deviation s = 4,500. (Hence, adapt the normal distribution
part of the template.)
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Risk Analysis
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Under Simulation Trials, copy formulas for the three
variables: C1, C2, X.
Define profit = (249 – C1 – C2)*X – 1,000,000.
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Risk Analysis
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Drag down row 21 to row 520 for 500 trials.
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Risk Analysis
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For this particular problem, we already precisely
computed the mean for the probability distribution of
profit, Ep = \$710,000.
The purpose of the simulation is to estimate other
features of the probability distribution of profit.
Compute Summary Statistics for the 500 simulation trials.
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Risk Analysis
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The central limit theorem predicts the averages of
simulated profit from many repeated simulations are most
likely close their expected value, Ep = \$710,000.
Since \$756,242 on the previous slide is far from
\$710,000, more simulations are needed.
Try 5,000 simulations. Even then, profit seems far from
\$710,000, so try more, …
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Risk Analysis
Recall: The simulation for
Hewlett Packard assumed that
demand x has a normal
distribution with a mean m =
15,000 and standard deviation s
= 4,500.
 There is no upper limit to profits
in the best-case scenario of the
true probability distribution of
profit because there is no upper limit to demand.
The maximum profits in the 500 scenario trials serve as a
practical upper bound to profits.
There is no lower limit to profits in the worst-case scenario
of the true probability distribution because demand x can
be negative. A more practical worst-case scenario is to
set demand x = 0, which makes profit p = -\$1,000,000
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Risk Analysis
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The true probability distribution of profit is not normal, so
we do not know with 95% probability that profits are
within 2 standard deviations of the mean.
We can use the simulation to estimate the probabilities
of being within two standard deviations of the mean.
The true mean is \$710,000. The estimated standard
deviation is about \$520,000. So profits are within 2
estimated standard deviations (\$1,040,000) when
\$710,000 – \$1,040,000 < p < \$710,000 +
\$1,040,000
– \$330,000
< p5,000
< \$1,750,000
 For
trials, the estimated
probability that profits are
– \$330,000 < p <
\$1,750,000 is 1 – 0.0498,