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Transcript
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
MORE ON VARIATION
AND DECISION MAKING
UNDER RISK
1
19. Learning Objectives
1. Understand certainty and risk.
2. Examine variables and distributions.
3. Relate to the context of random
variables.
4. Estimate expected value and standard
deviation from sampling.
5. Understand and apply Monte Carlo
techniques and simulation to
engineering economy problems.
2
19.1 Certainty, Risk, and Uncertainty
It is said, “There is nothing certain in
this world other than death and taxes.”
Situations and the passage of time
create change, variation, instability.
Engineering economy deals with
aspects of a very uncertain future.
3
19.1 A Parable worth remembering
Yesterday is history…
today is “now”…
and tomorrow is a mystery.
Putting this into perspective:


Accountants deal with yesterday.
Engineers deal with tomorrow…
 when it comes to engineering economy.
So, how can we deal with the
uncertainties of future estimation?
4
19.1 Certainty
Most of the chapters in this text have
presented problems where values were
given:


Assume certainty of occurrence;
Sorry, the real world is not like that!
About the only “certain” or near-certain
parameter in a problem might be the
purchase price of an asset; the rest of
the parameters will vary with time.
5
19.1 Decision Making under Risk
Risk is associated with knowing the
following about a parameter:
1. The number of observable values
and,
2. The probability of each value
occurring.
3. We know the “state of nature” of the
process at hand.
 Decision making under risk
6
19.1 Decision Making under Uncertainty
We will have two or more observable
values;
However, we find it most difficult to
assign the probability of occurrence of
the possible outcomes;
At times, no one is even willing to try to
assign probabilities to the possible
outcomes.
7
19.1 Discrete vs. Continuous Outcomes
If a parameter is “discrete,” then there
are a finite number of occurrences that
can occur, and we attempt to assign a
probability for each outcome.
If a parameter is continuous in nature,
then it can take on an infinite number of
values between two set limits, and we
would have to deal with continuous
functions.
8
19.1 Example 19.1
Two individuals assessing wedding
costs: (Charles and Sue)
Charles’ estimates are (subjective)
Estimated Costs
Prob(Cost)
$3,000
0.65
$5,000
0.25
$10,000
0.10
9
19.1 Ex. 19.1 Histogram Plot: Charles
Output produced
by Palisade’s
RiskView Excel
add-in software.
Note the mean
value for
Charles:
$4,200 for the
wedding
costs.
10
19.1 Example 19.1
Sue, on the other hand, estimates the
following estimated costs for the
Estimated Costs
Prob(Cost)
$8,000
0.333
$10,000
0.333
$15,000
0.333
You should feel sorry for Sue’s father:
He will have to pay for the wedding!
11
19.1 Sue’s Probability Distribution
Sue’s mean
value for the
wedding is
$11,000.
12
19.1 The Merged PDF Plots for Charles and Sue
After discussion, they
agree that the wedding
should cost from
$7,500 to no more than
$10,000 with equal
probability.
The agreed-to distribution
uniform from
7,500 to 10,000
with equal
probability.
13
19.1 Before a Study Is Started:
Must decide the following:


Analysis under certainty (point estimates);
Analysis under risk:
 Assign probability values or distributions
to the specified parameters;
 Account for variances;
 Which of the parameters are to be
probabilistic and which are to be treated
as “certain” to occur?
14
19.1 Two Ways to Account for Risk
1. Expected Value Analysis
1. Discrete or continuous?
2. Must assign or assume
probabilities/probability distributions.
2. Simulation Analysis
1. Assign relevant probability distributions:
2. Generate simulated data by applying
sampling techniques from the assumed
distributions.
15
19.1 Analysis under Uncertainty
This is the “worst” situation to be in.
Here, the states of nature may or may
not be known, or;

States of nature may be defined, but
assignment of probability distributions is at
best “a shot in the dark.”
What do you do?

Try to move from a degree to an improved
level of acceptable risk. (Hereon, we assume
decision making under risk.)
16
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
19.2
ELEMENTS IMPORTANT TO DECISION
MAKING UNDER RISK
17
19.2 Basic Probability and Statistics
RANDOM VARIABLE



A rule that assigns a numerical outcome to a
sample space.
Describes a parameter that can assume any
one of several values over some range.
Random Variables (RV’s) can be:
 Discrete or,
 Continuous.
18
19.2 Two Types
Discrete – assumes only
finite values.
Continuous –
can assume an infinite
number of values over
a defined range.
19
19.2 Probability
A number between 0 and 1.
Represents a “chance” of some event
occurring.
Notations:


P(Xi),
P(X = Xi): read as: “The probability that the
random variable, X, assumes a value of, say,
Xi”
20
19.2 Probability
For a given event and all of that event’s
possible outcomes:


The summation of the probabilities for all
possible outcomes must sum to 1.00.
A probability assignment of “0” means that
the event is impossible to occur.
21
19.2 Probability Distributions
A function that defines
how probability is
distributed over the
different values a random
variable can assume.
22
19.2 Probability Distributions
Individual probability values are stated as:




P(Xi) = probability that X = Xi
[19.1]
“X” represents the random variable or rule (math
function, for example)
Xi represents a specific value generated from the random
variable, X.
Remember, the random variable, X, is most likely a rule or
function that assigns probabilities.
23
19.2 Probability Distributions
Probabilities are developed two ways:
1. Listing the outcome and the
associated probability:
2. From a mathematical function that is
a proper probability function.
24
19.2 Cumulative Probability Distribution
Cumulative Probability Distribution, or
CDF – Cumulative distribution function:
Represents the accumulation for
probability over all values of the
variable.
Notation: F(Xi)
25
19.2 The CDF General Form
F(Xi)= P(X
 Xi) for all i in the domain of X.
26
19.2 Example 19.2 Data
No of
Treatments Per
Day
0
Prob. of
Infection
Reduction
0.07
1
0.08
2
0.10
3
0.12
4
0.13
5
0.25
6
0.25
Discrete data with
seven possible
outcomes. The
probabilities were
most likely developed
from experimental
observations.
27
19.2 Example 19.2 CDF Computed
No. of
Treatments Per
Day
0
Prob. of
Infection
Reduction
0.07
Cumulative
Probability
1
0.08
0.15
2
0.10
0.25
3
0.12
0.37
4
0.13
0.50
5
0.25
0.75
6
0.25
1.00
0.07
28
19.2 Discrete Probability Distributions
PDF and CDF from Example 19.2
29
19.2 Continuous Distribution: Uniform
If the distribution in question
represents outcomes that can assume
a continuous range of values, then:


The model is described by some
assignment of fitted continuous-type
distribution.
One common type of distribution is the:
UNIFORM DISTRIBUTION
30
19.2 Uniform Distribution
Discrete or continuous:
For the continuous case, new notation:


Let f(X) denote the PDF of the
random variable;
F(X) denotes the cumulative
density function (CDF) of the
random variable.
31
19.2 Equal Probability
For the uniform distribution:


All of the values from A to B are considered
equally likely to occur.
Thus, all values from A to B have the same
probability of occurring.
See example 19.3 for cash-flow
modeling.
32
19.2 Example 19.3 – Cash-Flow Modeling
Client 1



Estimated low cash flow: $10,000
Estimated high cash flow: $15,000
Distributed uniformly between $10,000 and
$15,000.
The cash flow is assumed to be best
described as a uniformly distributed
random variable between $10,000 and
$15,000.
33
19.2 Example 19.2: Client 1 Distribution
PDF –
Uniform{10k
–
15k}
CDF
from the PDF
34
19.2 Parameters for the Uniform
Density f(X):
1
1
f (X ) 

Max  Min B  A
Cumulative Density, F(X):
x  min
x A
F(X ) 

Max  Min B  A
35
19.2 Uniform Distribution
Mean:
Min  Max A  B


2
2
Variance:
( Max  Min)
( B  A)
Var ( ) 

12
12
2
2
2
X
36
19.2 Client 1: Question
What is the probability that the
monthly cash flow will be no more than
$12,000?




P(X  12,000) = F(12,000);
A = 10,000; B = 15,000
f(X) = 1/(15 – 10) = 1/5 = 0.200
F(12) = Cumulative of “12”;
 = (12 – 10)/5 = 4/5 = 0.80:
 80% chance that the CF will be $12,000
or less!
37
19.2 Example 19.3: Client 2
Parameters for Client 2:
Assumed Distribution – Triangular
Parameters:



Low =
Most Likely:
High:
20 ($ x 1000)
28 ($ x 1000)
30 ($ x 1000).
The triangular distribution is used to
model Client 2’s cash flow.
38
19.2 Triangular Distribution
Typical Triangular pdf and cdf:
PDF – Client 2
CDF – Client 2
39
19.2 Parameters for a Triangular
L = low value;
M = most likely;
H = High value.
f(X) is in two parts:
2( X  L)
f ( x) 
if L  X  M
( M  L)( H  L)
or,
2( M  X )
f(x) =
if M  X  H
( M  L)( H  M )
40
19.2 Example 19.2: Client 2, P(X  25,000)
“M” is the mode of the distribution;
Mode is the most frequently occurring
value.
For the Triangular:





f(mode) = f(M) = 2/(H-L);
(19.5)
Cumulative F(M) = (M-L)/(H-L)
(19.6)
f(28) = 2/(30-20)= 0.2 = 20%
The breakpoint is at the mode, M = 28;
F(28) = P(X  28) = (28-20)/30-20) = 0.8
41
19.2 Example 19.3: Client 2 Analysis
Given the CDF, F(X) one
locates 25 on the xaxis, projects up to the
curve and over to read
off 0.3125.
42
19.3 Random Samples
Assume an economic parameter can be
described by a random variable – X.
We assume that the random variable,
X, possesses a true mean and variance
denoted by:

 - the parameter’s true (but possibly
unknown) mean value and;

2 – the parameter’s true (but possibly
unknown) variance.
43
19.3 Random Samples – Population
A population is defined as a collection of
objects, elements, or all of the possible
outcomes a variable can assume.
A population may be:


Infinite in number, or
Finite.
A population is characterized numerically
by the population parameters.
44
19.3 Random Samples: Population Parameters
Just as a random variable, a population
possesses (numerically) a:

Mean ,

Variance 2.
In practice, we can define the
population, but probably do not know
the true mean and variance of the
population.
45
19.3 Random Samples: Inference
In the area of applied statistics, one
usually samples from the defined
population in order to make inferences
concerning the population.
There is always uncertainty present
when one samples from a parent
population.
This is why the area of probability is
usually studied before one studies
statistics.
46
19.3 Random Samples: Key Relationships
The diagram below presents an
overview of the relations between:

Populations, probability, a sample, and
statistics.
Probability
Sample
Population
Statistics
Ref: Probability and Statistics for Engineering and the Sciences, 4th Edition, Jay
Devore (Duxbury), p. 3.
47
19.3 Random Sample: Point Estimate
In many cases we assume certainty.
We supply a point estimate of the
parameter in question.
A point estimate is a sample of size 1
taken from the specified population.
An analysis under “certainty” is
essentially applying a point estimate,
which is a sample of size 1.
48
19.3 Random Sample
If we research the parameter of
interest and make another estimate,
then we have:


The original estimate and,
A second estimate:
 So that we now have a sample of size 2.
A point estimate is then:

The most likely value we perceive at the
time of the estimate, or a mean value
estimate.
49
19.3 Random Sample
A population is comprised of two or
more outcomes (values).
The population mean,
, is:
N

x
i 1
N
i
Often, we attempt
to estimate  from
a sample drawn
from the
population.
50
19.3 Random Sample
A random sample of size n is the
selection – in a random fashion – of n
values from the specified population.
Random means the sampling procedure
assumes or requires that every element
in the population has an equally likely
chance of being in the sample.
51
19.3 Random Sample: Variability
We assume that the values that make
up a population are not all the same
value.
The values perhaps are all different, or
mostly so.
Thus, a population possesses a concept
termed:

Variance
52
19.3 Variance of a Population
Variance of a population is a measure
of the dispersion of every element in the
population from the mean value of the
population.
If all of the values in the population are
numerically “close,” then the variance
will be “small.”
If the values are not all close, then the
associated variance will be “larger.”
53
19.3 Variance – Defined for a Population
Variance of a population – denoted as
Var(X) computationally, is:
N
 
2
 (x


)
i
2
i=1
N
Squaring the term in ( ) removes the effects of negative
signs in the summation.
54
19.3 Population Variance
N
 
2
 (x
i
 )
2
i=1
N
The variance is the sum of the squared
deviations about the population mean. Small
deviations generate a smaller variance: Large
deviations generate a larger variance.
55
19.3 Concept of Variance and Sampling
We sample to learn something about
the specified population.
We can use sample results (mean,
variance,…) as estimates of the random
variable in question.
Large sample sizes (say 30 or more)
yield more precision in an estimate as
opposed to small samples (under 30).
56
19.3 Random Sample: Simulation
One “samples” from a population to
either draw inferences about the
population, or:
To simulate the population.

Simulation is the art and science of
generating artificial data from one or more
random variables using statistical sampling
techniques.
57
19.3 Sampling and Variability
If a population possesses a high degree
of variance among its members, then:




Samples drawn from that population will
likely possess high variance:
High variance inhibits precise estimates;
The higher the variability of a population
the less reliable sample estimates will be.
Variability is a key factor is estimation.
58
19.3 Reviewing the Uniform Distribution
Assume a uniform distribution where A
= 0 and B = 1.
The mean of
the {0,1}
uniform is
0.5
59
19.3 Generating Uniform {0,1} Random Numbers
The Uniform Distribution – U{0,1} – is,
by far, the most widely used distribution
in all of statistics.
Numerous software programs in a
variety of languages have been written
to generate random numbers from
0 to 1.
Excel supports a random number
function generator.
60
19.3 Random Sample
Excel’s RAND Function
RAND( )
Remarks
To generate a random real number between a and b,
use: RAND()*(b-a)+a
If you want to use RAND to generate a random number,
but don't want the numbers to change every time the cell
is calculated, you can enter =RAND() in the formula bar,
and then press F9 to change the formula to a random
number. Ref. Microsoft’s Excel Help Data base.
61
19.3 Excel’s RAND Function
By marking and dragging the top
cell down, Excel will generate a
different random number in each cell.
62
19.3 Random Numbers Other Than {0,1}
Excel supports the RANDBETWEEN
function.
RANDBETWEEN(bottom,top)
Bottom is the smallest integer
RANDBETWEEN will return.
Top is the largest integer RANDBETWEEN
will return.
63
19.3 RANDBETWEEN Function
Formula Description:
(Result)=RANDBETWEEN (1,100)

Returns a uniformly distributed number
between 1 and 100
64
19.3 RANDBETWEEN Function
Formula Description:
(Result)=RANDBETWEEN (1,100)

Returns a uniformly distributed
number between 1 and 100
65
19.3 RANDBETWEEN Function – Example
66
19.3 Sampling from a Discrete Distribution
Our objective is to initiate simulating
outcomes from a discrete probability
distribution.
We apply a process known as the:

Inverse Transform Method.
This is part of a process known as:
 Monte Carlo Sampling

Derived from the development of the atomic
bomb in 1943 – 45 in the U.S.
67
19.3 Simulating from a Discrete Process
First, one defines the probability
distribution set: Use Example 19.4
Given a discrete probability set with
three possible outcomes:

{ 24, 30, and 36).
0.20 N = 24


P( X  Ni )  0.50 N = 30
0.30 N = 36


68
19.3 Example 19.4: Step 1
Generate the pdf and cdf.
PDF
CDF
69
19.3 Example 19.4: Step 2
Using the CDF, fix a sample size.
Assume “n” = 10.
Generate 10 U{0,1} random numbers.
Assume the 10 numbers are as shown
in Example 19.4
70
19.3 Example 19.4: 10 Uniform RN’s
Sample
No.
Random
Number
1
45
2
44
3
79
4
29
5
81
6
58
7
66
8
70
9
24
10
82
• Next, “map” the RN onto
the CDF to observe the
outcome for that random
number.
• The RN is located on the
y-axis: Project over to the
CDF curve, then down to
the x-axis and read the
associated outcome. This
is called the inverse
transform method.
71
19.3 Example 19.4: First RN = 45
First RN = 45
(Assume 0.45)
Observe the
outcome as N =
30 months
72
19.3 Example 19.4: Continue With Other RN’s
73
19.3 Summary of Outcomes for n = 10
From example 19.4, the 10
“simulated” outcomes are
shown on the next slide.
74
19.3 Summary of Outcomes for n = 10
Sample
No.
Random
Number
Simulated
Outcome
1
45
30
2
44
30
3
79
36
4
29
30
5
81
36
6
58
30
7
66
30
8
70
36
9
24
30
10
82
36
The simulated
outcomes are
proportional to
the originally
assigned
probabilities.
75
19.3 Sample Size Questions
A sample of size 10 is not sufficient to
perform a reasonable simulation
analysis for a real-world problem.
In simulation modeling of cash flows,
sample size is a critical issue.
Obviously, larger sample sizes are
better (provide more information and
better approximate the situation being
evaluated).
76
19.3 Sample Statistics for n = 10: Example 19.4
Sample
No.
1
2
3
4
5
6
7
8
9
10
Mean
Var
Std-Dev
Outcome
30
30
36
30
36
30
30
36
30
36
32.40
9.60
3.10
From a sample of size 10, the
sample mean is 32.40 with an
associated variance of 9.60.
77
19.3 Continuous Process
If the process under study is
considered “continuous,” then the
associated pdf and cdf must be
generated.
Assume a continuous uniform
distribution to illustrate.
Use Example19.3, Client 1’s data.



Est. low cash flow:
$10,000
Est. high value:
$15,000
Mean = (10 + 15)/2 = 12.5 ($12,500)
78
19.3 Sampling from a Continuous Uniform
Graphical Approach:
Fix the sample size (use 10 again).
Generate 10 U(0,1) random numbers.
Generate the CDF for the uniform and
plot.
Similar Procedure as the discrete case.
The pdf and cdf for the uniform are
presented on the next slide.
79
19.3 PDF and CDF for the Uniform (10,15)
We “sample” always from the CDF!
“Map” the RN over
to the CDF curve.
Generate a U{0,1}
random number.
Project down
to x-axis for
the outcome.
80
19.3 For Example 19.4: Client 1
Assume the RN is “45.”
This can be interpreted as 0.45.
Locate 0.45 on the y-axis;
Project to the right to intersect the CDF
curve;
Then project down to the x-axis to read
the associated outcome for y = 0.45.
81
19.3 Example 19.4: Using the Uniform CDF
RN = 0.45
outcome given
RN = 0.45
is
X = 12.5.
This would be
the first outcome
for the first
sample.
Repeat for all
other RN,s.
82
19.3 Random Number Generation
Excel and all other simulation software
support the generation of uniform
random numbers.
A U(0,1) random number generator
(software program) will generate a
sequence or stream of random numbers
between 0 and 0.99999.
“0” and “1” cannot be generated due to
software and floating point arithmetic
associated with digital computers.
83
19.3 Question of Sample Size
For simulation of cash flows using
computer software, the sample size can
be virtually as large as the analyst
desires (10,000 or more is not
uncommon).
Sampling is based upon:


The Law of Large Numbers, and
The Central Limit Theorem of statistics.
84
19.3 Sample Sizes
Small samples – less than 30

Apply the t-distribution from statistics.
Large Sample sizes – 30 or more.

Simulation studies – 1,000 to 20,000 per run
are not uncommon and take very little
computer time.
Obviously, the larger the sample the
greater the reliability of the use of the
values.
85
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
19.4
EXPECTED VALUE COMPUTATIONS
FOR ALTERNATIVES
86
19.4 Expected Value
Given a random variable – X.
Two important properties:


Expected Value of X; denoted E(X).
Variance of X. denoted Var(X).
IF the entire population was known
and available, then all parameters could
be calculated directly!
87
19.4 Why Sample?
One samples from a population in an
attempt to estimate some property or
value of a true but unknown population
parameter.
Thus, random samples are drawn or
generated from the population in
question to estimate, say, a population
parameter.
88
19.4 Common Estimators
Given a population with population
parameters:


 – the population’s mean value;
2 – the population’s variance.
We focus on estimating the mean
value –  – of a population.
89
19.4 Notation Concerns
We focus on two sets of notation:


Set 1: Population Parameters
 Denoted by Greek symbols
  - Population mean;
 2 - population variance;
  - population standard deviation.
Set 2: Sample Estimators
X
Mean of a sample of size “n.”
s2 – Sample variance
s – Sample standard deviation
90
19.4 Expected Value (Operation)
Expected Value is the long-run
expected average if the variable is
samples from a large number of times.
Estimating a population mean – .


 X i P( X i ) if X a discrete random variable.
E ( x)  





91
19.4 Mean of a Sample of Size “n”
Sample mean
Take or extract a sample of size “n”
from a population.
We require that the sample be
“random” – every element in the
population has an equally likely chance
of being selected:


If not, the sample is “biased.”
Compute the sample mean value…
92
19.4 Sample Mean Value: X-bar
The mean of a sample of size “n” is:
n
X
x
i
i
n
93
19.4 The Sample Mean
Measure of central tendency of a set of
values.
The sample mean X is termed an
unbiased estimator of the population
mean – ..
Unbiased means:
E( X )  
94
19.4 Sample Mean
Thus, a sample of sufficient size is
taken, say 30 or more:
The sample mean is calculated from the
sample data:
The sample mean, X, is then used as a
point estimate of the true but unknown
population mean value, .
95
19.4 Unbiased Estimator
An estimator or the process of
calculating the estimate should:




Provide a numerical result that is close to
the population parameter being estimated.
Not “biased” in any way.
The mean of a sample of size n is termed an
unbiased estimator of the population mean.
This is why a sample mean is often used as a
reliable “point” estimator.
96
19.4 Other Measures of Central Tendency
Mean – already been discussed.
Mode:


The value that most frequently occurs in a
population or in a sample.
The mode is a biased estimator of the
population mean.
Median:

The value where 50% of the observations
occur below the median value and 50%
occur above the median value (biased
estimator).
97
19.4 Measures of Variability
Variance of a population:
N
 
2
(x


)
 i
2
i=1
N
98
19.4 Standard Deviation of a Population
The standard deviation of a population;
Denoted as  is:
N
 
 (x
i
 )
2
i=1
N
99
19.4 Variance: Notes
2 = Var(X);
 X    Var(X)
2
X
100
19.4 Sample Variance and Standard Deviation
Sample of size “n.”
s 
2
(
X

X
)
 i
s s
n 1
2
101
19.4 Computational Form of the Variance
An alternative form for computations of
a sample variance is:
s
2
X


2
n
2

X
n 1 n 1
i
102
19.4 Sample Variance and Standard Deviation
Sample of size “n.”
s 
2
(
X

X
)
 i
n 1
Note: The variance is the sum of the squared distances
2
(deviations) that each variable in the sample is from the
mean value of the set. The divisor (n-1), is required to
permit s2 to be an unbiased estimator of 2 .
s s
103
19.4 Combining the Mean and the Variance
It is customary to refer to the following
values above and below a mean value.
X  (t )( s)
Where “t” is usually equal to { 1, 2, 3 }
104
19.4 t = 1, 2, 3:
In most engineering economy applications
we will find:

Virtually all of the sample values in the sample
will be within 3 standard deviations of the
mean value of the sample.
For the normal distribution (Example
19.10) we will find that well over 99% of
the normally distributed values will be
within 3 standard deviations of the mean
value of the sample.
105
19.4 Sample Distributions
Different means and standard
deviations illustrated.
106
19.4 Example 19.5
Utility bills from two different
populations (North American Data)
Sample
No.
1
2
3
4
5
6
7
Mean
X
40
66
75
92
107
159
275
116.29
(x-xbar)
-76.29
-50.29
-41.29
-24.29
-9.29
42.71
158.71
Sum
Sq'd
Dev.
5819.51
2528.65
1704.51
589.80
86.22
1824.51
25190.22
37743.43
107
19.4 Example 19.5
Utility Bills from two different
populations (Asian Data)
Sample
No.
1
2
3
4
5
Mean
X
84
90
104
187
190
131.00
(x-xbar)
-32.29
-26.29
-12.29
70.71
73.71
Sum
Sq'd
Dev.
1042.37
690.94
150.94
5000.51
5433.80
12318.55
108
19.4 Comparisons…
The dispersion of the data for the Asian
sample is smaller that that of the North
American sample.
Analysis from Excel is shown on the
next slide…
109
19.4 Comparing the Two Samples
North American
Sample
X
No.
1
40
2
66
3
75
4
92
5
107
6
159
7
275
Mean
116.29
Var
6,290.57
StdDev
79.313123
Max
275
Min
40
Range
235
• The variance of the
NA sample is larger
than the variance of
the Asian sample.
• Note: The Range
(Max – Min) for the
NA sample is also
higher.
• The NA sample
contains more
variability than the
Asian sample.
Asian
Sample
No.
1
2
3
4
5
Mean
Var
StdDev
Max
Min
Range
X
84
90
104
187
190
131.00
2,809.00
53
190
84
106
110
19.4 Deviation Ranges for Example 19.4
Examine:



 (1)(s) and (2)(s) “bands” for the NA data.
Standard deviation for NA: 79.31
Sample mean: 116.29
X  1s  116.29  (1)79.31  $36.98 to 195.60
Six out of seven data values fall in this range, or 85.7%.
X  2 s  116.29  (2)79.31  -42.33 to 274.91
111
19.4 Dispersion for Both Samples (Fig 19-9)
112
19.4 Continuous Functions: Summarization
Continuous PDF’s:


Replace summations with integrals;
Over the defined range of the random
variable in question.
Let R represent the defined range of the
specific pdf.
 P(X) is replaced by the differential element

f(x)dx.
113
19.4 Key Relationships for Continuous RV’s
If the random variable is described by a
continuous function, the following
relationships hold:
E(X)=  X(0.2)dX =
R
0.1X
2 15
10
= 0.1(225 - 100) = $12.5
E(X) =  = $12.5
114
19.4 Var(X): Continuous Uniform
The variance of the continuous uniform
is:
Var(X) =  X (0.2)dX-12.5 =
2
2
R
0.2
3
X
3
15
 (12.5)
2
10
= 0.06667(3375 - 1000) - 156.25 = 2.08
 X  2.08  $1.44
115
19.4 Summary
For a continuous Uniform{10,15};

Then mean  = $12.5;

The variance is 2 = Var(X) =
 2.08

The standard deviation  =
 $1.44
Thus, the mean for U{10,15} = $12.5
and the variance = 2.08.
116
19.4 Next…
The next section presents an
overview of a fundamental
simulation methodology
termed,
Monte Carlo Sampling
and…
introduces basic Simulation of
cash flows.
117
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
19.5
MONTE CARLO SAMPLING AND
SIMULATION ANALYSIS
118
19.5 Monte Carlo Sampling
Monte Carlo Sampling is a traditional
approach (method) for generating pseudorandom numbers to sample from a
prescribed probability distribution.
Pseudo-random refers to the fact that a
digital computer can generate
approximate random numbers due to fixed
word size, and can round off problems.
119
19.5 Monte Carlo Sampling
Requires:



The CDF of the assumed pdf;
A uniform random number generator;
Application of the inverse transform approach.
Why require the CDF?




The y-axis of a CDF is scaled from 0 to 1.
That is the same as the range of U(0,1).
Facilitates mapping of an RN to achieve the
outcome value on the x-axis.
The U(0,1) selects a X-value from the CDF.
120
19.5 Monte Carlo Sampling Illustrated
Cumulative
Probability
A U(0,1) R. N.
Xj
121
19.5 Simulation Using Monte Carlo Sampling
Formulate the economic analysis:






The alternatives – if more than one;
Define which parameters are “constants” and
which are to be viewed as random variables.
For the random variable(s), assign the
appropriate PDF:
 Discrete and or continuous.
Apply Monte Carlo sampling – a sample size of
“n” where it is suggested that n  30.
Compute the measure of worth (PW, AW,…)
Evaluate and draw conclusions.
122
19.5 Simulation Assumptions
It is assumed that:



All parameters are independent;
The value of a given parameter in no way
impacts or influences the value of another
parameter.
This is termed:
 The property of independent random
variables.
123
19.5 Example 19.7: Overview System 1
Two alternative systems ( i = 15%/yr.);

System 1
 P = $12,000, no salvage value.
 n1 = 7 years;
 No annual net revenue is offered.
 Considered a “high risk” venture.
 Assumption: NCF1-5 modeled by a
continuous U(-$4,000, +$6,000) – analyst's
best guess.
 Certainty: P = $12,000 and n = 7!
124
19.5 Example 19.7: Overview System 2
System 2
 P = $8,000, no salvage value.
 Annual net revenue = $1,000 for each of the
first 5 years;
 But after 5 years, no guarantee of future
revenues.
 Equipment updates may be useful up to 15
years – BUT, the exact number of years is
unknown.
 Cancel any time after 5 years.
125
19.5 Example 19.7: Overview System 2
Variable Assumptions:

P2 = $8,000;

NCFG = $1,000 first 5 years;

NCF2 years 6, 7, … assumed to follow:
 Discrete uniform($1,000 to $6,000) in
$1,000 increments.

N2 after t = 5 assumed to be continuous
U(6,15)
126
19.5 Define the Random Variables
System 1:
NCF1 = U(-4000,6000)
System 2: NCF2:
DiscreteU(1000,6000)
=1000
System 2: n2 U(6,15)
127
19.5 Systems 1 & 2 PW Formulations
PW1 = -P1 + NCF1 (P/A,15%,n1)
[19.17]
PW2 = -P2 + NCFG(P/A,15%,5) +
NCF2(P/A,15%,n2-5)(P/F,15%,5)
[19.18]
• Terms in
represent varying parameters.
• NCFG = the guaranteed cash flow for System2
128
19.5 Summarizing: Analytical Setup
PW1 = -12,000 + NCF1(P/A,15%,7)
= -12,000 + NCF1(4.1604)
[19.19]
PW2 = -8000 + 1000(P/A,15%,5)
+ NCF2(P/A,15%,n2-5)(P/F,15%,5)
= -4648 + NCF2(P/A,15%,n2-5)(0.4972)
[19.20]
At this point – Design a spreadsheet model to
conduct a simulation for, say, n = 30 samples.
129
19.5 Example 19.8: Spreadsheet Model
Simulate NCF1 First 5 Values Shown…
C
D
E
=RAND()*100
Sample No
6
7
8
9
10
1
2
3
4
5
RN1(100)
NCF1-$
12.5625
0.5158
12.0306
73.8529
30.3512
-$2,800
-$4,000
-$2,800
$3,300
-$1,000
=INT((100*$C74000)/100)*100
Important Note: The values shown in this simulation
may not match the values shown in Figure 19-12
due to inability to generate the same exact sequence
of U(0,1) used in the text.
130
19.5 Generating NCF2
NCF2 =
DiscUniform{1,000,6,000,=1,000}
Six possible outcomes;
{1,000,2,000,3,000,4,000,5,000,6,000}
P(each outcome) = 1/6 =0.1667
The categories will be as shown on the
next slide…
131
19.5 NCF-2 CDF Assignments
RN from-to
NCF2 will Equal….
00 – 16
$1,000
17-32
$2,000
33-49
$3,000
50-66
$4,000
67-82
$5,000
83-99
$6,000
132
19.5 Generating NCF-2 in Excel
In Excel this will require a complex IF
statement to slot the generated RN into
the appropriate outcome value.
We will apply a Nested IF construct.
F
G
Sample No. RN2(100)
6
7
8
9
10
1
2
3
4
5
83.6176
26.5754
18.6240
27.0229
13.4377
H
NCF2-$
$6,000
$2,000
$2,000
$2,000
$1,000
=IF($G7<=16,1000,
IF($G7<=32,2000,IF($
G7<=49,3000,IF($G7
<=66,4000,IF($G7<=8
2,5000,IF($G7<=100,
6000,6000)
133
19.5 Generating n2
The random variable – n2, represents the
time period in years AFTER t = 5.
We know that System 2 will have at least
a 5-year life, but could go up to as much
as 15 years.
We need to simulate the number of years
after t = 5, and then apply the (P/A, n2-5)
to come back to t = 5. See [19.18, 19.20]
134
19.5 “n2” Analysis: The PDF
The n2 random variable is defined as:

Continuous Uniform{6,15}
f(x) = 1/(16-9)
= 1/9 = 0.1111
Mean = 10.
135
19.5 The CDF for n2
X = 15
136
19.5 Generating n2 Life
We use the CDF for the uniform.
Specific problem of the CDF is:

F(X) = (x-Low)/(High – Low);

F(X) = (X -6)/(15-6);

F(X) = (X-6)/9.
Set F(X) = a U(0,1) say, RNj

Then, RNJ = (X-6)/9; (solve for X).

X = 9(RNJ) + 6. (returns the additional life
after 5 years)
137
19.5 Generating n2 Life…
X = 9(RNJ) + 6. (returns the additional
life after 5 years given the j-th random
number)
=RAND()
Cell Equation:
6
7
8
9
10
I
J
K
Sample No.
RN3
0.556
0.406
0.527
0.295
0.699
N-yrs
1
2
3
4
5
=INT(((9*J7)+1) +6)
12
10
11
9
13
138
19.5 Example 19.8: Computing the PV’s
If “n” = 30, then 30 individual present
values will be computed over the different
simulated lives (System 2) and the varying
cash flows.
From the 30 PV results, we determine
how many out of 30 generate a positive PV
and how many generate a negative PV.
Calculate the percentages out of 30 that
have a positive present value for each
alternative.
139
19.5 Excel’s PV Function Dialog Data Entry
The “PV” function will be used to compute
the associated present values.
140
19.5 The “Random Number” Worksheet
C
Sample No
6
7
8
9
10
1
2
3
4
5
D
E
F
G
H
I
RN1(100)
NCF1-$
RN2(100)
NCF2-$
N-yrs
12.5625
82.0129
37.4472
89.9367
89.7730
-$2,800
$4,200
-$300
$4,900
$4,900
83.6176
91.9930
84.3865
81.1570
64.2324
$6,000
$6,000
$6,000
$5,000
$4,000
RN3
0.556
0.513
0.429
0.741
0.088
12
11
10
13
7
The above shows the first 5 out of 30 samples for NCF1,
NCF2 and N for System 2. These values will not
compare to the values in Figure 19-12 due to a different
sequence of RN’s that were generated by Excel.
141
19.5 Spreadsheet Design
For this example:


One workbook with,
Two spreadsheets:
 Sheet 1 is named “Random Numbers”
 Sheet 2 is named “PW Values”
Sheet 2 summary calculations will refer
to certain values in the “Random
Numbers” spreadsheet and to values in
the “PW Values” spreadsheet.
142
19.5 Design Overview for the Model
Sheet 1
“Random Numbers”
contains the simulated
values as specified
by the problem
statement.
Sheet 2
“PW Values”
Calculates the 30 PW
values for the
two systems and
summarizes.
Sheet 2 contains the 30 calculated PW values by
referencing data in Sheet 1: Summary results are
then presented in Sheet 2.
143
19.5 Calculation of 30 PW’s
Input Parameters
Input Parameters for the Alternatives
System 1,P1 $ 12,000.00
System 2, P2 $
n
7
yrs
NCF
$
MARR
15%
MARR
8,000.00
1,000.00
15%
5
Yrs
144
19.5 Simulation Analysis: Results
Sample No. PW - Sys 1
1
-$23,649
2
-$9,504
3
-$351
4
-$21,153
5
-$26,561
29
$65
30
$481
Present Worth Calculations
+
Sample No.
0
1
1
0
1
2
0
1
3
0
1
4
0
1
5
1
0
29
1
0
30
11
19
36.67% 63.33%
(+)
(-)
PW - Sys. 2
$7,763
-$3,031
$2,045
$97
$2,163
$6,507
-$511
Counts
(+)
(-)
1
0
0
1
1
0
1
0
1
0
1
0
0
1
21
9
70.0% 30.0%
(+)
(-)
The first five samples and the last two are shown above
for a given sequence of RN’s.
145
19.5 Analysis for Current Sequence of RN’s
Sample No. PW - Sys 1
1
-$23,649
2
-$9,504
3
-$351
4
-$21,153
5
-$26,561
29
$65
30
$481
Present Worth Calculations
+
Sample No.
0
1
1
0
1
2
0
1
3
0
1
4
0
1
5
1
0
29
1
0
30
11
19
36.67% 63.33%
(+)
(-)
For this run, 36.67% of
the generated PW’s for
System 1 were positive
and 63.63% were
negative.
PW - Sys. 2
$7,763
-$3,031
$2,045
$97
$2,163
$6,507
-$511
Counts
(+)
(-)
1
0
0
1
1
0
1
0
1
0
1
0
0
1
21
9
70.0% 30.0%
(+)
(-)
System 2:
70% had a positive PW
and 30% were
negative.
146
19.5 Final Comments
The results shown on the previous slide
were from a given sequence of RN’s.
The spreadsheet can be recalculated and
a different sequence of RN’s will be
generated.
The % positive and % negative will
change from run to run – as expected.
147
19.5 Conclusions Based on Current Run
From a sample of size 30:



System 2 presented a greater percentage of
positive present worth over the percentage of
positive present worth for System 1.
This could lead one to believe that System 2,
over the long run, is the better alternative.
Suggest going with System 2!
148
19.5 The Normal Distribution
Popular Distribution.
Continuous pdf.
Often referred to as the “Bell-Shaped
Curve.”
Applicable to a large number of physical
applications.
See Figure 19-15…
149
19.5 Different Normal Distributions
Different means with equal
variance.
Different means with different
variances.
The Standard Normal with mean =
to 0 and variance = 1.
150
19.5 PDF for the Normal
The distribution function for the normal is:

PDF:
2 


1
 ( X  X ) 
f (X ) 
exp  

2
 X 2


  2 X

Note: There is no direct way to generate normally
distributed random numbers using the inverse
transform. Other approaches must be used.
151
19.5 The Standard Normal Distribution
Calculate a “Z” value as:
X X
Z
s
[19.22]
Where X-bar is the sample mean and s is the sample
standard deviation, Z is normally distributed with a
mean of 0 and a variance of 1.
152
19.5 Difficulties in Simulating
Generating random numbers that follow a
normal distribution are involved if one is
using Excel.
Most spreadsheets do not support a normal
random number generator.
Other methods must be used (refer to texts
on simulation and the generation of random
numbers from a prescribed normal).
153
19.5 Variation Ranges for the Normal
Variable X Range
Probability
Variable Z Range
 + 1
0.3413
0 to +1
  1
0.6826
-1 to +1
 + 2
0.4773
0 to +2
  2
0.9546
-2 to +2
 + 3
0.4987
0 to + 3
  3
0.9974
-3 to +3
154
19.5 Beware!
When using the normal distribution with a
mean close to 0, it is possible to generate a
negative outcome.
This may not be applicable to the problem
at hand.
Some software systems provide a
“truncated normal,” where you can specify
a lower and/or upper cutoff point.
155
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
Chapter Summary
156
19. Summary
To perform decision making under risk
implies that some parameters of an
engineering alternative are treated as random
variables.
Assumptions about the shape of the
variable's probability distribution are used to
explain how the estimates of parameter
values may vary.
Additionally, measures such as the expected
value and standard deviation describe the
characteristic shape of the distribution.
157
19. Summary
In this chapter, we learned several of the
simple, but useful, discrete and continuous
population distributions used in engineering
economy–uniform and triangular–as well as
specifying our own distribution, or assuming
the normal distribution.
158
19. Summary
Since the population's probability distribution
for a parameter is not fully known, a random
sample of size n is usually taken, and its
sample average and standard deviation are
determined.
The results are used to make probability
statements about the parameter, which help
make the final decision with risk considered.
159
19. Summary
The Monte Carlo sampling method is combined
with engineering economy relations for a
measure of worth such as PW to implement a
simulation approach to risk analysis.
The results of such an analysis can then be
compared with decisions when parameter
estimates are made with certainty.
160
19. Summary
Simulation analysis is a powerful tool when
properly applied.
One must be fully versed in applied statistics
and modeling.
Several Excel add-ins are on the market to
simplify sophisticated simulations of
engineering economy problems.
Example: Palisade Corporations “@RISK”
Excel add-in, and “Crystal Ball” are popular
additions to Excel.
161
19. Summary
Simulation modeling is a valuable tool
to assist in the analysis of engineering
economy problems where risk is an
important consideration.
Simulation is widely used in the
petrochemical, manufacturing,
aerospace, and other prominent
industrial sectors.
162
ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 19
End of Set
163