Download Probabilistic Risk Analysis

Probabilistic Risk Analysis
Part 1
Topics to be discussed




The definition of a random variable
The basic characteristics of probability
distributions
Evaluation of projects with discrete
random variables
Evaluation of projects with continuous
random variables
Risk



The chance that the cash flow will fall short
or exceed the estimate – chance of loss.
Sensitivity analysis questions the effect of
cash flow deviations and the cost of capital
when risk is considered.
Risk analysis is appropriate when significant
outcome variations are likely for different
future states and meaningful probabilities can
be assigned to those states.
Decision Making under
Uncertainty vs. Risk

Under uncertainty



There are only two or more observable
values;
However, it is 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.
Decision Making under
Uncertainty vs. Risk

Risk


The process of incorporating explicitly
random variation in the estimates of
measure of merit for an investment
proposal
Initial investments, Operating expenses,
Revenues, Useful life, and other economic
factors are seen as random variables
Risk Analysis

Risk is associated with knowing the
following about a parameter:



The number of observable values and,
The probability of each value occurring.
The “state of nature” of the process is known
at hand.
Risk Analysis

Approaches:

Expected value analysis



Discrete or continuous?
Must assign or assume probabilities/probability
distributions.
Simulation Analysis


Assign relevant probability distributions:
Generate simulated data by applying sampling
techniques from the assumed distributions
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?
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.
Discrete Probability Distributions
Continuous Distribution
Probability Distributions

Individual probability values are stated as:
P(Xi) = probability that X = Xi



“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.
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.
Probability Concepts

Expected value or mean

Discrete:
N
E ( X )   x i p ( x i )  X
i 1

Continuous:
E (X) 

 xf ( x ) dx

 X
Probability Concepts

Variance: the sum of squared deviations
about the population mean.


Var(X)= E{[X-E(X)]2}= E(X2)-[E(X)]2
Discrete:
N
N
i 1
i 1
Var ( X )   [ x i  E ( X )] 2 p ( x i )   x i2 p ( x i )   X2

Continuous:




Var ( X )   [ x  E ( X )]2 f ( x)dx   x 2 f ( x)dx   X2
Probability Concepts

Standard deviation:
SD ( X )  Var ( X )

Multiplication of a random variable by a constant

Expected value
E ( cX )  cE ( X )

Variance
Var ( cX )  c Var ( X )
2
Probability Concepts

Addition of two independent random
variables

Expected value
E ( X Y )  E ( X )  E (Y )

Variance
Var ( X Y )  Var ( X ) Var (Y )
Probability Concepts

Linear combination of two or more
independent variables.

Expected value
E(c1 X  c2Y )  c1E( X )  c2 E(Y )

Variance
Var (c1 X  c2Y )  c Var ( X )  c Var(Y )
2
1
2
2
Probability Concepts

Multiplication of two independent random
variables

Expected value
E ( XY )  E ( X ) E (Y )

Variance
Var ( XY )
 Var ( X )[ E (Y )] 2 Var (Y )[ E ( X )] 2 Var ( X )Var (Y )
Evaluation of Projects with
Discrete Random Variables


Factors are modeled as discrete random
variables.
Expected value and variance of
equivalent measure of merit are
investigated and used in the evaluation.
Example 1

A drainage channel in a community where flash floods
are experienced has a capacity sufficient to carry 700
cubic feet per second. Engineering studies produce the
following data regarding the probability that a given
water flow in any one year will be exceeded and the cost
of enlarging the channel
Water Flow
Ft3/sec
Probability of a greater flow
occurring in any one year
Capital investment to enlarge
channel to carry this flow
700
0.20
-
1,000
0.10
-$20,000
1,300
0.05
-$30,000
1,600
0.02
-$44,000
1,900
0.01
-$60,000
Example 1 (cont.)
The average property damage amounts to $20,000 when
serious overflow occurs. Reconstruction of the channel
would be financed by 40-year bonds bearing 8% interest per
year.
Solution:
If nothing is done and the drainage’s maximum capacity stays at 700
ft3/sec,
Capital Recovery Amount = 0
Expected AW(Property damage cost) = $20,000(0.2)+0(0.8)
= $4,000
Total expected AW of this “do nothing” choice would be $4,000
Example 1 (cont.)
If decide to enlarge the drainage’s maximum capacity to 1,000 ft3/sec,
Capital Recovery Amount = $20,000(A/P,8%,40)
= $20,000(0.0839)
= $1,678
Expected AW(Property damage cost) = $20,000(0.10)+0(0.9)
= $2,000
Total expected AW of this “do nothing” choice would be $3,678
Example 1 (cont.)
Enlarge the Capital Recovery
Amount
channel to
(ft3/sec)
Expected Annual
Property Damage
Total Expected
Equivalent Annual Cost
700
0
-$4,000
-$4,000
1,000
-$1,678
-2,000
-3,678
1,300
-2,517
-1,000
-3,517
1,600
-3,692
-400
-4,092
1,900
-5,034
-200
-5,234
Example 1 (cont.)
Enlarge the
channel to
(ft3/sec)
700
1000
1300
1600
1900
x
P(x)
Total Annual
Cost(x)
Expected Total
Annual Cost
Flood
0.20
-20,000
-4,000
No flood
0.80
0
Flood
0.10
-21,678
No flood
0.90
-1,678
Flood
0.05
-22,517
No flood
0.95
-2,517
Flood
0.02
-23,692
No flood
0.98
-3,692
Flood
0.01
-25,034
No flood
0.99
-5,034
-3,678
-3,517
-4,092
-5,234
Example 2

The Heating, Ventilating, and Air-Conditioning (HVAC)
system in a commercial building has become unreliable
and inefficient. Rental income is being hurt, and the
annual expenses of the system continue to increase.
Your engineering firm has been hired by the owners to

Perform a technical analysis of the system,

Develop a preliminary design for rebuilding the
system, and

Accomplish an engineering economic analysis to assist
the owners in making decision.
The estimated capital investment cost and annual
savings in O&M expenses, based on the preliminary
design, are shown in the following table.
Example 2 (cont.)
The estimated annual increase in rental income with a
modern HVAC system has been developed by the owner’s
marketing staff and is also provided in the following table.
These estimates are considered reliable because of the
extensive information available. The useful life of the rebuilt
system, however, is quite uncertain. The estimated
probabilities of various useful lives are provided. Assume
that the MARR = 12% per year and the estimated market
value of the rebuilt system at the end of its useful life is
zero. Based on this information, what is the E(PW), V(PW),
and SD(PW) of the project’s cash flows? Also, what is the
probability of the PW >= 0? What decision would you make
regarding the project, and how would you justify your
decision using the available information?
Example 2 (cont.)
Useful Life
Year (N)
p(N)
Economic Factor
Estimate
12
0.1
Capital
investment
-$521,000
13
0.2
Annual savings
48,600
14
0.3
15
0.2
Increases
annual revenue
31,000
16
0.1
17
0.05
18
0.05
Example 2 (cont.)

Solution:
The PW of the project’s cash flows, as a function of project
life (N), is
PW(N) = - $521,000 + $79,600(P/A,12%,N)
Example 2 (cont.)
Useful Life
Year (N)
(1)
p(N)
(2)
PW(N)
(3)
(1) X (2)
12
0.1
-27,926
-2,793
13
0.2
-9,689
-1,938
14
0.3
6,605
1,982
15
0.2
21,148
4,230
16
0.1
34,130
3,413
17
0.05
45,720
2,286
18
0.05
56,076
2,804
E(PW) = $9,984
Example 2 (cont.)
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-40,000
-20,000
0
20,000
40,000
60,000
80,000
E(PW) = $9,984
V(PW) = 477,847x106
SD(PW) = $21,859
Pr{PW >= 0} = 0.3+0.2+0.1+0.05+0.05 = 0.7
Evaluation of Projects with
Continuous Random Variables


Expected values and the variances of probabilistic
factors are computed mathematically.
Simplifying assumptions are made about the
distribution of the random variable and the
statistical relationship among the values it takes
on, for example,



Cash flow amounts are normal distribution.
Cash flow amounts are statistically independent.
When the situation is more complicated, Monte
Carlo simulation is normally used.
Example 3

For the following annual cash flow estimates, find the
E(PW), V(PW), and SD(PW) of the project. Assume that
the annual net cash flow amounts are normally
distributed with the expected values and standard
deviations as given and statistically independent, and
that the MARR = 15% per year.
End of
Year, k
Expected Value of
Net Cash Flow, Fk
SD of Net Cash
Flow, Fk
0
-$7,000
0
1
3,500
$600
2
3,000
500
3
2,800
400
Example 3 (cont.)

Solution:


Assume independency among cash flow amounts.
Expected PW are then linear combination of cash flow amounts.
3

E ( PW )  E  Fk ( P / F ,15%, k )
 k 0

3
  ( P / F ,15%, k ) E ( Fk )
k 0
 7,000  3,500( P / F ,15%,1)
 3,000( P / F ,15%,2)  2,800( P / F ,15%,3)
 153
Example 3 (cont.)
3

V ( PW )  V  ( P / F ,15%, k ) Fk 
k 0


3
V ( PW )   ( P / F ,15%, k ) 2V ( Fk )
k 0
 0212  6002 ( P / F ,15%,1) 2
 5002 ( P / F ,15%,2) 2  400 2 ( P / F ,15%,3) 2
 484,324
SD( PW )  V ( PW )
1/ 2
 696
Example 4

Refer to example 3. For this problem, what is the
probability that the IRR of the cash flow estimates is less
than the MARR, Pr{IRR<MARR}? Assume that the PW of
the project is a normally distributed random variable, with
its mean and variance equal to the values calculated in
example 3.
The probability that the IRR is less than the MARR
is the same as the probability that PW is less than zero
PW  E ( PW ) 0  153
Z

 0.22
SD( PW )
696
PrPW  0  PrZ  0.22  0.4129