Probabilistic Risk Analysis

Part 1Part 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:

Continuous:

X

N

iii xpxXE

1

)()(

XdxxxfXE

)()(

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:

Continuous:

N

iXii

N

iii xpxxpXExXVar

1

22

1

2 )()()]([)(

222 )()()]([)( XdxxfxdxxfXExXVar

Probability Concepts Standard deviation:

Multiplication of a random variable by a constant Expected value

Variance

)()( XVarXSD

)()( XcEcXE

)()( 2 XVarccXVar

Probability Concepts

Addition of two independent random variables Expected value

Variance

)()()( YEXEYXE

)()()( YVarXVarYXVar

Probability Concepts

Linear combination of two or more independent variables. Expected value

Variance

)()()( 2121 YEcXEcYcXcE

)()()( 22

211 2

YVarcXVarcYcXcVar

Probability Concepts

Multiplication of two independent random variables Expected value

Variance

)()()( YEXEXYE

)()()]()[()]()[(

)(22 YVarXVarXEYVarYEXVar

XYVar

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 FlowFt3/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 = 0Expected AW(Property damage cost) = $20,000(0.2)+0(0.8)

= $4,000Total 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)

= $2,000Total expected AW of this “do nothing” choice would be

$3,678

Example 1 (cont.)

Enlarge the

channel to

(ft3/sec)

Capital Recovery Amount

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)

x P(x) Total Annual Cost(x)

Expected Total Annual

Cost

700 Flood 0.20 -20,000 -4,000

No flood 0.80 0

1000 Flood 0.10 -21,678 -3,678

No flood 0.90 -1,678

1300 Flood 0.05 -22,517 -3,517

No flood 0.95 -2,517

1600 Flood 0.02 -23,692 -4,092

No flood 0.98 -3,692

1900 Flood 0.01 -25,034 -5,234

No flood 0.99 -5,034

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.)

Economic Factor Estimate

Capital investment

-$521,000

Annual savings

48,600

Increases annual

revenue

31,000

Useful LifeYear (N)

p(N)

12 0.1

13 0.2

14 0.3

15 0.2

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 LifeYear (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

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-40,000 -20,000 0 20,000 40,000 60,000 80,000

E(PW) = $9,984 V(PW) = 477,847x106 SD(PW) = $21,859Pr{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

0

)%,15,/()(k

k kFPFEPWE

153

)3%,15,/(800,2)2%,15,/(000,3

)1%,15,/(500,3000,7

FPFP

FP

)()%,15,/(3

0

k

kFEkFP

Example 3 (cont.)

)()%,15,/()(3

0

2k

k

FVkFPPWV

324,484

)3%,15,/(400)2%,15,/(500

)1%,15,/(600102222

2222

FPFP

FP

3

0

)%,15,/()(k

kFkFPVPWV

696)()( 2/1 PWVPWSD

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

22.0696

1530

)(

)(

PWSD

PWEPWZ

4129.022.0Pr0Pr ZPW