Dealing with Uncertainty
Probabilistic Risk Analysis
Introduction


Statistical and probability concepts can also be used
to analyze the economic consequences of some
decision situations involving risk and uncertainty.
The probability that a cost, revenue, useful life, NPW,
etc. will occur is usually considered to be the longrun relative frequency with which the value occurs.
Random Variables


Factors having probabilistic outcomes are called random
variables.
Useful information about a random variable is
– expected value (average, mean), denoted by E[X]
– variance, denoted by Var[X]
– standard deviation, denoted by SD[X]

When uncertainty is considered, the
– variability in the economic measures of merit and
– the probability of loss associated with the alternative are very
useful in the decision-making process.
Some Important Relationships

Discrete Random Variables
Probability: Pr{X=xi} = p(xi) for i = 1, ..., L
where p(xi)>0 and i p(xi)=1

Continuous Random Variables
d

Probability:Pr{c  X  d}   f (x)dx where  f (x)dx  1

Expected Value:
E[X] = i xi p(xi) or

Variance:
Var[X] = i (xi - E[X])


c
 xf (x)dx

2p(x )
i

or
2
(x

E(x))
f (x)dx


Some Important Relationships (cont’d)

Variance: Var[X] = E[X2] - (E[X])2

Standard Deviation: SD[X] = (Var[X]) 1/2

Expected value of a sum: E[X+Y] = E[X] + E[Y]
Variance of a sum or difference
Var[X+Y] = Var[X-Y] = Var[X] + Var[Y]
when X and Y are independent
Multiply by a constant: E[cX] = cE[X] and Var[cX] = c2 Var[X]
Expected Value of a function: E[h(X)] = i h(xi) p(xi) or




 h(x) f (x)dx

Evaluation of Projects with Random
Outcomes


We can use the expected value and variance concepts to
assess the project’s worth
We might be interested in
– the expected net present worth, E[NPW], or
expected net annual worth, E[NAW]
– the variance or standard deviation of the traditional measures,
Var[NPW], Var[NAW], SD[NPW], SD[NAW]
– the probability that the NPW or NAW is positive, i.e.,
Prob{NPW > 0} or Prob{NAW>0}
Example 7

A HVAC system has become unreliable and inefficient.
Rental income is being hurt and O&M continue to
increase. You decide to rebuild it.
Assume MARR = 12%
Economic Factor
Estimate
Capital Investment -$521,000
Annual Savings
$48,600
Increase Annual Rev. $31,000
Useful Life
Year (N)
12
13
14
15
16
17
18
p(N)
0.1
0.2
0.3
0.2
0.1
0.05
0.05
Example 7 (cont’d)

For year 12,
NPW = -$521,000 + ($48,600+$31,000) (P/A, 12%, 12)
= -$27,926
However, this useful life only has a 0.1 chance of occurring.

For year 13,
NPW = -$521,000 + ($48,600+$31,000) (P/A, 12%, 13)
= -$9,689
However, this useful life only has a 0.2 chance of occurring.
Example 7 (cont’d)

What is E[NPW] and Var[NPW] ?
(1)
N
12
13
14
15
16
17
18
(2)
NPW(N)
-$27,926
-$9,689
$6,605
$21,148
$34,130
$45,720
$56,076
Sum
(3)
p(N)
0.1
0.2
0.3
0.2
0.1
0.05
0.05
(4) = (2)x (3)
P(N)[NPW(N)]
-$2,793
-$1,938
$1,982
$4,230
$3,413
$2,286
$2,804
$9,984
(5) =(3)x(2)2
p(N)x NPW(N) 2
($ 2) 77.986 x 106
($ 2) 18.776 x 106
($ 2) 13.089 x 106
($ 2) 89.448 x 106
($ 2) 116.486 x 106
($ 2) 104.516 x 106
($ 2) 157.226 x 106
($2 ) 577.524 x 106
Example 7 (cont’d)






E[NPW] = $9,984
E[(NPW)2] = ($2) 577.524 x 106
Var[NPW] = E[(NPW)2] - (E[NPW])2
= ($2)477.847 x 106
SD[NPW] = (Var[NPW])1/2 = $21,859
Probability{NPW > 0} = 1- (0.1+0.2) = 0.7
The weakest indicator is SD(NPW) > 2E[NPW] !
Example 8

For the following cash flow estimates, find E[NPW],
Var[NPW], and SD[NPW]. Determine Prob{ ROR <
MARR}. Assume that the annual net cash flows are
normally distributed and independent. Use a MARR =
15%.
End of
Year, k
0
1
2
3
Net Cash Flow
Mean
Standard Dev.
-$7,000
0
$3,500
$600
$3,000
$500
$2,800
$400
Example 8 (cont’d)
The investment is known.
Year 0

1
0
-15000
-10000
-5000
0
5000
10000
15000
Example 8 (cont’d)

The cash flows for the years 1, 2 and 3 are not known.
Cash Flows
0.1
Ye ar 1
Ye ar 2
Ye ar 3
0.0
0
10 00
20 00
30 00
40 00
50 00
60 00
70 00
Example 8 (cont’d)

E[NPW] = -$7,000 + $3,500 (P/F,15%,1) +
$3,000 (P/F,15%,2) + $2,800 (P/F,15%,3)
= $153

Var[NPW] = 02 + ($600)2 (P/F,15%,1)2 +
($500)2 (P/F,15%,2)2 +($400 )2 (P/F,15%,3)2
= ($2 )484,324

SD[NPW] = $696
Example 8 (cont’d)

Prob{ ROR <= MARR} = ?
– Step 1: For a project having a unique ROR (simple
investments are such projects), the probability that the ROR is
less than the MARR is the same as the probability that the
NPW is less than 0. So
Prob{ ROR <= MARR} = Prob{ NPW <= 0}
– Step 2: Because the NPW is normally distributed, we can
normalize to a N(0,1) distribution. So
Z = (NPW - E[NPW])/SD(NPW) = (0-153)/696 = -0.22
– Step 3: Using Normal Tables, we get
Prob{NPW <=0} = Prob{Z <= -0.22} = 0.4129

Therefore Prob{ ROR <= MARR} = 0.4129
Decision Trees



Also called decision flow networks and decision
diagrams
Powerful means of depicting and facilitating the analysis
of problems involving sequential decisions and variable
outcomes over time
Make it possible to break down large, complicated
problems into a series of smaller problems
Diagramming


Square symbol depicts a decision node
Circle symbol depicts a chance outcome node
– All initial or immediate alternatives among which the decision
maker wishes to choose
– All uncertain outcomes and future alternatives that may
directly affect the consequences
– All uncertain outcomes that may provide information
Diagramming Example
Sales
Decision
Invest in new
product line
Status Quo
good
bad
Example 9

A new design is being evaluated as potential
replacement for a heavily used machine. The new design
involves major changes that have expected advantage,
but would be $8600 more expensive. In return, annual
expense savings are expected, but their extent depend on
the machine’s reliability.
Reliability
Excellent (E)
Good (G)
Standard (S)
Poor (P)

Probability
0.25
0.40
0.25
0.10
Annual Savings
$3,470
$2,920
$2,310
$1,560
Use MARR = 18%. Life = 6 years. Salvage = 0.
Example 9 (cont’d)
A = $3,470
NPW = $3,538
A = $2,920
NPW = $1,614
A = $2,310
NPW = -$520
25%
40%
25%
New Design
Current Design
10%
A = $1,560
NPW = -$3,143
Example 9 (cont’d)


Based on a before-tax analysis (MARR = 18%,
analysis period = 6 years, salvage value = 0),
is the new design economically preferable to the current
unit?
E[NPW] = - $8600 + 0.25 ($3470) (P/A,18%,6)
+ 0.4 ($2920) (P/A,18%,6)
+ 0.25 ($2310) (P/A,18%,6)
+ 0.10 ($1560)(P/A,18%,6)
= $1086
Example 9 (cont’d)
A = $3,470
NPW = $3,538
A = $2,920
NPW = $1,614
A = $2,310
NPW = -$520
25%
$1,086
40%
25%
New Design
10%
Current Design
$0
A = $1,560
NPW = -$3,143
Example 9 (cont’d)

Optimal decision based on perfect information
Reliability Probability
Decision with Perfect
Information
Decision Outcome
Excellent (E)
0.25
New
$3588
Good (G)
0.40
New
$1614
Standard (S)
0.25
Current
$0
Poor (P)
0.10
Current
$0
0.25($3588)+0.4($1614)=$1530

Expected Value of Perfect Information =
$1530 - $1086 = $444
Example 9 (cont’d)

Management is confident that data from an additional
comprehensive test will show whether future operational
performance will be favorable (excellent or good
reliability) or not favorable (standard or poor reliability).
The design team develop conditional probability
estimates.
Test Outcome
Favorable (F)
Not Favorable
(NF)
Conditional Probabilities that Test
Outcome is F or NF Given Reliability
E
G
S
P
0.95
0.85
0.30
0.05
0.05
0.15
0.70
0.95
Example 9 (cont’d)

We need to determine the joint probabilities of the
design goal being met at a particular level and a certain
test outcome occurring.
For example,
p(E, F) = p(F|E) p(E) = (0.95)(0.25) = 0.2375
p(E,NF) = p(NF|E) p(E) = (0.05)(0.25) = 0.0125
Test Outcome
Favorable (F)
Not Favorable (NF)
Marginal Prob. p(L)
E
0.2375
0.0125
0.25
Joint Probabilities
G
S
P
0.3400 0.0750 0.0050
0.0600 0.1750 0.0950
0.40
0.25
0.1
Example 9 (cont’d)

The revised probabilities of each outcome are obtained
from the joint probabilities and the marginal
probabilities
For example, when favorable
p(E) = p(E,F)/p(F) = 0.2375/0.6575 = 0.3612
When not favorable
p(E) = p(E,NF)/p(NF) = 0.0125/0.3425 = 0.0365
Example 9 (cont’d)
$1086
No test
Do test
Favorable
Unfavorable
New
Design
Current
Design
New
Design
Current
Design
E: 0.3612
$3538
G: 0.5171
$1614
S: 0.1141
-$520
P: 0.0076
-$3143
E: 0.0365
$3538
G: 0.1752
S: 0.5109
$1614
P: 0.2774
-$3143
$-520
Example 9 (cont’d)
$1086
$2029
No test
Do test
$2029
$2029 New
Design
Favorable
Current
Design
$0
-$726
Unfavorable $0
New
Design
Current
Design
$0