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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 long-run 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 VariablesProbability: Pr{X=xi} = p(xi) for i = 1, ..., L
where p(xi)>0 and i p(xi)=1
Continuous Random Variables
Probability: where
Expected Value: E[X] = i xi p(xi) or
Variance: Var[X] = i (xi - E[X]) 2p(xi) or
Pr{c ≤X≤d} = f (x)dxc
d∫ f (x)dx =1−∞
∞∫
€
xf (x)dx−∞
∞
∫
€
(x − E(x))2
−∞
∞
∫ 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 Useful Life p(N)Capital Investment -$521,000 Year (N)
Annual Savings $48,600 12 0.1Increase Annual Rev. $31,000 13 0.2
14 0.315 0.216 0.117 0.0518 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) (2) (3) (4) = (2)x (3) (5) =(3)x(2)2
N NPW(N) p(N) P(N)[NPW(N)] p(N)x NPW(N)2
12 -$27,926 0.1 -$2,793 ($2) 77.986 x 106
13 -$9,689 0.2 -$1,938 ($2) 18.776 x 106
14 $6,605 0.3 $1,982 ($2) 13.089 x 106
15 $21,148 0.2 $4,230 ($2) 89.448 x 106
16 $34,130 0.1 $3,413 ($2) 116.486 x 106
17 $45,720 0.05 $2,286 ($2) 104.516 x 106
18 $56,076 0.05 $2,804 ($2) 157.226 x 106
Sum $9,984 ($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 Net Cash FlowYear, k Mean Standard Dev.
0 -$7,000 01 $3,500 $6002 $3,000 $5003 $2,800 $400
Example 8 (cont’d)
0
1
-15000 -10000 -5000 0 5000 10000 15000
The investment is known.
Year 0
Example 8 (cont’d) The cash flows for the years 1, 2 and 3 are not known.
Cash Flows
0.0
0.1
0 1000 2000 3000 4000 5000 6000 7000
Year 1
Year 2
Year 3
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
DecisionInvest in new product line
Status Quo
Salesgood
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 Probability Annual SavingsExcellent (E) 0.25 $3,470Good (G) 0.40 $2,920
Standard (S) 0.25 $2,310Poor (P) 0.10 $1,560
Use MARR = 18%. Life = 6 years. Salvage = 0.
Example 9 (cont’d)
New Design
Current Design
25%
40%
25%
10%A = $1,560
A = $2,310
A = $2,920
A = $3,470 NPW = $3,538
NPW = $1,614
NPW = -$520
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)
$1,086
New Design
Current Design
25%
40%
25%
10%A = $1,560
A = $2,310
A = $2,920
A = $3,470 NPW = $3,538
NPW = $1,614
NPW = -$520
NPW = -$3,143
$0
Example 9 (cont’d)
Optimal decision based on perfect information
Expected Value of Perfect Information =
$1530 - $1086 = $444
Reliability Probability Decision with PerfectInformation
Decision OutcomeExcellent (E) 0.25 New $3588
Good (G) 0.40 New $1614Standard (S) 0.25 Current $0
Poor (P) 0.10 Current $00.25($3588)+0.4($1614)=$1530
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 Conditional Probabilities that TestOutcome is F or NF Given Reliability
E G S PFavorable (F) 0.95 0.85 0.30 0.05Not Favorable
(NF)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 Joint ProbabilitiesE G S P
Favorable (F) 0.2375 0.3400 0.0750 0.0050Not Favorable (NF) 0.0125 0.0600 0.1750 0.0950Marginal Prob. p(L) 0.25 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
Current Design
New Design
E: 0.3612
New Design
Current Design
$3538
G: 0.5171 $1614
S: 0.1141 -$520
P: 0.0076 -$3143
E: 0.0365 $3538
G: 0.1752 $1614
S: 0.5109 $-520
P: 0.2774 -$3143
Example 9 (cont’d)
$2029
-$726
$0
$2029
$1086
No test
Do test
Favorable
Unfavorable
Current Design
New Design
$2029
$0
New Design
Current Design
$0