dealing with uncertainty

28
Dealing with Uncertainty Probabilistic Risk Analysis

Upload: breanna-blevins

Post on 31-Dec-2015

38 views

Category:

Documents


4 download

DESCRIPTION

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. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Dealing with Uncertainty

Dealing with Uncertainty

Probabilistic Risk Analysis

Page 2: Dealing with Uncertainty

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.

Page 3: Dealing with Uncertainty

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.

Page 4: Dealing with Uncertainty

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

Page 5: Dealing with Uncertainty

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−∞

Page 6: Dealing with Uncertainty

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}

Page 7: Dealing with Uncertainty

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

Page 8: Dealing with Uncertainty

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.

Page 9: Dealing with Uncertainty

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

Page 10: Dealing with Uncertainty

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] !

Page 11: Dealing with Uncertainty

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

Page 12: Dealing with Uncertainty

Example 8 (cont’d)

0

1

-15000 -10000 -5000 0 5000 10000 15000

The investment is known.

Year 0

Page 13: Dealing with Uncertainty

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

Page 14: Dealing with Uncertainty

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

Page 15: Dealing with Uncertainty

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

Page 16: Dealing with Uncertainty

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

Page 17: Dealing with Uncertainty

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

Page 18: Dealing with Uncertainty

Diagramming Example

DecisionInvest in new product line

Status Quo

Salesgood

bad

Page 19: Dealing with Uncertainty

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.

Page 20: Dealing with Uncertainty

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

Page 21: Dealing with Uncertainty

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

Page 22: Dealing with Uncertainty

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

Page 23: Dealing with Uncertainty

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

Page 24: Dealing with Uncertainty

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

Page 25: Dealing with Uncertainty

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

Page 26: Dealing with Uncertainty

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

Page 27: Dealing with Uncertainty

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

Page 28: Dealing with Uncertainty

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