12 Uncertainty in Future Events

7/30/2019 12 Uncertainty in Future Events

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Kuliah Ekonomi Teknik Kimia

JTK FT UGM 2012


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Introduction

Probability estimation assumed that allvalues of costs, expenses, revenues,economic lives, and acceptable rates of

return were known with certainty. Sensitivity analysis is used to determine

the effect of technical and economicparameters on the profitability of a

project. The potential error of eachparametric variable is examined, as wellas its effect on the project.


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Sensitivity analysis

Break even analysis

The Strauss Plot

Relative profitability plot

Tornado plot


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Break even analysis


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Break-even plot for

pricing alternatives.


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A realistic break-even plot.


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The Strauss Plot


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Example: the Strauss plot


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TheStrauss

plot


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Relative probability plot


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Tornado plot


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Discussion of SensitivityAnalyses Only one variable at a time can be studied.

Interrelated variables such as fixed capitalinvestment, maintenance, and other

investment items in operating expensesand their effect upon one another cannotbe represented correctly.

Synergistic effects particularly with respect

to marketing variables such as salevolume, selling price, market share cannotbe taken onto account.


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Uncertainty analysis

To estimate the probability of a venture,one needs to take some assumptions.

Each assumption has its own degree of

uncertainty, and when taken togetherthese uncertainties can result in largepotential errors.


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Estimates and Their Use inEconomic AnalysisExample 10-1.Two alternatives are being considered. The best

estimates for the various consequences are asfollows:

A BCost $1000 $2000Net annual benefit $150 $250Useful life, in years 10 10End-of-useful-life-salvage value $100 $400

If interest is 3.5%,which alternative has the better netpresent worth (NPW)?


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Example 10-1.

Alternative ANPW = -1000+ 150(P/A, 3.5%,10) + 100(P/F, 3.5%,10)

= -1000 + 150(8.317) + 100(0.7089)= -1000 + 1248+ 71

=+$319Alternative BNPW = -2000 + 250(P/A, 3.5%, 10) + 400(P/F, 3.5%,

10)= -2000 + 250(8.317) + 400(0.7089)

= -2000 + 2079 + 284= +$363

Alternative B, with its larger NPW, would be selected.


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Example 10-2.

Suppose that at the end of 10years, the actual salvagevalue forB were $300 instead of the $400bestestimate. If all the other estimates were correct, is Bstill the preferred alternative?

Revised BNPW = -2000 + 250(P/A, 3.5%, 10) + 300(P/F, 3.5%,

10)= -2000 + 250(8.317) + 300(0.7089)= -2000 + 2079 + 213=+$292A is now the preferred alternative.


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A Range of Estimates

an optimistic estimate,

the most likely estimate

a pessimistic estimate


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Example 10-3

A firm is considering an investment. The most likelydata values were found during the feasibility study.

Analyzing past data of similar projects shows thatoptimistic values for the first cost and the annual

benefit are 5% better than most likely values.Pessimistic values are 15% worse. The firm's mostexperienced project analyst has estimated thevalues for the useful life and salvage value.

Optimistic Most Likely Pessimistic

Cost $950 $1000 $1150

Net annual benefit $210 $200 $175

Useful life, in years 12 10 8

Salvage value $100 $0 $0


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Example 10-3

Compute the rate of return for each estimate. If a10% before-tax minimum attractive rate of return isrequired, is the investment justified under all threeestimates? If it is only justified under some

estimates, how can these results be used.


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Example 10-3

Optimistic Estimate

PW of cost = PW of benefit

$950 = 21O(P/A, IRRopt,12) + 100(P/F,IRRopt, 12)

IRRopt = 19.8%

Most Likely Estimate

$1000 = 200(P / A, IRRmostlikely,10)

(P / A, IRRmostlikely,10) = 1000/200 = 5 IRRmostlikely =15.1%

Pessimistic Estimate

$1150 - 170(P/A, IRRpess,8)

(P / A, IRRpess,8) = 1150/170 = 6.76 IRRpess= 3.9%


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Example 10-3

From the calculations we conclude that the rate ofreturn for this investment is most likely to be15.1%, but might range from 3.9%to 19.8%.Theinvestment meets the 10%MARRcriterion for two

of the estimates. These estimates can beconsidered to be scenarios of what may happenwith this project. Since one scenario indicates thatthe project is not attractive, we need to have a

method of weighting the scenarios or consideringhow likely each is.


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Example 10-4

Compute the mean for each parameter:

Mean cost = [950 + 4 x 1000+ 1150]/6 = 1016.7

Mean net annual benefit = [210 + 4 x 200 + 170]/6 = 196.7

Mean useful life = [12 + 4 x 10+ 8]/6 = 10.0

Mean salvage life = 100/6 = 16.7

Compute the mean rate of return:

PW of cost = PW of benefit

$1016.7 = 196.7(P/A, IRRbeta,10) + 16.7(P/F,IRRbeta, 10)

IRRbeta= 14.2%


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Probability


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Example 10-5

What are the probability distributions for the annualbenefit and life for the following project? Theannual benefit's most likely value is $8000 with aprobability of 60%. There is a 30% probability that

it will be $5000, and the highest value that is likelyis $10,000. A life of 6 years is twice as likely as alife of 9 years.


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Example 10-5

Probabilities are given for only two of the possibleoutcomes for the annual benefit. The third value isfound using the fact that the probabilities for thethree outcomes must sum to 1,

1 = P(Benefit is $5000) + P(Benefit is $8000) +P(Benefit is $10,000)

P(Benefit is $10,000) = 1 - 0.6 - 0.3 = 0.1

The probability distribution can then be summarizedin a table.Annual benefit $5000 $8000 $10,000

Probability 0.3 0.6 0.1


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Example 10-5

The problem statement tells us:

P(life is 6 years) = 2P(life is 9 years)

P(6) + P(9)= 1

Combining these, we write2P(9) + P(9) = 1

P(9) = 1/3

P(6) = 2/3

The probability distribution for the life is P(6) = 66.7%and P(9) = 33.3%.


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Joint ProbabilityDistributionsThe project described in the previous example has a

first cost of $25,000. The firm uses an interest rateof 10%. Assume that the probability distributionsfor annual benefit and life are unrelated or

statistically independent. Calculate the probabilitydistribution for the PW.


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Example 10-6

Since there are three outcomes for the annual benefit andtwo outcomes for the life, there are six combinations.The first four columns of the following table show thesix combinations of life and annual benefit. Theprobabilities in columns 2 and 4 are multiplied to

calculate the joint probabilities in column 5. Forexample, the probability of a low annual benefit and ashort life is 0.3 x 2/3, which equals 0.2 or 20%.

The PW values include the $25,000 first cost and the

results of each pair of annual benefit and life. Forexample, the PW for the combination of high benefitand long life is:


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Example 10-6

Annual

BenefitProbability Life Probability

Joint

ProbabilityPW

$5,000 30% 6 66.70% 20.00% -3,224

8,000 60 6 66.7 40 9,842

10,000 10 6 66.7 6.7 18,553

5,000 30 9 33.3 10 3,795

8,000 60 9 33.3 20 21,072

10,000 10 9 33.3 3.3 32,590

100.00%


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Probability distribution function for PW


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Expected Value

The first cost of the project in Example 10-5 is $25,000.

Use the expected values for annual benefits and life toestimate the present worth. Use an interest rate of 10%.


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Example 10-7

EVbenefit = 5000(0.3) + 8000(0.6) + 10,000(0.1) =$7300

EVlife = 6(2/3) + 9(1/3) = 7 years

The PW using these values is

PW(EV) = -25,000 + 7300(P/A, 10%,7) = -25,000+6500(4.868) =$10,536

[Note: This is the present worth of the expectedvalues, PW(EV), not the expected value of thepresent worth, EV(PW). It is an easy value tocalculate that approximates the EV(PW), whichwill be computed using the joint probability


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Example 10-8

Use the probability distributionfunction of thePWthatwas derivedin Example 10-6 to calculate theEV(PW). Does this indicate an attractive project?

The table from Example 10-6 can be reused withone additional column for the weighted values ofthe PW (= PW x probability). Then, the expectedvalue of the PW is calculated by summing the

column of present worth values that have beenweighted by their probabilities.


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Example 10-8

Annual

Benefit

Probabilit

yLife

Probabilit

y

Joint

Probabilit

y

PW

PW xJoint

Probability

$5,000 30% 6 66.70% 20.00% -3,224 -$ 645

8,000 60 6 66.7 40 9,842 3,937

10,000 10 6 66.7 6.7 18,553 1,237

5,000 30 9 33.3 10 3,795 380

8,000 60 9 33.3 20 21,072 4,214

10,000 10 9 33.3 3.3 32,590 1,086100.00% EV(PW) = $ 10,209


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Example 10-8

With an expected PW of $10,209, this is an attractiveproject. While there is a 20% chance of a negative PW,the possible positive outcomes are larger and morelikely. Having analyzed the project under uncertainty,

we are much more knowledgeable about the potentialresult of the decision to proceed.

The 10,209 value is more accurate than the approximatevalue calculated in Example10-7. The values differbecause PW is a nonlinear function of the life. The

more accurate value of $10,209 is lower because theannual benefit values for the longer life are discountedby 1/(1 + i) for more year.


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Example 10-9

A dam is being considered to reduce river flooding.But if a dam is built, what height should it be?Increasing the dam's height will (1) reduce aflood's probability, (2) reduce the damage when

floods occur, and (3) cost more. Which dam heightminimizes the expected total annual cost? Thestate uses an interest rate of 5% for floodprotection projects, and all the dams should last

50 years.


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Example 10-9

Dam

Height (ft)First Cost

Annual

P (flood) x Height

Damages If

Flood Occurs

No dam $0 0.25 $800,000

20 700,000 0.05 500,000

30 800,000 0.01 300,000

40 900,000 0.002 200,000


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Example 10-9

The easiest way to solve this problem is to choosethe dam height with the lowest equivalent uniformannual cost (EUAC). Calculating the EUAC of thefirst cost requires multiplying the first cost by (A/P,

5%, 50). For example, for the dam 20ft high this is700,000(A/P,5%, 50)=$38,344.

Then the EUAC of the first cost and the expectedannual flood damage are added together to find

the total EUAC for each height. The 30 ft dam issomewhat cheaper than the 40 ft dam.


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Example 10-9

Dam height (ft)EUAC of first

cost

Expected

annual flood

damages

Total Expected

EUAC

No dam $0 $200,000 $200,000

20 38,344 25,000 63,344

30 43,821 3000 46,821

40 49,299 400 49,699


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Risk

Risk can be thought of as the chance of getting anoutcome other than the expected value with anemphasis on something negative.

One common measure of risk is the probability of

a loss (see Example 10-6).

The other common measureis the standarddeviation (a), which measures the dispersion ofoutcomes about the expected value


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Example 10-13

Using the probability distribution for thePW from Example 10-6, calculate thePW's standard deviation.


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Example 10-13

Annual

Benefit

Probab

ilityLife

Probabilit

y

JointProbabil

ity

PWPW x

Probability

PW2 xProbability

$5,000 30% 6 66.70% 20.00% -3,224 -$ 645 $ 2,079,480

8,000 60 6 66.7 40 9,842 3,937 38,747,954

10,000 10 6 66.7 6.7 18,553 1,237 22,950,061

5,000 30 9 33.3 10 3,795 380 1,442,100

8,000 60 9 33.3 20 21,072 4,214 88,797,408

10,000 10 9 33.3 3.3 32,590 1,086 35,392,740

100.00% 10,209 189,409,745


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Risk Versus Return

Example 10-14

A large firm is discontinuing an older product, so somefacilities are becoming available for other uses. Thefollowing table summarizes eight new projects that

would use the facilities. Considering expected returnand risk, which projects are good candidates? The firmbelieves it can earn 4% on a risk-free investmentinggovernment securities (labeed as project F).


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Example 10-14

Project IRR Standard Deviation1 13.10% 6.50%

2 12 3.9

3 7.5 1.5

4 6.5 3.55 9.4 8

6 16.3 10

7 15.1 7

8 15.3 9.4F 4 0


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Example 10-14

Answering the question is far easier ifwe use Figure 10-7.Since a largerexpected return is better, we want to

select projects that are as "high up" aspossible. Since a lower risk is better, wewant to select project that are as "farleft" as possible. The graph lets usexamine the trade-off of accepting morerisk for a higher return.


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Example 10-14

dominated projects.

efficient frontier.


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Simulation

Example 10-15 ShipM4U is considering installing a new, more

accurate scale, which will reduce the error incomputing postage charges and save $250 a year.

The scale's useful life is believed to be uniformlydistributed over 12, 13, 14, 15,and 16 years. Theinitial cost of the scale is estimated to be normallydistributed with a mean of $1500 and a standarddeviation of $150.

Use Excel to simulate 25 random samples of theproblem and compute the rate of return for eachsample. Construct a graph of rate of return versusfrequency of occurrence


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Example 10-15

250 Annual savings

Life First Cost

Min 12 1500 Mean

Max 16 150 Std dev

Iteration IRR

1 15 1510 14%

2 13 1217 18%

3 12 1391 14%

4 14 1297 17%

5 13 1621 12%

25 14 1408 15%


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Example 10-15

0

1

1

2

2

3

3

4

4

5

8% 9% 10% 11% 12% 13% 14% 15% 16% 17% 18% 19%

Frequency

IRR

IRR vs Frequency


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Example 10-16

Consider the scale described inExample 10-15. Generate 10,000iterations and construct a frequency

distribution for the scale's rate of return.


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10-16

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12 Uncertainty in Future Events