Evaluating Projects under

U ncerta inty

March 17, 2004

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• Project risk = possible variation in cash flow s

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Commonly used measure of project risk is the variability of the return

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M ethods of dealing w ith

uncertainty in project evaluations

• Sensitivity analysis

• Risk adjusted MARR

• Probability trees

• Monte Carlo simulations

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Exampl e showin g ho w us e o f ‘ risk-adjusted ’ MARR s can

lea d t o th e wron g decision (from Sullivan et al, Engineering Economy , (11th ed), p. 445)

The Atlas Corporation is considering two alternatives, both affected by uncertainty to different degrees, for increasing the recovery of a precious metal from its smelting process. The firm’s MARR f or its ri sk-free investments is 10% per year.

Alternative

End-of-year, k

P

Q

Because of te chnical considerations, Alternative P is thought to be m or e uncertain than

Alternative Q. Therefore, according to the Atlas Corporation’s “Engineering Economy Handbook”, the risk-adjusted MARR applied to P will be 20% per year and the risk-adjusted MARR f or Q has been set at 17% per year. Which alternative should be recommended?

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Solution

At the risk-free MARR, bot h alternative s hav e th e s am e PW of $39, 659. What to do?

All else equal, choose Q, because it is less uncertain (hence less riskier) than P.

But now, do a PW analysis, using Atlas Corporation’s prescribed ri sk-adjusted MARRs for the two options:

PW P = -160 ,000 + 120,000 (P/F, 20%, 1) + 60,000 (P/F , 20%, 2) + 60,000 (P/ F, 20% , 4)

= $ 10,602

PW Q = -160 ,000 + 20,827 (P/F,17%,1) + 60,000 (P/F,17%, 2) + 120,000 (P/F,17%,3) +

+ 60,000 (P/F,17%, 4)

= $ 8575

Hence according to this method we would choose P. In other words, using the risk-adjusted MARRs makes the more uncertain project, P, look MORE attractive than Q!!

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Example of probabilistic analysis

Consider th e simple decision whether to make a new investment, when there is uncertainty about the duration of demand.

I 0 = 6800

R = 7000/yr

M = 2000 + (n-1)1000 i = 20%

The probability of demand fo r the service provided by this asset persisting for:

Question: Should this investment be made?

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Solution (contd.)

From Sullivan et al, p.447

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Example (continued)

Question: Should this investment be made?

LAC  I o ( A / P ,20%, N )  I N ( A / F ,20 %, N )  2000  1000 ( A / G , 20 %, N )

1yr p =0.1 PW = -2132 px PW = -213.2

2yr p =0.2 PW=-828 p x PW = -166

Yes

3yr p =0.3 PW = 6 p x PW = 2

Expected Value

= -$236

4 yr p = 0.4 PW = 352 p x P W = 141

No

0

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The Problem of Investment Timing

Exampl e 1 : Uncertaint y ove r Prices Widget factory

Initial cost = I = $1600 Annual operating cost = 0

Production rate = 1 widget per year Current widget p rice = $200

Price next year (and forever after):

$300 with probability 0.5

$100 with probability 0.5

t =0

t =1

t =2

0.5

P 1 = $300

P 2 = $300

P o = $200

0.5

P 1 = $100

P 2 = 100

Assume interest rate of 10%/yr

Question : Should the f irm invest now, or should it w ait for 1 year and see whether the price of widgets goes up or down?

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Example 1 (contd.)

Since expected price of widgets is always $200, th e NPV of an investment now is

NPV   1600 





0

200

( 1  0 . 1 )

t

  1600  2200  $600

Thus it might seem sensible to go ahead.

But what if we wait until next year? Then we would decide to invest only if the price goes up. If the price falls, i t would make no sense to invest.

The NPV i n this case is given by:

  1600



NPV  (0.5 ) 

 1.1 ( 1  0. 1 )

 t  1  



300 ˘   1600 3300 ˘ 850

t



 0.5    $ 773



1 . 1 1.1 1 . 1





So if we wait a year before de ciding whether to invest in the factory, the project NPV today is $773. Clearl y i t i s bette r t o wai t tha n t o inves t righ t away .

If we had n o choice, and either ha d to invest now or never, we would obviously choose to invest, since this would have a positive NPV of $600. But the flexibility to choose to postpone the de cision and invest next year if the market price is right is w orth something. Specifically, it is worth 77 3-600 = $173.

In other words, we should be willing to p ay up to $173 more for an investment o pportunity that is flexible than one that only allows us the choice of investing now or never. This is the value of flexibility in this case.

Still another way of saying this is that there is an opportunit y cost of investing now, rather than waiting.

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Example 2 -- Uncertainty over costs

We can consider two different kinds of cost uncertainty

a. Suppose that I = $1600 today, but th at next year it will increase to $2400 or decrease to $800, each with a probability of 0.5. ( The cause of this uncertainty could be stochastic fluctuations in input prices, or regulatory uncertainties.) The interest rate is again 10% per year

Question: Should we invest today or wait to decide until next year? As before, if we invest today the NPV is given by:

NPV   1600 





0

200

( 1  0 . 1 )

t

  1600  2200  $600

If we wait until next year, it will be sensible to invest only if the investment cost falls to

$800, which happens with a probability of 0.5. In th is case the NPV is given by:

NPV  (0.5 ) 

   80 0 200 ˘   800 2200 ˘



 ( 1  0. 1 ) ( 1  0 . 1 )





t  1 

t

 0.5   $636

 

1.1 1.1

 

so once again i t is better to wait than to invest immediately.

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Example 2 -- contd.

a. Suppose, alternatively, that there are uncertainties over how much it is going to cost to com plete the project that can only be resolved by actually doing it. You don’t know for certain how much it is going to cost until you complete it. Let’s say that th is uncertainty takes the following form: To build the widget fa ctory you first have to spend $1000, and that there is a 50% probability that the f actory will then be complete, and a 50% probability that you will have to spend another $3000 to complete it. Assume that the widget price remains constant at $200, an d that the interest rate is 10%.

At first blush, the investment would make no sense. The expected cost of the factory is: 1000 + 0.5.(3000) = 2500.

 200

And since the value of the factory =



t  0

1.1

t

 2200

, we might conclude that it makes no

sense to proceed. But this ignores the add itional information that is generated b y completing the f irst phase of the project, and that we can choose to abandon t he pr oject if completion requires an extra $3000. The true NPV is:

NPV   1000  (0 . 5 )  1 . 1 t   $ 100

 200

t  0

Since the NPV is positive, one should invest in the first stage of the project.

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