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Uncertainty in Future
Events
Chapter 10
Mechanical Engineering 431
Engineering Economics
Chapter 10 …
 Uses the expected values of variables to evaluate
projects and make decisions.
 Describes uncertain outcomes using probability
distributions.
 Combines the probability distributions of single
variables in joint probability distributions.
 Measures/assesses risk in decision making.
 Uses simulations for decision making.
2009-2010 Term 2
MECH 431 — Engineering Economics
10-2
Precise estimates are still
estimates
All estimates are inherently uncertain.
 Far-term estimates are almost always more uncertain than nearterm estimates.
Minor changes in any estimate(s) may alter the results of
an economic analysis.
Using breakeven and sensitivity analysis yields an
understanding of how changes in variables will affect the
economic analysis.
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10-3
Decision-making and uncertainty
of future outcomes
In the left box, one cash flow in Project B is uncertain
and NPVA > NPVB (i = 10%).
In right box, the cash flow estimate has changed and
now NPVA < NPVB.
Year
Project A
Project B
Year
Project A
Project B
0
$1000
$2000
0
$1000
$2000
1
$400
$700
1
$400
$700
2
$400
$700
2
$400
$700
3
$400
$700
3
$400
$700
4
$400
$700
4
$400
$800
$267.95
$218.91
$267.95
$287.21
NPV
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NPV
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10-4
Decision-making and uncertainty
of future outcomes …
It is good practice to examine the effects of variability in
the estimates on the outcomes.
 By how much and in what direction will a measure of merit
(e.g., NPV, EACF, IRR) be affected by variability in the
estimates?
But, this does not take the inherent variability of
parameters into account in an economic analysis.
We need to consider a range of estimates.
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MECH 431 — Engineering Economics
10-5
A range of estimates
Usually, we consider three scenarios:
 optimistic
 most likely
 pessimistic
Compute the internal rate of return for each scenario.
Compare each IRR to the MARR.
What if IRR < MARR for one or more scenarios?
 Are scenarios equally important?
 We need a weighting scheme.
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A range of estimates …
Assigning probabilities
 Assign a probability to each of the three scenarios.
 The largest probability is assigned to the most likely scenario.
The optimistic and pessimistic scenarios may have equal or
unequal probabilities.
There are two possible approaches:
 Calculate the average of the measure of merit (e.g., IRR) for all
scenarios.
 Calculate the measure of merit (e.g., IRR) from the averages of
each parameter (annual benefit and cost; first cost; salvage
value …).
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A range of estimates …
Use the example below to:
 explore the impact on the IRR of uncertainty in future results;
 explore the effect on the IRR of changing the variability of the
parameters.
Alternative: Optimistic Most likely Pessimistic Average value
Cost
-$950
-$1,000
-$1,150
-$1,026.50
Net annual benefit
$210
$200
$180
$197.10
Useful life, in years
12
10
8
9.9
Salvage value
$150
$100
$50
$97.50
IRR =
19.97%
15.72%
5.98%
14.564%
Probability:
19%
57%
24%
Average IRR =
14.192%
Std Dev IRR =
4.885%
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MECH 431 — Engineering Economics
10-8
Probability and risk
Probabilities of future events can be based on theory,
empirical data, judgement, or a combination, e.g.
 games of chance,
 weather and climate data, or
 expert judgement on events.
Most data has some level of uncertainty.
 Small uncertainties are often ignored.
Variables can be known with certainty (deterministic) or
with uncertainty (random or stochastic).
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MECH 431 — Engineering Economics
10-9
Probability and risk …
0 ≤ probability ≤ 1
The sum of probabilities for all possible outcomes = 1 or
100%.
It is usual in engineering economics to use between two
and five outcomes with discrete probabilities.
 Expert judgement limits the number of outcomes.
 Each additional outcome requires more analysis.
Probability can be considered as the long-run relative
frequency of an outcome’s occurrence.
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MECH 431 — Engineering Economics
10-10
Joint probability distributions
Random variables are assumed to be statistically
independent.
 For instance, the project lifetime and the annual benefit are
assumed to be independent (unrelated).
Project criteria, e.g., NPV, IRR, depend on the
probability distributions of input variables.
We need to determine the joint probability distributions
of different combinations of input parameters.
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MECH 431 — Engineering Economics
10-11
Joint probability distributions …
If A and B are independent, P(A and B) = P(A)  P(B).
 “A and B” means that A and B occur simultaneously.
Suppose there are three values for the annual benefit and
two values for the life. This leads to six possible
combinations that represent the full set of outcomes and
probabilities.
Joint probability distributions are burdensome to
construct when there is a large number of variables or
outcomes.
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MECH 431 — Engineering Economics
10-12
Expected value
The expected value of a random variable, X, is its mean
or average based on the values of the variable and their
probabilities. We can also determine the variance of X.
 Expected value: E[X] = μ = (pj)(xj) for all j.
 Variance: Var[X] = ² = (pj)(E[X] – (xj))² for all j.
 p = probability; x = discrete value of the variable.
The expected value and variance can be determined
when two or more possible outcomes and their
associated probabilities are known.
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MECH 431 — Engineering Economics
10-13
Expected value …
Example: a firm is considering an investment that has
annual net revenue and lifetime with the probability
distributions shown below. Find E(NPV) and SD(NPV).
Project Cost:
MARR:
Net Revenue
$10,000
$12,800
$15,000
Joint PD
$10,000
$12,800
$15,000
E(NPV)=
2009-2010 Term 2
$45,000
12%
Prob
Lifetime Prob
25%
5 65%
55%
7 35%
20%
100%
100%
5
7
NPVs
5
7
16.25%
8.75% 25% $10,000 -$8,952.24
$637.57
35.75%
19.25% 55% $12,800 $1,141.14 $13,416.08
13.00%
7.00% 20% $15,000 $9,071.64 $23,456.35
65%
35% 100%
$4,412.86
Std Dev(NPV)= $8,826.76
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10-14
Decision analysis
A decision tree is a logical structure of a problem in
terms of the sequence of decisions and outcomes of
chance events.
 e.g. demand for a new product will depend on economic factors
or conditions (“states of nature”).
When decisions depend on the outcomes of random
events, decision makers are forced to anticipate what
those outcomes might be as part of the analysis process.
 This analysis is suited to decisions and events that have a
natural sequence in time or space.
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Decision analysis …
A decision tree grows from left to right and usually
begins with a decision node.


Represents a decision required by the decision maker.
Branches extending from a decision node represent
decision options available to the decision maker.

A chance node represents events for which outcomes are
uncertain.

Branches extending from a chance node represent
possible outcome factors, sometimes called “states of nature”.
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Decision analysis …
The decision analysis procedure:
1. Develop the decision tree.
2. Execute the rollback procedure on the decision tree from right
to left.
 Compute the expected value (EV) of each possible outcome at each
chance node.
 At each decision node, select the option with best EV.
 Continue the rollback process until the leftmost node is reached.
3. Select the optimal decision and note its expected value at the
final node.
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MECH 431 — Engineering Economics
10-17
Decision analysis …
Example: the engineers of a firm must decide whether
the company will build a new product or have it built
under contract (“build or buy” decision).
 The initial cost of building the product is $800,000 and net
sales will be $200,000 in the first year.
 The initial cost of buying the product is $175,000 and net sales
will be $120,000 in the first year.
After one year of operation, the company has the choice
either to continue with the product, expand operations, or
abandon the product.
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Decision analysis …
The expand/continue/abandon decision will depend on
whether the economy is good or bad. The probability of
a good economy in one year is 60%.
If they build the product, in one year:
 the cost of expanding will be $450,000;
 they could receive $600,000 if they abandon;
 if they continue, net annual sales will be $300,000 if the
economy is good and $150,000 if it is bad;
 if they expand, net annual sales will be $400,000 if the
economy is good and $180,000 if it is bad.
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Decision analysis …
If they buy the product, in one year:
 the cost of expanding will be $80,000;
 they will receive $0 if they abandon;
 if they continue, net annual sales will be $160,000 if the
economy is good and $96,000 if it is bad;
 if they expand, net annual sales will be $185,000 if the
economy is good and $110,000 if it is bad.
They use a MARR of 14% for decisions like this and the
expected lifetime is ten years. Use a decision analysis to
make a recommendation to the engineers.
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MECH 431 — Engineering Economics
10-20
Decision analysis …
Build:
Buy:
Start net sales:
Good net sales:
Bad net sales:
MARR:
Lifetime (yrs):
Build=
Buy=
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Cost (t=0) Expand (t=1) Abandon (t=1)
$800,000
$450,000
$600,000
$175,000
$80,000
$0
Build
Buy
$200,000
$120,000
Build/Continue Build/Expand Buy/Continue Buy/Expand
Probability
$300,000
$400,000
$160,000
$185,000
60%
$150,000
$180,000
$96,000
$110,000
40%
14%
10
at t= 0
at t= 1
at t= 1
Good= $1,728,548.73
Expand= $1,528,548.73
Continue= $1,483,911.55
Abandon=
$600,000.00
Bad=
$941,955.78
Expand=
$440,346.93
Continue=
$741,955.78
Abandon=
$600,000.00
$440,273.29
Decision is to buy.
$536,392.94
Good=
$955,078.79
Expand=
$835,078.79
Continue=
$791,419.49
Abandon=
$0.00
Bad=
$594,851.70
Expand=
$464,100.90
Continue=
$474,851.70
Abandon=
$0.00
MECH 431 — Engineering Economics
10-21
Simulation
When any of the project components is a random
variable, the outcome of the project, e.g. the NPV, is also
a random variable.
If we want to assess a project with uncertain parameters,
we estimate the probability distribution of the outcome
using the relative frequency approach.
We can do this by repeatedly sampling from the
distributions of the project’s parameters. Spreadsheets
are helpful in this procedure.
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Simulation …
Monte Carlo simulation procedure:
1. Formulate the model for determining the project outcome from
the project components.
2. Determine the probability distributions of all project
components that are random variables.
3. Use a random number generator to produce values for the
project components that are random variables and calculate the
project outcome using the model.
4. Repeat step 3 until a large enough sample has been taken (250
is usually a sufficient number).
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Simulation …
Monte Carlo simulation procedure (cont’d):
5. Produce a frequency distribution and a histogram to estimate
the probability distribution of the project outcome.
6. Produce summary statistics of the project outcome, e.g. mean,
median, standard deviation, range, minimum, maximum, …
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Simulation …
Example: A machine initially costs $19,000. After a
lifetime of five years, its salvage value is expected to be
$4000. The annual operating profit of the machine has
the probability distribution shown in the table below.
Perform a simulation of the NPV of the project and make
a recommendation for a cost of capital of 13 percent.
Annual profit $3500 $5000 $6000 $8000
0.25 0.50 0.20 0.05
Probability
 See the spreadsheet on the second page following.
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MECH 431 — Engineering Economics
10-25
Simulation …
Example: A machine’s initial cost is uniformly
distributed between $18,000 and $20,000 and its salvage
value is expected to be $4000. The annual profit of
operating the machine has a normal distribution with a
mean of $6000 and a standard deviation of $2500. The
machine’s lifetime is five years. Use simulation to
analyze the project for a cost of capital of 13 percent.
 A spreadsheet for both discrete and continuous distributions is
on the following page (click on the tabs when it is open).
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Simulation …
Simulation Example (Discrete)
Annual profit
$3,500
$5,000
$6,000
$8,000
Initial cost:
Salvage value:
Cost of capital:
NPV mean=
NPV median=
NPV std. dev.=
NPV maximum=
NPV minimum=
Prob(NPV < 0)=
Probability
0.25
0.50
0.20
0.05
$19,000
$4,000
13%
$760.91
$730.36
$1,810.38
$6,033.97
-$3,704.51
35.60%
Net Present Value
$757.20
-$2,664.94
$4,888.26
$738.55
$601.08
$1,102.23
-$570.24
-$419.21
$757.20
$5,384.29
-$949.36
$2,233.39
$122.81
Year 1
5000
3500
8000
5000
6000
5000
3500
5000
5000
8000
6000
5000
3500
NPV Frequency Distribution for Machine
60
50
40
$2,171 (PV)
30
20
10
0
-$3,500 -$2,500 -$1,500 -$500
$500
$1,500 $2,500 $3,500 $4,500 $5,500 $6,500
NPV
Annual profit
2009-2010 Term 2
Year 2
5000
3500
6000
3500
5000
5000
5000
6000
5000
8000
3500
6000
5000
Year 3
5000
5000
6000
5000
6000
8000
5000
3500
5000
5000
3500
6000
6000
Year 4
5000
3500
5000
6000
3500
3500
5000
3500
5000
3500
3500
5000
5000
Year 5
5000
5000
5000
6000
3500
3500
5000
5000
5000
6000
6000
5000
5000
MECH 431 — Engineering Economics
Lowest=
-4000
Highest=
7000
Lo limit Hi Limit
NPV
-4000
-3000 -$3,500
-3000
-2000 -$2,500
-2000
-1000 -$1,500
-1000
0
-$500
0
1000
$500
1000
2000 $1,500
2000
3000 $2,500
3000
4000 $3,500
4000
5000 $4,500
5000
6000 $5,500
6000
7000 $6,500
Freq
2
13
24
50
52
50
31
18
6
3
1
250
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Suggested problems — Chapter 10
10-12, 10-13, 10-26, 10-31, 10-32, 10-34 (for problem
10-13 only), 10-37, 10-38, 10-39, 10-42 (for problem
10-13 only).
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MECH 431 — Engineering Economics
10-28