Dealing with
Uncertainty
Uncertainty


Decisions under Certainty vs. Decisions
under Uncertainty (risk)
Sources of Uncertainty
– possible inaccuracy of the estimates
– the type of business involved in relation to the
future health of the economy
– the type of physical plant and equipment
involved
– length of the assumed study period
HKUST, IEEM, Dr. C. J. Su
Nonprobabilistic Methods
for Dealing with Uncertainty
 Breakeven
analysis
 Sensitivity
Analysis
HKUST, IEEM, Dr. C. J. Su
Breakeven Analysis
When the selection among alternatives is
heavily dependent on a single factor, such as
capacity utilization, energy utilization, etc. that
is uncertain.
 A breakeven point for the factor is determined
such that two alternatives are equally desirable.
 Estimating the most likely value of the
uncertain factor then choose between the
alternatives by and comparing this estimate to
the breakeven value.

HKUST, IEEM, Dr. C. J. Su
Breakeven Analysis
EWA (y) = EWB (y)
where EWA and EWB = an equivalent worth value
for the net cash flow of Alternatives A & B .
y = a common factor of interest affecting the
equivalent worth of Alternative A and B .
The common factor y can be :
–
–
–
–
Annual revenue and expenses
Rate of return
Market (or salvage) value
Equipment life
HKUST, IEEM, Dr. C. J. Su
Example

Two alternative electric motors that provide 100 hp output
are under consideration. An Alpha motor can be purchased
for $12,500 and has an efficiency of 74%, an estimated
useful life of 10 years, and maintenance expenses of $500
/year. A Beta motor will cost $16,000 and has an efficiency
of 92%, a life of 10 years, and annual maintenance expenses
of $250. Annual taxes and insurance expenses on either
motor will be 1.5 % of the investment. If the MARR is
15%, how many hours per year would the motors have to
be operated at full load for the annual expenses to be
equal? Assume that market values for both are negligible
and that electricity costs $0.05 per kilowatt-hour
HKUST, IEEM, Dr. C. J. Su
Alpha Machine
1 hp = 0.746 kW. If X = # of hours of operation /year
Capital recovery amount (depreciation and minimum profit):
-12,500(A/P,15%,10) = -12,500(0.1993) = -2,490 / year
Operating expense for power:
- (100)(0.746) X ($0.05) /0.74 = - 5.04X per year
Maintenance expense: - 500 /year
Taxes and insurance: -12,500(0.015) = -187 /year
Beta Machine
Capital recovery amount:
- 16,000(A/P,15%,10) = - 16,000(0.1993) = - 3,190 /year
Operating expense for power:
- (100) (0.746) X($0.05) /0.92 = - 4.05X /year
Maintenance expense: - 250 /year
Taxes and insurance:
-16,000 (0.015) = - 240 /year
At the breakeven point, AW (X) = AW (X)
-2,490 - 5.04X - 500 - 187 = - 3,190 - 4.05X- 250 - 240
-5.04X - 3,177 = - 4.05X- 3,680, X = 508 hours/year
Graphical Solution
At the breakeven point, AW (X) = AW (X)
-2,490 - 5.04X - 500 - 187 = - 3,190 - 4.05X- 250 - 240
AW (X) = -5.04X - 3,177 & AW (X) = - 4.05X- 3,680
HKUST, IEEM, Dr. C. J. Su
Example
The Universal Postal Service is considering the possibility of
putting wind deflectors on their long-haul tractors. Three
types of deflectors, with the following characteristics, are
being considered (MARR = 10% per year):
Windshear Blowbv
Air-vantage
Capital investment
- 1,000
- 400
- 1,200
Drag reduction
20%
10%
25%
Maintenance/year
- 10
-5
-5
Useful life
10 years
10 years
5 years
If 5% in drag reduction means 2% in fuel savings per mile, how
many miles do tractors have to be driven per year before the
Windshear deflector is favored over the other deflectors? Over
what range of miles driven per year is Air-vantage the best
choice? (Fuel cost is $1.00 per gallon and average fuel
consumption is 5 miles per gallon without the deflectors). State
any assumptions you make
HKUST, IEEM, Dr. C. J. Su
Assumption: Repeatability (the deflector will
continuously)
Let X be the mileage driven per year then
be
Windshear: Total 8% reduction in fuel, operating cost/year :
- (X/5) (0.92)(1.00) = -0.184 X /year
AW = - 1,000(A/P,10%,10) - 10 - 0.184 X = - 172.75 - 0.184 X
Blowby: Total 4% reduction in fuel, operating cost/year :
- (X/5) (0.96)(1.00) = -0.192 X /year
AW = - 400(A/P,10%,10) - 5 - 0.192 X = - 70.1 - 0.192 X
Air-vantage: Total 10% reduction in fuel, operating cost/year :
- (X/5) (0.90)(1.00) = -0.180 X /year
AW = -1200(A/P, 10%, 5) - 5 - 0.18 X = -321.56 - 0.18 X
used
Breakeven points:
Windshear vs. Blowby deflector
- 172.75 - 0.184X = - 70.1 - 0.192X, X = 12,831 miles/year
Windshear vs. Air-vantage deflector
- 172.75 - 0.184X = -321.56 - 0.18 X, X = 37,203 miles/year
Blowby vs. Air-vantage deflector
- 70.1 - 0.192X = -321.56 - 0.18 X, X = 20,955 miles/year
Graphically ...
HKUST, IEEM, Dr. C. J. Su
Example

In planning 2-story office building, the architect has submitted two
designs. The first provides foundation and structural details so that two
additional stories can be added to the required initial two stories at a
later date and without modifications to the original structure. This
building would cost $1,400,000. The 2nd design, without such
provisions, would cost only $1,250,000. If the first plan is adopted, an
additional two stories could be added at a later date at a cost of
$850,000. If the second plan is adopted, however, considerable
strengthening and reconstruction would be required, which would add
$300,000 to the cost of a 2-story addition. Assuming that the building is
expected to be needed for 75 years, by what time would the additional
two stories have to be built to make the adoption of the first design
justified? (The MARR is 10% per year.)
Let T be the time when the additional 2-stories is needed
1st Design
2nd Design
PW cost:
First unit
- 1,400,000
Second unit - 850,000(P/F, 10%, T)
PW(1st design) = PW(2nd Design)
- 1,400,000 - 850,000 (P/F, 10%. T) =
- 1,250,000 - 1,150,000 (P/F, 10%,T)

=> (P/F, 10%. T) = 0.5 => T = 7 years
- 1,250,000
- 1,150,000(P/F, 10%, T)
If the additional space will be required < 7 years then
choose the 1st design (with the foundation);
Else choose the 2nd design (without the foundation)
PW
- 2,250,000
1st Design
- 2,400,000
2nd Design
7 years
T
HKUST, IEEM, Dr. C. J. Su
Sensitivity Analysis
Sensitivity - the relative magnitude of
change in the measure of merit (such as PW)
caused by one or more changes in estimated
study factor values.
 provide information about the potential
impact of uncertainty in selected factor
estimates.
 Its routine use is fundamental to achieving
sound results useful in the decision process.

HKUST, IEEM, Dr. C. J. Su
Example






A machine for which cash flow estimates are given
in the following list is being considered for
immediate installation. Because of the new
technology built into this machine, it is desired to
investigate its PW over a range of  40% in (a)
capital investment, (b) annual net cash flow, (c)
market value, and (d) useful life. Based on these
estimates, how much can the initial investment
increase without making the machine an
unattractive venture? MARR = 10%
Capital investment, I -$11,500
Revenues/yr
5,000
Expenses/yr
-2,000
Market value, MV
1,000
Useful life, N
6 years
HKUST, IEEM, Dr. C. J. Su
Sensitivity to Capital Investment (I)
PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6)
= 2,130
(a) When the capital investment varies by  p% the
PW(10%) = - (1  p%)(11,500) + 3,000(P/A,10%,6)
+ 1,000(P/F,10%,6)
If p% = 0 => PW(10%) = 2,130
If p% = 10%
PW(10%) = - ( 1.1 )( 11,500 ) + 3,000(P/A,10%,6)
+ 1,000(P/F,10%,6)
= - (1.1 )(11,500) + 3,000 (4.3553) + 1,000( 0.5645) = 1060.4
HKUST, IEEM, Dr. C. J. Su
PW (10%)
I
4,000
3,000
2,130
2,000
1,060.4
1,000
-40
-30
-20
-10
% Change
in Parameter
0
10
-1,000
-2,000
-3,000
20
30
40
Sensitivity to
Net Annual CF (A)
PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6)
= 2,130
(a) When the net annual CF varies by  a% the
PW(10%) = - 11,500 + (1  a%) 3,000(P/A,10%,6)
+ 1,000(P/F,10%,6)
If a% = 0 => PW(10%) = 2,130
If a% = 10%
PW(10%) = - 11,500 + ( 1.1 ) 3,000(P/A,10%,6)
+ 1,000(P/F,10%,6)
= - 11,500 + ( 1.1 ) 3,000 (4.3553) + 1,000( 0.5645) = 3886.99
HKUST, IEEM, Dr. C. J. Su
PW (10%)
A 3886.99
4,000
3,000
2,130
2,000
1,000
-40
-30
-20
-10
% Change
in Parameter
0
10
-1,000
-2,000
-3,000
20
30
40
Sensitivity to
Market Value (MV)
PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6)
= 2,130
(a) When the Market Value (MV) varies by  s % the
PW(10%) = - 11,500 + 3,000(P/A,10%,6)
+ (1  s%) 1,000(P/F,10%,6)
If s% = 0 => PW(10%) = 2,130
If s% = 10%
PW(10%) = - 11,500 + 3,000(P/A,10%,6)
+ ( 1.1 ) 1,000(P/F,10%,6)
= - 11,500 + 3,000 (4.3553) + ( 1.1 ) 1,000( 0.5645) = 2,186.85
HKUST, IEEM, Dr. C. J. Su
PW (10%)
4,000
3,000
2,130
MV
Low Slope => Insensitive
2,000
2,186.85
1,000
-40
-30
-20
-10
% Change
in Parameter
0
10
-1,000
-2,000
-3,000
20
30
40
Sensitivity to Useful Life (n)
PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6)
= 2,130
(a) When the Useful Life (n) varies by  n% the
PW(10%) = - 11,500 + 3,000(P/A,10%, 6  n% )
+ 1,000 (P/F,10%, 6  n%)
If n % = 0 => PW(10%) = 2,130
If n % = 10%
PW(10%) = - 11,500 + 3,000(P/A,10%, 6.1)
+ ( 1.1 ) 1,000(P/F,10%,6.1 )
= - 11,500 + 3,000 (4.3553) + ( 1.1 ) 1,000( 0.5645) = 2,560.35
HKUST, IEEM, Dr. C. J. Su
PW (10%)
4,000
N
3,000
2,560.35
2,000
1,000
-40
-30
-20
-10
% Change
in Parameter
0
10
-1,000
-2,000
-3,000
20
30
40
Which factor is
the most
sensitive to PW ?
I
PW (10%)
A
4,000
N
MV
3,000
2,000
1,000
-40
-30
-20
-10
% Change
in Parameter
0
10
-1,000
-2,000
-3,000
20
30
40
END
HKUST, IEEM, Dr. C. J. Su