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© 2014, McGraw-Hill Education All rights reserved
8 - 1
Basics of Engineering Economy
byby
Leland Blank and Anthony TarquinLeland Blank and Anthony Tarquin
Slides to accompanySlides to accompany
Chapter 8
Breakeven, Sensitivity, and
Payback Analysis
Slides to accompany Blank and Tarquin Basics of Engineering Economy, 2nd
ed.

Chapter 8 – Breakeven, Sensitivity, Payback
PURPOSES
Determine a parameter’s value to
breakeven or payback for one or more
alternatives
Evaluate
parameter sensitivity
to variation in estimates
TOPICS
Breakeven value for single
project
Breakeven between two
alternatives
Sensitivity of ≥ 1 parameter
to estimate variation
Sensitivity of selection
between ME alternatives
Payback period for i = 0%
and i > 0%; plus cautions
Spreadsheet usage for
sensitivity and breakeven
© 2014, McGraw-Hill Education All rights
reserved8 - 2

Sec 8.1 – Breakeven for Single Project
DEFINITION
• Breakeven point is the value of a parameter or decision variable
that makes two relations equal
• Breakeven is point of indifference between two projects
• Breakeven analysis can be performed considering time value of
money (i > 0%) or without it (i = 0%) without is very common, bu
has real shortcomings
Example breakeven points
Sales volume necessary to just cover costs
Price of gas to consumers to just recover drilling,
processing and delivery costs
Interest rate such that two alternatives’ cash flows are equivalent
8 - 3

• Two relations equated may be
PWA = PWB or Revenue = Cost
• Popular application of breakeven involves
cost-revenue-volume relationships with linear or nonlinear
relations with no interest considered
8 - 4
DIFFERENT REVENUE CURVES
Linear and nonlinear revenue
curves
1 – increasing cost per unit
2 – decreasing cost per unit2

COST ELEMENTS
• Fixed costs (FC) - Costs incurred regardless of volume
level to stay in business
• Examples are:
• Building costs - Minimum labor costs
• Insurance - Utilities
• Capital recovery of equipment
• Variable costs (VC) – Costs that vary with level of
activity
• Examples are:
• Materials costs - Labor costs
• Advertisement - Warranty
• Legal costs - Subcontractor costs
8 - 5
Costs can
be
linear or
nonlinear

Sample cost curves forTC = FC +VC
Q = quantity in units
8 - 6
Linear FC and VC Linear FC; VC with
decreasing unit cost

Linear Breakeven AnalysisVariables and Equations
Linear variable cost: VC = vQ
where v = unit cost, $/unit
Q = quantity, units
Breakeven quantity or point: Q = QBE
Linear revenue: R = rQ
where r = unit revenue, $/unit
Profit when Q > QBE: P = R –TC
= rQ – (FC + vQ)
8 - 7

8 - 8
Breakeven occurs
at P = 0
Solve for Q to
obtain breakeven
point

Sec 8.1 – Breakeven Quantity –Example
Water vending machine: FC = $900 per month
per site
r = 30¢ per gallon (~4 liters) v = 18¢ per gallon
8 - 9
Must sell 7,500 gallons per month per site to breakeven
Selling more means a profit is realized
What if the variable cost per unit is reduced through
improved efficiency methods? (see graph on next slide)

Sec 8.1 – Breakeven Quantity
8 - 10
As v decreases,
VC = vQ has a
lower slope
Therefore, TC line
has a lower slope
Breakeven point
decreases
Result: Larger profit
for same amount of
revenue

Sec 8.1 – Example - Breakeven Quantity
Assume v is lowered from 18¢ to 15¢ per gallon
Determine profit at sales of Q = 8,000 gallons/site
QBE = FC/(r-v) Profit = (r-v)Q – FC
= 900/(0.30-v) = (0.30-v)8,000 - 900
8 - 11
Big increase in profit for reduced variable cost

Additional considerations
Breakeven on a per unit basis:
Has the same QBEformula
Average cost per unit is Cu
Cu= FC/Q + v
Nonlinear R and/orTC relations:
May have more than 1 breakeven points and there
may be a profit range
For 2 breakeven points, maximum profit occurs
where separation between R andTC curves is largest
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8 - 13

For PW, AW and FW relations when a (one) parameter
estimate is not reliable or known
♣Given P, F, A, i, n
♣If all these parameters above are reliably estimated except
one, the unknown parameter can be calculated or
approximated as a breakeven value
♣To determine it, set PW, FW, or AW equivalency relation =
0 and solve for the unknown parameter
♣This approach was used to find ROR (Sec 6.2); the i* value
is a breakeven value
8 -
14

Sec 8.1 - Example-Breakeven Using AW
P = $-20,000 S = 10% of P = $2,000 n = 5 years
r = $300 per event FC = $-100 per event X = ?
Find X, number of events per year to breakeven at i = 5%
Solution: Set AW relation = 0 and solve for X
0 = -20,000(A/P,5%,5) + 2,000(A/F,5%,5) – 100X + 300X
200X = 4257.46
X = 21.3 events per year
© 2014, McGraw-Hill Education All rights
reserved8 - 15
A total of 22 events per year will recover
investment and costs plus slightly more
than a 5% rate of return

Sec 8.2 - Breakeven Between 2 Alternatives
Breakeven point is one of indifference between alternatives
Involves finding value of parameter common to the two
relations
Can use:
PW or AW relation at i%, or
TC relations without time value of money considered
8 - 16
Solution
Set relations equal and
solve for parameter value

Sec 8.2 - Breakeven Between 2 Alternatives
Selection guideline
Compare the common
parameter expected
level to its breakeven
quantity
If expected level is <
breakeven, select higher
variable cost (larger slope on
TC line)
(Alternative 1 in graph)
If expected level is >
breakeven, select lower
variable cost (smaller slope on
TC line)
(Alternative 2 in graph)
8 - 17

Sec 8.2-Two Alternative Breakeven-Example
1
Alternative A M
P, $ -23,000 -8,000
S, $ 4,000 0
Life, years 10 5
M&O, $/year -3,500 -1,500
Labor rate $/hr 24 12
People 1 3
Output, tons/hr 8 6
8 -
18
Find breakeven in tons per year (x) to justify
the higher initial cost alternative (A) at i = 10%
cont →

1
Solution approach:
1. DetermineVC expression for both A and M using the common variable x = tons per
year
2. Set up AW relations for both alternatives
3. Equate AW relations and solve for x
4. Determine when A or M is selected
A: variable cost expression
M: variable cost expression
8 -
19
cont →

1
A: AW relation
M: AW relation
Equate AWA= AWM and solve for x
-6,992 – 3x = -3,610 - 6x
x = 1127 tons per year
8 -
20
cont →

1
Select smaller variable
cost if tonnage
exceeds 1127 per year
A has smaller variable
cost (3x)
Conclusion: Choose A if more than 1127
tons per year
Spreadsheet plot of AW curves
shows this graphically
8 -
21

2
8 - 22
Select make or buy at 125,000 units/year and MARR = 15%
cont →
Buy alternative: 60¢ per unit

2 VC AND AW FOR MAKE ALTERNATIVE
Breakeven: x = number of units per year
Variable cost relation:
AWA+ AWB–VC = -18,000(A/P,15%,6)+2000(A/F,15%,6)
-6000 -3000(P/F,15%,3)(A/P,15%,6)
-12,000(A/P,15%,4)-500(A/F,15%,4)-5000 -0.4x
Equate AW relations for buy and make; solve for x
© 2014, McGraw-Hill Education All rights reserved 8 - 23
cont →

2
-0.60x = AWA+ AWB-VC
-0.20x = -20,352
x = 101,762 units per year
At level of 125,000 > 101,762 breakeven quantity
Select lower variable cost alternative (0.4x and 0.6x)
Select to make inhouse
8 -
24

Sec 8.3 – Sensitivity Analysis of Parameters
• Sensitivity analysis is all about reaction to variation in
estimates
• Sensitivity to estimate variation can change the economic
decision
• Sensitivity analysis helps determine which parameters matter
the most to the economic evaluation
• It determines sensitivity of an entire alternative’s
accept/reject decision or of a particular measure of worth
(PW, AW, FW, ROR, B/C) to estimate variation
• Breakeven analysis is a form of sensitivity analysis
8 -
25

Variation in some parameters causes
more sensitivity than others
Examples:
Variation in salvage value S estimate –- little effect on
AW over 20 year life
Variation in revenue per unit – major changes in
breakeven quantity to cover costs
Variation in life estimate of asset – large impact on
capital recovery (CR) each year to recover P plus i% return
Variation in MARR – No real effect on alternative
accept/reject decision if all alternatives have i* >>
MARR
8 -
26

Types of sensitivity analysis
• One parameter for single project
• One parameter and selecting between 2
alternatives
• Variation of several parameters for single
project
Note: Sensitivity of entire ME alternative selection (3rd
bullet) is covered in next section
8 -
27

General procedure of sensitivity analysis
1. Determine parameter(s) that might vary and are of
interest
2. Select the probable range (# or %) and increment of
variation
3. Select the measure of worth (e.g., PW, AW, ROR, B/C)
4. Compute the measure of worth for each parameter
5. Graphically display the results by plotting the parameter
vs. measure of worth
8 -
28
Evaluate one parameter at a time and
assume independence of parameters

→It is best to consistently use the same measure -- PW,
AW, etc.
→Don’t use the ROR measure for multiple alternative
sensitivity analysis
→Suggestion: Use a spreadsheet to perform the analysis
8 -
29
Low High
AW
Parameter Value

Sec 8.3 – Example: Sensitivity of 1 Parameter
P = $-800,000 n = 10 S = 0
Revenue estimate is $250,000 per year and will decrease
each year (best estimate is G = $-20,000 per year).
Determine impact on PW at i = 15%
Use steps in procedure
Step 1. Gradient G may vary
Step 2. Consider range of G from $-10,000 to
$-30,000 in $5,000 increments
Step 3. Use PW relation
Step 4. Relation for PW has G as the variable
8 -
30
cont →

Sec 8.3 – Example: Sensitivity of 1 Parameter
PW = - 800,000 + 250,000(P/A,15%,10) +G(P/G,15%,10)
8 -
31
PW,$1000
Step 5. Plot PW vs. G values
PW is sensitive to variation in G

Sec 8.3 – Sensitivity of Several Parameters
Evaluate sensitivity of > 1 parameter using a single
measure
Graph measure vs. percent change in estimate
Flat curve implies little sensitivity
Sloping or arched curve means the measure is sensitive
to estimate variation
Example:
‘Spider graph’ on next slide plots project ROR vs. %
variation from most likely estimated value for 4
parameters
ROR is quite sensitive to sales price and materials
cost (larger negative slopes) compared to labor and
maintenance cost (relatively flat curves)
8 -
32

Sec 8.3 – Sensitivity of Several Parameters
8 -
33

Sec 8.4 – Sensitivity of Multiple Parameters and Multiple
Alternatives
Purpose:To compare ME alternatives when more than 1
parameter may vary
Approach:
• Make 3 estimates (high, medium, low) for each
parameter of interest
• Perform evaluation via PW, AW, etc.
• Select alternative with best overall measure
(compromise may be necessary)
• The three estimates are called
 Pessimistic  Most likely  Optimistic
8 -
34

Sec 8.4 – Sensitivity of Multiple
Parameters and Multiple Alternatives
Nature of parameter dictates whether pessimistic estimate is
a high or low value
8 -
35
Example:
• Estimates made for first cost, AOC, S, n for 3 methods
• First cost values are known for each alternative (A, B, C)
• AOC, S and n may vary
• Choose the best alternative at i = 12% per year
• Use P = pessimistic; O = optimistic; ML = most likely
cont →

Sec 8.4 – Example: Sensitivity for ME Alternatives
8 - 36
cont →

Establish 3 AW relations and substitute P, ML
and O
estimates (shown in red) for each alternative
AWA = -20,000(A/P,12%,n) – AOC
AWB = -15,000(A/P,12%,n) – AOC +
S(A/F.12%,n)
AWc = -30,000(A/P,12%,n) – AOC +
3000(A/F,12%,n)
8 - 37
B is clear choice here, but this
can change as estimates change
cont →

Sec 8.5 – Payback (Payout) Period
Purpose: Determine time (in years, usually) to recover
the first cost or initial investment, P
Payback period is termed np
Two types of payback analysis
i = 0% no return expected; recover P only
i > 0% recover P plus return of i%
(also called discounted payback)
8 -
39
There are some important cautions to observe
when using this technique
Discussed throughout slides

Sec 8.5 – Discounted Payback Period
At i% > 0, find np such that annual net cash flow (NCF)
estimates just offset P
where NCF = cash outflows – cash inflows
If NCF forms a uniform series, find npusing
8 - 40

Sec 8.5 – Example: Discounted Payback
Find payback using PW at 10% return
0 = -18M +6M(P/A,10%,x)(P/F,10%,5)
x = 6.9
np = 5 + 6.9 = 12 years (approx)
8 -
41
0 1 2 3 4 5 6 7 8
np
P = $18M
A = $6M
x

Sec 8.5 – Discounted Payback Period
Caution when using discounted payback
• Cash flows that occur after payback time (np) are neglected
• May favor shorter-lived alternatives when longer-lived ones provide a higher
ROR
• Different than PW, AW, ROR, and B/C method where all estimates are used in
the evaluation
What to do: Use payback only as a supplemental or first- cut evaluation tool in
conjunction with a primary measure of worth technique (PW, etc.)
8 - 42

Sec 8.5 – No-return Payback Analysis
Determines npwhen i = 0%; no return expected
If NCF forms a uniform series, npis simply
Remember the cautions on using payback
8 -
43

Sec 8.5 – No-return Payback Analysis
Cautions when using no-return payback
Cash flows that occur after payback time (np) are
neglected
No return is anticipated on the initial investment since
time value of money is neglected
What to do: Again, use payback only as a supplemental or
first- cut evaluation tool in conjunction with a primary
technique
8 -
44

Sec 8.5 – Example: No-return Payback
8 -
45
0 1 2 3 4 5 6 7 8
np
Find payback using PW at 0% (no return)
0 = -18M +6M(x)
x = 3
np = 5 + 3 = 8 years
Requiring a 10% return increases payback
by 50% from 8 to 12 years
P = $18M
A = $6M
x

Sec 8.6 – Spreadsheet Usage – Sensitivity Example 1
PML = $-100,000 AOC = $-20,000
R = $50,000 n = 5 years
Analyze sensitivity of ROR to changes in P of ± 25%
8 - 46 cont →

Sec 8.6 – Spreadsheet Usage - Sensitivity Example 1
First cost variation of ± 25% causes P to range from
$75,000 to $125,000
• All other parameters are fixed at most likely
estimates
• Resulting ROR ranges from 28.6% high to 6.4%
low
• From the most likely value of 15.2%, this ROR
variation ranges from +88% to - 58%
8 - 47

Sec 8.6 – Spreadsheet Usage – Sensitivity Example 2
P = $-100,000 AOC = $-20,000
RML = $50,000 n = 5 years
Analyze sensitivity of ROR to changes in R of +20% with a rock-
bottom low of $25,000
8 - 48
cont →

Sec 8.6 – Spreadsheet Usage -
Sensitivity Example 2
Revenue variation causes R to range from $25,000 to
$60,000
• ROR range is 28.6% high at R = $60,000
to
• Revenue variation to as low as $25,000 (-50% below
ML estimate) causes ROR to decrease significantly to
-33.5%
• ROR is very sensitive to revenue variation
8 - 49

Breakeven Example 1
P = $-8 M AOC = $-100,000 n = 5 years
r = $1600 per testv = $800 per test
Use GOAL SEEK to determine breakeven number of tests
Set up profit relation. Let i = 0% and Q = # tests per year
Profit = R – TC
= 1600Q – [800Q + 8,000,000(0.2) + 100,000]
= 800Q – 1,700,000
8 - 50
cont →

Breakeven Example 1
8 - 51
BEFORE GOAL SEEK AFTER GOAL SEEK
cont →

Breakeven Example 1
Breakeven quantity is 2,125 test per year
GOAL SEEK parameters to find breakeven
8 - 52
• Set cell (B9) -- Change profit from $-1.7 M to 0
• Changing cell (C7) -- This displays breakeven value

Breakeven Example 2
P = $-8M AOC = $-100,000 n = 5 years r = $1600
v = varies from $800 to $1400 per test
Determine breakeven values for variation in cost per test
Set up spreadsheet for different v amounts
Repeatedly use GOAL SEEK to get breakeven for each
v estimate; use $200 increment
Results for breakeven quantity:
Quite sensitive to cost per test
Varies from 2125 tests (test cost is $800) up to 8500 (test cost is
$1400)
8 - 53
cont →

Breakeven Example 2
8 - 54
GOAL SEEK template for
breakeven at v of $1400/test

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