CE 231 – ENGINEERING ECONOMY
BREAK-EVEN ANALYSIS
This chapter covers the basics of break-even analysis, the simplest analytical tool in
management. It details what break-even analysis is, what it is used for, what definitions are
used in break-even analysis and how break-even analysis can be helpful in decision- making
of professionals in construction industry.
In construction industry, break-even analysis can be a handy tool to find answers to questions
such as: How many years should I operate the facility to recover the initial investment and
annual operating costs? How much does our company need to sell to reach the desired
profitability? What should be the toll rate to cover my costs?
Break-even analysis for a single project
Definitions: Basically, break-even analysis determines the “break-even point”, at which
operations neither make money nor loose money (Paek 2000, Blank and Tarquin 2008). At the
break-even point, there is no gain or loss; hence costs or expenses are equal to
revenues/incomes.
Break-even analysis utilizes two types of inputs for calculation of costs as: fixed costs and
variable costs:
 Fixed cost represents the expenses that are not related with the volume of production
(or activity level) over a feasible range of operations. Examples include buildings,
insurance expenses, depreciation, overheads, cost of information systems (Blank and
Tarquin 2008). It is the sum of all costs to produce the first unit of a product. Another
example could be the cost of an excavation equipment regardless of the excavation work
performed on different projects.

Variable cost represents the cost items that change with the volume of production or
construction. Input materials and time to produce a unit affect variable costs.
Examples include direct labor costs, fuel costs, material types (e.g., a certain type of paint
used for painting a facility), and marketing costs (Blank and Tarquin 2008, Paek 2000).
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Total cost is the sum of the fixed and total variable costs for any production or
construction. Total revenue is the product of expected unit sales and the unit price of each
unit.
Cost/expense and revenue/income relations are commonly assumed as linear; however
non-linear relations are more realistic with more revenue for larger volumes (Blank and
Tarquin 2008). Examples of different revenue cost relations are presented in Figure 1.
(a) Revenue relations – linear, increasing
and decreasing per unit of production
(b) Linear cost relations
Figure 1. Linear and nonlinear revenue cost relations (copyright © Blank and Tarquin
2008).
Mathematically, the formula for break-even point can be shown as:
TR = TC
or
Profit = 0
where;
TR represents the total revenues and TC represents total costs or expenses for an
operation.
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TR = TC
Expected unit sales (Q) x Unit price (P) = Fixed cost (FC) + Total variable cost (VC)
Q x P = FC +Variable unit cost (V) x Expected unit sales(Q)
QxP = FC + (VxQ)
(QxP) – (VxQ) = FC
Q(P-V) = FC
Q = FC / (P-V)
Here, Q (expected unit sales) is break-even point in sales. As seen from the above
formulation, break-even analysis depends on fixed costs, variable costs, unit price of a
product and expected unit sales (volume of sales). Graphical depiction of break-even
point is provided in Figure 2.
Figure 2. Graphical depiction of break-even point (copyright © Blank and Tarquin 2008).
Example: Assume that as an investor, you are planning to enter the construction industry
as a panel formwork supplier. Given the size of the construction industry in Turkey and
the potential number of forthcoming projects, you forecasted that within two years, your
fixed cost for producing formworks is 300.000 TL. The variable unit cost for making one
panel is 15 TL. The sale price for each panel will be 25 TL. If you charge 25 TL for
each panel, how many panels you need to sell in total, in order to start making money?
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Solution:
Variable unit cost = 15 TL/panel
Total fixed cost = 300.000 TL
Price per unit = 25 TL
TC = TR
VC + FC = TR
15 x Q + 300.000 = Q x 25 (Q refers to the number of panels)
Q = 300.000 / (25-15) = 30.000 panels
Example: A manufacturing company supplies its products to construction job sites. The
average monthly fixed cost per site is 4.500 TL, while each unit costs 35 TL to produce,
and selling price is 50 TL. (a) determine the monthly volume of supplies to job sites in
order to break-even; (b) the company has to modify the selling prices due to severe
competition. In this case, the fixed cost and production costs will be the same, but the
sales price per unit will be 50 TL for the first 200 units, and 40 TL for all above this
threshold level. Determine the monthly breakeven volume.
Solution:
(a) Q = 4.500 / (50-35) = 300 units, where Q refers to the number of units per month
(b) At 200 units, the profits is negative at - 1500 TL, as determined by
Profit = Revenue – cost
Profit at 200 units production = 200 x 50 – (4.500 + 35 x 200)
= 10.000 – 4.500 – 7.000 = - 1.500 TL
The revenue curve has a lower slope above this threshold production.
50 x 200 + 40 x Q = 4.500 + 35x (200 + Q)
Q = (4.500 + 7.000 -10.000) / (40 – 35) = 1.500 / 5 = 300 units per month
Hence the required volume is 500 units per month, the point at which revenue and total
cost break even at 22.000 TL, as shown in Figure 3.
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Figure 3. Break-even graph for the manufacturing company
Break-even analysis between two alternatives
Break-even analysis can also be used to select among alternatives (e.g., projects or
construction processes). In order to perform break-even analysis between alternatives,
there needs to be a parameter (e.g., cost or revenue variables) that is common in both
alternatives. When two alternatives are compared, the break-even point represents the
point of indifference between the alternatives (i.e., the point at which two alternatives are
equally desirable) (Badiru 1996).
The steps to find the point of indifference between alternatives:

Find the common variable between the alternatives

Express the total cost of each alternative as a function of the common variable

Equate expressions and solve for the point of indifference

Select the alternative with higher variable cost (larger slope) if the expected level is
below the point of indifference, and select the alternative with lower variable cost if
the level is above the point of indifference.
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For the example provided in Figure 4, profit functions are graphed. This graph shows that
Project B is favorable over the other alternatives if the production is between 0 and 100
units, Project A is favorable if the production is between 100 and 178 units, and Project C
is favorable if the production is larger than 178 units.
Figure 4. Plot of profit functions (copyright © Badiru 1996)
Example: There exist two alternative locations for an asphalt mixing plant to transport
materials from. Characteristics of these two locations and associated costs are tabulated
below. Which location is best for the asphalt mixing plant, the cheaper Location A or
closer Location B? (This example has been adopted from MIT Engineering Economics
lecture notes, copyright © MIT, Civil and Env. Engineering Department).
Transportation distance
Transportation expense
Monthly rental expense
Set-up cost
Workmanship costs
Total volume available
Time to use the location
Location A
Location B
6 km
1,15 TL/m3-km
1.000 TL/month
15.000 TL
0
50.000 m3
4 months (85 days)
4,3 km
1,15 TL/m3-km
5.000 TL/month
25.000 TL
96 TL/day
50.000 m3
4 months (85 days)
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Solution:
First obtain the total cost functions for all alternatives
Location A
Fixed Costs
Rental expense
Set-up cost
Workmanship costs
Variable costs
Transportation
Total Cost
Location B
4 month x 1.000 TL/month =
4.000 TL
15.000 TL
0
4 month x 5.000 TL/month =
20.000 TL
25.000 TL
85 days x 96 TL/day = 8.160 TL
6 km x 1,15 TL x Q
19.000 TL + 6,9 Q
4,3 km x 1,15 TL x Q
TL 53.160 + 4,945 Q
Equate the total cost functions to solve for volume to be transported for break-even point
19.000 TL+ 6,9 Q = 53.160 TL + 4,945 Q
Q = 17.473 m3.
At 17.473 m3 of material usage, both sites are equally desirable. If less material is
transported than 17.473 m3, then selecting location A is favorable, and if more volume is
expected to be transported than 17.473 m3, then selecting location B is more favorable
with less variable cost.
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Break-even analysis between three alternatives
For the break-even analysis between more than two alternatives, the pairwise comparison
of the alternatives is required. This analysis may lead several break-even points and the
graphical representation of the alternatives is essential to evaluate the ranges of economical
solutions. In case of existence of break-even points that are dominated by the other breakeven points, the irrelevant ones should be eliminated (Tockey 2004). The following graphs
exemplify the evaluation process of break-even analysis between three alternatives.
 three break-even points, one
irrelevant
break-even
point
between Plan A and Plan C
 for hours less than 15 select
Plan A, for hours between 15 and
25 select Plan B, and for hours
more than 25 select Plan C
Figure 5. Plot of three break-even points with one irrelevant (Tockey 2004)
In this example the break-even point between Plan A and Plan C is dominated with the
existence of Plan B (Figure 5). There may also be one break-even point between three
alternatives as in the following example (Figure 6).
 the point of indifference is 15
hours
 Plan B is irrelevant except the
point of indifference
 for hours less than 15 select
Plan A, and for hours more than 15
select Plan C
Figure 6. Plot of one break-even point between three alternatives (Tockey 2004)
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As a final example, an alternative may also be dominated by existence of the other
alternatives. The irrelevant alternative should be eliminated from the evaluation when it
exists as in the following figure (Figure 7). One alternative may also dominate the other
alternatives if their functions are parallel to each other (Tockey 2004).
 the Plan C is dominated by Plan
A and Plan B
 Plan C is eliminated from the
analysis
 for hours less than 15 select
Plan A, and for hours more than 15
select Plan B
Figure 7. Plot of one alternative dominated by others (Tockey 2004)
SUPPLEMENTARY EXAMPLES
Example 1:
A contractor finds that he can buy from a concrete manufacturer, components for 800 TL
per unit. Alternatively, he can manufacture the same size and quality components for a
variable cost of 400 TL/unit. It is estimated that the additional fixed cost in the plant
would be 1.200.000 TL per year, if the components are manufactured by himself. What is
the break-even point?
Solution 1:
Total annual cost as a function of the number of units for the make alternative:
TCMake = 1.200.000 TL + 400 Q (Q refers to the number of units)
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Total cost for the buy alternative:
TCBuy = 800 Q
Break-even volume occurs when TCBuy = TCMake
i.e.,
800 Q = 1.200.000 TL + 400 Q, solving for Q will give Q = 3.000 units/year.
For the requirements in excess of 3.000 units/year, the make alternative would be more
economical. If the rate of use is less than 3.000 units/year, the buy alternative should be
chosen. This is shown in the following graph.
Example 2:
A special equipment is needed for a specific construction activity at a job site. This
equipment can be leased for 5.000 TL/day including its cost of maintenance.
Alternatively, the equipment can be purchased for 2. 500. 000 TL. The equipment is
estimated to have a useful life of 12 years with a salvage value of 400.000 TL at the end
of its useful time. It is estimated that annual maintenance costs will be 300.000 TL. If the
interest rate is 6% and it costs 5.000 TL per day to operate the equipment, how many days
per year are required for the two alternatives to break-even?
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Solution 2:
Annual costs of the equipment if leased:
TCL = 5.000 Q + 5.000 Q (lease + operation costs)
= 10.000 Q (Q refers to the number of days per year)
Annual costs of the equipment if purchased:
TCP = initial cost – salvage value + maintenance cost + operating costs
TCP = 2.500.000(A/P,6,12) – 400.000(A/F,6,12) + 300.000 + 5.000 Q
TCP = 2.500.000 x 0,1193 – 400.000 x 0,0593 + 300.000 + 5.000 Q
TCP = 298.250 – 23.720 + 300.000 + 5.000 Q
TCP = 574.30 + 5.000 Q
Break-even occurs when TCL= TCP
10.000 Q = 574.530 + 5.000 Q
Q = 115 days/year
For all the levels of use exceeding 115 days/year it would be more economical to
purchase the equipment. If the level of use is anticipated to be below 115 days/year, the
computer should be leased, as shown in the following graph.
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Example 3:
A hydraulically operated equipment can be fabricated for 140 000 TL and it will have a
salvage value of 20. 000 TL at the end of 4 years. Maintenance cost will be 12. 000
TL/year and the cost of operation will be 85 TL/hr. As an alternative, a mechanically
operated equipment can be fabricated for 55. 000 TL. This equipment will have no
salvage value at the end of a 4 year service life. The cost of operation and maintenance is
estimated to be 140 TL/hr.
With an interest rate of 10%, what is the break-even number of service hours per year?
Solution 3:
Annual equivalent total cost for the hydraulic equipment:
TCH = initial cost – salvage value + maintenance cost + operating costs
TCH = 140.000(A/P,10,4) – 20.000 (A/F,10,4) +12.000 + 85 Q
(Q refers to the hours of use per year)
TCH = 140.000 x 0,3155 – 20.000 x 0,2155 +12.000 + 85 Q
TCH = 51.860 + 85 Q
Annual equivalent total cost for the mechanical equipment:
TCM = initial cost + operating and maintenance costs
TCM = 55.000(A/P,10,4) + 140 Q
TCM = 17.353 + 140 Q
Hydraulic
Mechanical
Break –even occurs when TCH = TCM , or
51.860 + 85 Q = 17.353 +140 Q
Q = 627 hrs/year
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For rates of use exceeding 627 hrs/yr, the hydraulic equipment would be economical.
However, if it is anticipated that the rate of use will be less than 627 hrs/yr, then the
mechanically operated equipment should be used, as shown in the following graph.
Example 4:
A company manufacturing prefabricated reinforced concrete building elements is
operating with an annual cost of 40.000.000 TL, a revenue (income) of 1.100 TL/element,
and a variable cost of 600 TL/element. The company hopes to produce 100.000 elements
per year. What is the break-even point? Would the company make a profit or loss?
Assuming that the market is ready for an increase in selling price to 1.600 TL/element,
how would this affect the yearly profit?
Solution 4:
Break-even point occurs when;
Revenue = Cost
1.100 Q = 40.000.000 + 600 Q, where Q is the total elements sold in a year.
Q = 80.000 elements/year
If 100.000 elements per year are made and sold, then annual profit will be;
Profit = Revenue – Cost
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Profit = 1.100 x 100.000 – 600 x 100.000 – 40.000.000
P = 10.000.000 TL/yr
If selling price is increased from 1.100 TL/element to 1.600 TL/element, then annual
profit will be:
P = (1.600 – 600) x 100.000 – 40.000.000
P = 60.000.000 TL/yr
In other words, if the selling price is increased about 40 to 50%, the annual profit
increases by 600%.
Example 5:
The company in Example 4 is considering usage of a new production equipment which
will save 3.000.000 TL/year in supervision and related fixed costs. With a revenue of
1.100 TL/element, and a variable cost of 600 TL/element, at what volume of production
the company starts making profit? What is the expected profit per year for 100.000 units
of expected production?
Solution 5:
Revenue = Cost
1.100 Q = 40.000.000 – 3.000.000 + 600 Q (Q refers to the number of elements per year)
500 Q = 37.000.000
Q = 74.000 elements/yr
At 100.000 elements per year, the profit will be;
P = (1.100- 600) x 100.000 – (40.000.000 – 3.000.000)
P = 13.000.000 TL/yr which is greater than 10.000.000 TL/yr when the fixed cost was
40.000.000 TL/yr.
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Example 6:
The company in example 4 considers an advertisement campaign which will make it
possible to sell elements for 1.125 TL/element. What is the new volume of production at
which revenues will be equal to costs? What will annual profits become? How much can
the company spend on the advertisement campaign?
Solution 6:
Break even volume will be at:
40.000.000 + 600 Q = 1.125 Q (Q refers to the number of elements per year)
Q = 76.190 elements/yr
At 100.000 elements per year, profits will be:
P = (1.125-600) x 100.000 – 40.000.000
P = 12.500.000 TL/yr > 10.000.000 TL/yr when the selling price was 1.100
TL/element. The company can spend for the advertisement:
12.500.000 – 10.000.000 = 2.500.000 TL/yr.
Example 7:
A toy company currently purchases the metal parts which are required in the manufacture
of certain toys, but there has been a proposal that the company make these parts themselves.
Two machines will be required for the operation: Machine A will cost 18.000 TL, have a
life of six years, and a 2.000 TL salvage value; Machine B will cost 12.000 TL, have a life
of four years, and a -500 TL salvage value. Machine A will require an overhaul after three
years costing 3.000 TL. The annual operating cost of Machine A is expected to be 6.000
per year and for Machine B 5.000 TL per year. A total of four laborers will be required for
the two machines at a cost of 2,50 TL per hour per worker. In a normal eight-hour day, the
machines can produce parts sufficient to manufacture 1.000 toys. If the company’s present
price for these parts is 0,50 TL per toy, how many toys must be manufactured each year in
order to justify the purchase of the machines? Use a MARR of 15% per year.
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Solution 7:
Variable cost/year = (cost/unit)(unit/year) =
4(2,50)(8)
x  0,08 x
1.000
Where x = number of units per year
For Machine A, the fixed costs per year:
AEA = 18.000 (A/P,15%,6) – 2.000 (A/F,15%,6) + 6.000
+ 3.000(P/F,15%,3)(A/P,15%,6)
AEA = 18.000*0,2642 – 2.000*0,1142 + 6.000 + 3.000*0,6575*0,2642
AEA = 11.048 TL
For Machine B, the fixed costs per year:
AEB = 12.000 (A/P,15%,4) +500 (A/F,15%,4) + 5.000
AEB = 12.000*0,3503 + 500*0,2003 + 5.000
AEB = 9.304 TL
Equating the annual costs of the purchase option (0,50x) and the manufacture option:
0,50x = 11.048 +9.304 + 0,08x
0,42x = 20.352
x = 48.457 units
Example 8:
A construction company is planning to convert a plant from manufacturing steel
components for railways (lines, axes, etc.) to manufacturing for buildings (beams, columns,
etc.) The initial cost for equipment conversion will be 60 million TL with a 15% salvage
value anytime within a 7-year period. The cost of producing each component will be 100
TL, and they will be sold for 250 TL. The production capacity for the first year will be
50.000 components. At a MARR of 9% per year, by what uniform amount will production
have to increase each year in order for the company to recover its investment in 4 years?
Solution 8:
Let X = gradient increase per year
Total costs =
-60.000.000 (A/P,9%,4) + (0,15)(60.000.000)(A/F,9%,4) – [50.000 + X (A/G,9%,4)](100)
Revenue = [50.000 + X (A/G,9%,4)](250)
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At breakeven, revenue + costs* = 0, then
[50.000 + X (A/G,9%,4)] (250-100) = 60.000.000 (A/P,9%,4) – (0,15) (60.000.000)
(A/F,9%,4)
50,000+ X(1,3925) (150) = 60.000.000(0,30867) – 9.000.000(0,21867)
7.500.000 + 208,875X = 18.520.200 – 1.968.030
X = 43.338 units
Example 9:
(Translated, source: Akbıyıklı 2014)
There is a woodworks workshop that is producing doors. With the current capacity, the
workshop can produce 10.000 doors per year. For this production amount, the fixed costs
are 200.000 TL, variable costs are 120 TL/door and selling price of each door is 200 TL.
The owner of the workshop is considering two strategies for production growth:
i.
Increasing the production to 20.000 doors
ii.
Increasing the production to 25.000 doors
The fixed costs would increase to 420.000 TL for the first strategy and 535.000 TL for the
second one.
Evaluate all three alternatives for the workshop owner, and comment on them regarding
their desirability in terms of profit per unit and breakeven production amounts.
a. No growth
b. Production amount of 20.000 doors
c. Production amount of 25.000 doors
Solution 9:
a. No growth
FC = 200.000TL
VC = 120 TL/door
Unit price = 200 TL/door
Total fixed cost = 200.000TL
Total VC = 120 x 10.000 = 1.200.000 TL
Total revenue = 200 x 10.000 = 2.000.000 TL
Profit = total revenue – total cost = 600.000 TL
Profit/unit = 600.000/10.000 = 60 TL/door
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b. Production amount of 20.000 doors
Total fixed cost = 420.000TL
Total VC = 120 x 20.000 = 2.400.000 TL
Total revenue = 200 x 20.000 = 4.000.000 TL
Profit = total revenue – total cost = 1.180.000 TL
Profit/unit = 1.180.000/20.000 = 59 TL/door
c. Production amount of 25.000 doors
Total fixed cost = 535.000TL
Total VC = 120 x 25.000 = 3.000.000 TL
Total revenue = 200 x 25.000 = 5.000.000 TL
Profit = total revenue – total cost = 1.465.000 TL
Profit/unit = 1.465.000/25.000 = 58,6 TL/door
Breakeven amounts for each alternative:
a. No growth
Q = 200.000 / (200-120) = 2.500 doors
b. Production amount of 20.000 doors
Q = 420.000 / (200-120) = 5.250 doors
c. Production amount of 25.000 doors
Q = 535.000 / (200-120) = 6.688 doors
Profitability is decreasing as the capacity and revenue increases.
Example 10:
A product currently sells for 20 TL per unit. The variable costs are 8 TL/unit, and this
specific design (Design A) sells 10.000 units annually with a profit of 40.000 TL acquired
per year. The company is trying to make a 10-year plan to decide if any improvements are
can be initiated for the production line. As an initial idea, company is considering to change
the existing design of the product. This new design (Design B) will increase the variable
costs by 25% and fixed costs by 10% for this production machine (Machine I). Sales are
forecasted to increase 2.500 units per year with Design B. Alternatively, without changing
the design of the product (using Design A), the company can change the production
machine (Machine II) for 43.625 TL initial cost, no salvage value, and additional annual
maintenance cost of 5.000 TL. The variable costs are expected to be 8 TL/unit in that case.
The last alternative is integrating Design B with Machine II, then the associated variable
cost will be 10 TL/unit. MARR is 10%. Given this information please answer the following
questions:
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a. If the company decides to go with Design B, at what selling price does this alternative
break-even?
b. If the selling price is to be kept same (20 TL/unit) what will the annual profit be for
the design update (Machine I, Design B)?
c. For the new machine, the selling quantities are estimated as 13.500 units for Design
A and 15.000 units for Design B. If the selling price is kept the same, which
alternative should be selected as the 10-year investment plan?
d. If 12.000 products will be produced annually, please calculate the selling price for
each of these alternatives separately so that each of them breaks-even individually.
Solution 10:
Profit = revenue – costs
40.000 = 10.000(20) – [10.000(8) + FC]  FC = 80.000 TL
Design change:
New variable cost = 8(1,25) = 10 TL/unit
New fixed costs = 80.000(1,1) = 88.000 TL
a. Breakeven analyses for Design B
Let X = breakeven selling price per unit, then
12.500X = 88.000 + 12.500(10)
X = 17,04 TL/unit
b. Profit analysis for Design B
Profit = 12.500(20) – 12.500(10) – 88.000 = 37.000 TL
c. New machine, Design A
FC = 43.625(A/P, 10%, 10) + 5.000 + 80.000
FC = 43.625*0,16275 + 85.000 = 92.100 TL
Profit = 20*13.500 – 8*13.500 – 92.100 = 69.900 TL
New Machine, Design B
FC = 43.625 (A/P, 10%, 10) + 5.000 + 88.000
FC = 43.625*0,16275 + 93.000 = 100.100 TL
Profit = 20*15.000 – 10*15.000 – 100.100 = 49.900 TL
New Machine, Design A should be selected.
d. Machine 1, Design A: 12.000X = 80.000 + 12.000*8  X = 14,67 TL/unit
Machine 1, Design B: 12.000X = 88.000 + 12,000*10  X = 17,33 TL/unit
Machine 2, Design A: 12.000X = 92.100 + 12.000*8  X = 15,68 TL/unit
Machine 2, Design B: 12.000X = 100.100 + 12.000*10  X= 18,34 TL/unit
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Example 11:
Rephrased, costs, numbers and units changed, source: (NPTEL Lecture 16)
A construction company is planning to purchase an excavator to be used in one of its
construction sites, and has three proposals as detailed below:
Alternative-1: Initial purchase cost = 325.000 TL
Salvage value = 83.300 TL
Useful life = 12 years
Operating cost: excavating 1m3 of earth is 1 TL
This excavator can excavate 50 m3 of earth in one hour.
Alternative-2: Initial purchase cost = 360.000 TL
Salvage value = 94.000 TL
Useful life = 12 years
Operating cost: The operating cost for excavating 1m3 of earth is 0,55 TL
This excavator can excavate 60 m3 of earth in one hour.
Alternative-3: Initial purchase cost = 400.000 TL
Salvage value = 100.000 TL
Useful life = 12 years
Operating cost: excavating 1m3 of earth is 0,4 TL
This excavator can excavate 70 m3 of earth in one hour.
The company identified MARR as 10,5% per year. Determine the economical alternative
by plotting the cost of these three alternatives at different values of annual operating hours
on the chart provided to you.
a. If the excavator is expected to be used for only 500 hours, which alternative should
be used?
b. If the excavator is expected to be used for only 1.500 hours, which alternative should
be used?
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Solution 11:
Pair-wise comparison between the alternatives is performed to determine the breakeven
point.
Let ‘X’ be the number of operating hours per year.
Annual operating cost (TL) for Alternative-1:
1 TL/m3 * 50 m3/hr * X hr/year = 50X TL/year
AEAlternative-1:
AE1 = -325.000 (A/P, 10,5%, 12) – 50X + 83,300 (A/F, 10,5%, 12)
AE1 = -325.000* 0,1504 – 50X + 83.300* 0,0454 = -45.098 – 50X
Annual operating cost (TL) for Alternative-2:
0,55 TL/m3 * 60 m3/hr * X hr/year = 33X TL/year
AEAlternative-2:
AE2 = -360,000(A/P, 10,5%, 12) –3X + 94,000 (A/F, 10,5%, 12)
AE2 = -360.000*0,1504 – 33X + 94.000* 0,0454 = -49.876 – 33X
Annual operating cost (TL) for Alternative-3:
0,4TL/m3 * 70 m3/hr * X hr/year = 28X TL/year
AEAlternative-3:
AE3 = -400.000 (A/P, 10,5%, 12) – 28X + 100.000 (A/F, 10,5%, 12)
AE3 = -400.000* 0,1504 – 28X + 100.000* 0,0454 = -55,620 – 28X
The equivalent uniform annual
worth (cost) of all the
alternatives are determined at
different values of annual
operating hours and are shown
in the figure.
AE1 = AE2
-45.098 – 50X = -49.876 – 33X
 X = 281 hours/year
AE1 = AE3
-45.098 – 50X = -55.620 – 28X
 X = 478 hours/year
AE2 = AE3
-49.876 – 33X = -55.620 – 28X
 X = 1.149 hours/year
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Alternative-1 should be selected, if the expected annual operating hours are less than the
281 hours, Alternative-2 should be selected, if the expected annual operating hours are
between 281 and 1.149 hours. Alternative-3 should be selected, if the expected annual
operating hours are greater than 1.149 hours.
a. Alternative 2 should be selected.
b. Alternative 3 should be selected.
Example 12:
An investor is going to get a multi-story apartment constructed. They have four alternatives
as building materials: steel, concrete, stone masonry or brick. The costs associated with
each alternative is presented in the following table. Please carry out breakeven analyses to
determine which material should be used for what ranges of building area so that cheapest
alternative is selected. In your calculations, consider MARR as 15% per year and plot your
findings on the chart.
Initial cost
Maintenance
Steel
Concrete
30 TL/m2
40.000
TL/year
20 TL/m2
25.000
TL/year
1,5
TL/m2/year
90% of initial
cost
25 years
Heating
1 TL/m2/year
Salvage
Value
Life span
75% of initial
cost
25 years
Stone
Masonry
32 TL/m2
15.000
TL/year
2,2
TL/m2/year
100%
of
initial cost
25 years
Brick
25 TL/m2
18.000
TL/year
2 TL/m2/year
100%
of
initial cost
25 years
Solution 12:
Let X = area in m2
Using AE values:
AES = -(30X)(A/P,15%,25) + (0,75)(30X)(A/F,15%,25) – 40.000 –X
= -(30X)(0,15470) + (0,75)(30X)(0,00470) – 40.000 –X
= -4,641X + 0,10575X – 40.000 – X
= -5,535X – 40.000
AEC = -(20X)(A/P,15%,25) + (0,9)(20X)(A/F,15%,25) – 25.000 – 1,5x
= -(20X)(0,15470) + (0,9)(20X)(0,00470) – 25.000 – 1,5x
= -3,094X + 0,0846X – 25.000 – 1,5X
= -4,509X – 25.000
22
AESM = -(32X)(A/P,15%,25) + (1,0)(32X)(A/F,15%,25) – 15.000 – 2,2X
= -(32X)(0,15470) + (1,0)(32X)(0,00470) – 15.000 – 2,2X
= -4,950X + 0,1504X – 15.000 – 2,2X
= -7X – 15.000
AEB = -(25X)(A/P,15%,25) + (1,0)(25X)(A/F,15%,25) – 18.000 – 2X
= -(25X)( 0,15470) + (1,0)(25X)( 0,00470) – 18.000 – 2X
= -3,8675X + 0,1175X – 18.000 – 2X
= -5,75X – 18.000
Breakeven Analyses:
Steel vs Concrete:
-5,535X – 40.000 = -4,509X – 25.000
X < 0 negative
Steel vs Brick:
-5,535X – 40.000 = -5,75X – 18.000
x =102.326 m2
Steel vs Stone Masonry
-5,535X – 40.000 = -7X – 15.000
x= 17.065 m2
Concrete vs Brick:
-4,509X – 25.000 = -5,75X – 18.000
x= 5.641 m2
Concrete vs Stone Masonry
-4,509X – 25.000 = -7X – 15.000
x= 4.015 m2
180000
160000
140000
120000
100000
80000
60000
40000
20000
0
0
5000
steel
10000
concrete
15000
stone
20000
25000
brick
Brick vs Stone Masonry
-5.750x -18,000 = -7X – 15.000
x= 2.400 m2
for X < 2.400  stone
for 2.400 < X < 5.641  brick
for X > 5.641  concrete
Following examples are adapted from Lecture Notes provided by Department of Industrial
Engineering, Eastern Mediterranean University (“EMU”).
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Example 13:
The ABC Company is faced with three alternatives for their fabrication work. Alternative
A involves the purchase of a machine for 5.000 TL. It will have a seven-year life, with a
zero salvage value at that time. Alternative A involves additional costs of TL0,20 per unit
of product produced per year. Alternative B involves the purchase of a machine for 10.000
TL. It will also have a seven-year life, with 2.000 TL salvage value at that time. Alternative
B involves additional costs of 0,15 TL per unit of product produced per year. Alternative C
involves the purchase of a machine for 8.000 TL. It will have a 2.000 TL salvage value
when disposed of in seven years. Additional costs of 0,25 TL per unit of product per year
arise when Alternative C is used. An 8% interest rate is used by the ABC Company in
evaluating investment alternatives. For what range of annual production volume values is
each method preferred?
Solution 13:
Let X = number of units per year
AEA = -5.000 (A/P,8%,7) – 0,2X = -5.000 (0,19207) – 0,2X
AEA = -960,35 – 0,2X
AEB = -10.000(A/P,8%,7) + 2.000(A/F,8%,7) – 0,15X = -10.000(0,19207) +
2.000(0,11207) – 0,15X
AEB = -1.696,56 – 0,15X
AEC = -8.000(A/P,8%,7) + 2.000(A/F,8%,7) – 0,25X
AEC = -1.312,42 – 0,25X
A vs B:
-960,35 – 0,2X = -1.696,56 – 0,15X
XAB = 14.724,2
B vs C:
-1.696,56 – 0,15X = -1.312,42 – 0,25X
XBC = 3.841,4
A vs C:
No intersection for X > 0.
24
C
A
B
Total Cost
4.000
X  14.724, select A
X > 14.724, select B
2.000
X
3.841
10.000
Number of units per year
14.724
20.000
Alternative solution (without graph):
Check minimum cost at each unit interval by picking random values at each interval (break
even points represent the points of possible change in cost trend).
0
3.841
3.000
Units
CA
CB
CC
14.724
10.000
3000
-1.560,35 TL
-2.146,96 TL
-2.062,42 TL
10000
-2.960,35 TL
-3.196,56 TL
-3.812,42 TL
16.000
16000
-4.160,35 TL
-4.096,56 TL
-5.312,42 TL
Therefore Select A for units<14.724, Select B for units>14.724.
Example 14:
Three types of design proposals for a commercial one - storey building is to be evaluated
details given below:
25
STEEL
CONCRETE
BRICK
First cost
72 TL/m2
76 TL/m2
81 TL/m2
Annual maintenance
14.000 TL
9.000 TL
6.000 TL
Annual heating cost
3 TL/m2
3,4 TL/m2
3,9 TL/m2
SV (%of first cost)
%80
%100
%110
20
20
20
Life (years)
For what range of building area (m2) which type of design is the most suitable (cheapest) to
select? Carry out break-even analysis using an interest rate of %18 per year and plot your
ranges to illustrate.
Solution 14:
Let X = area in m2
First cost of steel = 72x
and its salvage value = (0,8)72x
Using AE values:
AES = -(72x)(A/P,18%,20) + (0,8)(72x)(A/F,18%,20) – 14.000 – 3x
AES = -(72x)(0,18682) + (0,8)(72x)(0,00682) – 14.000 – 3x
AES = -16,058x – 14.000
Similarly,
AEC = -(76x)(A/P,18%,20) + (1,0)(76x)(A/F,18%,20) – 9.000 – 3,4x
AEC = -(76x)(0,18682) + (1,0)(76x)(0,00682) – 9.000 – 3,4x
AEC = -17,08x – 9.000
AEB = -(81x)(A/P,18%,20) + (1,1)(81x)(A/F,18%,20) – 6.000 – 3,9x
AEB = -(81x)(0,18682) + (1,1)(81x)(0,00682) – 6.000 – 3,9x
AEB = -18,425x – 6.000
Break-evens:
Steel vs Concrete:
-16,058x – 14.000 = -17,08x – 9.000
x = 4.892 m2
Steel vs Brick:
-16,058x – 14.000 = -18,425x – 6.000
x = 3.380 m2
Concrete vs Brick:
-17,08x – 9.000 = -18,425x – 6.000
x = 2.231 m2
26
B
C
Total Cost
0 < x  2.231 Select Brick
2.231 < x  4.892 Select Concrete
x > 4.892
Select Steel
S
2.231 3.380
Area, m2
4.892
Alternative solution (without graph):
Check minimum cost at each area interval by picking random values at each interval (break
even points represent the points of possible change in cost trend).
0
2.231
1.000
3.380
3.000
4.892
4.000
6.000
Area (m2)
1.000
3.000
4.000
6.000
CSTEEL -30.058 TL -62.174 TL -78.232 TL -110.348 TL
CCONCRETE -26.080 TL -60.240 TL -77.320 TL -111.480 TL
CBRICK -24.425 TL -61.275 TL -79.700 TL -116.550 TL
Therefore, Select Brick for area<2.231 m2, Select Concrete for 2.231 m2<area<4.892 m2,
and Select Steel for area>4.892 m2.
Example 15:
Three options are considered for an engine part:
A – complete in-house manufacturing, with initial equipment cost of 50.000 TL, labor cost
of 26.000 TL per year, and material cost of 10 TL per engine part.
B – partial manufacture, (i.e. partially finished engine parts are purchased), with initial
equipment cost of 35.000 TL, labor cost of 10.000 TL per year, material cost of 3 TL per
engine part, and an additional cost of 40 TL per the partially finished engine part.
C – purchase from outside at a cost of 120 TL per engine part.
Any equipment purchased will have a life of 6 years. If the MARR is 10% per year,
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determine the number of engine parts that must be manufactured to justify (a) complete inhouse manufacture and (b) partial manufacture. (c) Plot the total cost lines for all three
options, and state the ranges of engine parts for which each option will have the lowest cost.
Solution 15:
Let x = number of engine parts per year.
Using AE values:
AEIN = -50.000(A/P,10%,6) – 26.000 – 10x = -50.000(0,22961) – 26.000 – 10x =
- 37.480,5 – 10x
AEPM = -35.000(A/P,10%,6) – 10.000 – 3x – 40x = -35.000(0,22961)– 10.000– 43x
= - 18.036,35 - 43x
AEOUT = -120x
(a) Complete in-house manufacturing vs purchase from outside:
AEIN = AEOUT
or,
-37.480,5 - 10x = -120x
x = 341 parts per year
(b) Partial manufacture vs purchase from outside:
AEPM = AEOUT
-18.036,35 - 43x = -120x
x = 234 parts per year
(c) Complete in-house vs partial manufacture
AEIN = AEPM
-37.480,5 - 10x = -18.036,35 - 43x
x = 589 parts per year
from outside
Ranges for the lowest total cost are:
0 < x  234 select purchase from outside
partial
Total
cost
in-house
234 < x  589 select partial manufacture
589 < x
select in-house manufacture
234 341
589
Number of engine parts
28
Alternative solution (without graph):
Check minimum cost at each engine parts interval by picking random values at each interval
(break even points represent the points of possible change in cost trend).
0
234
200
341
300
589
400
600
Parts
200
300
400
600
CIN -39.480,50 TL -40.480,50 TL -41.480,50 TL -43.480,50 TL
CPM -26.636,35 TL -30.936,35 TL -35.236,35 TL -43.836,35 TL
COUT -24.000,00 TL -36.000,00 TL -48.000,00 TL -72.000,00 TL
Therefore, Select Purchase from Outside for engine parts<234, Select Partial Manufacture
for engine parts 234<area<589, and Select In-House Manufacture for Engine Parts>589.
References:
Akbıyıklı, R. (2014). Mühendislik Ekonomisi: Temel Prensipleri ve Uygulamaları, Birsen Yayınevi,
İstanbul.
Badiru, A. B. (1996). Project Management in Manufacturing and High Technology Operations,
Wiley-IEEE, USA.
Blank, L. T., and Tarquin, A. J. (2008). Engineering Economy, 5th Edition, McGraw-Hill, USA.
Eastern Mediterranean University, Lecture Notes of Industrial Engineering, retrieved on 22nd
February 2014, from http://ie.emu.edu.tr/development/dosyalar/{o2M-or_-Mco}Break.doc
Massachusetts Institute of Technology, OpenCourseWare, 2006 Lecture Notes of 1.011 Project
Evaluation Course.
NPTEL Lecture 16, Lecture Notes on Breakeven analysis for two and more than two alternatives,
retrieved on 3rd January 2017, from http://nptel.ac.in/courses/105103023/33
Paek, J. H. (2000). “Running a profitable construction company: Revisited break-even analysis.”
Journal of Management in Engineering, 16(3), 40-46.
Tockey, S. (2004). Return on Software: Maximizing the Return on Your Software Investment,
Addison-Wesley Professional, USA.
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