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Engineering Economy 82 Dr. Emad Elbeltagi

CHAPTER 4

APPLICATIONS IN THE CONSTRUCTION INDUSTRY

This chapter introduces some topics that are related to the economic evaluation of

alternatives such as the depreciation, breakeven analysis and the sensitivity analysis.

These topics are very important for the economic evaluation of projects. Also, the use of

investment laws and economic evaluation are presented and applied for the construction

projects such as: calculation of the renting cost of equipment, equipment replacement

analysis and the evaluation of bids.

4.1 Depreciation

The depreciation in defined as “the decrease in market value of an asset over time”

through wear, deterioration or obsolescence. A machine may depreciate (decline in value)

because it is wearing out and no longer performing its function as well as when it was

new. Many kinds of machinery require increased maintenance as they age, reflecting a

slow but continuing failure of individual parts. Also, the quality of outputs may decline

due to wear in components. Another aspect of depreciation is that caused by

obsolescence. Obsolescence occurs when the asset is no longer technologically superior

to available alternatives. A machine is described as obsolete when the function it

performs can be done in some better manner. A machine may be in excellent working

condition, yet may still be obsolete. For example, electronic machines, computers, etc.

As asset always has different values: initial value (P), book value (BV), salvage value (F)

and market value (MV). The initial value represents the purchase price of an asset.

Salvage value represents the expected price for selling the asset at the end of its useful

life. The book value represents the current value in the accounting systems. The book

value equals the initial value of the asset minus all the depreciation costs till given time.


The book value is always calculated at the end of each year. The market value, on the

other hand, represents the value of the asset if it is sold in the free market. It is not

necessary that the book value equals the market value.

Depreciation is an accounting charge that allows for the recuperation of capital that was

used to procure equipment or other physical assets. There are three common methods for

calculating depreciation expense for financial accounting purposes: straight-line, sum-of-

years digits and the sinking fund method. Each method involves the spreading of the

amount to be depreciated over the recovery life of an asset in a systematic manner.

Each depreciation method selected produces different patterns of depreciation expense

per period. The straight-line method assumes linear depreciation or the depreciation cost

is allocated equally over the asset useful life. The sum-of-years digits assumes high rate

of depreciation at the early age of an asset and decreasing rate at its aged life. The Third

method assumes lower rate at the early ages and faster rate at the late age.

4.1.1 Straight-Line method

The simplest and best known of the various depreciation methods is the straight-line

depreciation method. In this method a constant depreciation charge is made. To obtain

the annual depreciation charge at any year, Dn, the total amount to be depreciated (initial

value, P – salvage value, F) is divided by the useful life in years, N.

(Annual depreciation charge) Dn = (P – F) / N (4.1)

Where:

Dn = Annual depreciation

P = Initial value of the asset (include purchase price, delivery cost,

installation cost and other equipment related costs).

P = Salvage value of the asset (the net value after dismantling or removal

costs have been subtracted from the actual monetary value).

N = Expected depreciable life of the asset.


The book value at any time, n, could be calculated as given in the following equation:

BV(n) = P – nDn (4.2)

Example 4.1: If an asset has a initial value of LE50,000 with LE10,000 salvage value

after five years. Calculate the annual depreciation and calculate the book

value of the asset after each year.

Solution:

Annual depreciation:

Dn = (P - F) / N = 50,000 - 10,000 / 8 = LE8,000 per year

Book value of the asset after each year:

BV(n) = P – nDn (n = 1, 2, 3, 4, 5)

BV(1) = 50,000 – (1) 8,000 = LE42,000

BV(2) = 50,000 – (2) 8,000 = LE34,000

BV(3) = 50,000 – (3) 8,000 = LE26,000

BV(4) = 50,000 – (4) 8,000 = LE18,000

BV(5) = 50,000 – (5) 8,000 = LE10,000 = F

Example 4.2: If the purchase price of an equipment is LE60,000 and its salvage value

after 8 years is LE6,000, calculate the annual depreciation and the book

value of the equipment each year.

Solution:

P = 60,000; F = 6,000; N = 8;

Total depreciation = 60000 – 6000 = LE54,000

Annual depreciation = 54000 / 8 = LE6,750

Notice that the book value of the equipment equals its salvage value at the

end of its useful life as shown in Table 4.1.


Table 4.1: Straight-line depreciation of Example 4.2

Year Annual depreciation Book value

0 0 60,000

1 6,750 53,250

2 6,750 46,500

3 6,750 39,750

4 6,750 33,000

5 6,750 26,250

6 6,750 19,500

7 6,750 12,750

8 6,750 6,000

4.1.2 Sum-of-years digits method

Another method for allocating the cost of an asset minus its salvage value over its useful

life is called sum-of-years digits depreciation method. The sum of the year digits is a

rapid write-off technique by which most of the value of the asset is written off in the first

one-third of its life. This method results in faster depreciation at the early life of an asset.

Larger depreciation charges than straight-line depreciation during the early years of an

asset and smaller charges as the asset nears the end of its estimated life. Each year, the

depreciation charge is computed as the remaining useful life at the beginning of the year

divided by the sum of the years digits for the total useful life, with this ratio multiplied by

the total amount of depreciation (P – F). The mechanics of the method involve finding

the sum of the year digits from 1 through N. The depreciation charge for any given year is

then obtained by multiplying total amount of depreciation by a ratio of the number of

years remaining in the life of the asset to the sum of year digits. Thus means that the

depreciation is calculated as the percentage of the remaining life to the original life.

For an asset with useful life N, to obtain the annual depreciation charge, Dn, at any year

n, can be calculated as follows:

Dn = (Remaining useful life at beginning of a year / Sum of years digits) × (P – F)


Sum of years digits = N (N + 1) / 2

Example 4.3: Resolve Example 4.2 using the straight-line depreciation method.

Solution:

P = 60,000; F = 6,000; N = 8

Sum-of-years digits = 8 (8 + 1) / 2 = 36 years

The calculations are shown in the following table (Table 4.2).

Table 4.2: Sum-of-years depreciation of Example 4.3

Year Remaining life /

sum-of-years

Annual

depreciation Book value

0 0 0 60,000

1 8/36 12,000 48,000

2 7/36 10,500 37,500

3 6/36 9,000 28,500

4 5/36 7,500 21,000

5 4/36 6,000 15,000

6 3/36 4,500 10,500

7 2/36 3,000 7,500

8 1/36 1,500 6,000

Example 4.4: Calculate the depreciation charge for the first three years and the book

value for year three for an asset which had a first cost of LE25,000, and a

salvage value of LE4,000 and a life of eight years.

Solution:

The sum of years digits must be calculated first:

Sum of Years Digits = 8 (8 + 1) / 2 = 36

D1 = [(8 – 1 + 1) / 36] (25,000 – 4,000) = LE4667

)3.4()(1)/2 (

1 - FP

NN

nΝDn


D2 = [(8 – 2 + 1) / 36] (25,000 – 4,000) = LE4083

D3 = [(8 – 3 + 1) / 36] (25,000 – 4,000) = LE3500

The book value for year three:

BV(3) = 25,000 – [4667 + 4083 + 3500] = LE12,750

4.1.3 Sinking fund method

This method assumes that a uniform series of end-of-payments are deposited into an

imaginary sinking fund at a given interest rate i. The amount of the annual deposit is

calculated so that the accumulated sum at the end of the asset life, and at the stated

interest rate, will just equal the value of the asset depreciated (i.e., P – F). The amount of

yearly depreciation is invested in a compound manner for the remaining period as a

uniform series of payments using Eq. 4.4 as follows:

Then the depreciation value, Dn, at any year n is calculated using the following equation.

Dn = A × (1 + i)n-1

; n = 1, 2, 3, …….. ……., N (4.5)

Example 4.5: Resolve Example 4.2 using the sinking fund depreciation method,

assuming that the interest rate is 10%.

Solution:

P = 60,000; F = 6,000; N = 8; i =10%

A = (60000 – 6000) × [(0.1) / (1.18 – 1)] = LE4,722

Accordingly, the annual depreciation could be calculated as follows:

At the first year: D1 = LE4,722

At the second year: D2 = 4722 × (1.1) = LE5,194

At the third year: D3 = 4722 × (1.1)2 = LE5,714

……………..

At the eighth year: D8 = 4722 × (1.1)7 = LE9,202

The results of the depreciation calculations are summarized in Table 4.3.

)4.4(1)1(

)(

ni

iFPA


Table 4.3: Sinking fund depreciation of Example 4.5

Year Annual

depreciation Book value

0 0 60,000

1 4,722 55,278

2 5,194 50,084

3 5,714 44,370

4 6,285 38,085

5 6,913 31,172

6 7,605 23,567

7 8,365 15,202

8 9,202 6,000

After studying depreciation calculations from the previous listed three methods,

Figure 4.1 illustrates the difference between the three methods. The figure shows that

the sum-of-year digits method gives an accelerated depreciation compared to the

straight-line method. On the other hand, the sinking fund is a decelerated method

compared with the straight-line method. However, the straight-line method is the

commonly used for calculating asset depreciation.

Figure 4.1: Comparison among the three depreciation methods

Age

Bo

ok

val

ue

Salvage

value

Initial

value

Sinking

fund

Sum-of

years

Straight-

line


4.2 Estimating Equipment Costs (Rentals)

The cost per unit of time of owning an item of equipment has to be determined. Costs

associated with owing equipment called the ownership costs. Estimating equipment cost

involves identifying the ownership and operating costs. Ownership costs include: initial

cost, financing (investment) costs, depreciation costs and taxes and insurance costs. The

operating costs include: maintenance and repair costs, storage costs and fuel and

lubrication costs.

4.2.1 Initial cost

The initial cost is the total cost required to purchase a piece of equipment. This initial

cost is the basis for determining other costs related to ownership as well as operating

costs. Generally, initial cost is made up of: price at the factory or used equipment price,

extra options and accessories, sales tax, freight and assembly or setup charges. The initial

cost is very straight forward, whereas the other costs require more analysis and

computation. Te annual depreciated cost of the equipment should be calculated as

described in the previous section.

4.2.2 Investment cost

The purchase of construction equipment requires a significant investment of money. This

money either be borrowed from a lender, or it will be taken from reserve fund of the

contractor. Either the lender will charge an interest rate for the borrowed money, or the

contractor will lose any interest money that could be gained if the contractor invest that

amount of money used to purchase a piece of equipment.

In order to calculate the cost of finance (or investment cost) of an equipment, both the

purchase price, P, and the salvage value, F, should be converted into uniform annual

values. In this situation, the purchase price is considered as a present value invested for n

yeas as a series of uniform payments (equipment useful life) and the salvage value is

considered as a future sum of money to be discounted for n years as a series of uniform

payments.


Example 4.6: An excavator purchase price is LE460,000 and its salvage value is

LE40,000 after 10 years of useful life. Find the annual cost of finance of

this excavator if the annual interest rate is 15%.

Solution:

P = 460,000; F = 40,000; n = 10; i = 15%

Annual cost of finance =

= LE47,684/year

4.2.3 Operating costs

Operating cost accrue only when the unit of equipment is used, whereas ownership costs

accrue whether or not the equipment is used. Operating costs include maintenance and

repairs, fuel, oil and lubricants. The amounts consumed by a piece of equipment vary

with the type and size of equipment, the conditions under which it is operated. An

equipment is seldom used its total horse power and also it is seldom to work for 60

minute/hour. Thus, the fuel consumed should be based on the actual operating conditions.

Perhaps the average demand on an engine might be 50 percent of its maximum power for

an average 45 minutes/hour.

Maintenance and repair costs: The cost for maintenance and repairs include the

expenditures for replacement parts and the labor required to keep the equipment in

good working condition. Historical cost records of maintaining and servicing

equipment are the most reliable guide in estimating maintenance and repair cost.

The manufacturers of construction equipment provide information showing

recommended costs for maintenance and repairs for the equipment they

)6.4(1)1(1)1(

)1(cos

n

F

i

iF

n

P

i

iiPfinanceoftAnnual

nn

n

10

40000

1)15.1(

15.040000

10

460000

1)15.1(

)15.1(15.0460000

1010

10


manufacture. The annual cost of maintenance and repairs is often expressed as a

percentage of purchase prices or as a percentage of the straight-line depreciation

costs.

Fuel consumption: When operating under standard conditions, a gasoline engine

will consume approximately 0.06 gallon of fuel for each horsepower-hour

developed. A diesel engine will consume approximately 0.04 gallon of fuel for

each horsepower-hour developed.

Lubricating oil consumption: The quantity of lubricating oil consumed by an

engine varies with the size of the engine, the capacity, the equipment condition

and the number of hours between oil changes.

Cost of rubber tires: Many types of construction equipment use rubber tires,

whose life usually will not be the same as the equipment on which they are used.

For example, a unit of equipment may have an expected useful life of six years,

but the tires on the equipment may last only for two years. Therefore, a new set of

tires must be placed on the equipment every two years, which would require three

sets of tires during the six years the equipment will be used. Thus, the cost of

depreciation and repairs for tires should be estimated separately from the

equipment.

Example 4.7: Calculate the ownership cost per hour for an excavator powered by a

250-hp engine based on the following data:

- Purchase price (P) = LE420,000

- Salvage value (F) = LE250,000

- Operation factor = 50%

- Useful life (N) = 6 years

- Working hours per year = 2000

- Maintenance and repair costs = 110% of annual depreciation

- Diesel fuel price = 3.8/gallon


- Fuel consumption = 0.04 gallon/hp/hr

- Lube oil cost = 10% of fuel

- Interest rate (i) = 10%

Solution:

Depreciation (assume straight-line) = (420000 – 250000) / 6

= LE28333.33/year

Investment annual cost is calculated as follows:

Annual investment = (420000 × 0.2296 – 70000) – (250000 × 0.1296 –

41666.67) = 26432 – (- 9264.82) = LE35696.82/year

Maintenance and repair cost = 1.1 × 28333.33 = LE31166.67/year

Then, the total yearly costs = 28333.33 + 35696.82 + 31166.67

= LE95196.81/year

Accordingly, the hourly cost = 95196.81 / 2000 = LE47.6/hr

Fuel consumption = 250 × 0.04 × 0.5 = 5 gallon/hr

Fuel cost = 5 × 3.8 = LE19/hr

Lubricate oil cost = 19 × 0.1 = LE1.9/hr

Finally, the total hourly cost = 47.6 + 19 + 1.9 = LE68.5/hr

Example 4.7: Calculate the hourly rate of equipment based on the following data:

- Purchase price (P) = LE460,000

- Salvage value (F) = LE40,000

- Useful life (N) = 10 years

- Working hours per year = 2000 years

- Annual maintenance costs = 10% of purchase price

- Annual operating costs = LE47,000


6

250000

1)1.1(

1.0250000

6

420000

1)1.1(

)1.1(1.0420000

66

6


Solution:

Depreciation (assume straight-line) = (460000 – 40000) / 10 =

LE42000/year

Investment annual cost is calculated as follows:

Annual investment = LE47684/year

Maintenance and repair cost = 0.1 × 460000 = LE46000/year

Operating costs = LE47000/year

Then, the total annual costs = 42000 + 47684 + 46000 + 47000

= LE182684/year

Accordingly, the hourly cost = 182684/ 2000 = LE91.34/hr

4.2.4 Calculate the equipment rentals using cash flows

In this method, the cash flow for both the costs and the revenues are calculated and the

net present worth is calculated so that the net present worth equal zero taking into

account the time value of money.

Example 4.8: Resolve example 4.7 using the cash flows method.

Solution:

Let’s assume that the annual return from renting the equipment is x,

accordingly, the cash flows could be summarized as given in Table 4.4.

Annual costs = annual maintenance + operating cost

= 0.1 × 460000 + 47000 =

= 46000 + 47000 = LE93,000

10

40000

1)15.1(

15.040000

10

460000

1)15.1(

)15.1(51.0460000

1010

10


Table 4.4: Cash flows of Example 4.8

Year Initial

cost

Salvage

value

Annual

costs Cash in Cash out Net cash

0 460000 0 0 -460000 -460000

1 -93000 x -93000 x - 93000

2 -93000 x -93000 x - 93000

3 -93000 x -93000 x - 93000

4 -93000 x -93000 x - 93000

5 -93000 x -93000 x - 93000

6 -93000 x -93000 x - 93000

7 -93000 x -93000 x - 93000

8 -93000 x -93000 x - 93000

9 -93000 x -93000 x - 93000

10 40000 -93000 x + 40000 -93000 x - 53000

By equating the PV for both the costs and revenues (i.e., NPV =0), as

follows:

0 = -460000 + (x – 93000) (P/A, 15%, 10) + 40000 (P/F, 15%, 10)

= -460000 + (x – 93000) (5.0188) + 40000 (0.2472)

x – 93000 = 89685

Then, x = LE182,685 / year

The hourly rent = 182685 / 2000 = LE91.34/hr

4.3 Sensitivity Analysis

All the previous analysis assumes that the data used for an economic decision is

deterministic and certain. Also, since many data gathered in solving a problem represents

projections of future consequences, there may be considerable uncertainty regarding the

accuracy of that data. As the desired results of the analysis is decision making, as

appropriate question is, “to what extent do variations in the data affect my decision?”

What variations in a particular estimate would change selection of the alternative? In

such case, the decision is said to be sensitive to the estimate.


The sensitivity analysis, as such, aims to check if the economic decision changes as one

or more of the data used change such as the MARR, service life, annual maintenance

cost, etc. This section is dedicated to show how the change of these values may change

the economic decisions.

To perform sensitivity analysis, follows the following steps:

- Identify the factors(s) that values may change than that has/have been identified.

- Identify the range of values change for this/these factor(s).

- Identify the economic decision criterion (e.g., NPW, EUAW, IRR, etc.) that will

be used to perform the economic analysis.

- Calculate the values of this economic criterion at different levels of the

changeable factor.

- Draw a diagram for the obtained results from the previous step to study the

effect of these changes on the economic criterion under study.

Example 4.9: The purchasing cost of a given equipment is LE90,000 and its return

for the first year is LE30,000 decreasing annually by LE3,000. If the

investment rate changes between 10% and 25% and the equipment age

ranges from 8 to 12 years. It is required to study the sensitivity of the

decision considering the effect of the change of the investment rate and the

age using the EUAW method. Neglect the equipment salvage value.

Solution:

The cash flow could be represented as follows:

G = 3,000

LE30,000

LE90,000


The sensitivity of the decision to the change is the investment rate:

In this case, let’s assume that the average age of the equipment is 10 years.

Then, calculate the EUAW at different values for i, (10, 15, 20 and 25%)

i = 10%

EUAW = -90000 (A/P, 10%, 10) + 30000 – 3000(A/G, 10%, 10)

= -90000 (0.16274) + 30000 – 3000 (3.7255)

= LE4177

Similarly, The EUAW could be calculated at different i as follows:

i = 15% EUAW = LE1919

i = 20% EUAW = - LE687

i = 25% EUAW = - LE3600

The sensitivity of the decision to the change is the equipment age:

In this case, let’s assume that the average investment rate is 15%.

Then, calculate the EUAW at different age values (8, 10 and 12 years).

n = 8 EUAW = LE1601

n = 10 EUAW = LE1919

n= 12 EUAW = LE1673

The previous results show that all the values of the EUAW remain positive

for all values of n (in the specified range). Accordingly, the decision does

not affected with the change of the equipment life.

On the other hand, the change of the investment rate changes the values of

the EUAW from being positive to negative. Accordingly, the decision

changes with the change of the i values.

Example 4.10: A test vehicle has a cost of LE100,000 with a life of 10 years.

Additional revenues of LE17,500 per year are expected and the required

MARR is 15%. If the additional revenues are only estimates and might

fluctuate 15%. Evaluate the sensitivity of the analysis.


Solution:

EUAW= -100,000(A/P, 15%, 10) + Revenue

= - 19,925 + Revenue

Then, calculate the EUAW at different values for the revenue within the

range of ±15% (14875 to 20125).

Example 4.11: A company wants to purchase a new core driller for information

gathering. It is expected that the machine can be purchased for LE275,000

it will last 8 years, expenses will total LE50,000 per year and that the

revenues will be LE100,000 per year. The company requires a MARR of

10%.

Revenue EUAW

14,875 -5,050

15,750 -4,175

16,625 -3,300

17,500 -2,425

18,375 -1,550

19,250 -675

20,125 200


a. Determine the sensitivity of the present worth to a change in the

MARR.

b. What would be the effects of a 10% range to all other parameters?

Solution:

a. PW = -275,000 - 50,000(P/A, i%, 8) + 100,000(P/A, i%, 8)

= -275,000 + 50(P/A, i%, 8)

b. By changing the cost, expenses and revenues by 10% (-10% to +10%)

at the interval of 5%, the effect on the present worth would be as

shown in the following table.

MARR (P/A, i%, 8) PW

6 6.46321 48160.5

7 5.9713 23565

8 5.74664 12332

9 5.53482 1741

10 5.33493 -8253.5


Range Cost PW Cost Exp. PW Exp. Rev. PW Rev.

-10 247500 39832 45000 41065.2 90000 -45134.4

-5 261250 26082 47500 26698.6 95000 -16401.2

0 275000 12332 50000 12332 100000 12332

5 288750 -1418 52500 -2034.6 105000 41065.2

10 302500 -15168 55000 -16401 110000 69798.4

4.4 Breakeven Analysis

A fundamental of accounting is that all revenues and costs must be accounted for and the

difference between the revenues and costs is the profit, or loss, of the business. Costs can

be classified as either a fixed cost or a variable cost. A fixed cost is one that is

independent of the level of sales; rather, it is related to the passage of time. Examples of

fixed costs include rent, salaries and insurance. A variable cost is one that is directly

related to the level of sales, such as cost of goods sold and commissions. In planning and

managing your business it is important to know what level of sales must be achieved in

order to meet total costs. Every LE of sales above this will contribute to profits.

Advantages of Break-even Analysis

- It is simple to conduct and understand.

- It shows profit and loss at different levels of output.

- It can cope with changing circumstances, e.g. the following changes in the

business environment can be shown in a break even chart.

Disadvantages of Break-even Analysis

- It assumes that all output is sold at the given price (this may well be untrue).

- Although it can cope with changes in circumstances, these factors change

regularly reducing its usefulness as a forecasting tool.

- The model assumes that costs increase constantly and do not benefit from

economies of scale. If the firm obtains purchasing economies of scale then its

total cost line will no longer be straight.


- Break-even analysis is only as good as the data upon which it is based. Poor

quality data will lead to inaccurate conclusions being drawn.

Example 4.12: A factory produces concrete blocks units. Each unit sells for LE15

and costs LE5. The annual maintenance and operation costs are LE75,000.

Calculate the number of blocks that should be produced to justify keep this

business running.

Solution:

Let’s assume that number of units produced is x. Accordingly, the cost of

blocks equal 5x and the annual revenue is 15x.

The annual cost = 75000 + 5x

At breakeven point when total costs equal total revenue:

75000 + 5x = 15x

Then, x = number of units produced = 7500 blocks

The investment in this project would be acceptable when the production

increases than 7500 units per year, otherwise it is rejected.

Total cost

Revenue

Fixed cost

Units

Cost


In case of evaluating alternatives using the breakeven analysis, the breakeven point

represents the point at which the total cost for alternatives are equal considering that all

alternatives have equal revenue. In case of having alternatives with different revenues,

the breakeven for each alternative is identified first, then the breakeven between each two

alternatives is identified as follows:

- Identify the common criterion (IRR, NPW, etc.) among alternatives.

- Calculate the cost in one of these criteria.

- For each two alternatives, find the value at which the two alternatives have

equal cost.

- Represent the results graphically and compare between alternatives on the

breakeven point.

Example 4.13: There are two proposals to buy a new production line for brick

manufacturing. The information related to both alternatives is shown in the

following table. If the interest rate is 10%, find:

a. The volume of production that justifies the purchase of Alternative A.

b. If the demand on the production of this factory is 2000 ton annually,

which alternative do you recommend for purchasing?

Alternative A B

Initial cost LE230,000 LE80,000

Annual costs LE35,000 LE15,000

Salvage value LE40,000 -

Labor cost (EL/ton) 15 40

Age (years) 10 5

Solution:

First, let’s calculate the EUAW for both alternatives:

EUAWA = 230000(A/P, 10%, 10) + 35000 – 40000(A/F, 10%, 10)

= LE69,922

EUAWB = 80000(A/P, 10%, 5) + 15000


= LE36,103

Assume that x represents the annual production (ton/year), the direct costs

for both alternatives could be represented as:

Direct costA = 15x

Direct costB = 40x

Accordingly, the total cost for both alternatives are:

Total costA = 69922 + 15x

Total costB = 36103 + 40x

By equating the total cost for both alternatives, then, the breakeven point

equals; x = 1353 ton/year

a. Thus means that, if the annual production is more than 1353 ton, then

alternative A is better.

b. In case of the required production is 2000, then alternative A is better.

As shown in the figure below, if the production is less than 1353 ton

annually then it is better to use alternative B. While alternative A is

better in case of production is more that 1353 ton annually.

A

B


Example 4.14: It is required to calculate the breakeven point for the three options

listed in the table below. Which alternative do you recommend?

Alternative Annual fixed

cost

Variable cost

(LE/ unit)

Sell price

(LE/unit)

A LE30,000 17 27

B LE50,000 12 35

C LE60,000 10 40

Solution:

First, calculate the breakeven point for each alternative (the point at which

the total costs and the revenue are equal).

Alternative A:

30000 + 17x = 27x; then x = 3000 units

Alternative B:

50000 + 12 x = 35x; then x = 2174 units

Alternative C:

60000 + 10 x = 40z; then x = 2000 units

Second, draw the relation between the total cost and the number of units

produced.

B

A

C


To find the points of intersection, you may solve each two equations

together or you may find it from the graph shown above.

The above shows the following:

- If the production is less than 4000 units, then alternative A is the

best.

- If the production is greater than 4000 and less than 5000, then

alternative B is the best.

- Finally, if the production is greater than 5000 units, then alternative

C is the best.

4.5 Exercises

1. A company purchased a piece of equipment 3 years ago with an initial value of

LE15,000, salvage value of LE3,000, annual operating cost of LE2,000, and

estimated life of 10 years. Calculate the book value of the machine now using the

straight-line, sum-of years digits and sinking fund depreciation method. Assume

interest rate 10%.

2. A backhoe will be purchased for a cost of LE109,750. After a useful life of 5

years, it is assumed the equipment will be sold for LE35,000. Assume interest of

8% for borrowing money, 4% for risk and 2% for taxes, insurance and storage.

Calculate the annual ownership cost and the cost per hour assuming the

equipment will be used 1800 hr/year.

3. Calculate the ownership cost per hour for a dump truck powered by a 120-hp

gasoline engine based on the following data:

- Purchase price = LE175,000

- Freight charges = LE2,000

- Estimated salvage value = LE57,500

- Operation factor = 40%

- Useful life = 5 years

- Hours used per year = 1800


- Maintenance and repair = 130% of annual depreciation

- Tire cost = LE5,000

- Tire life = 4,000 hours

- Maintenance and repairs (tires) = 15% of tire depreciation

- Gasoline fuel price =LE4.0/gallon

- Fuel consumption = 0.06 gallon/hp/hr

- Lube oil cost = 10% of fuel


4. When studying the different alternative air conditioning systems for a building,

there were two available systems with their information as shown in the table

below. It is required to calculate the decision of selecting any of the two systems

to if the interest rate ranges from 8% to 15%. Use the EUAW method.

Alternative A B


Annual costs LE2,000 LE1,500

Salvage value LE5,000 LE10,000

Maintenance at mid-age LE20,000 -

Age (years) 8 12

5. Find the breakeven point for the following two alternatives. Assume that the

investment rate is 10%. Which one do you select if the expected production is

2000 m3/year?

Alternative Equipment A Equipment B


Maintenance cost/year LE3,500 LE1,500

Salvage value LE4,000 -

Labor cost/hr LE12 24

Age (years) 10 5

Production (m3/hr) 8 6


REFERENCES

Newman, D.G., and Lavelle, J.P., Engineering Economic Analysis, 7th

edition,

Engineering Press, Austin, Texas, 1998.

Ammar, M., Principles of Engineering Economy, Lecture Notes, Tanta University, 2008.

Griffis, F.H., Farr, J.V., and Morris, M.D., Construction Planning for Engineers,

McGraw-Hill, Inc., New York, 2000.

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