DEPRECIATION

Depreciation is the decrease in the value of a property, such as machinery,

equipment, building or other structure, due to the passage of time.

An asset starts to lose value as soon as it is purchased. For example, a car

bought for Php 1 200 000 today may be worth Php 1 000 000 next week, Php 950
000 next year, and Php 500 000 in 10 years. This loss in value, called depreciation,
occurs for several reasons.

Use-related physical loss: As something is used, parts wear out. An

automobile engine has a limited life span because the metal parts within it wear
out. This is one reason why a car diminishes in value over time. Often, use-related
physical loss is measured with respect to units of production, such as thousands of
kilometers for a car, hours of use for a light bulb, or thousands of cycles for a punch
press.

Time-related physical loss: Even if something is not used, there can be a

physical loss over time. This can be due to environmental factors affecting the
asset or to endogenous physical factors. For example, an unused car can rust and
thus lose value over time. Time-related physical loss is expressed in units of time,
such as a 10-year-old car or a 40-year-old sewage treatment plant.

Functional loss: Losses can occur without any physical changes. For
example, a car can lose value over time because styles change so that it is no
longer fashionable. Other examples of causes of loss of value include legislative
changes, such as for pollution control or safety devices, and technical changes.
Functional loss is usually expressed simply in terms of the particular unsatisfied
function.

 Value of an Asset

Models of depreciation can be used to estimate the loss in value of an asset over
time, and also to determine the remaining value of the asset at any point in time.
This remaining value has several names, depending on the circumstances.

Terms and Definitions:

V. Market Value (MV) - is what will be paid by a willing buyer to a willing

seller for a property where each has equal advantage and is under no
compulsion to buy or sell.
VI. Salvage or residual value (SV) - is the price that can be obtained from the
sale of the property after it has been used. It is the best estimate of an
asset’s net market value at the end of its useful life. If used in depreciation
 calculations, it is referred to as estimated salvage (ES), representing an
asset’s terminal value.
VII. Useful life sometimes referred to as depreciable life, is the expected
period of time that a property will be used in a trade or business or to
produce income. It is not how long the property will last but how long the
owner expects to productively use it.
VIII. Book Value (BV) is the worth of a property as shown on the accounting
records of a company. Ordinarily, it means the original cost of the
property less all amounts that have been charged as depreciation
expense.

It is desirable to be able to construct a good model of depreciation in order to

state a book value of an asset for a variety of reasons:

1. In order to make many managerial decisions, it is necessary to know the

value of owned assets. For example, money may be borrowed using the
firm’s assets as collateral. In order to demonstrate to the lender that there is
security for the loan, a credible estimate of the assets’ value must be made.
A depreciation model permits this to be done. The use of depreciation for
this purpose is explored more thoroughly in the second part of this chapter.

2. One needs an estimate of the value of owned assets for planning purposes.
In order to decide whether to keep an asset or replace it, you have to be
able to judge how much it is worth. More than that, you have to be able to
assess how much it will be worth at some time in the future.

3. Government tax legislation requires that taxes be paid on company

profits. Because there can be many ways of calculating profits, strict rules
are made concerning how to calculate income and expenses. These rules
include a particular scheme for determining depreciation expenses.

 Purpose of Depreciation

a) To provide for the recovery of capital which have been invested in

physical property.

b) To enable the cost of depreciation to be changed to the cost of

producing products or services that results from the use of the property.
  Types of Depreciation

1. Normal Depreciation

o Physical Depreciation is the result of deterioration of an asset due to

age and wear. It results from use, decay and the action of the
elements. Physical depreciation is a constant factor. It begins as
soon as an asset is exposed to the action of the elements or is put
into use.

o Functional depreciation is the depreciation provided as a result of

lack of adaptation of an asset to function. It results from change of
conditions and surroundings which render the asset ill adapted to its
work, from the growth of business which renders the asset
inadequate, or to the decline of business.

2. Depreciation due to changes in price levels is atmost impossible to

predict and therefore is not considered in economy studies.

3. Depletion refers to the decrease in the value of a property due to the

gradual extraction of its contents.

 Physical and Economic Life

Physical Life of a property is the length of time during which it is capable of

performing the function for which it is designed and manufactured.

Economic Life is the length of time during which the property may be
operated at a profit.

Requirement of a Depreciation Method

1. It should be simple
2. It should recover capital

3. The book value will be reasonably close to the market value at any
time.

4. The method should be accepted by the Bureau of Internal Revenue

DEPRECIATION METHODS:

 Time-Independent Methods
1. Unit Production Method
2. Working Hours Method

 Time-Dependent Methods
1. Straight Line Method
2. Sinking Fund Method
3. Sum-of-years-digits
4. Declining Balance Method

We shall use the following symbols for the different depreciation methods.

FC = First cost

Dn = total depreciation ant the end of n years

Dr = accumulated depreciation at the end of r periods.

BVr = Book value at the end of periods.

dr = annual depreciation

SV= Salvage Value

 TIME-INDEPENDENT METHODS

1. UNIT PRODUCTION METHOD

Example:
A Xerox machine costing P75,000 with a salvage value of P10,000 at the end
of its life can print 3 million sheets in its entire life. If it has already printed
500,000 sheets, find the total depreciation at that time.

Solution:

FC=75,000
= 10,000

0.5M 3M

Note: x represents the number of printed sheets.

Dn = FC-SV = 65,000

By Ration and Proportion:

𝐷𝑟 65,000
=
500,000 3,000,000

Dr = P10833.33
 2. WORKING HOURS METHOD

Example:
A lathe machine costing 3M has a salvage value of P0.1M at the end of 20
years. It can be used for 5000 hours in its entire life. Find the accumulated
depreciation after using it for 1000 hours.

Solution:

= P2.9M
FC= P3M

= P0.1M

1000 hrs 5000 hrs

By Ration and Proportion:

𝐷𝑟 2.9𝑀
=
1000 ℎ𝑟𝑠 5000

Dr = P0.58M

 TIME-DEPENDENT METHODS

1. STRAIGHT LINE METHOD (SLM)

IX. It is the simplest and most widely used depreciation method.
X. It assumes that the loss in value is directly proportional to the age of the
property.
XI. In the straight line method the depreciation base is evenly allocated over
the lifetime of the asset, resulting in equal annual depreciation

Where:

FC = First cost
Dn = total depreciation ant the end of n years
Dr = accumulated depreciation at the end of r
periods.
BVr = Book value at the end of periods.
SV= Salvage Value
d = annual depreciation/depreciation charge
Dn
d=
n
Example:
A company has purchased an equipment whose first cost is
P100,000 with an estimated life of eight years. The estimated salvage
value of the equipment at the end of its lifetime is P20,000. Determine
the depreciation charge and book value at the end of various years
using the straight line method of depreciation.

Solution:

Given: FC = P100,000
SV = P20,000
n = 8 years

Dn = FC-SV = P80,000
𝐃𝐧 𝟖𝟎,𝟎𝟎𝟎
dr= = =P10,000 - Ans. depreciation charge
𝐧 𝟖

In this method of depreciation, the value of d is the same for all the years.
The calculations pertaining to BVr for different values of r are summarized in the
table below.

Depreciation charge, d Book Value

100,000
BVr = BVr-1 - d
 2. SINKING FUND METHOD (SFM)
 This method depreciates an asset as if the firm were to make a series of
equal deposits whose value at the end of the asset’s useful life just equal
the cost of replacing the asset.
 It is assumed that a sinking fund is established in which funds will
accumulate for replacement purposes and will bear interest.
 The total depreciation which has occurred up to any given time is
assumed to equal the amount in the sinking fund at that time.

Where:

FC = First cost
Dn = total accumulated depreciation at the
end of n years
Dr = accumulated depreciation at the end of r
periods.
BVr = Book value at the end of periods.
SV= Salvage Value
d = annual depreciation/depreciation charge

Formula:
(𝟏+𝒊)𝒏−𝟏
Dn = 𝒅 ( )
𝒊

(𝟏+𝒊)𝒓 −𝟏
Dr = 𝒅 ( )
𝒊

Example:
A company has purchased an equipment whose first cost is P100,000 with
an estimated life of eight years. The estimated salvage value of the equipment
at the end of its lifetime is P20,000. Determine the book value of the equipment
after 5 years using Sinking Fund Method. i = 12%

Solution:
Given: FC = P100,000
SV = P20,000 = P80,000
n = 8 years FC= P100,000

Dn = FC-SV = P80,000 = P20,000

r n=8
 First, determine depreciation charge:

(1+𝑖)𝑛 −1
Dn = 𝑑 ( )
𝑖
8
(1+0.12) −1
80,000 =𝑑 ( ), d = 6504.23
0.12

 Compute for the accumulated depreciation after 5 years:

(1+𝑖)𝑟−1
Dr = 𝑑 ( )
𝑖
5
(1+0.12) −1
D5 = 6504.23 ( ) , D5 = 41320.38
0.12
 Solve for the Book value at r=5
BVr = FC- Dr
BV5 = 100,000 – 41,320.38
BV5= P58679.62

3. SUM-OF-YEARS-DIGITS METHOD (SYD)

 It provides very rapid depreciation during the early years of life of the
property, and therefore enables faster recovery of capital.
 The basic assumption for this method is that the value of the property
decreases at a decreasing rate.
 To compute:
1. List the digits corresponding to the number for each permissible year
of life in reverse order
2. Determine the sum of these digits
3. For any year, the depreciation factor is the number from the
reversed-order listing for that year divided by sum of the digits.
4. The depreciation deduction for any year is the product of the SYD
depreciation factor that year and the difference between the cost
basis/FC and the SV.

Where:
d1 = depreciation charge during first year
only.
d2 = depreciation charge during second
year only.
Formula:

dr = (SYD factor)(Dn) Or

(𝑛−𝑟+1)
dr = 𝐷𝑛
1+2+3…..+𝑛
Example 1:
A new electrical saw for cutting small pieces of lumber in a furniture
manufacturing plant has a cost basis of $4,000 and a 10-year depreciable life.
The estimated SV of the saw is zero at the end of 10 years. Determine the annual
depreciation amounts using the straight-line method. Tabulate the annual
depreciation amounts and the book value of the saw at the end of each year.
Use SYD.

Solution:
(Using table)
1. Find for the depreciation factor each year.

2. Solve for depreciation charge and BV for each year using the formula
Ex: dr = (SYD factor)(Dn), Dn = FC-SV = 4000 – 0 = 4000
10
d1 = (55) (4000) = $ 727.27
9
d2 = ( ) (4000) = $654.55
55

The depreciation and book value amounts for each year are shown below.
EOY, r dr BVr
 4. DECLINING BALANCE METHOD (DBM)
 It is assumed that the annual cost of depreciation is a fixed percentage of
the BV at the beginning of the year.
 The ratio of the depreciation is a fixed percentage of the BV at the
beginning of the year is constant throughout the life of the asset and is
designated by k (0 ≤ k ≤ 1).
 The annual depreciation charges, different each year, decrease from
year to year, greatest during the 1st year and least in the last year of life of
the property.
 With this method, a property can never depreciate to zero value.

Formula:
BVr = FC(1 − k)𝑟 𝑟 𝐵𝑉𝑟 Where: k = annual rate of
Or 𝑘 =1− depreciation
𝑟 𝐹𝐶
𝑆𝑉 𝑛
BVr = FC (𝐹𝐶 ) Or

Sv = FC(1 − k)𝑛 𝑛 𝑆𝑉
𝑘 = 1−
𝐹𝐶
dr = k FC(1 − k)𝑟−1

Example:

A company has purchased an equipment whose first cost is P100,000 with an

estimated life of eight years. Determine the (a) depreciation charge in its 3rd
year, (b) accumulated depreciation after 3 years, (c) Book value of the
equipment after 3 years using Declining balance method (DBM) of depreciation
if the annual rate of depreciation k is 0.2.

Solution:

FC = P100,000
n = 8 years
k = 0.2

(a) Depreciation charge @ 3rd year.

dr = k FC(1 − k)𝑟−1
d3= 0.2(100,000)(1-0.2)3-1
d3 = P12,800

(b) Total accumulated depreciation after 3 years, D3.

 D3 = d1 + d2 + d3

 Solving for d1, d2 and d3 using the formula dr = k FC(1 − k)𝑟−1 , we get:
d1 =0.2(100,000)(1-0.2)1-1 = P20,000
d2 = 0.2(100,000)(1-0.2)2-1 = P16,000
d3 = P12,800
Therefore,
D3 = 20,000 + P16,000 + 12,800
D3 = P48,800

(C) Book value at year 3.

BV3 = FC – D3
= 100,000 – 48,800
BV3 = P51,200
Or by using the formula,
BVr = FC(1 − k)𝑟

BV3 = 100,000(1 − 0.2)3

BV3 = P51,200

5. DOUBLE DECLINING BALANCE METHOD

In this method, k = 2/N when a 200% declining balance is being used.

Formula:

BVr = FC(1 − k)𝑟 Where: k = 2/N annual rate of depreciation

Or

Sv = FC(1 − k)𝑛

dr = k FC(1 − k)𝑟−1
Example:

Determine the rate of depreciation, the total depreciation up to the end of

the 8th year and the book value at the end of 8 years for an asset that costs Php
15, 000 new and has an estimated scrap value of Php 2, 000 at the end of 10 years
by (a) the declining balance method and (b) the double declining balance
method.

Solution:
𝐹𝐶 = Php 15, 000 𝑆𝑉 = Php 2, 000 𝑛 = 10 r= 8
(a) Declining Balance Method
𝑛 𝑆𝑉 10 𝑃ℎ𝑝 2, 000
k=1− √ =1− √ = 0.1825 or 18.25%
𝐹𝐶 𝑃ℎ𝑝 15, 000

𝐵𝑉8 = 𝐹𝐶 (1 − 𝑘 )8 = 𝑃ℎ𝑝 15, 000 (1 − 0.1825) 8 = 𝑃ℎ𝑝 2, 992

𝐷8 = 𝐹𝐶 − 𝐵𝑉8 = 𝑃ℎ𝑝 15, 000 − 𝑃ℎ𝑝 2, 992 = 𝑷𝒉𝒑 𝟏𝟐, 𝟎𝟎𝟖

(b) Double Declining Balance Method

2 2
Rate of depreciation = = = 0.20 𝑜𝑟 20%
𝐿 20

2 8 2 8
𝐵𝑉8 = 𝐹𝐶 (1 − ) = 𝑃ℎ𝑝 15, 000 (1 − ) = 𝑃ℎ𝑝 2, 517
𝑁 10
𝐷8 = 𝐹𝐶− 𝐵𝑉8 = 𝑃ℎ𝑝 15, 000 − 𝑃ℎ𝑝 2, 517 = 𝑃𝒉𝒑 𝟏𝟐, 𝟒𝟖𝟑
AFTER– TAX ECONOMIC ANALYSIS

Taxes are fees paid by individuals or businesses to support their government.

Taxes are generally the primary source of government revenue, and are used to
provide public goods and services such as highways and dams, police services,
the military, water treatment, education, health care, and other social programs.

Taxes can have a significant impact on the economic viability of a project

because they change the actual cash flows experienced by a company. A vital
component of a thorough economic analysis will therefore include the tax
implications of an investment decision.

Income tax is the amount of the payment (taxes) on income or profit that
must be delivered to a federal (or lower-level) government unit. Taxes are real
cash flows; however, for corporations tax computation requires some noncash
elements, such as depreciation. Corporate income taxes are usually submitted
quarterly, and the last payment of the year is submitted with the annual tax return.

The required after-tax MARR is established using the market interest rate,
the corporation’s effective tax rate, and its weighted average cost of capital. The
CFAT (cash flow after taxes) estimates are used to compute the PW or AW at the
after-tax MARR. When positive and negative CFAT values are present, a PW or AW
< 0 indicates the MARR is not met.

One project. PW or AW ≥ 0, the project is financially viable because the after-

tax MARR is met or exceeded.

Two or more alternatives. Select the alternative with the best (numerically
largest) PW or AW value.

If only cost CFAT amounts are estimated, calculate the after-tax savings
generated by the operating expenses and depreciation. Assign a plus sign to
each saving and apply the guidelines above.

Remember, the equal-service assumption requires that the PW analysis be

performed over the least common multiple (LCM) of alternative lives. This
requirement must be met for every analysis—before or after taxes.
 Example 4.14

Paul is designing the interior walls of an industrial building. In some places, it is

important to reduce noise transmission across the wall. Two construction options—
stucco on metal lath (S) and bricks (B)—each have about the same transmission
loss, approximately 33 decibels. This will reduce noise attenuation costs in
adjacent office areas. Paul has estimated the first costs and after-tax savings
each year for both designs. Use the CFAT values and an after-tax MARR of 7% per
year to determine which is economically better.

PLAN S PLAN B

Year CFAT $ Year CFAT $

0 -28, 000 0 - 50, OOO

1–6 5, 400 1 14, 200

7 – 10 2, 040 2 13, 300

10 2, 792 3 12, 400

4 11, 500

5 10, 600

Solution:
𝑃 𝑃 𝑃
𝐴𝑊𝑆 = [−28, 000 + 5400 ( , 7%, 6) + 2040 ( , 7%, 4) ( , 7%, 6)
𝐴 𝐴 𝐹
𝑃 𝐴
+ 2792 ( , 7%, 10)] ( , 7%, 10)
𝐹 𝑃

AW𝑆 = $ 422

𝑃 𝑃
𝐴𝑊𝐵 = [−50, 000 + 14, 200 ( , 7%, 1) + 13, 300 ( , 7%, 2)
𝐹 𝐹
𝑃 𝑃
+ 12, 400 ( , 7%, 3) + 11, 500 ( , 7%, 4)
𝐹 𝐹
𝑃 𝐴
+ 10, 600( , 7%, 5)] ( , 7%, 5)
𝐹 𝑃

AW𝐵 = $ 327

Both plans are financially viable; select plan S because AW𝑆 is larger
REPLACEMENT STUDY

The need for a replacement study can develop from several sources:

Reduced performance. Because of physical deterioration, the ability

to perform at an expected level of reliability (being available and
performing correctly when needed) or productivity (performing at a given
level of quality and quantity) is not present. This usually results in increased
costs of operation, higher scrap and rework costs, lost sales, reduced
quality, diminished safety, and larger maintenance expenses.

Altered requirements. New requirements of accuracy, speed, or

other specifications cannot be met by the existing equipment or system.
Often the choice is between complete replacement or enhancement
through retrofitting or augmentation.

Obsolescence. International competition and rapidly changing

technology make currently used systems and assets perform acceptably
but less productively than equipment coming available. The ever-
decreasing development cycle time to bring new products to market is
often the reason for premature replacement studies, that is, studies
performed before the estimated useful or economic life is reached.

Replacement studies use some terminology that is closely related to terms in

previous chapters.

Defender and challenger are the names for two mutually exclusive
alternatives. The defender is the currently installed asset, and the
challenger is the potential replacement. A replacement study compares
these two alternatives. The challenger is the “best” challenger because it
has been selected as the best one to possibly replace the defender. (This is
the same terminology used earlier for incremental ROR and BC analysis, but
both alternatives were new).

Market value is the current value of the installed asset if it were sold
or traded on the open market. Also called trade-in value , this estimate is
obtained from professional appraisers, resellers, or liquidators familiar with
the industry. As in previous chapters, salvage value is the estimated value
at the end of the expected life. In replacement analysis, the salvage value
at the end of one year is used as the market value at the beginning of the
next year.
 AW values are used as the primary economic measure of
comparison between the defender and challenger. The term equivalent
uniform annual cost ( EUAC) may be used in lieu of AW, because often only
costs are included in the evaluation; revenues generated by the defender
or challenger are assumed to be equal. (Since EUAC calculations are
exactly the same as for AW, we use the term AW.) Therefore, all values will
be negative when only costs are involved. Salvage or market value is an
exception; it is a cash inflow and carries a plus sign.

Economic service life (ESL) for an alternative is the number of years

at which the lowest AW of cost occurs . The equivalency calculations to
determine ESL establish the life n for the best challenger and the lowest cost
life for the defender in a replacement study. The next section explains how
to find the ESL.

Defender first cost is the initial investment amount P used for the
defender. The current market value ( MV ) is the correct estimate to use for
P for the defender in a replacement study. The estimated salvage value at
the end of one year becomes the market value at the beginning of the
next year, provided the estimates remain correct as the years pass. It is
incorrect to use the following as MV for the defender fi rst cost: trade-in
value that does not represent a fair market value , or the depreciated book
value taken from accounting records. If the defender must be upgraded
or augmented to make it equivalent to the challenger (in speed, capacity,
etc.), this cost is added to the MV to obtain the estimated defender fi rst
cost.

Challenger first cost is the amount of capital that must be recovered

(amortized) when replacing a defender with a challenger. This amount is
almost always equal to P , the fi rst cost of the challenger.

If an unrealistically high trade-in value is offered for the defender

compared to its fair market value, the net cash fl ow required for the
challenger is reduced, and this fact should be considered in the analysis.
The correct amount to recover and use in the economic analysis for the
challenger is its fi rst cost minus the difference between the trade-in value
(TIV) and market value (MV) of the defender. In equation form, this is P –
(TIV – MV).
 A sunk cost is a prior expenditure or loss of capital (money) that
cannot be recovered by a decision about the future. The replacement
alternative for an asset, system, or process that has incurred a
nonrecoverable cost should not include this cost in any direct fashion; sunk
costs should be handled in a realistic way using tax laws and write-off
allowances.

The nonowner’s viewpoint , also called the outsider’s viewpoint or

consultant’s viewpoint, provides the greatest objectivity in a replacement
study. This viewpoint performs the analysis without bias; it means the analyst
owns neither the defender nor the challenger. Additionally, it assumes the
services provided by the defender can be purchased now by making an
“initial investment” equal to the market value of the defender.

The fundamental assumptions for a replacement study parallel those of an AW

analysis. If the planning horizon is unlimited , that is, a study period is not specified,
the assumptions are as follows:

1. The services provided are needed for the indefinite future.

2. The challenger is the best challenger available now and in the future to
replace the defender. When this challenger replaces the defender
(now or later), it will be repeated for succeeding life cycles.
3. Cost estimates for every life cycle of the defender and challenger will
be the same as in their first cycle.

Example 4.15

Only 2 years ago, Techtron purchased for $275,000 a fully loaded SCADA
(supervisory control and data acquisition) system including hardware and
software for a processing plant operating on the Houston ship channel. When it
was purchased, a life of 5 years and salvage of 20% of fi rst cost were estimated.
Actual M&O costs have been $25,000 per year, and the book value is $187,000.
There has been a series of insidious malware infections targeting Techtron’s
command and control software, plus next-generation hardware marketed only
recently could greatly reduce the competitiveness of the company in several of
its product lines. Given these factors, the system is likely worth nothing if kept in
use for the fi nal 3 years of its anticipated useful life.

Model K2-A1, a new replacement turnkey system, can be purchased for $300,000
net cash, that is, $400,000 first cost and a $100,000 trade-in for the current system.
A 5-year life, salvage value of 15% of stated first cost or $60,000, and an M&O cost
of $15,000 per year are good estimates for the new system. The current system
was appraised this morning, and a market value of $100,000 was confirmed for
today; however, with the current virus discovery, the appraiser anticipates that
the market value will fall rapidly to the $80,000 range once the virus problem and
new model are publicized.

Using the above values as the best possible today, state the correct defender
and challenger estimates for P, M&O, S, and n in a replacement study to be
performed today.

Solution:

Defender: Use the current market value of $100,000 as the fi rst cost for the
defender. All others—original cost of $275,000, book value of $187,000, and trade-
in value of $100,000—are irrelevant to a replacement study conducted today.
The estimates are as follows:

First cost P = $100,000

M&O cost A = $25,000 per year

Expected life n = 3 years

Salvage value S=0

Challenger: The $400,000 stated first cost is the correct one to use for P, because
the trade-in and market values are equal.

First cost P = $400,000

M&O cost A = $15,000 per year

Expected life n = 5 years

Salvage value S = $60,000

If the replacement study is conducted next week when estimates will have
changed, the defender’s first cost will be $80,000, the new market value
according to the appraiser. The challenger’s first cost will be $380,000, that is, P –
(TIV – MV) = 400,000 – (100,000 – 80,000).
 o Performing a Replacement Study

Replacement studies are performed in one of two ways: without a study period
specified or with one defined. Figure 4–4 gives an overview of the approach
taken for each situation. The procedure discussed in this section applies when no
study period (planning horizon) is specified. If a specific number of years is
identified for the replacement study, for example, over the next 5 years, with no
continuation considered after this time period in the economic analysis.

A replacement study determines when a challenger replaces the in-place

defender. The complete study is finished if the challenger (C) is selected to
replace the defender (D) now. However, if the defender is retained now, the
study may extend over a number of years equal to the life of the defender 𝑛𝐷 ,
after which a challenger replaces the defender. Use the annual worth and life
values for C and D determined in the ESL analysis in the following procedure.
Assume the services provided by the defender could be obtained at the AW D
amount.

The replacement study procedure is:

New replacement study:

1. On the basis of the better 𝐴𝑊𝐶 or 𝐴𝑊𝐷 value, select the challenger C or
defender D. When the challenger is selected, replace the defender now,
and expect to keep the challenger for 𝑛𝐶 years. This replacement study is
complete. If the defender is selected, plan to retain it for up to 𝑛𝐷 more
years. (This is the leftmost branch of Figure 4–4.) Next year, perform the
following steps.

One-year-later analysis:

2. Determine if all estimates are still current for both alternatives, especially first
cost, market value, and AOC. If not, proceed to step 3. If yes and this is year
𝑛𝐷 , replace the defender. If this is not year 𝑛𝐷 , retain the defender for
another year and repeat this same step. This step may be repeated several
times.
3. Whenever the estimates have changed, update them and determine new
𝐴𝑊𝐶 and 𝐴𝑊𝐷 values. Initiate a new replacement study (step 1).
 Figure 4 – 4. Overview of replacement study approaches.

If the defender is selected initially (step 1), estimates may need updating after 1
year of retention (step 2). Possibly there is a new best challenger to compare with
D. Either significant changes in defender estimates or availability of a new
challenger indicates that a new replacement study is to be performed. In
actuality, a replacement study can be performed each year or more frequantly
to determine the advisability of replacing or retaining any defender, provided a
competitive challenger is available.