Chapter 6: Depreciation Analysis

 What is Depreciation?
Depreciation is a decrease in value of an
asset each year.
 Any equipment which is purchased today
will not work for ever.
 This may be due to wear and tear of the
equipment or obsolescence of technology.
 Therefore, it is to be replaced at the proper
time for continuance of any business.
 The replacement of the equipment at the end of
its life span involves financial expenditure.
 This must be internally generated from the
earnings of the equipment.
 The recovery of money from the earnings of an
equipment for its replacement purpose is called
depreciation fund since we make an assumption
that the value of the equipment decreases with
the passage of time.
 Thus, the word “depreciation” means decrease
in value of any physical asset with the passage of
time.
6.1: Causes of Depreciation
 Deterioration …..wear on parts/portions that affect its
functionality
 Obsolescence….product becomes outdated
 A business asset can be depreciated if:
1) It is used for business purposes to produce income
(revenue)
2) It has a determinable useful life that is longer than one
year
3) It is an asset that decays, gets used up, wears out,
becomes obsolete, or loses value to the owner from
natural causes.
 Types of Depreciation
1.Economic Depreciation: Economic
depreciation is a decrease in market value or
value to the owner.
It contains:
A) Physical depreciation…Loss of originality
B) Functional depreciation…loss of activity
2.Accounting depreciation: The systematic
allocation of an asset’s value in portions over its
depreciable life—often used in engineering
economic analysis.
• Accounting depreciation is defined as the
systematic allocation of the cost of an asset over
its depreciable life. This period may differ from
the useful life of equipment.
• Accountant definition is used for determining
taxable income.
• Accounting depreciation has two parts: Book
depreciation & Tax depreciation.
 6.2: Depreciable Properties
 Business Property can be classified as either:
1. Tangible property, which can be seen, touched,
and felt – Real property (think “real estate”)
includes land, buildings, and all things growing on,
built on, constructed on, or attached to the land.
Personal property includes equipment, furnishing,
vehicles, office machinery, and anything that is
tangible excluding those assets defined as real
property.
2. Intangible property, which includes all
property that has value to the owner but
cannot be directly seen or touched.
• Examples include patents, trademarks,
trade names, and franchises and etc
 Assets to Depreciate
Examples of depreciable business assets:
 Copy machines, computer networks, pc's, ..
 Buildings and interior furnishings
 Production equipment
 Many different types of properties that
wear out, decay, or lose value can be
depreciated as business assets
 Non-depreciable business assets:
• Land……….. Fertility…..
• Leased/rented property (only the owner may
claim depreciation expenses)
 Land does not wear out, lose value, or have a
determinable useful life. In fact, it often
increases in value
 Depreciation on tangible property used for both
business and personal activities (e.g., home,
office, equipment, materials) can be taken only
in proportion to the use for business expenses
Depreciation costs.
 6.3: Methods of Depreciation
 There are several methods of accounting
depreciation fund. These are as follows:
1. Straight line method of depreciation
2. Declining balance method of depreciation
3. Sum of the years—digits method of depreciation
4. Sinking-fund method of depreciation
5. Service output method of depreciation
1 . Straight line method of depreciation

 In this method of depreciation, a fixed sum

is charged as the depreciation amount
throughout the lifetime of an asset such that
the accumulated sum at the end of the life of
the asset is exactly equal to the purchase
value of the asset.
 Here, we make an important assumption that
inflation is absent.
Let:
P = First cost of the asset,
F = Salvage value of the asset,
n = Life of the asset,
Bt = Book value of the asset at the end of the
period t,
Dt = Depreciation amount for the period t.
 The formulae for depreciation and book value are
as follows:
Dt = (P – F)/n
Bt = Bt–1 – Dt = P – t [(P – F)/n]
Example 1: A company has purchased an
equipment whose first cost is $100,000 with an
estimated life of eight years. The estimated salvage
value of the equipment at the end of its lifetime is
$20,000.
Determine the depreciation charge and book value
at the end of various years using the straight line
method of depreciation.
Solution:
P = $100,000
F = $ 20,000
n = 8 years
Dt = (P – F)/n
= (100,000 – 20,000)/8
= $10,000
 In this method of depreciation, the value of Dt
is the same for all the years.
Dt and Bt Values under Straight line Method of Depreciation

End of year (t) Depreciation(Dt) Book value((Bt = Bt–1 – Dt)

0 10,000 100,000
1 10,000 90,000
2 10,000 80,000
3 10,000 70,000
4 10,000 60,000
5 10,000 50,000
6 10,000 40,000
7 10,000 30,000
8 10,000 20,000
 If we are interested in computing Dt and Bt for a specific period (t),
the formulae can be used. In this approach, it should be noted that the
depreciation is the same for all the periods.

Example 2: Consider Example 1 and compute the depreciation and the

book value for period 5.
Solution
P = $100,000
F = $20,000
n = 8 years
Dt = (P – F)/n
= (1,00,000 – 20,000)/8
= $10,000 (This is independent of the time period.)
Bt = P – t (P – F)/n
B5 = 100,000 – 5 (100,000 – 20,000)/8
= $50,000
2. Declining Balance Method of Depreciation
 In this method of depreciation, a constant percentage
of the book value of the previous period of the asset
will be charged as the depreciation amount for the
current period.
 This approach is a more realistic approach, since the
depreciation charge decreases with the life of the asset
which matches with the earning potential of the asset.
 The book value at the end of the life of the asset may
not be exactly equal to the salvage value of the asset.
 This is a major limitation of this approach.
 Let
P = first cost of the asset,
F = salvage value of the asset,
n = life of the asset,
Bt = book value of the asset at the end of the period t,
K = a fixed percentage, and
Dt = depreciation amount at the end of the period t.
 The formulae for depreciation and book value are as follows:
Dt = K Bt-1
Bt = Bt–1 – Dt = Bt–1 – K Bt–1
= (1 – K) Bt–1
 The formulae for depreciation and book value in terms of P are as
follows:
Dt = K(1 – K)t–1 x P
Bt = (1 – K)t x P
• While availing income-tax exception for the
depreciation amount paid in each year, the rate
K is limited to at the most 2/n. If this rate is
used, then the corresponding approach is
called the double declining balance method of
depreciation.
Example 2: Consider example 1 above and
demonstrate the calculations of the declining
balance method of depreciation by assuming
0.2 for K.
 Solution:
P = $100,000
F = $ 20,000
n = 8 years
K = 0.2
• The calculations pertaining to D t and Bt for
different values of t are summarized in table 2
using the following formulae:
Dt = K Bt–1
Bt = Bt–1 – Dt
Table 2: Dt and Bt according to Declining Balance Method of
Depreciation

End of year(n) Depreciation (Dt) Book value (Bt)

0 100,000.00
1 20,000.00 80,000.00
2 16,000.00 64,000.00
3 12,800.00 51,200.00
4 10,240.00 40,960.00
5 8,192.00 32,768.00
6 6,553.60 26,214.40
7 5,242.88 20,971.52
8 4,194.30 16,777.22
• If we are interested in computing Dt and Bt for a
specific period t, the respective formulae can be
used.
For example calculate the depreciation and the
book value for period 5 using the declining balance
method of depreciation by assuming 0.2 for K.
Solution:
P = $ 100,000
F = $ 20,000
n = 8 years
K = 0.2
 Dt = K(1 – K)t –1 x P
D5 = 0.2(1 – 0.2)4 x 100,000
= $ 8,192
Bt = (1 – K)t x P
B5 = (1 – 0.2)5 x 100,000
= $32,768
3. Sum-of-the-Years-Digits Method of Depreciation

• In this method of depreciation also, it is assumed that

the book value of the asset decreases at a decreasing
rate. If the asset has a life of eight years, first the sum
of the years is computed as Sum of the years = 1 + 2
+ 3 + 4 + 5 + 6 + 7 + 8 = 36 = n(n + 1)/2
• The rate of depreciation charge for the first year is
assumed as the highest and then it decreases. The
rates of depreciation for the years 1–8, respectively
are as follows: 8/36, 7/36, 6/36, 5/36, 4/36, 3/36,
2/36, and 1/36.
• For any year, the depreciation is calculated by
multiplying the corresponding rate of depreciation
with (P – F).
Dt = Rate x (P – F)
Bt= Bt–1 – Dt
• The formulae for Dt and Bt for a specific year t are
as follows:
Dt = n-t + 1(P – F)
n(n+1)/2
Bt = (P – F) (n-t) (n-t+1) + F
n(n +1 )
• Consider Example1 and demonstrate the
calculations of the sum-of-the-years-digits
method of depreciation.
Solution
P = $ 100,000
F = $20,000
n = 8 years
Sum = n(n + 1)/2 = 8 9/2 = 36
• The rates for years 1–8, are respectively 8/36,
7/36, 6/36, 5/36, 4/36, 3/36, 2/36 and 1/36.
The calculations of Dt and Bt for different
values of t are summarized in table 3 using the
following formulae:
Dt = Rate x (P – F)
Bt = Bt–1 – Dt
Table 3: Dt and Bt under Sum-of-the-years-
digits Method of Depreciation
End of year (n) Depreciation (Dt) Book value(Bt )

0 100,000.00
1 17,777.77 82,222.22
2 15,555.55 66,666.68
3 13,333.33 53,333.35
4 11,111.11 42,222.24
5 8,888.88 33,333.36
6 6,666.66 26,666.70
7 4,444.44 22,222.26
8 2,222.22 20,000.04
• If we are interested in calculating D t and Bt for
a specific t, then the usage of the formulae
would be better.
Example 3: Consider example 1 above and
find the depreciation and book value for the 5 th
year using the sum-of-the-years-digits method
of depreciation.
Solution:
P = $100,000
F = $20,000
n = 8 years
Dt = n-t + 1(P – F)
n(n+1)/2
= Dt = 8-5 + 1(100,000 – 20,000)
8(8+1)/2
= $8,888.88
Bt = (P – F) (n-t) (n-t+1) + F
n(n +1 )
B5 = (1,00,000 – 20,000) (8-5) (8-5+1) + 20,000
8(8 +1 )
= 80,000 x (3/8) x (4/9) + 20,000
=$33,333.33
4. Sinking Fund Method of Depreciation

In this method of depreciation, the book value decreases

at increasing rates with respect to the life of the asset.
Let: P = first cost of the asset,
F = salvage value of the asset,
n = life of the asset,
i = rate of return compounded annually,
A = the annual equivalent amount,
Bt = the book value of the asset at the end of the period t,
Dt = the depreciation amount at the end of the period t.
• The loss in value of the asset (P – F) is made available
an the form of cumulative depreciation amount at the
end of the life of the asset by setting up an equal
depreciation amount (A) at the end of each period
during the lifetime of the asset.
A = (P – F) [A/F, i, n]
• The fixed sum depreciated at the end of every time
period earns an interest at the rate of i% compounded
annually, and hence the actual depreciation amount
will be in the increasing manner with respect to the
time period. A generalized formula for Dt is
Dt = (P – F) x (A/F, i, n) x(F/P, i, t – 1)
 The formula to calculate the book value at the end of period t is Bt = P –
(P – F) (A/F, i, n) (F/A, i, t)
• The above two formulae are very useful if we have to calculate D t and
Bt for any specific period. If we calculate Dt and Bt for all the periods,
then the tabular approach would be better.
Example 4: Consider example1 and give the calculations regarding
the sinking fund method of depreciation with an interest rate of 12%,
compounded annually.
Solution:
P = $100,000
F = $ 20,000
n = 8 years
i = 12%
A = (P – F) x [A/F, 12%, 8]
= (100,000 – 20,000) x 0.0813
= $ 6,504
• In this method of depreciation, a fixed amount of
$6,504 will be depreciated at the end of every year
from the earning of the asset. The depreciated
amount will earn interest for the remaining period
of life of the asset at an interest rate of 12%,
compounded annually. For example, the
calculations of net depreciation for some periods
are as follows:
• Depreciation at the end of year 1 (D1)
= $ 6,504.
• Depreciation at the end of year 2 (D2) = 6,504 +
6,504 x 0.12 = $7,284.48
• Depreciation at the end of the year 3 (D3)
= 6,504 + (6,504 + 7,284.48)x 0 .12
= $ 8,158.62
• Depreciation at the end of year 4 (D4)
= 6,504 + (6,504 + 7,284.48 + 8,158.62)x 0.12
= $9,137.65
 These calculations along with book values are
summarized in table 4.
Table 4: Dt and Bt according to Sinking Fund Method of
Depreciation Bt (Bt = Bt–1 – Dt)
End of year (t) Fixed depreciation($) Net depreciation (Dt) Book value (Bt$)

0 6,504 - 1,00,000.00
1 6,504 6,504.00 93,496.00
2 6,504 7,284.48 86,211.52
3 6,504 8,158.62 78,052.90
4 6,504 9,137.65 68,915.25
5 6,504 10,234.17 58,681.08
6 6,504 11,462.27 47,218.81
7 6,504 12,837.74 34,381.07
8 6,504 14,378.27 20,002.80
 Example 4: Consider example 1 above and compute D 5
and B7 using the sinking fund method of depreciation
with an interest rate of 12%, compounded annually.
Solution:
P = $ 100,000
F = $ 20,000
n = 8 years
i = 12%1.574
Dt = (P – F) (A/F, i, n) (F/P, i, t – 1)
D5 = (P – F) (A/F, 12%, 8) (F/P, 12%, 4)
= (100,000 – 20,000)x 0.0813x 1.574
= $10,237.30
• This is almost the same as the corresponding
value given in the table. The minor difference
is due to truncation error.
Bt = P – (P – F) (A/F, i, n) (F/A, i, t)
B7 = P – (P – F) (A/F, 12%, 8) (F/A, 12%, 7)
= 100,000-(100,000 -20,000) x 0.0813 x 10.08
= $34,381.10
5. Service Output Method of Depreciation

• In some situations, it may not be realistic to compute

depreciation based on time period. In such cases, the
depreciation is computed based on service rendered by an asset.
• Let
P = first cost of the asset
F = salvage value of the asset
X = maximum capacity of service of the asset during its lifetime
x = quantity of service rendered in a period.
Then, the depreciation is defined per unit of service rendered:
Depreciation/unit of service = (P – F)/X
• Depreciation for x units of service in a period = (P – F )X
X
 Example 5: The first coat of a road laying machine is $ 8,000,000.
Its salvage value after five years is $50,000. The length of road that
can be laid by the machine during its lifetime is 75,000 km. In its
third year of operation, the length of road laid is 2,000 km. Find the
depreciation of the equipment for that year.
Solution:
P =$ 8,000,000
F = $ 50,000
X = 75,000 km
x = 2,000 km
Depreciation for x units of service in a period == (P – F )X
X
Depreciation for year 3 = (8,000,000 -50,000) x 2000
75,000
= $ 212,000
 Exercise
1. Define the following:
A) Depreciation
B) Book value
2. Distinguish between declining balance
method of depreciation and double declining
balance method of depreciation.
6.4: Break-Even Analysis
• Break-even analysis(profit contribution
analysis) is an important analytical tool used to
study the relationship between the total costs,
total revenue and the total profit, and the
losses over the whole range of stipulated
output.
• The relationship between cost and output and
between price and output may be linear or
non-linear in nature.
• Break-even point is the level at which total
revenue and total cost equates (TR=TC)
• The difference between total cost and total
revenue (TC>TR) is known as Operating Loss.
• The difference between total revenue and total
cost (TR>TC) is called Operating Profit.
• Contribution is the difference between total
revenue and variable costs.
• The contribution is the difference between the
sales and the variable costs. The margin of
safety (M.S.) is the sales over and above the
break-even sales. The formulae to compute
these values are Contribution = Sales –
Variable costs Contribution/unit = Selling
price/unit – Variable cost/unit
M.S. = Actual sales – Break-even sales
= Profit x Sales
Contribution
 M.S. as a per cent of sales = (M.S./Sales)x 100
 Example 1: Alpha Associates has the following details:
• Fixed cost = $ 20,00,000
• Variable cost per unit = $ 100
• Selling price per unit = $200
Find:
(a) The break-even sales quantity,
(b) The break-even sales
(c) If the actual production quantity is 60,000, find:
(i) contribution; and
(ii) margin of safety by all methods.
Solution:
Fixed cost (FC) = $20,00,000
Variable cost per unit (v) = $ 100
Selling price per unit (s) = $200
 Margin of Safety (M.S)
• Margin of safety represents the difference between sales at
break-even point and the total actual sales. Three
measures of M.S
a) Margin of safety = Profit x Sales
PV ratio
b) Margin of safety = Profit
PV ratio
c) Margin of safety = Sa –Sb x100
Sa
Where
Sa=Actual sales, Sb= Sale at BEP
 Numerical example:
• Suppose that TR & TC functions are given as,
 TR = 10Q, and TC = 50 + 5Q
 Sa = 20
Calculate the value of SB:
Sb = 10Q = 50 + 5Q
= 10Q- 5Q = 50
= 5Q = 50
 Sb = 10
Margin of Safety = 20 -10 x100 = 50%
20
Profit/Volume Ratio (P/V Ratio)
Example 1.Consider the following data of a
company for the year 2016
• Sales = $ 1,20,000
• Fixed cost = $ 25,000
• Variable cost = $ 45,000
Find the following:
(a) Contribution
(b) Profit
(c) BEP
(d) M.S.
 Solution:
(a) Contribution = Sales – Variable costs
= $ 1,20,000 – $ 45,000
= $ 75,000
(b) Profit = Contribution – Fixed cost
= $ 75,000 – $ 25,000
Profit = $ 50,000
(c) BEP
P/V ratio = Contribution
Sales
 = 75,000 x100 = 62.50%
120,000
• BEP = Fixed cost = 25,000 x 100 = $40,000
P/V ration 62.50
M.S = Profit = 50,000 x100 = $80,000
P/V ration 62.50
 PV= Profit Volume Ratio
 Example 2: Consider the following data of a company for the year 2015:
• Sales = $ 80,000
• Fixed cost = $ 15,000
• Variable cost =$ 35,000
Find the following:
(a) Contribution
(b) Profit
(c) BEP
(d) M.S.
Solution
(a) Contribution = Sales – Variable costs
= $ 80,000 – $35,000
Contribution = $45,000
(b) Profit = Contribution – Fixed cost
= $45,000 – $ 15,000
Profit = $ 30,000
 (c) BEP:
P/V ratio = Contribution = 45,000 x100
=56.25%
Sales 80,000
BEP = Fixed cost = 15,000 x100 = 26,667
P/V ratio 56.25
(d) M.S = Profit = 30,000 x 100 =$53,333.33
P/V ratio 56.25
 Use of Break-even Analysis
 Sales volume can be determined to earn a
given amount of return on capital
 Profit can be forecasted if estimate of revenue
& cost are available
 Effect of change in volume os sales, sale price,
cost of production, can be appraised
 Choice of products or processes can be made
from the alternatives available….product-mix
can be determined
 Impact of increase or decrease in fixed and
variable costs can be highlighted
 Effect high fixed costs and low variable costs
to the total cost can be studied
 Valid inter-firm comparisons of profitability
can be made
 Cash break-even helps proper planning of cash
requirements
 Break-even analysis emphasis the importance
of capacity utilization for achieving economy