Chapter: Mechanical : Engineering Economics & Cost Analysis : Depreciation
Depreciation
DEPRECIATION
Introduction
ü 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.
ü Hence, it is to be replaced at the proper time for continuance
of any business.
ü The replacement of the equipment at the end of its life involves
money. This must be internally generated from the earnings of the equipment.
ü 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.
ü Thus the word “depreciation” means decrease in value of any physical asset with the passage
of time.
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
A company has
purchased an equipment whose first cost is Rs. 1,00,000 with an estimated life
of eight years. The estimated salvage value of the equipment at the end of its
lifetime is Rs. 20,000. Determine the depreciation charge and book value at the
end of various years using the straight line method of depreciation.
Solution
P = Rs.
1,00,000 F = Rs. 20,000 n = 8 years
Dt = (P – F)/n
= (1,00,000 – 20,000)/8
= Rs. 10,000
In this method of depreciation, the value of Dt
is the same for all the years. The calculations pertaining to Bt
for different values of t are summarized in Table .
Table Dt and Bt
Values under Straight line Method of Depreciation
If we are interested in computing Dt and Bt for a specific
period (t), In the formulae can be used.
this approach, it should be noted that the all the depreciation is the
same for periods.
EXAMPLE
Consider Example and compute the depreciation and the book value
for period 5.
P = Rs. 1,00,000
F = Rs. 20,000
n = 8 years
D5 = (P – F)/n
= (1,00,000 – 20,000)/8
= Rs. 10,000 (This is independent of the time period.)
Bt = P – t
(P – F)/n
B5 = 1,00,000 – 5 (1,00,000 – 20,000)/8
= Rs. 50,000
2.Declining Balance Method of
Depreciation
ü 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.
ü 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.
ü 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
P Bt = (1 – K)t
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
Consider Example 9.1 and demonstrate the calculations of the
declining balance method of depreciation by assuming 0.2 for K.
Solution
P = Rs.
1,00,000
F = Rs.
20,000
n = 8
years
K = 0.2
The calculations pertaining to
Dt and Bt
for different values
of t
are summarized in Table 9.2 using the following formulae:
Dt = K
Bt–1
Bt = Bt–1 – Dt
Table Dt
and Bt according to Declining
Balance Method of Depreciation
If we are
interested in computing Dt and Bt for a
specific period t, the respective formulae can be used.
EXAMPLE Consider Example 9.1 and 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 = Rs. 1,00,000
F = Rs. 20,000
n = 8 years
K = 0.2
= Rs. 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 (P – F)
Bt = Bt–1 – Dt
The formulae for Dt and Bt
for a specific year t are as follows:
EXAMPLE
Consider Example 9.1 and demonstrate
the calculations of the sum-of-the-years-digits method of depreciation.
Solution
P = Rs.
1,00,000
F = Rs.
20,000 Sum = n (n + 1)/2 = 8 9/2 = 36
n = 8 years
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 using the following formulae:
Dt = Rate (P – F)
Bt = Bt–1 – Dt
Table Dt and Bt under
Sum-of-the-years-digits Method of Depreciation
If we are
interested in calculating Dt and Bt for a
specific t, then the usage of the formulae would be better.
EXAMPLE 9.6 Consider Example 9.1 and find the depreciation and book value
for the 5th year using the sum-of-the-years-digits method of depreciation.
Solution
P = Rs. 1,00,000
F = Rs. 20,000
n = 8 years
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,
and
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) (A/F, i, n)
(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 Dt and Bt for
any specific period.
If we calculate Dt and Bt
for all the periods, then the tabular approach would be better.
EXAMPLE
Consider Example and give the calculations regarding the sinking
fund method of depreciation with an interest rate of 12%, compounded annually.
In this method of
depreciation, a fixed amount of Rs. 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) = Rs. 6,504.
Depreciation at the end of year 2 (D2)
= 6,504 + 6,504 0.12 = Rs. 7,284.48
Depreciation at the end of the year 3 (D3)
= 6,504 + (6,504 + 7,284.48) .12
= Rs. 8,158.62
Depreciation at the end of year 4 (D4)
= 6,504 + (6,504 + 7,284.48 + 8,158.62) 0.12 = Rs. 9,137.65
These calculations along with book values are summarized in
Table .
Table Dt and Bt
according to Sinking Fund Method of Depreciation
EXAMPLE
Consider Example and compute D5 and B7 using the sinking fund
method of depreciation with an interest rate of 12%, compounded annually.
Solution
P = Rs. 1,00,000
F = Rs. 20,000 n = 8 years
i = 12%
Dt = (P – F) (A/F, i,
n) (F/P, i, t – 1)
D5 = (P – F) (A/F, 12%,
8) (F/P, 12%, 4)
= (1,00,000 –
20,000) 0.0813 1.574
= Rs. 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)
= 1,00,000 –
(1,00,000 – 20,000) 0.0813 10.089
= 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
The first coat of a road laying
machine is Rs. 80,00,000. Its salvage value after five years is Rs. 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 = Rs. 80,00,000
F = Rs. 50,000
X = 75,000 km
Evaluation Of
Public Alternatives
In evaluating alternatives of private
organizations, the criterion is to select the alternative with the maximum
profit.
ü The profit maximization is the main goal of private
organizations while providing goods/services as per specifications to their
customers.
ü But the same criterion cannot be used while evaluating public
alternatives. Examples of some public alternatives are
ü constructing bridges,
ü roads, dams,
ü establishing public utilities, etc.
The main objective of any public
alternative is to provide goods/services to the public at the minimum cost. In
this process, one should see whether the benefits of the public activity are at
least equal to its costs.
Inflation Adjusted Decisions
Inflation is the rate of increase in
the prices of goods per period. So, it has a compounding effect. Thus, prices
that are inflated at a rate of 7% per year will increase 7% in the first year,
and for the next year the expected increase will be 7% of these new prices.
If economic decisions are taken
without considering the effect of inflation into account, most of them would
become meaningless and as a result the organizations would end up with
unpredictable return.
Procedure To Adjust Inflation
A procedure to deal with this situation is
summarized now.
1. Estimate all the costs/returns associated with an investment
proposal in terms of today’s rupees.
2. Modify the costs/returns estimated in step 1 using an assumed
inflation rate so that at each future date they represent the costs/returns at
that date in terms of the rupees that must be expended/received at that time,
respectively.
3. As per our requirement, calculate either the annual equivalent
amount or future amount or present amount of the cash flow resulting from step
2 by considering the time value of money.
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