Depreciation12

Lecture Note-12, ME-430: Industrial Management (OE), Autumn 2015-16
DEPRECIATION
Dr. S.K. Patel, Mech Engg, NIT, Rourkela 8-Sep-15 Page 1 of 6
Depreciation is defined as the reduction in value or worth of a physical asset with time or use.
At the end of its life period, it is replaced with a new one when they no longer function
economically. Assets like land and gold are however exception. The causes of depreciation
may be wear and tear due to usage, aging, obsolescence due to availability of better product
in the market, reduction in effectiveness due to chemical reaction, damage in accidents etc. It
is charged under overhead in accounting records.
TYPES OF DEPRECIATION
1. Physical depreciation
This type of depreciation is the reduction in value of an item because of its physical
impairment or increasing the ability of a physical asset to render its intended service.
It occurs because of wear and tear from use. When in use an item is subjected to
abrasion, shock, vibration, impact etc. Even if an item is not used and simply kept idle
also, it gradually gets deteriorated. The most common phenomenon of deterioration is
that it gets rusted. Others factors responsible for its deterioration when it is not in use
may be corrosion, chemical decomposition, bacterial action etc..
2. Functional depreciation
It is the reduction in functional value because of change in demand for the services it
can render. The primary causes for this are
i) Obsolescence: An item is no longer needed when another item of better quality
can do the same job more efficiently and more economically e.g. computers,
diodes, slide rule, T-square, floppy disk etc.
ii) Inadequacy: The product may not be adequate for the purpose for which it is used.
3. Accident
Items get damaged because of unforeseen causes like fire, flood, earthquake etc. But,
it’s not unpredictable.
4. Amortization
This is the type of depreciation applicable to intangible assets like patents,
copyrights, goodwill, trade marks etc..
5. Depletion
The term depletion refers to measure the rate of exhaustion of the natural resources or
assets such as mines, iron ore, oil wells, quarries etc.
DEPRECIATION MEASUREMENT METHODS
1) Straight line method
It is also called as fixed instalment method or proportional method. In this method,
depreciation in every year is constant throughout its life.
Annual depreciation = (P-S)/n
where P = initial value , S = scrap value = value after the expiry of life, and n =
service life period in years.

DEPRECIATION
The total depreciation amount accumulated since beginning is called depreciated
fund. The market value of this asset at any time is called book value B. It is the value
of an item at any instant of time. The depreciation calculation should be such that
book value is obtained by subtracting the total cumulative depreciation from the initial
value. This is also called as unamortised value or undepreciated value. The years-wise
calculation of depreciation, depreciation fund and boom value are tabulated below.
The book values at the beginning and end of life period are equal to initial value and
scrap vale respectively.
Year
r
Depreciation
in year r
Depreciation fund
collected after the year r
Book value
after the year r
1 (P-S)/n (P-S)/n P-(P-S)/n
2 (P-S)/n 2(P-S)/n P-2(P-S)/n
3 (P-S)/n 3(P-S)/n P-2(P-S)/n
- - - -
r (P-S)/n r(P-S)/n P-r(P-S)/n=B
- - - -
N (P-S)/n n(P-S)/n P-n(P-S)/n = S
Depreciation in each year = (P-S)/n
Book Value at the end of r year = B = P-r(P-S)/n
Total depreciation during life time = P-S
Advantages :
i) Easy and simple to calculate
ii) Passage of time is major part
Disadvantages
Wear is usually not uniform over the life period. It is less during initial period and
goes on increasing with the increase in its life. Repair cost is more in later years of life
So, the rate of depreciation should not be uniform
2) Declining balance method
It is also called as reducing balance method, diminishing balance method,
minimising method, percentage on book value method or constant ratio method.
Depreciation during the year r is at a fixed rate i on the book value at the end of (r-1)
years.
Year
r
Depreciation
during the year r
Depreciation fund
collected after the year r
Book value at the end of
the year r
0 - - P
1 Pi Pi P-Pi = P(1-i)
2 P(1-i).i Pi+Pi(1-i) = Pi(2-i) P(1-i)-P(1-i)i = P(1-i)2
3 P(1-i)2
.i
- - - -
r P(1-i)r-1
.i Pi(r-i) P(1-i)r
=B
- - - -

DEPRECIATION
n P(1-i)n-1
.i Pi(n-i) P(1-i)n
= S
Depreciation in year r = Pi(1-i)r-1
Book value at the end of r year = B = P(1-i)r
Salvage value = Book value at the end of life = S = P(1-i)n
Total depreciation = P-S = (1-i)n
Here, asset decreases its value at a faster rate in early portion of its service life than the
latter portion of its life
3) Sinking fund method
This is also called as depreciation fund method. It is a method for depreciating an asset
in bookkeeping records while also generating money to purchase a replacement for the
asset when it reaches the end of its useful life. Under the sinking fund method, the
business sets aside an amount of money to invest annually so that the principal plus the
interest earned in the fund will be enough to replace the asset. In this method, there is a
separate sinking fund where money is allowed to get accumulated. Depreciation is at an
increasing rate. Here, a series of equal amount of A is assumed to be deposited into a
sinking fund at the end of each year of the asset’s life, which in turn gets accumulated to
an amount at compound interest at the end of the estimated life of asset. This amount is
equal to the total depreciation of the asset or the difference between first cost and
salvage value. The special feature of this method is that the sum required to buy the new
asset is available from depreciation or sinking fund. As a result, the working capital of
business is preserved. Sinking fund method is specially applicable to costly machines in
large scale industries. Thus the annual depreciation in any year has two components. The
first component is the fixed sum that is deposited into the sinking fund and the second
component is the interest earned on the amount accumulated in sinking fund till the
beginning of that year. The variable component of annual depreciation increases every
year.
Fixed component of depreciation in each year = A = constant
So that total depreciation over a life of n years = P-S = (A/i){(1+i)n
- 1}
Let variable component of annual depreciation during the year r = Vr = cumulative
sum of interests generated on all investments made in sinking fund during the past
years from the beginning till end of year r-1.
Thus, total depreciation during year r = A+ Vr. This amount is again kept in sinking
fund as investment for next year.
The concept of sinking fund can be clarified further through an example given below.
Example-1
A machine has initial cost Rs. 45000. It can be salvaged for Rs. 5000 after a life of 5
years. Find out year-wise depreciation provision @ 5%.
Solution-1

DEPRECIATION
Data given: P = Rs. 45000, S = Rs. 5000, n = 5 years, i = 5% = 0.05
If, fixed component of annual depreciation = A, then at compound interest
𝑃 − 𝑆 =
𝐴
𝑖
{(1 + 𝑖) 𝑛
− 1} => 45000 − 5000 =
𝐴
0.05
{(1 + 0.05)5
− 1}
 A = Rs. 7239
Year,
r
Sinking
fund
investment,
F
Annual
interest on
F = I
Cumulative
interest =
∑ I
Fixed
annual
depreciation
= A
Total
Depreciation
during the
year r = Dr =
A+ ∑ I
1 0 0 0 7239 7239
2 7239 7239x0.05
= 362
0+362 =
362
7239 7239 +362 =
7601
3 7601 7601x0.05
=380
362+380 =
742
7239 7239 + 742
= 7981
4 7981 7981x0.05
= 399
742+399 =
1141
7239 7239 +1141
= 8380
5 8380 8380x0.05
= 419
1141+419
= 1560
7239 7239 + 1560
= 8799
Total 40000
It can be observed from last column in the above table that depreciation amount
increases with time every year as D1 < D2 < D3 < D4 < D5 .
4) Sum of the years digits (SYD) method
For almost all items, there is maximum drop in book value during the first year and then
it gradually reduces with time. It is the highest in the first year and the least in the last
year. It has some sort of relationship with the digits of the year. This is also called fixed
base diminishing rate method.
Depreciation during the year r = (n-r+1).(P-S)/(1+2+.....+n).
For example, consider an item having 5 years of life has initial value Rs. 5000 and
estimated salvage value Rs. 1000. Sum of digits of all the years during its life
=1+2+3+4+5 = 5*6/2 = 15. Then, depreciations during
1st
year = (5000-1000)x5/15 =Rs. 1333.33
2nd
year = (5000-1000)x4/15 = Rs. 1066.67
5) Production Output method
This method is applicable to productive machines for which the production quantity is
clearly measurable. Here, rate of depreciation of an asset = (P-S)/ total number of
units already produced using this asset. If Qr is output quantity from the machine

DEPRECIATION
during the year r, then total quantity produced during its life span of n years = Q1 +
Q2 + …. + Qn = ∑Qr.
Depreciation during 1st
year = (P – S) (Q1/∑Qr)
Depreciation during 2nd
year = (P – S) (Q2/∑Qr)
In general, Depreciation during the year r = (P – S) (Qr/∑Qr)
6) Repair Provision
It is equal to the amount of money kept aside for the purpose of repair of the asset
every year.
7) Annuity method: This method is most suitable for a firm where capital is invested in
the least hold properties. Under this method, while calculating the amount of
depreciation, a fixed amount of depreciation is charged for every year of the estimated
useful life of the asset in such a way that at a fixed rate the interest calculated on the
same amount had been invested in some other form of capital investment. In other
words, depreciation charged for every year refers to interest losing or reduction in the
original cost of the fixed assets. Under the annuity method where the loss of interest is
due to the investment made in the form of an asset is considered while calculating the
depreciation.
8) Revaluation method: This method is specially designed to revalue the assets in the
case of livestock, loose tools, patents etc. This method also termed as Appraisal
Method. The calculation of depreciation of these assets is valued at the end of the
accounting year by comparing the opening value of the asset of the additional if any,
the difference is treated as depreciation.
9) Insurance policy method: Under this method an asset to be replaced by taking
required amount of insurance policy from an Insurance Company. A fixed premium is
paid which is equal to the amount of depreciation for every year. At the end of the
agreed sum, i.e., on the maturity of the policy, the amount will be used for replacing
the existing assets.
10) Depletion method: It is mostly used for natural resources such as mines, quarries, oil
and gas etc. from which certain quantity of the resources can be obtained on the basis
of the availability of minerals. The rate of depreciation is determined on the basis of
the quantity obtained for every year. The relevant formulae used are :
i) Rate of Depreciation = cost of mines / estimated cost of minerals to be
extracted
ii) Depreciation = Annual Quantity x Rate of Depreciation
11) Machine hour rate method : This method is similar to the Depletion Method but
instead of taking estimated available quantities in advance, the working life of the
machine is estimated in terms of hours. The hourly rate of depreciation is determined
by dividing the cost of the machine minus scrap value of the machine by the estimated
total number of hours utilized every year.

DEPRECIATION
The rates of depreciation with respect to time for various methods of depreciation calculation
are as given below.
i) Straight line method --- constant
ii) Declining balance method----decreasing
iii) Sinking fund method --- increasing
iv) Sum of the years digits method --- decreasing
Example-2
An equipment cost Rs 100000 has an estimated life of 9 years and an estimated salvage value
of Rs. 10000. Find out the following:
(1) Book value at the end of 7th
year using a straight line depreciation
(2) Annual depreciation in 5th
year following double declining method of depreciation
(3) Annual depreciation in the 9th
year, following sum of year’s digits method.
Answer-2
Cumulative depreciation up to 7th
yr =
  7
9
10000100000
*

= Rs. 70,000
Book Value at the end of 7th
yr = 30000.7*
9
90000
100000 Rs
(2) For 5th
year R =   5
1
1
1011 






r
S
P
Depreciation = P(1-R)4
*R = 100000 [10](4/5)
*[1-10(1/5)
]
(4) sum of digits = n(n+1)/2 = 10*9/2 = 45
Depreciation in 9th
year =   2000
45
90000
45
1
 SP* Rs.
Exercise Problem
1. An equipment with initial cost of Rs.55000 has an estimated life of 10 years and a
salvage value of Rs 5000. Calculate the annual depreciation for the third year of its
life and the book value at the end of the third year by all methods of depreciation
accounting.

Depreciation12

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