- The fixed depreciation charge is made every year and interest compounded calculated on it annually.
- The constant depreciation charge is calculated such that the sum of the total annual installment and interest of it is equal to cost of replacement of equipment after its useful life.

Let

P = Initial cost
of equipment / plant

S = Salvage /
Scrap value of equipment / plant

n = Life of
equipment

r = Annual rate
of interest

q = Annual
depreciation

Cost of
replacement for new equipment = P – S

Let us consider
that the q amount is set as depreciation charge every year and r will be
interest of it.

**After first year
this amount **

= q + qr

= q ( 1 + r )

**After second
year this amount **

= q ( 1 + r ) +
rq ( 1 + r )

= q + qr + qr +
r^{2}q

= q ( 1 + 2r + r^{2}
)

= q ( 1 + r )^{2}

**After nth year
this amount **

= q ( 1 + r )^{n}

Now, amount q
deposited at the end of first year will earn compound interest for ( n – 1 )
years and it is equal to

q ( 1 + r )^{n
– 1 }

Amount q
deposited at the end of 2^{nd} year becomes

q ( 1 + r )^{n
– 2}

Amount q
deposited at the end of 3^{rd} year becomes

q ( 1 + r )^{n
– 3}

Amount q
deposited at the end of ( n – 1 )th year becomes

q ( 1 + r )^{n
– n + 1 }

q ( 1 + r )^{
1 }

Total fund
deposited after n years = [ q ( 1 + r )^{n – 1} ]^{ } + [ q ( 1 + r )^{n – 2 }] +^{ } [ q ( 1 + r )^{n – 3 }] + ……+ q ( 1 +
r ) ]

**Solution of Sum
of Geometric Series = a1 ( 1 – r ^{n} ) / ( 1 – r )**

**Where **

**a1 = first term
= [ q ( 1 + r ) ^{n – 1} ]^{ } **

**r = Common ratio
= ( 1 + r )**

**n = Number of
terms = Here n**

Total fund = [ q
( 1 + r )^{n – 1} ]^{ } {( 1 – ( 1 + r )^{n} } / { 1 – ( 1 + r
) }

= [ q ( 1 + r )^{n
– 1} – q ( 1 + r )^{n – 1} ( 1 + r )^{n} } / – r

= q [ ( 1 + r )^{n
– 1} – ( 1 + r )^{2n – 1} } / – r

= q [ ( 1 + r )^{2n
– 1} – ( 1 + r )^{n – 1} } / r

Dividing
numerator by ( 1 + r )^{n – 1}

= q { ( 1 + r )^{n
}– 1 } / r

Now total fund
deposited = P – S

Therefore

P – S = q { ( 1
+ r )^{n }– 1 } / r

Depreciation
every year q = ( P – S ) [ r / { ( 1 + r
)^{n }– 1 ]

Where

**Sinking fund
factor = [ r / { ( 1 + r ) ^{n }– 1 ]**

**Example **

**The cost of an electrical equipment is Rs.75,000 and its useful life is 10 years. The salvage value of equipment is Rs.5,000.****Calculate annual depreciation charges using Sinking fund method if the annual rate of interest is 5% .****Calculate the cost of equipment after 5 years.**

**Solution **

P = Rs. 75,000

S = Rs. 5,000

n = 10 years

q = ( P – S ) {
r / ( 1 + r )^{n} – 1 }

= ( Rs. 75,000 – Rs. 5000 ) { 0.05 / ( 1 +
0.05 )^{10} – 1 }

= Rs. 70,000 { 0.05 / 1.6288 – 1 }

= Rs. 70,000 { 0.05 / 0.6288 }

= Rs. 70,000 { 0.0795 }

= Rs. 5566

**Sinking fund
after 5 ^{th} year**

= q { ( 1 + r )^{n}
– 1 / r }

= Rs. 5566 { ( 1
+ 0.05 )5 – 1 / 0.05 }

= Rs. 5566 { ( 1
+ 0.05 )5 – 1 / 0.05 }

= Rs. 5566 { 1.2762
– 1 / 0.05 }

= Rs. 5566 { 0.2762
/ 0.05 }

= Rs. 30746.60

The value of equipment after 5 years = Rs. 75,000 – Rs. 30746 = Rs. 44,254

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