Electrical Revolution

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 + r²q

= q ( 1 + 2r + r² )

= 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 – rⁿ ) / ( 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 }