# Sinking Fund Method of Depreciation

ElectronElectronics & ElectricalDigital Electronics

## Depreciation of Power Station Equipment

The reduction in the value of the equipment and other property of the power station every year is known as depreciation. Therefore, a suitable amount, called depreciation charge, must be set aside annually so that by the time the life span of the power plant is over, the collected amount equals to the cost of the replacement of the power plant.

## Sinking Fund Method of Depreciation

In the sinking fund method of depreciation, a fixed depreciation charge is made every year and the interest is compounded on it annually. The constant depreciation charge is such that the sum of annual investment and the interest accumulations is equal to the cost of replacement of equipment after its useful life.

Explanation

Let,

• X = Initial Value of Equipment

• S = Scrap value after useful life

• n = Useful life of equipement in years

• r = Annual interest rate

Therefore, the cost of replacement of the equipment is,

$$\mathrm{\mathrm{Cost\: of\: replacement}\:=\:\mathit{X-S}}$$

Let an amount of p is set aside as depreciation charge every year and interest compounded on it so that an amount of (X-S), i.e. cost of replacement is available after n years. Therefore, the amount p at annual interest rate of r at the end of n years is given by,

At the end of first year,

$$\mathrm{\mathrm{Amount\:=\:}\mathit{p\:\mathrm{+}\:rp}\:=\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$

At the end of second year,

$$\mathrm{\mathrm{Amount\:=\:}\mathrm{\left ( \mathit{p}+\mathit{rp} \right )}\:+\:\mathit{r}\mathrm{\left ( \mathit{p}+\mathit{rp} \right )}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{2}}}$$

Similarly, at the end of n years,

$$\mathrm{\mathrm{amount\:=\:}\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}}$$

Now, the amount p deposited at the end of first year will earn compound interest for (n-1) years and it becomes,

$$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:first\: year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n-\mathrm{1}}}}}$$

And the amount p deposited at the end of second year becomes,

$$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:second\: year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n-\mathrm{2}}}}}$$

Similarly, amount p deposited at the end of (n-1) year becomes,

$$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:}\mathrm{\left ( \mathit{n} -1\right )}\mathrm{year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}\mathrm{\left ( \mathit{n}-1 \right )}}}\:=\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$

Therefore, the total fund after n years is given by,

$$\mathrm{\mathrm{Total\: fund\: after\: n\: years}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-1}}\:+\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-2}}\:+\:...\:+\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$

$$\mathrm{\Rightarrow \mathrm{Total\: fund\: after\: n\: years}\:=\:\mathit{p}\mathrm{\left [ \mathrm{\left ( 1\:+\mathit{r} \right )^{\mathit{n}-1}}\:+\:\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-2}}\:+\:...\:+\mathrm{\left ( 1\:+\mathit{r} \right )} \right ]}}$$

As it is a geometric progression series and its sum is given by,

$$\mathrm{\mathrm{Total \:fund\: after\: n \:years}\:=\:\frac{\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}-1}}{\mathit{r}}}$$

This total fund must be equal to the cost of replacement of the equipment, i.e.,

$$\mathrm{\mathit{X-S}\:=\:\frac{\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}-1}}{\mathit{r}}}$$

Therefore, the amount of sinking fund is,

$$\mathrm{\mathrm{Sinking \:fund,}\:\mathit{p}\:=\:\mathrm{\left ( \mathit{X-S} \right )}\mathrm{\left[ \frac{\mathit{r}}{\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}-1}\right ]}}$$

Where,

$$\mathrm{\mathrm{Sinking \:fund\:factor}\:=\:\mathrm{\left[ \frac{\mathit{r}}{\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}-1}\right ]}}$$

## Numerical Example

A transformer costs Rs 150000 and has a useful life of 25 years. If the scrap value of the transformer is 10000 and the rate of annual compound interest is 7$\%$. Then, calculate the amount to be saved annually for the replacement of the transformer after the end of 25 years.

Solution

Given data is −

• Initial cost of transformer, X = Rs.150000

• Scrap value of transformer, S = Rs.10000

• Useful life of transformer, n = 25 years

• Annual rate of interest, r = 7$\%$

Then, the annual amount for sinking fund is given by,

$$\mathrm{\mathrm{Sinking\: fund,}\mathit{p}\:=\:\mathrm{\left ( \mathit{X-S} \right)}\mathrm{\left[\frac{\mathit{r}}{\mathrm{\left(1\:+\:\mathit{r}\right)^{\mathit{n}}-1}} \right ]}}$$

$$\mathrm{\mathit{p}\:=\:\mathrm{\left (150000-10000 \right )}\mathrm{\left[ \frac{0.07}{\mathrm{\left( 1\:+\:0.07 \right )^{25}}-1}\right ]}\:=\:\mathrm{Rs\:2212.18}}$$

Updated on 15-Feb-2022 09:23:42