Methods to Calculate Annual Depreciation Cost of a Power Plant
Straight Line Method
In this method, provision is made from setting aside a certain and fixed amount of money every year. This value remains fixed for every year and depends upon the useful lifespan of the plant. It can be given as, total depreciation value divided by the useful life of the plant. The total depreciation value is calculated by subtracting the 'salvage (scrap) value after the lifespan' from the 'initial cost'.
Annual Depreciation = (Initial or capital cost of plant  Scrap value) / Useful life of the plant
For example, if a plant initially costs Rs. 500,000 and its scrap (salvage) value is Rs. 20,000 after 40 years of useful life, then, Annual depreciation charges = (500,000  20,000) / 40 = Rs. 12,000. let us take one another example: Initial cost = USD 62,500; Salvage value: USD 2500; and life time = 3, then
Advantages of straight line method

Simple method

Easy to apply
Disadvantages

In actual practice, depreciation of equipment is not constant every year.

It does not consider the amount of interest earned by the annual depreciation amount set aside annually.
Graphically, it can be represented as follows:
Equipment or plant value after k (≤ n) years,
V_{k} = P  qk
Diminishing Value Method
In this method, a fixed rate of depreciation value is set. This rate is first applied to the capital cost (P) and then to the diminishing value. The rate is decided according to the useful lifespan of the plant. Yearly depreciation value can be calculated as follows:
Let, P = Capital cost of the equipment,
x = Annual unit depreciation (if annual depreciation is 10%, then annual unit depreciation is 0.1)
After one year, the value of equipment = P  Px = P (1  x)
The annual depreciation after 1st year is equal to = (P  Px).x = Px  Px^{2}
Hence, value of the equipment after 2 years = Diminishing value  Annual depreciation
= P  Px  Px + Px^{2} = P(1  x)^{2}
After n years (useful lifespan), value of the equipment = P(1  x)^{n}
Graphically,
We can see that depreciation charges are higher in the early years and reduce with time.
Equipment or plant value after k (≤ n) years,
V_{k} = P (1  x)^{k}
Advantage

Better distribution of charges: In the early years, depreciation charges are more while maintenance and repair charges are less. In the later years, depreciation charges are less while maintenance and repair charges are higher.
Disadvantage

In the early years, the plant is supposed to collect money and then collect interest on it as time passes. But in this method, the amount of interest is not taken into account.
Sinking Fund Method
In this method, the arrangement is made such that a fixed amount is set aside annually and then invested at a certain interest rate which is compounded yearly. This fixed depreciation charges will be such that sum of these charges and the interest collected must be equal to the cost of replacement of the equipment.
Let, P = initial value of plant
n = useful life span
S = Salvage value after n years of useful life
r = Annual rate of interest
After n years, the salvage value will be received. Therefore, Cost of equipment replacement = P  S
If the depreciation amount set aside every year is q, then interest will be received on this amount till n years are completed. Also, after n years, the total amount must be equal to the cost of replacement (i.e. P  S)
If we deposit amount q for the first year, then,
Interest after 1 year = rq
Amount after 1 year = q + rq = q(1 + r)
Similarly, after n years, amount = q(1 + r)^{n}
But, amount q is added at the end of 1st year. Hence, it will collect interest only for (n1) years.
Therefore, amount q (deposited at the end of 1st year) = q(1+r)^{n1}
Similarly, amount q (deposited at the end of 2nd year) = q(1+r)^{n2}
Amount q (deposited at the end of (n1)^{th} year) = q(1+r)^{n(n1)} = q(1+r)
Therefore, the total sum (after n years) = q[(1+r)^{n1} + (1+r)^{n2} + ... + (1+r)]
The terms in the brackets constitute a geometric progression with a ratio (1+r)^{1}
Therefore, Total funds collected = Sum of G.P = q[(1+r)^{n}  1]/r
But, total funds must be equal to the cost of replacement, i.e. PS. Therefore,
P  S = q[(1+r)^{n}  1]/r
So, Sinsking Fund, q = (P  S)[r/(1 + r)^{n}  1]
Equipment or plant value after k (≤ n) years,
V_{k} = P  q[(1+r)^{k}  1]/r
This method is not widely used in practice. However, it does find application in economic assessments.