Featured post

What Is Fluid and Its Properties?

Science helps to understand things in a better way. We know about fluid but science helps to understand different aspects of it in a better ...

11/03/2021

Depreciation : Diminishing Method

  • The depreciation charge is first applied to the initial cost of equipment, then its value goes on diminishing. 
  • This method is more accurate than the straight line method. 
  • The depreciation charges are more in the early years and its value goes on decreasing after some years.



diminishing-method-of-depreciation.png


Let

P = Initial cost of equipment / plant

S = Salvage / Scrap value of equipment / plant

n = Life of equipment

r = Annual rate of interest

x = Annual depreciation

Value of equipment after one year

S = P – Px

S = P ( 1 – x )

Value of equipment after second year

= Diminished value after first year – Annul Depreciation

= P ( 1 – x ) – [ P ( 1 – x )x ]

= P ( 1 – x – x + x2 )

= P ( 1 – 2x + x )2

= P ( 1 – x )2

Value of equipment after 10th year

S = P ( 1 – x )10

Value of equipment after nth years

S = P ( 1 – x )n

Annual depreciation charge

( 1 – x )n = S / P

( 1 – x ) = ( S / P )1/n

x = 1 – ( S / P )1/n

Depreciation for the first year = xP

                                                = P { 1 – ( S / P ) }

Depreciation for the nth year = xP

                                               = P {( 1 – ( S / P )n}

Disadvantages

  • The depreciation charge is not depending upon rate of interest on annual depreciation. 
  • The depreciation charges are more during early years and its value goes on decreases in late years when the maintenance cost of equipment is quite high.

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 Diminishing method. Calculate the cost of equipment after 5 years.

Solution

P = Rs 75,000

S = Rs. 5,000

n = 10 years

Annual Depreciation x = 1 – ( S / P )1/n

   = 1 – ( 5,000 / 75,000 )1/10

   = 1 – ( 0.0667 )0.1

   = 1 – ( 0.762 )

   = 0.238

Value of equipment after 5 years

    = P ( 1 – x )5

    = Rs, 75,000 ( 1 – 0.238 )5

    = Rs. 75,000 ( 0.762 )5

    = Rs. 75,000 ( 0.2569 )

    = Rs. 19268

You may also like :

Importance of load curve

Function of shunt reactor

What do you mean by AC Drive?

How to reduce mitigation of third harmonic?


No comments:

Post a Comment