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1.1 Compound depreciation

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Compound depreciation

The second method of calculating depreciation is to assume that the value of the asset decreases at a certain annual rate, but that the initial value of the asset this year, is the book value of the asset at the end of last year.

For example, if our second hand car has a limited useful life of 5 years and it has an initial value of R60 000, then the interest rate of depreciation is 20% (100%/5 years). After 1 year, the car is worth:

Book value after first year = P ( 1 - n × i ) = R 60 000 ( 1 - 1 × 20 % ) = R 60 000 ( 1 - 0 , 2 ) = R 60 000 ( 0 , 8 ) = R 48 000

At the beginning of the second year, the car is now worth R48 000, so after two years, the car is worth:

Book value after second year = P ( 1 - n × i ) = R 48 000 ( 1 - 1 × 20 % ) = R 48 000 ( 1 - 0 , 2 ) = R 48 000 ( 0 , 8 ) = R 38 400

We can tabulate these values.

End of first year R 60 000 ( 1 - 1 × 20 % ) =R 60 000 ( 1 - 1 × 20 % ) 1 = R48 000,00
End of second year R 48 000 ( 1 - 1 × 20 % ) =R 60 000 ( 1 - 1 × 20 % ) 2 = R38 400,00
End of third year R38 400 ( 1 - 1 × 20 % ) =R 60 000 ( 1 - 1 × 20 % ) 3 = R30 720,00
End of fourth year R30 720 ( 1 - 1 × 20 % ) =R 60 000 ( 1 - 1 × 20 % ) 4 = R24 576,00
End of fifth year R24 576 ( 1 - 1 × 20 % ) =R 60 000 ( 1 - 1 × 20 % ) 5 = R19 608,80

We can now write a general formula for the book value of an asset if the depreciation is compounded.

Initial value - Total depreciation after n years = P ( 1 - i ) n

For example, the book value of the car after two years can be simply calculated as follows:

Book value after 2 years = P ( 1 - i ) n = R 60 000 ( 1 - 20 % ) 2 = R 60 000 ( 1 - 0 , 2 ) 2 = R 60 000 ( 0 , 8 ) 2 = R 38 400

as expected.

Note that the difference between the compound interest calculations and the compound depreciation calculations is that while the interest adds value to the principal amount, the depreciation amount reduces value!

The Flamingo population of the Bergriver mouth is depreciating on a reducing balance at a rate of 12% p.a. If there are now 3 200 flamingos in the wetlands of the Bergriver mouth, how many will there be in 5 years' time ? Answer to three significant numbers.

  1. P = R 3 200 i = 0 , 12 n = 5 A is required
  2. A = 3 200 ( 1 - 0 , 12 ) 5
  3. A = 3 200 ( 0 , 88 ) 5 = 3 200 × 0 , 527731916 = 1688 , 742134
  4. There would be approximately 1 690 flamingos in 5 years' time.

Farmer Brown buys a tractor for R250 000 and depreciates it by 20% per year using the compound depreciation method. What is the depreciated value of the tractor after 5 years ?

  1. P = R 250 000 i = 0 , 2 n = 5 A is required
  2. A = 250 000 ( 1 - 0 , 2 ) 5
  3. A = 250 000 ( 0 , 8 ) 5 = 250 000 × 0 , 32768 = 81 920

  4. Depreciated value after 5 years is R 81 920

Compound depreciation

  1. On January 1, 2008 the value of my Kia Sorento is R320 000. Each year after that, the cars value will decrease 20% of the previous year’s value. What is the value of the car on January 1, 2012?
  2. The population of Bonduel decreases at a rate of 9,5% per annum as people migrate to the cities. Calculate the decrease in population over a period of 5 years if the initial population was 2 178 000.
  3. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses 3% of its water each day. How much does it weigh after a month of 31 days ?
  4. A computer depreciates at x % per annum using the reducing-balance method. Four years ago the value of the computer was R10 000 and is now worth R4 520. Calculate the value of x correct to two decimal places.
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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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