Question

A depreciable item is purchased for $500,000. The salvage value at the end of 25 years is

estimated at $100,000. What is the depreciation in each of the first three years using the (a)

straight line, (b) sum-of-the-years' digits, and (c) double-declining balance methods?

Answer 1

a) straight line methods

purchase value = 500000

salvage value = 100000

useful life = 25

Depreciation per year = (500000 - 100000)/25 = 400000/25 = 16000

Depreciation in three years = 3*16000 = 48000

b) Sum of the years digit method

purchase value = 500000

salvage value = 100000

useful life, n = 25

sum of digits over useful life = n*(n+1)/2 = 25*26/2 = 325

Depreciaiton in year m = [(total life - m +1)/ sum of digits ] * (purchase value - salvage value)

Depreciation in first year = (25-1+1)/325 * (500000-100000) = 25/325 * 400000 = 30769.23

Depreciation in second year = (25-2+1)/325 * 400000 = 29538.46

Depreciation in third year = (25-3+1)/325 * 4000000 = 28307.69

Total depreciation = 30769.23+29538.46+28307.69 = 88615.38

c) double declinign methods

purchase value = 500000

salvage value = 100000

useful life = 25

d = 2/25 = 0.08

Depreciation in any year m = purchase value *(1-d)^m-1*d

Depreciation in first year = 500000 * (1-0.08)^1-1 * 0.08= 500000 * 0.08 = 40000

Depreciation in second year = 500000 * (1-0.08)^2-1 * 0.08= 500000 * 0.08*0.92 = 36800

Depreciation in third year = 500000 * (1-0.08)^3-1 * 0.08= 500000 * 0.08 * 0.92² = 33856

Total depreciation = 40000 + 36800 + 33856 = 110656