Question

. A dump truck is purchased for $110,000 and has an estimated salvage value of $10,000 at the end of the recovery period. Prepare a depreciation schedule for the dump truck with a recovery period of five years using: a) the straight-line method b) the sum-of-the-years method c) the declining-balance method

Answer 1

solution (a): according to straight line method :

annual depreciation rate in this method is same for all years and it is given by the formula = (P - F) / N

where P is purchase price, F is salvage value at the end of its design life and N is recovery period

therefor, annual depreciation = (110,000 - 10,000) / 5 = $20,000

now, book value at the end of each year is given by BV_i = P - m(D_i)

where BV_i is book value for any year i, P is purchase price, m is no. of year starting from the purchase year, and D_i is depreciation value of any year i

BV_{1 =} 110,000 - 1$\times$(20,000) = $90,000

BV₂ = 110,000 - 2$\times$(20,000) = $70,000

BV₃ = 110,000 - 3$\times$(20,000) = $50,000

BV₄ = 110,000 - 4$\times$(20,000) = $30,000

BV₅ = 110,000 - 5$\times$(20,000) = $10,000

(b) sum of the years method:

sum of year is calculated using formula: N(N + 1) / 2

where N is recovery period , therefor sum of years = 5(5+1) / 2 = 15

first year annual depreciation rate (ADR) is given by: (N - m + 1) / Sum Of Years

where m is number of year, therefor ADR = (5 - 1 + 1 ) / 15 = 1/3

now, depreciation is given by: (P - F) / ADR = (110,000 - 10000) $\times$ (1/3) = $33,333

book value at the end of first year is given by purchase price - depreciation of that year = 110,000 - 33,333 = $76,667

similarly for second year ADR = (5 - 2 + 1) / 15 = 4/15

and annual depreciation for second year is given by (110,000 - 10,000) $\times$ (4/15) = $26,667

book value = 76,667 - 26,667 = $50,000

for third year, ADR = (5 - 3 + 1) / 15 = 3/15

annual depreciation = (110,000 - 10,000) $\times$ (3/15) = $20,000

book value = 50,000 - 20,000 = $30,000

for fourth year, ADR = (5 - 4 + 1) / 15 = 2/15

annual depreciation = (110,000 - 10,000) $\times$ (2/15) = $13,333

book value = 30,000 - 13,333 = $16,667

for fifth year, ADR = (5 - 5 + 1) / 15 = 1/15

annual depreciation = (110,000 - 10,000) $\times$ (1/15) = $6,667

book value = 16,667 - 6,667 = $10,000

(c) declining-balance method: considering 200% declining-balance,

ADR for each year = 2 / 5 = 0.40

first year annual depreciation (AD)₁ = 110,000 $\times$ 0.40 = $44,000 and in straight line method (AD)_{2 =} $20,000

greater value is taken

Book value = 110,000 - 44,000 = $66,000

for second year, (AD)_{1 =} (66,000)0.40 = $26,400, (AD)₂ = (66,000 - 10,000) / 4 = $14,000

therefor, book value = 66,000 - $26,400 = $39,600

for third year, (AD)₁ = (39,600)0.40 = $15,840, (AD)₂ = (39,600 - 10,000) / 3 = $9,867

therefor, book value = 39,600 - 15,840 = $23,760

for fourth year, (AD)₁ = (23,760)0.40=$9,504, (AD)₂ = (23,760 - 10,000)/2 = $6,880

therefor, book value = (23,760 - 9,504) =$14,256

for fifth year,  (AD)₁ = (14,256)0.40=$5,702, (AD)₂ = (14,256 - 10,000)/1=$4,256

therefor, book value = 14,256 - 5,702=$8,554