DDB METHOD

Universal Arm (a robot) costs $25,000. At the end of its 4-year useful life its salvage value is estimated to be $5,000.

(a) Prepare its DDB depreciation schedule.

(b) Discuss any discrepancy in the schedule, especially between book value at the end of useful life and the salvage value.

Question

Universal Arm (a robot) costs $25,000. At the end of its 4-year useful life its salvage value is estimated to be $5,000.

(a) Prepare its DDB depreciation schedule.

(b) Discuss any discrepancy in the schedule, especially between book value at the end of useful life and the salvage value.

Accepted Answer

The given data are: [latex]P=\$25,000[/latex] , [latex]S=\$5,000[/latex] , [latex]N=4 years[/latex]

(a) For DDB depreciation the rate is 2/N. Since the entire depreciation schedule is being asked for, we can either follow the first-principles or use Equation (10.4). Let us follow the first-principles, especially since N is small[latex]^{14}[/latex]. This involves following the steps used in deriving Equation (10.4), but with the given data. Thus,

[latex] ext{jth year depreciation}=({2P}/{N})({1-2}/{N})^{j-1}[/latex]

.

[latex] ext{Nth year depreciation}=({2P}/{N})({1-2}/{N})^{N-1}[/latex] (10.4)

Year 1

Book value at the [latex]beginning[/latex] of the 1st yea

= First ( initial ) cost

[latex]=\$25,000[/latex]

1st year depreciation[latex]^{15}[/latex] =[latex]({2}/{N})[/latex] × 1st year book value

[latex]=({2}/{4}) imes \$25,000[/latex]

[latex]=\$12,500[/latex]

Book value at the [latex]end[/latex] of 1st year

= Book value at the beginning of 1st year

–1 st year depreciation

[latex]=\$25,00-\$12,500[/latex]

[latex]=\$12,500[/latex]

Year 2

Book value at the [latex]beginning[/latex] of the 2nd year

=Book value at the end of 1st year

[latex]=\$12,500[/latex]

2nd year depreciation = (2/N) × 2nd year book value

[latex]=({2}/{4}) imes \$12,500[/latex]

[latex]=\$6,250[/latex]

Book value[latex]^{16}[/latex] at the [latex]end[/latex] of 2nd year

= Book value at the beginning of 2nd year - 2nd year depreciation

[latex]=\$12,500-\$6,250[/latex]

[latex]=\$6,250[/latex]

Year 3

Book value at the [latex]beginning[/latex] of the 3rd year

= Book value at the end of 2nd year

[latex]=\$6,250[/latex]

3rd year depreciation = (2/N) × 3rd year book value

[latex]=({2}/{4}) imes \$6,250[/latex]

[latex]=\$3,125[/latex]

Book value at the [latex]end[/latex] of 3rd year

= Book value at the beginning of 3rd year –3rd year depreciation

[latex]=\$6,250 -\$3,125[/latex]

[latex]=\$3,125[/latex]

Year 4

Book value at the [latex]beginning[/latex] of the 4th year

= Book value at the end of 3rd year

[latex]=\$3,125[/latex]

4th year depreciation =(2/N) × 4th year book value

[latex]=({2}/{4}) imes \$3,125[/latex]

[latex]=\$1,563[/latex]

Book value at the [latex]end[/latex] of 4th year

= Book value at the beginning of 4th year

–4th year depreciation

[latex]=\$3,125-\$1,563[/latex]

[latex]=\$1,562[/latex]

The DDB depreciation schedule can be summarized in a table as:

(b) We notice a discrepancy between the expected salvage value $5,000 and the book value $1,562 at the end of useful life. These two should have been the same. This discrepancy is also reflected in the total depreciation cost $23,438 being greater than the maximum allowable of $20,000 (P-S = $25,000 – $5,000). Thus, there is something wrong with the results.

[latex]^{14}[/latex]For large values of N, the first-principles approach becomes lengthy.

[latex]^{15}[/latex]One can evaluate 2/N as 2/4 = 0.5 at this stage, and use this 0.5, instead of 2/N, throughout the calculations to achieve simplicity.

[latex]^{16}[/latex]Since the given data, rather than symbols, are being used as the calculation proceeds, the results get simplified; we do not face any lengthy equation as while deriving Equation (10.4). This illustrates that the first-principles approach can be simpler.

Question 10.4: DDB METHOD Universal Arm (a robot) costs $25,000. At the end...

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