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Sum-of-Years Digits

An asset often loses more of its value early in its lifetime. A method that exhibits this dynamic is desirable.

Assume an asset depreciates from price $P$ to salvage value $S$ in $N$ years. First compute the sum-of-years as $T = 1+2+\cdots +N$. The depreciation for the years after the asset’s purchase is:

Table 59.1: Sum-of-Years General Example

Year Number

Annual Depreciation

first

$\frac{N}{T}(P-S)$

second

$\frac{N-1}{T}(P-S)$

third

$\frac{N-2}{T}(P-S)$

$\vdots $

$\vdots $

final

$\frac{1}{T}(P-S)$


For the ith year of the asset’s use, the annual depreciation is:

\[ \frac{N+1-i}{T}(P-S) \]

For our example, $N=5$ and the sum of years is $T=1+2+3+4+5=15$. The depreciation during the first year is

\[ (\$ 20,000-\$ 5,000)\frac{5}{15}=\$ 5,000 \]

Table 59.2 describes how Declining Balance would depreciate the asset.

Table 59.2: Sum-of-Years Example

Year

Depreciation

Year-End Value

1

$(\$ 20,000-\$ 5,000)\frac{5}{15}=\$ 5,000$

$\$ 15,000.00$

2

$(\$ 20,000-\$ 5,000)\frac{4}{15}=\$ 4,000$

$\$ 11,000.00$

3

$(\$ 20,000-\$ 5,000)\frac{3}{15}=\$ 3,000$

$\$ 8,000.00$

4

$(\$ 20,000-\$ 5,000)\frac{2}{15}=\$ 2,000$

$\$ 6,000.00$

5

$(\$ 20,000-\$ 5,000)\frac{1}{15}=\$ 1,000$

$\$ 5,000.00$


As expected, the value after $N$ years is $S$.

\begin{align*} S & = P - (\text {5 years’ depreciation}) \\ & = P - \left( \frac{5}{10}(P-S) + \frac{4}{10}(P-S) + \frac{3}{10}(P-S) + \frac{2}{10}(P-S) + \frac{1}{10}(P-S) \right) \\ & = P - (P-S) \end{align*}