Sum-of-Years Digits :: SAS/ETS(R) 13.1 User's Guide

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 58.1: 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 $i$ th 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 58.2 describes how Declining Balance would depreciate the asset.

Table 58.2: 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 $S$ .

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