Straight line: Difference between revisions
imported>Doug Williamson (Standardise appearance of page) |
imported>Doug Williamson (Standardise calculs) |
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The total expected accounting cost: | The total expected accounting cost, in $m: | ||
= | = 12 - 2 | ||
= | = 10 | ||
Allocated on a straight line basis over 4 years, the depreciation charge in each year would be: | Allocated on a straight line basis over 4 years, the depreciation charge in each year, in $m, would be: | ||
= | = 10 / 4 | ||
= | = 2.5 | ||
The net book value of the fixed asset would be | The net book value of the fixed asset in $m would be in each successive year: | ||
Year 1: | Year 1: | ||
| Line 33: | Line 33: | ||
= 12.0 - 2.5 | = 12.0 - 2.5 | ||
= | = 9.5 | ||
| Line 40: | Line 40: | ||
= 9.5 - 2.5 | = 9.5 - 2.5 | ||
= | = 7.0 | ||
| Line 47: | Line 47: | ||
= 7.0 - 2.5 | = 7.0 - 2.5 | ||
= | = 4.5 | ||
| Line 54: | Line 54: | ||
= 4.5 - 2.5 | = 4.5 - 2.5 | ||
= | = 2.0 | ||
Revision as of 15:25, 18 March 2015
1.
A basis of allocating total costs or income equally across successive time periods.
Example
A fixed asset has a cost of $12m,
an expected disposal value of $2m,
and an expected useful life of 4 years.
The total expected accounting cost, in $m:
= 12 - 2
= 10
Allocated on a straight line basis over 4 years, the depreciation charge in each year, in $m, would be:
= 10 / 4
= 2.5
The net book value of the fixed asset in $m would be in each successive year:
Year 1:
= 12.0 - 2.5
= 9.5
Year 2:
= 9.5 - 2.5
= 7.0
Year 3:
= 7.0 - 2.5
= 4.5
Year 4:
= 4.5 - 2.5
= 2.0
Using a straight line basis of depreciation, the net book value of a retained asset will often fall to zero.
(But it would never be depreciated to a negative value of course.)
2.
An estimation method which assumes a straight line relationship between the items under review.
Sometimes known as Linear interpolation.
