Straight line: Difference between revisions
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'''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 | |||
The total expected accounting cost: | |||
= $12m - $2m | = $12m - $2m | ||
= $10m. | = $10m. | ||
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 would be: | ||
= $10m / 4 | |||
= $2.5m. | |||
The net book value of the fixed asset would be (at the end of each year): | The net book value of the fixed asset would be (at the end of each year): | ||
Year 1 = 12.0 - 2.5 = $9.5m. | Year 1 | ||
= 12.0 - 2.5 | |||
= $9.5m. | |||
Year 2 | |||
= 9.5 - 2.5 | |||
= $7.0m. | |||
Year 3 | |||
= 7.0 - 2.5 | |||
= $4.5m. | |||
Year | Year 4 | ||
= 4.5 - 2.5 | |||
= $2.0m. | |||
Revision as of 13:00, 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:
= $12m - $2m
= $10m.
Allocated on a straight line basis over 4 years, the depreciation charge in each year would be:
= $10m / 4
= $2.5m.
The net book value of the fixed asset would be (at the end of each year):
Year 1
= 12.0 - 2.5
= $9.5m.
Year 2
= 9.5 - 2.5
= $7.0m.
Year 3
= 7.0 - 2.5
= $4.5m.
Year 4
= 4.5 - 2.5
= $2.0m.
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.
