AMORLINC

AMORLINC(cost, date_purchased, first_period, salvage, period, rate [,basis])

Returns the depreciation of an asset in a single period (straight-line).

cost	The cost of the asset.
date_purchased	The date the asset was purchased.
first_period	The date of the end of the first period.
salvage	The salvage value at the end of the life of the asset.
period	The period to calculate the depreciation.
rate	The rate of depreciation.
basis	(Optional) The type of day counting to use: 0 = 30/360 (default) 1 = Actual/Actual 2 = Actual/360 3 = Actual/365 4 = 30E/365

REMARKS

* For an illustrated example refer to the [[Depreciation]] page.
* This function is used with the French accounting system.
* This function is identical to the AMORDEGRC function except that there is no depreciation coefficient applied in the calculation.
* The value of an asset at the end of the depreciation is also known as the salvage value.
* Dates must be entered as text strings within quotation marks or as serial numbers.
* If "cost" < "salvage", then #NUM! is returned.
* If "basis" is left blank, then 0 is used.
* If "basis" is not an integer, it is truncated.
* If "basis" < 0, then #NUM! is returned.
* If "basis" > 4, then #NUM! is returned.
* If an asset is purchased in the middle of an accounting period then the pro-rated depreciation is taken into account.
* You can use the AMORDEGRC function to return the depreciation of an asset in a single period.
* For the Microsoft documentation refer to support.microsoft.com
* For the Google documentation refer to support.google.com

	A
1	=AMORLINC(300, "01/01/2003", "31/12/2005", 50, 1, 0.15, 1) = 45
2	=AMORLINC(300, "01/01/2003", "31/12/2005", 50, 1, 0.15, 0) = 45
3	=AMORLINC(3000, "01/01/2003", "31/12/2005", 50, 1, 0.15, 4) = 450
4	=AMORLINC(50000, "01/01/2003", "31/12/2003", 500, 1, 0.15, 1) = 7500
5	=AMORLINC(50000, "01/01/2003", "31/12/2003", 500, 2, 0.15, 1) = 7500
6	=AMORLINC(50000, "01/01/2003", "31/12/2003", 500, 3, 0.15, 1) = 7500
7	=AMORLINC(50000, "01/01/2003", "31/12/2003", 500, 3, 0.15, 0) = 7500
8	=AMORLINC(50000, "01/01/2003", "31/12/2003", 500, 3, 0.15) = 7500
9	=AMORLINC(100, "01/01/2003", "31/12/2005", 500, 1, 0.15, 1) = #NUM!

1 - What is the depreciation of an asset worth £300.