Puzzled by values returned by the function depsoyd(Cost, Salvage, Life) from Financial Toolbox when Life is fractional

1 Ansicht (letzte 30 Tage)
Antonio Seabra
Antonio Seabra am 31 Jan. 2021
Bearbeitet: Antonio Seabra am 31 Jan. 2021
I am puzzled by the values returned by depsoyd(Cost, Salvage, Life) from Financial Toolbox when Life is fractional.
>> dep = depsoyd(13000,1000,2.5)
dep =
1.0e+03 *
6.857142857142857 4.114285714285714
Here the depreciation amounts do not sum to 13000 - 1000 = 12000 as I was expecting, summing instead 10971.42857142857. Would appreciate very much any clarification about how these depreciation values are been computed. Many thanks in avance!

Melden Sie sich an, um diese Frage zu beantworten.

Antworten (0)

Kategorien

Mehr zu Financial Toolbox finden Sie in Help Center und File Exchange

Tags

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by