depreciationDb¶
Evaluates the depreciation of an asset using the fixed-declining balance method.
Synopsis¶
depreciationDb (cost, salvage, life, period, month)
Required Arguments¶
- float
cost
(Input) - Initial value of the asset.
- float
salvage
(Input) - The value of an asset at the end of its depreciation period.
- int
life
(Input) - Number of periods over which the asset is being depreciated.
- int
period
(Input) - Period for which the depreciation is to be computed.
period
cannot be less than or equal to 0, and cannot be greater thanlife
+1. - int
month
(Input) - Number of months in the first year.
month
cannot be greater than 12 or less than 1.
Return Value¶
The depreciation of an asset for a specified period using the fixed-declining balance method. If no result can be computed, NaN is returned.
Description¶
Function depreciationDb
computes the depreciation of an asset for a
specified period using the fixed-declining balance method. Routine
depreciationDb
varies depending on the specified value for the argument
period
, see table below.
period | Formula |
---|---|
period = 1 |
\[\mathrm{cost} \times \mathrm{rate} \times \frac{\mathrm{month}}{12}\]
|
period = life |
\[(\mathrm{cost} - \text{total depreciation from periods}) \times
\mathrm{rate} \times \frac{12 - \mathrm{month}}{12}\]
|
period other than 1 or life |
\[(\mathrm{cost} - \text{total depreciation from prior periods}) \times
\mathit{rate}\]
|
where
\[\mathit{rate} = 1 -
\left(\frac{\mathrm{salvage}}{\mathrm{cost}}\right)^
{\left(\frac{1}{\mathit{life}}\right)}\]
NOTE: rate is rounded to three decimal places. |
Example¶
In this example, depreciationDb
computes the depreciation of an asset,
which costs $2,500 initially, a useful life of 3 periods and a salvage value
of $500, for each period.
from __future__ import print_function
from numpy import *
from pyimsl.math.depreciationDb import depreciationDb
cost = 2500
salvage = 500
life = 3
month = 6
for period in range(1, life + 2):
db = depreciationDb(cost, salvage, life, period, month)
print("For period %i, db = $%.2f." % (period, db))
Output¶
For period 1, db = $518.75.
For period 2, db = $822.22.
For period 3, db = $481.00.
For period 4, db = $140.69.