Purchase
Price :
|
£5,000
|
|||||
Life
in Years :
|
5
|
|||||
Salvage
value :
|
£200
|
|||||
Year
|
Deprecation
|
|||||
1
|
£2,375.00
|
=DB(E3,E5,E4,D8)
|
||||
2
|
£1,246.88
|
=DB(E3,E5,E4,D9)
|
||||
3
|
£654.61
|
=DB(E3,E5,E4,D10)
|
||||
4
|
£343.67
|
=DB(E3,E5,E4,D11)
|
||||
5
|
£180.43
|
=DB(E3,E5,E4,D12)
|
||||
Total
Depreciation :
|
£4,800.58
|
* See example 4 below.
|
||||
What Does It Do ?
|
||||||
This function calculates deprecation
based upon a fixed percentage.
|
||||||
The first year is depreciated by
the fixed percentage.
|
||||||
The second year uses the same
percentage, but uses the original value of the item less
|
||||||
the first years depreciation.
|
||||||
Any subsequent years use the same
percentage, using the original value of the item less
|
||||||
the depreciation of the previous
years.
|
||||||
The percentage used in the
depreciation is not set by the user, the function calculates
|
||||||
the necessary percentage, which
will be vary based upon the values inputted by the user.
|
||||||
An additional feature of this
function is the ability to take into account when the item was
|
||||||
originally purchased.
|
||||||
If the item was purchased part way
through the financial year, the first years depreciation
|
||||||
will be based on the remaining
part of the year.
|
||||||
Syntax
|
||||||
=DB(PurchasePrice,SalvageValue,Life,PeriodToCalculate,FirstYearMonth)
|
||||||
The FirstYearMonth is the month in
which the item was purchased during the
|
||||||
first financial year. This is an
optional value, if it not used the function will assume 12 as
|
||||||
the value.
|
||||||
Formatting
|
||||||
No special formatting is needed.
|
||||||
Example 1
|
||||||
This example shows the percentage
used in the depreciation.
|
||||||
Year 1 depreciation is based upon
the original Purchase Price alone.
|
||||||
Year 2 depreciation is based upon
the original Purchase Price minus Year 1 deprecation.
|
||||||
Year 3 deprecation is based upon
original Purchase Price minus Year 1 + Year 2 deprecation.
|
||||||
The % Deprc has been calculated
purely to demonstrate what % is being used.
|
||||||
Purchase
Price :
|
£5,000
|
|||||
Salvage
value :
|
£1,000
|
|||||
Life
in Years :
|
5
|
|||||
Year
|
Deprecation
|
%
Deprc
|
||||
1
|
£1,375.00
|
27.50%
|
||||
2
|
£996.88
|
27.50%
|
||||
3
|
£722.73
|
27.50%
|
||||
4
|
£523.98
|
27.50%
|
||||
5
|
£379.89
|
27.50%
|
||||
=DB(E47,E48,E49,D56)
|
||||||
Total
Depreciation :
|
£3,998.48
|
|||||
Example 2
|
||||||
This example is similar to the
previous, with the exception of the deprecation being calculated
|
||||||
on a monthly basis. This has been
done by multiplying the years by 12.
|
||||||
Purchase
Price :
|
£5,000
|
|||||
Life
in Years :
|
£5
|
|||||
Salvage
value :
|
100
|
|||||
Month
|
Deprecation
|
|||||
56
|
£8.79
|
|||||
57
|
£8.24
|
|||||
58
|
£7.72
|
|||||
59
|
£7.23
|
|||||
60
|
£6.78
|
|||||
=DB(E66,E68,E67*12,D75)
|
||||||
Example 3
|
||||||
This example shows how the length
of the first years ownership has been taken into account.
|
||||||
Purchase
Price :
|
£5,000
|
|||||
Life
in Years :
|
5
|
|||||
Salvage
value :
|
£1,000
|
|||||
First
Year Ownership In Months :
|
6
|
|||||
Year
|
Deprecation
|
%
Deprc
|
||||
1
|
£687.50
|
13.75%
|
||||
2
|
£1,185.94
|
27.50%
|
||||
3
|
£859.80
|
27.50%
|
||||
4
|
£623.36
|
27.50%
|
||||
5
|
£451.93
|
27.50%
|
||||
=DB(E74,E76,E75,D84,E77)
|
||||||
Total
Depreciation :
|
£3,808.54
|
|||||
Why Is The Answer
Wrong ?
|
||||||
In all of the examples above the
total depreceation may not be exactly the expected value.
|
||||||
This is due to the way in which
the percentage value for the depreceation has been calculated
|
||||||
by the =DB() fumction.
|
||||||
The percentage rate is calculated
by Execl using the formula = 1 - ((salvage / cost) ^ (1 / life)).
|
||||||
The result of this calculation is
then rounded to three decimal places.
|
||||||
Although this rounding may only
make a minor change to the percentage rate, when applied
|
||||||
to large values, the differnce is
compounded resulting in what could be considered as
|
||||||
approximate values for the the
depreceation.
|
||||||
Example 4
|
||||||
This example has been created with
both the Excel calculated percentage and the 'real'
|
||||||
percentage calculated manually.
|
||||||
The Excel Deprecation uses the
=DB() function.
|
||||||
The Real Deprecation uses a manual
calculation.
|
||||||
This
is the 'real' deprecation percentage, calculated manually :
|
27.522034%
|
|||||
=1-((E117/E116)^(1/E118))
|
||||||
Purchase
Price :
|
£5,000
|
=
1 - ((salvage / cost) ^ (1 / life)).
|
||||
Salvage
value :
|
£1,000
|
|||||
Life
in Years :
|
5
|
|||||
Year
|
Excel
Deprecation |
Real
Depreciation |
Excel
% Deprc |
|||
1
|
£1,375.0000
|
£1,376.1017
|
27.500%
|
|||
2
|
£996.8750
|
£997.3705
|
27.500%
|
|||
3
|
£722.7344
|
£722.8739
|
27.500%
|
|||
4
|
£523.9824
|
£523.9243
|
27.500%
|
|||
5
|
£379.8873
|
£379.7297
|
27.500%
|
|||
Total
Depreciation :
|
£3,998.48
|
£4,000.00
|
||||
Error
difference :
|
£1.52
|
You will learn the following features | Using Formulae and Functions | Formatting Spreadsheet | Printing and Layout | Creating Charts and Graphs | Generation of Charts | Workbook Management | Cursor Management | excel formulas with examples |
Thursday, 2 February 2017
How to use Excel DB formula
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