Depreciation

[eddy2017]

There's a link in the forum guidelines to purplemath's four-page explanation about how to text math using a keyboard.

If you're going to teach math, then you'll be communicating with machines via keyboard (eg: graphing calculator, CAS). Most software is not coded to interpret missing or mismatched grouping symbols, so you need to force yourself to practice using grouping symbols.

Also, I find it awkward and annoying to type anything here (other than English words) when using a cell phone. Typing proper grouping symbols is not tricky at all, when using a computer. I'll show how I do it.

Obviously, there's no issue typing something like a/(b-c) directly.

If I were to enter something like this into a computer

\(\displaystyle \frac{(2x)(3x+5)}{(x-7)(2x+1)}\)

then I could type all the parentheses first and then insert the expressions. I know that I need ()() both on top and bottom. I also need parentheses around numerator and denominator.

(()())/(()())

Then I move the cursor and type the expressions, like

((2x)())/(()())

and so forth. With enough experience through practice, you will be able to type complex expressions without having to type the grouping symbols separately. You'll learn to read and type such expressions naturally.

I'll definitely give that a try.

 

[eddy2017]

The formula for annual linear depreciation is [imath]\dfrac{a - s}{y}[/imath], where a is acquisition cost, s is expected salvage value, and y is number of years of expected useful life for items of that type. In many cases, s is assumed to be 0.

The formula for annual exponential depreciation after n years is [math]\left ( \dfrac{y - 1}{y} \right )^n * a.[/math]

If I were to apply the formula for annual exponential depreciation,
a= acquisition cost= $17000
n= amount of years the asset has been losing value or depreciating.
What would be the value of y? According to you, y stands for the number of years of expected useful life for items of the type in question.
How am I supposed to know that for certain when it is not given in the problem?.

 

[Deleted member 4993]

If I were to apply the formula for annual exponential depreciation,
a= acquisition cost= $17000
n= amount of years the asset has been losing value or depreciating.
What would be the value of y? According to you, y stands for the number of years of expected useful life for items of the type in question.
How am I supposed to know that for certain when it is not given in the problem?.

Do you remember the original problem:

A new car is purchased for 17000 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be, to the nearest cent, after 12 years?

 

[eddy2017]

n = 12
How about y?.

 

[Otis]

n = 12
How about y?

Eddy, did you forget that you and Jeff are each using a different set of variable names?

By the way, S =salvage value in the formula you posted is the same as residual value (R) in the one I posted. Same concept different names.
Your a stands for my C = cost
Your y stands for expected useful value, which in mine us n.

It would make sense for everyone to agree on a single set of variable names, before continuing.

 

[JeffM]

Eddy

Otis is correct. My description of my y was not helpful.

In straight-line depreciation, annual depreciation is constant fraction of a constant amount, namely a - s. The denominator of that fraction is expected years of useful life. I called that y.

In exponential depreciation, annual depreciation is a constant fraction of a declining balance. Supposedly, that fraction is computed with some reference to salvage value and expected useful life, but the denominator of the fraction is not expected useful life. I should have changed variables.

 

[eddy2017]

Any way we slice it, my question was this:
Your y stands for expected useful value, which in mine is n. BUT,
what is the expected useful value of the thing in question, its expected life span?? .That info is not provided in the problem. that was I what and why I asked.

Oh, I think JeffM answered that when he posted this::
quoting 'n straight-line depreciation, annual depreciation is constant fraction of a constant amount, namely a - s. The denominator of that fraction is expected years of useful life.

I asked this question #27 in reply to two formulas Jeff posted. (to formulas Jeff posted at #20)
Could this formula be used to solve the problem in this post? that was my question
Now, after what Jeff posted at 46, are there any changes in any of those two formulas, are they still okay like that>. I am sorry but i stopped following you at post 42. I need a little bit of clarification because I am taking notes of everything.

Any way we slice it, my question was this:
Your y stands for expected useful value, which in mine is n. BUT,
what is the expected useful value of the thing in question, its expected life span?? .That info is not provided in the problem. that was what and why I asked. It seems to me that piece of info is not there. It has to be calculated somehow.

 

[JeffM]

Eddy

It has been decades since I studied accounting theory. And practicing accountants frequently take shortcuts and have their own jargon. So this may not be perfect in terms of nomenclature and may not reflect actual day to day practice exactly.

Exponential (or declining balance) depreciation uses a depreciation factor.

[math]a = \text {acquisition cost of asset (including installation).}\\ s = \text {expected salvage value or } 0.01, \text { whichever is higher.}\\ e = \text {expected useful life.}\\ d = \text {depreciation factor.}\\ b_n = \text {net book value at end of year } n.\\ x_n = \text {depreciation expense for year n.}[/math]
In theory,

[math]a(1 + d)^e = s \implies ln(a) + e * ln(1 + d) = ln(s) \implies[/math]
[math]ln(1 - d) = \dfrac{1}{e} * \{ln(s) - ln(a)\} = ln \left ( \sqrt[e]{\dfrac{s}{a}} \right ) \implies d = 1 - \sqrt[e]{\dfrac{s}{a}}.[/math]
[math]b_n = (1 - d)b_{n-1} \implies x_n = b_n - b_{n-1} \text { and } b_n = a(1 - d)^n.[/math]
So you are right. The depreciation factor is (supposedly) the result of a calculation involving acquisition cost, salvage value, and expected life. In practice, this method is almost never used because it is a computational nightmare. Over the decades, I have read thousands of corporate financial statements and cannot remember one that used this method.

 

[eddy2017]

Eddy

It has been decades since I studied accounting theory. And practicing accountants frequently take shortcuts and have their own jargon. So this may not be perfect in terms of nomenclature and may not reflect actual day to day practice exactly.

Exponential (or declining balance) depreciation uses a depreciation factor.

[math]a = \text {acquisition cost of asset (including installation).}\\ s = \text {expected salvage value or } 0.01, \text { whichever is higher.}\\ e = \text {expected useful life.}\\ d = \text {depreciation factor.}\\ b_n = \text {net book value at end of year } n.\\ x_n = \text {depreciation expense for year n.}[/math]
In theory,

[math]a(1 + d)^e = s \implies ln(a) + e * ln(1 + d) = ln(s) \implies[/math]
[math]ln(1 - d) = \dfrac{1}{e} * \{ln(s) - ln(a)\} = ln \left ( \sqrt[e]{\dfrac{s}{a}} \right ) \implies d = 1 - \sqrt[e]{\dfrac{s}{a}}.[/math]
[math]b_n = (1 - d)b_{n-1} \implies x_n = b_n - b_{n-1} \text { and } b_n = a(1 - d)^n.[/math]
So you are right. The depreciation factor is the result of a calculation involving acquisition cost, salvage value, and expected life. In practice, this method is almost never used because it is a computational nightmare. Over the decades, I have read thousands of corporate financial statements and cannot remember one that used this method.

Great. Thanks so much for taking the time to explain. Very interesting and it is always good to know. Thanks again.