I have recently encountered the problem of annuities. That is to say, paying off a loan with equal size payments each month/year. The example problem that is posed, which is the only thing containing information about annuities (it's a subtopic), is the following:
John took out a loan in the beginning of 2013 of $100 000. He will pay interest and amortization on the loan with 10 payments equal in size (annuities) at the end of each year, beginning with the end of 2013. The annual interest rate is 6%. How large is the annuity?
(Note that this is more or less directly translated from swedish, so I do not know whether the terms I'm using correspond to the same concepts as their swedish counterparts.)
Now the proposed solution:
We will draw a timeline and state what each payment, $x, has grown to at the end of the 10th year.
|-2013-|-2014-|-2015-|-2016-|-2017-|-2018-|-2019-|-2020-|-2021-|-2022-|-->
100k____x_______x______x_______x_______x_______x______x______x_______x_______x
----------|_______|____________________________________________________ |--------x*1.06
----------|_______|-------------------------------------------------------------- x*1.06^9
----------|---------------------------------------------------------------------- x*1.06^9
The value of the 10 payments (annuities) with interest together are to be equal to the whole loan with 10 years of interest. This give the equation:
[MATH]x + x * 1.06 + x * 1.06^2 ... + x * 1.06^9 = 100000 * 1.06^{10}[/MATH]
The sum of a geometric series give the sum of the left side to be:
[MATH]x*(1.06^{10}-1)/(1.06-1)=100000*1.06^{10}[/MATH] -> [MATH]x=13586.80[/MATH]
Now, what I do not understand is why expressing the solution like this works and why it makes sense.
If the payment of $x each month contains some varying amount of amortization (paying off a part of the loan, so that less of the loan remains for interest to apply to), then the loan will not have been the size of 100000*1.06^{10} at the end of the 10 years. There is only $100000 for interest to apply to at the end of 2013. Subsequent years there is less of the loan remaining. Also, the payments only pay off the loan, which means it isn't stored somewhere, which means there is no reason why it should increase in value due to interest. As I understand the real situation, what happens each year, say starting with 2013, is that 100000*1.06 is decreased by 13600, and remaining is 92400. At the end of 2014, 92400*1.06 is decreased by 13600, and so on. I have done this calculation and it actually checks out. After ten such payments, the loan is perfectly payed off, none left, no excess.
However, when I add up the amount of interest that you have payed for the loan, it is not the same as if you calculate with [MATH]1.06^10[/MATH]. The interest that you pay each month is [whatever is left of the loan * 1.06] - [whatever is left of the loan]. In other words, the amount the loan increased with. Applying that to the entire loan to get the cost of the loan is the same as taking the entire payment, 13600, times the amount of payments, 10, and subtracting the loan itself.
[MATH]13586.8*10-100000=35868[/MATH]. I did the above ^ calculation for all ten payments and it is indeed the same, see the attachment. This is different from [MATH]100000*1.06^{10}-100000=79085[/MATH].
What the problem seems to assume and calculate is if you create an account where you make a deposit of $x each year for ten years, never paying off the loan until 10 years later, when the loan has grown to $179000.
So why would their solution, which seemingly does not portray the problem at all, give you the correct answer?
John took out a loan in the beginning of 2013 of $100 000. He will pay interest and amortization on the loan with 10 payments equal in size (annuities) at the end of each year, beginning with the end of 2013. The annual interest rate is 6%. How large is the annuity?
(Note that this is more or less directly translated from swedish, so I do not know whether the terms I'm using correspond to the same concepts as their swedish counterparts.)
Now the proposed solution:
We will draw a timeline and state what each payment, $x, has grown to at the end of the 10th year.
|-2013-|-2014-|-2015-|-2016-|-2017-|-2018-|-2019-|-2020-|-2021-|-2022-|-->
100k____x_______x______x_______x_______x_______x______x______x_______x_______x
----------|_______|____________________________________________________ |
----------|_______|
----------|
The value of the 10 payments (annuities) with interest together are to be equal to the whole loan with 10 years of interest. This give the equation:
[MATH]x + x * 1.06 + x * 1.06^2 ... + x * 1.06^9 = 100000 * 1.06^{10}[/MATH]
The sum of a geometric series give the sum of the left side to be:
[MATH]x*(1.06^{10}-1)/(1.06-1)=100000*1.06^{10}[/MATH] -> [MATH]x=13586.80[/MATH]
Now, what I do not understand is why expressing the solution like this works and why it makes sense.
If the payment of $x each month contains some varying amount of amortization (paying off a part of the loan, so that less of the loan remains for interest to apply to), then the loan will not have been the size of 100000*1.06^{10} at the end of the 10 years. There is only $100000 for interest to apply to at the end of 2013. Subsequent years there is less of the loan remaining. Also, the payments only pay off the loan, which means it isn't stored somewhere, which means there is no reason why it should increase in value due to interest. As I understand the real situation, what happens each year, say starting with 2013, is that 100000*1.06 is decreased by 13600, and remaining is 92400. At the end of 2014, 92400*1.06 is decreased by 13600, and so on. I have done this calculation and it actually checks out. After ten such payments, the loan is perfectly payed off, none left, no excess.
However, when I add up the amount of interest that you have payed for the loan, it is not the same as if you calculate with [MATH]1.06^10[/MATH]. The interest that you pay each month is [whatever is left of the loan * 1.06] - [whatever is left of the loan]. In other words, the amount the loan increased with. Applying that to the entire loan to get the cost of the loan is the same as taking the entire payment, 13600, times the amount of payments, 10, and subtracting the loan itself.
[MATH]13586.8*10-100000=35868[/MATH]. I did the above ^ calculation for all ten payments and it is indeed the same, see the attachment. This is different from [MATH]100000*1.06^{10}-100000=79085[/MATH].
What the problem seems to assume and calculate is if you create an account where you make a deposit of $x each year for ten years, never paying off the loan until 10 years later, when the loan has grown to $179000.
So why would their solution, which seemingly does not portray the problem at all, give you the correct answer?