Calculating the size of an annuity, what is going on??

[Fredrik]

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?

 

[nasi112]

I think that the problem is assuming No Payment will be made during the term of the loan, but John is required to establish a sinking fund, and make annual deposits into this fund in order to accumulate the fully maturity value.

And remember this, if the annual payments are not fixed, you cannot call this problem an annuity.

 

[JeffM]

I would not explain the process that way at all. First, let's use a simpler example.

You can get a loan for 100,000 at 5% compounded annually and payable at the end of two years. You will make a payment of 110250 at the end of two years. Alternatively, you can get two loans, each of 50000 at 5% compounded annually, but one due in one year and the other due after two years. On the first loan, you will make a payment of 52500 on the due date. On the second loan, you will make a payment of 55125. THE PAYMENTS ARE NOT EQUAL.

If you want the payments to be equal, you are essentially asking the banker to make two loans, both at the same interest rate and compounding period, that sum to 100000 but have equal payments due at maturity. Those loan amounts cannot be equal. In our simple example, we can solve through a system of simultaneous equations, where x and y are the amounts of the two loans.

[MATH]x + y = 100000 \text { and } x(1.05) = y(1.05)^2 \implies[/MATH]
[MATH]x = 100000 - y \text { and } x = 1.05y \implies y = \dfrac{100000}{2.05} \approx 48780.49[/MATH]
[MATH]x = \dfrac{205000 - 100000}{2.05} \approx 51219.51.[/MATH]
They add, to the penny, to 1000000. How about the payments?

[MATH]51219.51 * 1.05 \approx 53780.49 \text { and } 48780.49 * 1.05^2 \approx 53780.49.[/MATH]
Equal payments.

Now from the bank's standpoint, this can also be viewed as a sequence of equal payments such that the sum of the present values must equal the initial amount lent.

So, for the general case, we want to solve for c where

[MATH]L= \text {initial amount of loan;}[/MATH]
[MATH]i= \text {interest rate per COMPOUNDING PERIOD;}[/MATH]
[MATH]n= \text {number of compounding periods; and}[/MATH]
[MATH]c = \text{constant payment AT THE END of each compounding period.}[/MATH]
[MATH]L = \sum_{j=1}^n \dfrac{c}{(1 + i)^j}.[/MATH]
That should be quite comprehensible. We have n equal payments of c, one each at the end of one of the n compounding periods. The bank wants the present value of those payments to equal the amount of the loan. Now admittedly the ALGEBRA to solve for c gets a bit ugly, but it is just algebra working from a very common-sense initial equation.

[MATH]L = \sum_{j=1}^n \dfrac{c}{(1 + i)^j} \implies L(1 + i)^n = c * \sum_{j=1}^n (1 + i)^{(n-j)} \implies[/MATH]
[MATH]L(1 + i)^n = c * \sum_{j=0}^{n-1} (1 + i)^j = c * \dfrac{1 - (1 + i)^n }{1 - (1 + i)} = c * \dfrac{1 - (1 + i)^n}{-i} \implies [/MATH]
[MATH]c = -Li * \dfrac{(1 + i)^n}{1 - (1 + i)^n} = L * \dfrac{i(1 + i)^n}{(1 + i)^n - 1} * \dfrac{(1 + i)^{-n}}{(1 + i)^{-n}} \implies[/MATH]
[MATH]c = L * \dfrac{i}{1 - (1 + i)^{-n}},[/MATH] which is the standard formula.

Let's stick that into our example.

[MATH]c = 100000 * \dfrac{0.05}{1 - 1.05^{-2}} = 5000 * \dfrac{1}{1 - \dfrac{1}{1.1025}} \approx 53780.49. \checkmark[/MATH]
Let's stick it into your problem.

[MATH]c = 100000 * \dfrac{0.06}{1 - 1.06^{-10}} \approx 13586.80 \checkmark[/MATH]
They went through their explanation in order to avoid some algebra.

 

[Fredrik]

I would not explain the process that way at all. First, let's use a simpler example.

You can get a loan for 100,000 at 5% compounded annually and payable at the end of two years. You will make a payment of 110250 at the end of two years. Alternatively, you can get two loans, each of 50000 at 5% compounded annually, but one due in one year and the other due after two years. On the first loan, you will make a payment of 52500 on the due date. On the second loan, you will make a payment of 55125. THE PAYMENTS ARE NOT EQUAL.

If you want the payments to be equal, you are essentially asking the banker to make two loans, both at the same interest rate and compounding period, that sum to 100000 but have equal payments due at maturity. Those loan amounts cannot be equal. In our simple example, we can solve through a system of simultaneous equations, where x and y are the amounts of the two loans.

[MATH]x + y = 100000 \text { and } x(1.05) = y(1.05)^2 \implies[/MATH]
[MATH]x = 100000 - y \text { and } x = 1.05y \implies y = \dfrac{100000}{2.05} \approx 48780.49[/MATH]
[MATH]x = \dfrac{205000 - 100000}{2.05} \approx 51219.51.[/MATH]
They add, to the penny, to 1000000. How about the payments?

[MATH]51219.51 * 1.05 \approx 53780.49 \text { and } 48780.49 * 1.05^2 \approx 53780.49.[/MATH]
Equal payments.

Now from the bank's standpoint, this can also be viewed as a sequence of equal payments such that the sum of the present values must equal the initial amount lent.

So, for the general case, we want to solve for c where

[MATH]L= \text {initial amount of loan;}[/MATH]
[MATH]i= \text {interest rate per COMPOUNDING PERIOD;}[/MATH]
[MATH]n= \text {number of compounding periods; and}[/MATH]
[MATH]c = \text{constant payment AT THE END of each compounding period.}[/MATH]
[MATH]L = \sum_{j=1}^n \dfrac{c}{(1 + i)^j}.[/MATH]
That should be quite comprehensible. We have n equal payments of c, one each at the end of one of the n compounding periods. The bank wants the present value of those payments to equal the amount of the loan. Now admittedly the ALGEBRA to solve for c gets a bit ugly, but it is just algebra working from a very common-sense initial equation.

[MATH]L = \sum_{j=1}^n \dfrac{c}{(1 + i)^j} \implies L(1 + i)^n = c * \sum_{j=1}^n (1 + i)^{(n-j)} \implies[/MATH]
[MATH]L(1 + i)^n = c * \sum_{j=0}^{n-1} (1 + i)^j = c * \dfrac{1 - (1 + i)^n }{1 - (1 + i)} = c * \dfrac{1 - (1 + i)^n}{-i} \implies [/MATH]
[MATH]c = -Li * \dfrac{(1 + i)^n}{1 - (1 + i)^n} = L * \dfrac{i(1 + i)^n}{(1 + i)^n - 1} * \dfrac{(1 + i)^{-n}}{(1 + i)^{-n}} \implies[/MATH]
[MATH]c = L * \dfrac{i}{1 - (1 + i)^{-n}},[/MATH] which is the standard formula.

Let's stick that into our example.

[MATH]c = 100000 * \dfrac{0.05}{1 - 1.05^{-2}} = 5000 * \dfrac{1}{1 - \dfrac{1}{1.1025}} \approx 53780.49. \checkmark[/MATH]
Let's stick it into your problem.

[MATH]c = 100000 * \dfrac{0.06}{1 - 1.06^{-10}} \approx 13586.80 \checkmark[/MATH]
They went through their explanation in order to avoid some algebra.

Well thank you very much for the thorough explanation. Your own solution I understand, I believe. However, I still can't wrap my head around why it would be the same as my textbook's solution.
Your solution means that c, which I called x, is a completed payment, so to speak. A certain amount was borrowed in 2013 and and at maturity it has reached $13586, no matter which year it reached maturity. But that sub-loan is then payed off. It never increases beyond this. Yet in the equation [MATH]x+x∗1.06+x∗1.06^2...+x∗1.06^9=100000∗1.06^{10}[/MATH], the 13586 increases every year until 2022. It isn't a completed sub-loan, as far as I'm able to interpret the math.

 

[JeffM]

Well thank you very much for the thorough explanation. Your own solution I understand, I believe. However, I still can't wrap my head around why it would be the same as my textbook's solution.
Your solution means that c, which I called x, is a completed payment, so to speak. A certain amount was borrowed in 2013 and and at maturity it has reached $13586, no matter which year it reached maturity. But that sub-loan is then payed off. It never increases beyond this. Yet in the equation [MATH]x+x∗1.06+x∗1.06^2...+x∗1.06^9=100000∗1.06^{10}[/MATH], the 13586 increases every year until 2022. It isn't a completed sub-loan, as far as I'm able to interpret the math.

Yes, I think you understand my reasoning. We are treating the loan as a series of sub-loans, each of which accrues interest at a compounded rate and each of which is paid at a different maturity date in a one-time payment that is the same amount for each loan. When a sub-loan is paid, it no longer accrues interest.

Now you can get to your book's equation quite easily. We had the equation

[MATH]L = \sum_{j=1}^n \dfrac{c}{(1 + i)^j}[/MATH]. And I presume that you understand that.

And from there we did just a bit of algebraic manipulation and got

[MATH]L(1 + i)^n = c * \sum_{j=1}^n (1 + i)^{(n-j)}.[/MATH]
Almost trivial algebra.

Now consider that series. Expanded, it looks like

[MATH]c * \{((1 + i)^{(n-1)} + (1 + i)^{(n-2)} + ... (1 + i)^2 + (1 + i)^1 + (1 + i)^0\}.[/MATH]
But we can reverse the order of the terms to

[MATH]c * \{(1 + i)^0 + (1 + i)^1 + (1 + i)^2 + ... + (1 + i)^{(n-2)} + (1 + i)^{(n-1)} \}.[/MATH]
In fact, I IMPLICITLY did that myself to get things into a form where the geometric series was easily recognizable. I showed that RESULT as

[MATH]L(1 + i)^n = c * \sum_{j=0}^{n-1} (1 + i)^j.[/MATH]
But what your book does is to multiply out the expanded form.

[MATH]L(1 + i)^n = c * \{(1 + i)^0 + (1 + i)^1 + (1 + i)^2 + ... + (1 + i)^{(n-2)} + (1 + i)^{(n-1)} \} \implies[/MATH]
[MATH]L(1 + i)^n = c * \{1+ (1 + i) + (1 + i)^2 + ... + (1 + i)^{(n-2)} + (1 + i)^{(n-1)} \} \implies[/MATH]
[MATH]L(1 + i)^n = c + c(1 + i) + c(1 + i)^2 + ... + c(1 + i)^{(n-2)} + c(1 + i)^{(n-1)} \}.[/MATH]
The math is exactly the same although they do not bother to go through the algebraic manipulations needed to solve for c. As I said in my first post, I find the approach unintuitive, and skipping the algebra makes the whole thing look like magic. I do not understand that method of teaching a simple formula and the logic behind it.