How to Calculate the Future Value of Geometric Annual Annuity?

[waptrick]

The geometric annual annuity runs for
a period of 45 years which starts now.

• The first payment at the end of the
first year from now amounts to R2 000.

• The growth rate of the annual payments
equals 7% per year during the first
30 years (of the period of 45 years).

• The growth rate of the annual payments
equals 11% per year during the last
15 years (of the period of 45 years).

• The interest rate during the first
20 years (of the period of 45 years)
amounts to 10% per annum.

• The interest rate during the last
25 years (of the period of 45 years)
is equal to 14% per annum.

Calculate the future value of the first 20 annual
payments (of the total period of 45 years)
at the end of the period, 45 years from now

 

[MarkFL]

Hello and welcome to MHB, waptrick! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[waptrick]

I've used the geometric annuity formula: PMT [((1+i)^n-(1+g)^n)/(i-g)]

with

PMT=2000
i=10%
g=7%
n=20

The answer I got was 190 521.03

 

[waptrick]

The main problem I am having is that the interest rates and growth rates differ and have different periods.

 

[MarkFL]

If I were going to work this, without being given a formula, I would begin by finding the $n$th payment, $P_n$ which grows at the rate $g$. We can do so with the following difference equation:

\(\displaystyle P_{n}=(1+g)P_{n-1}\)

We see the characteristic root $\lambda$ is:

\(\displaystyle \lambda=1+g\)

And so the general form of $P_{n}$ is:

\(\displaystyle P_n=k_1(1+g)^n\)

Now, using:

\(\displaystyle P_1=k_1(1+g)\)

we obtain:

\(\displaystyle k_1=\frac{P_1}{1+g}\)

And so we have:

\(\displaystyle P_n=P_1(1+g)^{n-1}\)

Now, to find the future value $F_n$ of the annuity at the end of year $n$, where $i$ interest is compounded annually, we may use the difference equation:

\(\displaystyle F_{n}=(1+i)F_{n-1}+P_{n}\)

I would arrange this as (and put in our closed-form for $P_n$):

\(\displaystyle F_{n}-(1+i)F_{n-1}=P_1(1+g)^{n-1}\)

We see that the homogeneous solution will have the form:

\(\displaystyle h_n=k_1(1+i)^n\)

Now, we may assume the particular solution will have the form:

\(\displaystyle p_n=A(1+g)^{n-1}\)

And by substitution, there results:

\(\displaystyle A(1+g)^{n-1}-(1+i)A(1+g)^{n-2}=P_1(1+g)^{n-1}\)

And this reduces to:

\(\displaystyle A(1+g)-A(1+i)=P_1(1+g)\implies A=\frac{P_1(1+g)}{g-i}\)

Hence, the particular solution is:

\(\displaystyle p_n=\frac{P_1}{g-i}(1+g)^n\)

And so the general solution is:

\(\displaystyle F_{n}=h_{n}+p_{n}=k_1(1+i)^{n}+\frac{P_1}{g-i}(1+g)^n\)

Now, given that we must have:

\(\displaystyle F_1=P_1\)

We find:

\(\displaystyle F_{1}=k_1(1+i)+\frac{P_1}{g-i}(1+g)=P_1\)

Thus:

\(\displaystyle k_1=\frac{P_1}{i-g}\)

And so our solution becomes:

\(\displaystyle F_{n}=\frac{P_1}{i-g}(1+i)^{n}-\frac{P_1}{i-g}(1+g)^n=\frac{P_1}{i-g}\left((1+i)^{n}-(1+g)^{n}\right)\)

This agrees with the formula you gave, and so using the given data for the first 20 years, we find (in dollars, to the nearest penny):

\(\displaystyle F_{20}\approx190521.03\)

This agrees with the result you gave. (Yes)

Now, we are asked to find just the value of this amount after the total period of 45 years, so we have 25 years of interest to account for. The interest rate for the remaining 25 years is 14%, so this become a simple interest problem, where the principle is $F_{20}$. Can you proceed?

 

[waptrick]

Proceeding with your logic, would the formula be 190521.03(1+0.14*25)= $857344.64 to the nearest penny

 

[MarkFL]

I would wait to round only the final result. I would use:

\(\displaystyle F_{45}=F_{20}(1.14)^{25}\approx5,041,551.52\)

 

[waptrick]

Would that not be compounding interest as opposed to simple interest?

 

[MarkFL]

Would that not be compounding interest as opposed to simple interest?

My use of the word "simple" was unfortunate there...I should have said it would be compounded annually, which is a simple problem (compared to what you had already correctly done). Sorry for the confusion.

 

[waptrick]

Perfect I understand now. I also have another related question. What would the value of the payment that occurs at the end of the 23rd year be?

 

[MarkFL]

Perfect I understand now. I also have another related question. What would the value of the payment that occurs at the end of the 23rd year be?

You can use the formula I derived for the $n$th payment $P_n$:

\(\displaystyle P_n=P_1(1+g)^{n-1}\)

 

[waptrick]

I got $8860.80 by 2000(1+0.07)^22

 

[MarkFL]

I got $8860.80 by 2000(1+0.07)^22

I get the same value. (Yes)

 

[waptrick]

My next question relates to the future value of the 10 annual payments that are received from the end of Year 21 until the end of Year 30 at the end of the period of 45 years?

Do you use PMT=2000, or do you have to apply the growth rate first?

 

[MarkFL]

My next question relates to the future value of the 10 annual payments that are received from the end of Year 21 until the end of Year 30 at the end of the period of 45 years?

Do you use PMT=2000, or do you have to apply the growth rate first?

I would first determine the value of the 21st payment $P_{21}$, and then work it in the same way as your first question. It looks like $g$ and $i$ remain constant during these years. :)

 

[waptrick]

So in effect 2000(1+0.07)^20= 7739.37

But, the interest rate would change from 0.10 to 0.14 because of the initial question?

Then, 7739.37[[(1+0.14)^10-(1+0.07)^10]/0.14-0.07]

 

[MarkFL]

Yes:

\(\displaystyle P_{21}=P_1(1.07)^{20}\approx7739.37\)

Then we determine the value of the 10 payments in question using the formula (so yes, you are correct):

\(\displaystyle F_{n}=\frac{P_{21}}{i-g}\left((1+i)^{n}-(1+g)^{n}\right)\)

where $i=0.14,\,g=0.07,\,n=10$ to get:

\(\displaystyle F_{10}=\frac{2000(1.07)^{20}}{0.07}\left(1.14^{10}-1.07^{10}\right)\approx192386.33\)

Now we have 15 years of annually compounded interest at 14% to compute:

\(\displaystyle A=F_{10}\cdot1.14^{15}\approx?\)

 

[waptrick]

I get $ 1373241.69. If the initial question was changed so that it was asked what is the future value of the last 15 annual payments at the end of the period of 45 years? Would you compound the annuity or just calculate the annuity for the last 15 payments?

 

[MarkFL]

I get $ 1373241.69.

I get:

\(\displaystyle A\approx1,373,241.72\)

Difference is most likely due to my not rounding until the very end. :)

If the initial question was changed so that it was asked what is the future value of the last 15 annual payments at the end of the period of 45 years? Would you compound the annuity or just calculate the annuity for the last 15 payments?

You would first determine $P_{35}$, and then work it the same as the previous part, where there is no period of years to just compound interest as the last step.

 

[waptrick]

Would it be 2000(1+0.11)^34= $69,504.24

Then, 69504.24[[(1+0.14)^15-(1+0.11)^15]/0.14-0.11]= $5,452,256.30

Then, 5,452,256.30(1.14)= $6,215,572.18

 

[MarkFL]

Getting $P_{35}$ is a bit trickier now, because the growth rate changes after the first 30 years...:)

 

[waptrick]

So do you get the value of Payment 30 and then use the new growth rate to get payment 35?

Such as 2000(1+0.07)^29 = $14,228.51

Then, 14228.51(1+0.11)^4 = $21,599.89

 

[MarkFL]

You have the right idea, although I get:

\(\displaystyle P_{35}=(2000(1.07)^{29})(1.11)^5\approx23975.87\)

Do you see why the exponent is 5 on the last factor?

 

[waptrick]

No why is the exponent 5 and not 4?

 

[MarkFL]

No why is the exponent 5 and not 4?

Well, once we have $P_{30}$, then we need to apply the new growth rate an additional $n$ times to get the payment at the end of year $y=30+n$. The reason we had an exponent of $n-1$ before is because we were given the value of $P_1$ and that didn't grow until the end of year 2. :)

 

[waptrick]

Ahh that makes sense now. :)So...23975.87[[(1+0.14)^15 -(1+0.11)^15)/0.14-0.11] = $1,880,785.921880785.92(1.14)= 2,144,095.94

 

[MarkFL]

Ahh that makes sense now. :)So...23975.87[[(1+0.14)^15 -(1+0.11)^15)/0.14-0.11] = $1,880,785.921880785.92(1.14)= 2,144,095.94

Why did you add 14% at the end?

 

[jonah1]

Beer soaked ramblings follow.

If the initial question was changed so that it was asked what is the future value of the last 15 annual payments at the end of the period of 45 years? Would you compound the annuity or just calculate the annuity for the last 15 payments?

You would first determine $P_{35}$, and then work it the same as the previous part, where there is no period of years to just compound interest as the last step.

Nay Sir Mark, I say nay.
Me beer goggles, courtesy of my alcohol enhanced awareness (chemically impaired to many a frowning soul), tells me that you may have meant to write $P_{30}$

All told, I be almost certain that we should get something like $1,238,932.13
Maybe. Maybe not.
Let's see if Sir Denis will agree with his caffeine soaked calculations.

 

[MarkFL]

Yes, it should have been $P_{30}$. :)

 

[Wilmer]

Agree (just had a strong green tea!).
In case this helps, pertinent transactions in Bank statement format:

year      payment         interest           balance
  0                                              .00
  1      2,000.00              .00          2,000.00
  2      2,140.00(7%)       200.00(10%)     4,340.00
...
 19      6,759.86        14,533.23        166,625.43
 20      7,233.06        16,662.54        190,521.03
 21      7,739.37        26,672.94(14%)   224,933.34
...
 30     14,228.51       108,618.07        898,689.97
 31     15,793.65(11%)  125,816.60      1,040,300.22
...
 45     68,077.60       931,570.78      7,653,725.37

 

[MarkFL]

Since I botched part of one of the questions, I am going to amass everything into one post as a follow-up. The questions asked are as follows:

---

A geometric annual annuity runs for a period of 45 years which starts now.

• The first payment at the end of the first year from now amounts to R2 000.
• The growth rate of the annual payments equals 7% per year during the first 30 years (of the period of 45 years).
• The growth rate of the annual payments equals 11% per year during the last 15 years (of the period of 45 years).
• The interest rate during the first 20 years (of the period of 45 years) amounts to 10% per annum.
• The interest rate during the last 25 years (of the period of 45 years) is equal to 14% per annum.

a) Calculate the future value of the first 20 annual payments (of the total period of 45 years) at the end of the period, 45 years from now.

b) What would the value of the payment that occurs at the end of the 23rd year be?

c) Find the future value of the 10 annual payments that are received from the end of Year 21 until the end of Year 30 at the end of the period of 45 years?

d) Find the future value of the last 15 annual payments at the end of the period of 45 years.

---

In post #5, I derived the following formulas:

\(\displaystyle P_{n}=P_1(1+g)^{n-1}\tag{1}\)

\(\displaystyle F_{n}=\frac{P_1}{i-g}\left((1+i)^{n}-(1+g)^{n}\right)\tag{2}\)

a) For the first 20 payments, we have the following data:

\(\displaystyle P_1=2000,\,i=0.10,\,g=0.07,\,n=20\)

Hence, using (2), we find:

\(\displaystyle F_{20}=\frac{2000}{0.03}\left(1.1^{20}-1.07^{20}\right)\approx190521.03\)

And then applying the interest rate of 14% compounded annually for the remaining 25 years, we have:

\(\displaystyle F_{45}=F_{20}(1.14)^{25}\approx5,041,551.52\)

b) For the first 23 years, we have:

\(\displaystyle P_1=2000,\,g=0.07,\,n=23\)

And so, using (1), we get:

\(\displaystyle P_{23}=2000\cdot1.07^{22}approx8860.80\)

c) From the end of year 21 until the end of year 30, we have the following data:

\(\displaystyle P_21=2000\cdot1.07^{20},\,i=0.14,\,g=0.07,\,n=10\)

And so, using (2), we find:

\(\displaystyle F_{10}=\frac{2000(1.07)^{20}}{0.07}\left(1.14^{10}-1.07^{10}\right)\)

And then the 15 years of annually compounded interest at 14% gives:

\(\displaystyle A=F_{10}\cdot1.14^{15}\approx1,373,241.72\)

d) For the last 15 annual payments at the end of the period of 45 years, we have:

\(\displaystyle P_{31}=2000\cdot1.07^{29}\cdot1.11,\,i=0.14,\,g=0.11,\,n=15\)

And so, using (2) we have:

\(\displaystyle A=\frac{2000\cdot1.07^{29}\cdot1.11}{0.03}\left(1.14^{15}-1.11^{15}\right)\approx1,238,932.13\)