- #31
MarkFL
Gold Member
MHB
- 13,284
- 12
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.
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:
$$P_{n}=P_1(1+g)^{n-1}\tag{1}$$
$$F_{n}=\frac{P_1}{i-g}\left((1+i)^{n}-(1+g)^{n}\right)\tag{2}$$
a) For the first 20 payments, we have the following data:
$$P_1=2000,\,i=0.10,\,g=0.07,\,n=20$$
Hence, using (2), we find:
$$F_{20}=\frac{2000}{0.03}\left(1.1^{20}-1.07^{20}\right)\approx190521.03$$
And then applying the interest rate of 14% compounded annually for the remaining 25 years, we have:
$$F_{45}=F_{20}(1.14)^{25}\approx5,041,551.52$$
b) For the first 23 years, we have:
$$P_1=2000,\,g=0.07,\,n=23$$
And so, using (1), we get:
$$P_{23}=2000\cdot1.07^{22}approx8860.80$$
c) From the end of year 21 until the end of year 30, we have the following data:
$$P_21=2000\cdot1.07^{20},\,i=0.14,\,g=0.07,\,n=10$$
And so, using (2), we find:
$$F_{10}=\frac{2000(1.07)^{20}}{0.07}\left(1.14^{10}-1.07^{10}\right)$$
And then the 15 years of annually compounded interest at 14% gives:
$$A=F_{10}\cdot1.14^{15}\approx1,373,241.72$$
d) For the last 15 annual payments at the end of the period of 45 years, we have:
$$P_{31}=2000\cdot1.07^{29}\cdot1.11,\,i=0.14,\,g=0.11,\,n=15$$
And so, using (2) we have:
$$A=\frac{2000\cdot1.07^{29}\cdot1.11}{0.03}\left(1.14^{15}-1.11^{15}\right)\approx1,238,932.13$$
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.
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:
$$P_{n}=P_1(1+g)^{n-1}\tag{1}$$
$$F_{n}=\frac{P_1}{i-g}\left((1+i)^{n}-(1+g)^{n}\right)\tag{2}$$
a) For the first 20 payments, we have the following data:
$$P_1=2000,\,i=0.10,\,g=0.07,\,n=20$$
Hence, using (2), we find:
$$F_{20}=\frac{2000}{0.03}\left(1.1^{20}-1.07^{20}\right)\approx190521.03$$
And then applying the interest rate of 14% compounded annually for the remaining 25 years, we have:
$$F_{45}=F_{20}(1.14)^{25}\approx5,041,551.52$$
b) For the first 23 years, we have:
$$P_1=2000,\,g=0.07,\,n=23$$
And so, using (1), we get:
$$P_{23}=2000\cdot1.07^{22}approx8860.80$$
c) From the end of year 21 until the end of year 30, we have the following data:
$$P_21=2000\cdot1.07^{20},\,i=0.14,\,g=0.07,\,n=10$$
And so, using (2), we find:
$$F_{10}=\frac{2000(1.07)^{20}}{0.07}\left(1.14^{10}-1.07^{10}\right)$$
And then the 15 years of annually compounded interest at 14% gives:
$$A=F_{10}\cdot1.14^{15}\approx1,373,241.72$$
d) For the last 15 annual payments at the end of the period of 45 years, we have:
$$P_{31}=2000\cdot1.07^{29}\cdot1.11,\,i=0.14,\,g=0.11,\,n=15$$
And so, using (2) we have:
$$A=\frac{2000\cdot1.07^{29}\cdot1.11}{0.03}\left(1.14^{15}-1.11^{15}\right)\approx1,238,932.13$$