Motivation
1. If you have some amount of unused money, what do you do
with it? What are type of investments that provide you a known expected
outcome?
If I have some amount of unused money, I will investments it,
example:
- Saving Account
- Goverment Bonds
- Corporate Bonds
Tuition fee preparation
When a child was born in 2017, her parents decided to invest
in her college education. This was motivated by a forecast that 4 years
of tuition will be about $300, 000 when she will attend college. Suppose
that the parents want to accumulate that amount by their child’s 17th
birthday. They open a bank account into which they make a deposit on
each birthday of the child up to the 17th birthday. Assume that the
first deposit is for the amount P and thereafter the parents increase
the deposited amount by 8% annually. Suppose that the bank fixed
interest rate at 6% per annum compounded monthly. What should the
minimum annual deposits be in order for the amount in the fund to reach
at least $300, 000 after the 17th deposit?
\[ r = 6\% \]
Future value of the first saving
\[ P(1+\frac{6\%}{12}) \]
Future value of the second saving:
\[P(1+8\%)(1+\frac{6\%}{12})^{15\times12}
\]
Future value of the third saving:
\[
P(1+8\%)^2(1+\frac{6\%}{12})^{14\times12} \]
.
.
.
Future value of the final saving:
\[P(1+8\%)^{16}\]
The amount in the fund to reach
\[
P(\sum_{i=1}^{17}(1+8\%)^{i-1}(1+\frac{6\%}{12})^{17\times12-12i}) =
$300.000 \]
\[ \to P\times 50.9689 = $300.000
\] \[ \to P=$5885.94 \]
Monthly mortgage loan payments
1. A person receives a loan of 800 Million VND in return for
fixed monthly payments K over the next 20 years. Calculate the payment K
if the interest rate is 8% per year compounded annually?
The first period:
\[ \frac{K}{(1+8\%)^{\frac{1}{12}}}
\]
The second periods:
\[ \frac{K}{(1+8\%)^{\frac{2}{12}}}
\]
The present value of the ith payment is
\[ \frac{K}{(1+8\%)^{\frac{i}{12}}}
\]
The present value of payment
\[ P =
\sum_{i=1}^{240}\frac{K}{(1+8\%)^{\frac{i}{12}}} = K\times
\sum_{i=1}^{240}\frac{K}{(1+8\%)^{\frac{i}{12}}}\]
\[ \to 800\times 10^6 = K\times 122.078
\] \[ \to K = 6,553,203.411
\]
2. A person receives a loan of 500 Million VND from a local
bank with loan tenor is 13 years and the interest rate is 9% per year
compounded annually. Due to the Covid 19, the customer only pays a fixed
monthly interest I in the first 2 years and then pay fixed monthly
payments K over the next 11 years. Compute the payment K and I. Note
that the first monthly payment is exactly due at one month after the
loan principal is disbursed.
The interest rate (r) per year compounded daily
\[ P(1+\frac{r}{12}^{2\times12}) =
P(1+9\%)^2 \to r = 8.65\%\]
Interest rate (I):
\[ I = 500\times10^{6}\times\frac{r}{12} =
$3,604,166.667 \]
Present value for all payment in the next 11 years:
\[ P =
\sum_{i=1}^{11\times12}\frac{K}{(1+8.65\%)^{\frac{i}{12}}} \to K =
5,883,884.81\]
Mortgage payments toward capital and interest
1. A couple bought their house 10 years ago for $500000 and
put down 50% on the house. On the balance, they took out a 20-year
mortgage at 9% per annum with annually compounding. To payoff the
mortage, the couple paid a constant amount $P every year. Suppose that
the current net market value is now $650000 and the couple wants to sell
their house. Note that the first yearly payment is exactly due at one
year after the loan principal is disbursed.
a. Caluculate P.
The loan amount for the mortgage:
\[ F = 500,000\times(1-0.5) = 250.000
\]
The present value of the loan amount for 20 next-year:
\[ F = P\times
\sum_{i=1}^{20}\frac{1}{(1+9\%)^{i}} \to P = 27,386.61875 \]
b. How much equity (to the nearest dollar) is in the house
today? Equity
\[ B_{10} = F(1+r_{p})^{10} -
\frac{P}{9\%}((1+9\%)^{10}-1) = 175,757.9448 \]
The equity is:
\[ 650.000 - 175,757.9448 = 474,242.0552
\]
c. What are the 1st and 10nd interest
payments?
The first interest payment:
\[ I_{1} = F\times9\% = 22.500 \]
\[B_{9} = \frac{B_{10}+P}{1+9\%} =
186,371.1592 \]
The tens interest payment:
\[I_{10} = B_{9}\times0.09 = 16,773.40433
\]
d. What are the 1st and 10nd payments toward
capital?
The first payment towards capital is:
\[ P - I_{1} = 273,86.61875 - 22.500 =
4886.61875 \]
The tens payment:
\[ P - I_{10} = 273,86.61875 -
16,773.40433 = 10,613.21442 \]
Daily simple interest method
1. A person receives a loan of 25 Million VND on 20/05/2021
from a finance consumer company with the interest rate is 34% per year
with daily simple interest. On 16th day of every month, starting from
16/06/2021, the person has to pay a fixed payment K. The contract closed
date is on 16/10/2022. Compute the payment K.
From 20/5/2021 to 16/6/2021 : 27 days
From 16/6/2021 to 16/7/2021 : 30 days
Present value of the first payment
\[ r = 34\% daily \]
\[ K = A_{1}(1+\frac{34\%}{365}\times27)
\to A_{1} = \frac{K}{(1+\frac{34\%}{365})\times27} \]
Present value of the second payment:
\[ A_{2} = (1+\frac{34\%}{365}\times27) =
B_{2} \] \[ B_{2} =
(1+\frac{34\%}{365}\times30) = K \] \[
A_{2}(1+\frac{34\%}{365}\times27)(1+\frac{34\%}{365}\times30) = K
\] \[ \to A_{2} =
\frac{K}{(1+\frac{34\%}{365}\times27)(1+\frac{34\%}{365}\times30)}
\]
Using Excel, we have
\[ 25\times10^{6} = \frac{K}{13.3825} \to
K = 1.868111339 \] ``
2. A person receives a loan of 25 Million VND on 20/05/2021
from a finance consumer company with the interest rate is 34% per year
with daily simple interest. Due to the Covid 19 impact, the customer has
to pay a fixed monthly interest I in the first 6 months. In particular,
on 16th day of every month, starting from 16/06/2021 to to 16/11/2021,
the person only pays interest on the outstanding principal. On 16th day
of every month, from 16/12/2021 to 16/11/2022, the person has to pay a
fixed monthly payment K. The contract closed date is on 16/11/2022.
Compute the payment K and I
The future value of the original capital
\[ F =
25(1+\frac{34\%}{365}\times27)(1+\frac{34\%}{365}\times30)(1+\frac{34\%}{365}\times31)(1+\frac{34\%}{365}\times31)(1+\frac{34\%}{365}\times30)(1+\frac{34\%}{365}\times31)
= 29.496 \]
The total accumulated interest: \(F-25\)
Future value of first I:
\[
I(1+\frac{34\%}{365}\times30)^2(1+\frac{34\%}{365}\times31)^3
\]
Future value of the second interest payment:
\[
I(1+\frac{34\%}{365}\times30)(1+\frac{34\%}{365}\times31)^3
\]
\[ F-25 =
I(1+\frac{34\%}{365}\times30)^2(1+\frac{34\%}{365}\times31)^3 +
I(1+\frac{34\%}{365}\times30)(1+\frac{34\%}{365}\times31)^3 +...+ I
\]
\[\to I = 0.33 \]
Apply Using Excel
\[ \to K = 559,600 \]
Future value of the second interest payment:
\[
I(1+\frac{34\%}{365}\times30)^2(1+\frac{34\%}{365}\times31)^3
\]
Personal financial plan
1. How much money that you need to have so that you can live
without working? How would you obtain that monthly amount?
I think I need 10 billions to live without working. I will hard
working in 40 year and after that I using salary every month to
investment.
Inflation effect
1. Suppose a family’s monthly expense is 20 million VND and
the inflation rate is constant at 4% per year. To maintain the same
current living quality, how much does the family expect to pay per month
in the next 10 years?
The payment per month in the next 10 years
\[ 20\times(1+4\%)^{10} = 29.60
\]
2. Suppose the inflation rate is constant at 4% per year.
Compute the present used value of 100 million in the next 2 years (i.e.,
what is the amount of money that is equivalent in buying power with 100
million VND in the next 2 years)?
Present used value of I
\[ \frac{100}{(1+4\%)^{10}} = 92.46
\]
Forward planning
1. A person decides to deposit 120 MVND at the begin of each
of the next 30 years, with a constant interest rate 7% per year
compounded yearly.
a) At the begin of the next 31 years, the person has
obtained an accumulated amount X. What is X?
The first year:
\[ P_{1} = 120(1+7\%) \]
The second year:
\[ P_{2} = 120(1+7\%)^2 \]
The thirty year:
\[ P_{30} = 120(1+7\%)^{30} \]
The accumulated amount is:
\[ X_{31} =
\sum_{i=1}^{30}120\times10^{6}\times(1+7\%)^{i} = 1.21\times10^{10}
\]
b) What is the amount of interest I that X produces in a
year
Interest rate I is:
\[ I = X_{31}\times7\% =
1.21\times10^{10}\times7\% = 847M\]
c) If the inflation rate is fixed at 4%, what is the present
used value of I discounted by the inflation rate?
Present used value of I:
\[ PV = \frac{847}{(1+4\%)^{32}} = 241.444
\]
d) At year 32th, the person desires to take an amount D.
From year 33th on ward, the yearly amount withdrew increases by 4% to
avoid the effect of inflation. In particular, the yearly amount withdrew
in year 33th, 34th are 1.04D and 1.042D, respectively. What is the value
of D such that the remaining amount also increases 4% each
year?
| X |
\(X(1+7\%)\) |
\(X(1+7\%)(1+4\%)\) |
\(X(1+7\%)(1+4\%)^2\) |
|
-D |
\(-1.04D\) |
\(-1.04^2D\) |
| Total |
\(1.04X\) |
\(1.04^2X\) |
\(1.04^3X\) |
In the \(32^{th}\) year:
\[X(1+7\%) - D = 1.04X \to D = 3\%X
\]
In the \(33^{th}\) year:
\[ X(1.07)(1.04) - 1.04D = 1.04^2X
\] \[\to D = 0.03X \to D = 363M
\]
e) At year 32th, the person desires to take an amount E.
From year 33th on ward, the yearly amount withdrew increases by 4% to
avoid the effect of inflation. In particular, the yearly amount withdrew
in year 33th, 34th are 1.04E and 1.042E, respectively. What is the value
of E such that no money left over after year 72th?
|
E |
\(1.04E\) |
\(1.04^{40}E\) |
| X |
\(1.07\) |
\(1.07^2\) |
\(1.07^{41}\) |
In the \(1^{st}\) period : \(\frac{E}{1.07}\)
In the \(2^{nd}\) period : \(\frac{1.04E}{1.07^2}\)
.
.
.
In the \(41^{th}\) period : \(\frac{1.04^{40}E}{1.07^{41}}\)
PV at year \[ \to PV at year 31 =
E\times\sum_{i=0}{40}\frac{(1+4\%)^i}{(1+7\%)^{i+1}} \]
\[ X_{31} =
E\times\sum_{i=0}{40}\frac{(1+4\%)^i}{(1+7\%)^{i+1}} \to E = 528.6
\]