1.1 Problem 2.3.1

A person decides to deposit $100 at the end of each of the next 18 years, with a constant interest rate 12% p.a. compounded annually. How much this person would have at the end of the final year?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{18}100\cdot(1+12\%)^{18-i}\approx5574.971.\]

1.2 Problem 2.3.2

A person decides to deposit $100 at the end of each month during 20 years, with a constant interest rate 6% p.a. compounded monthly. How much this person would have at the end of the final month?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{240}100\cdot\left(1+\frac{6\%}{12}\right)^{240-i}\approx46204.09.\]

1.3 Problem 2.3.3

Suppose you have two options to choose:

1. You can receive a fixed amount 20,000,000 VND per month at the end of each of the next 12 months;
2. You can receive 250,000,000 VND at the end of 12 months.

If the interest rate is fixed at 7% p.a. compounded monthly, which option you should choose and why?

Solution. The future value of Option 1 is \[\textrm{FV}_1=\sum_{i=1}^{12}2\cdot10^7\cdot\left(1+\frac{7\%}{12}\right)^{12-i}\approx2.479\cdot10^8<2.5\cdot10^8=\textrm{FV}_2,\] so Option 2 should be chosen.

1.4 Problem 2.3.4

A person want to have $500000 at the end of 12 years. To obtain that aim, this person deposit $P at the end of each month during 12 years, with a constant interest rate 6% p.a. compounded monthly. What is P?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{144}P\cdot\left(1+\frac{6\%}{12}\right)^{144-i}\approx210.15P,\] yielding the monthly deposit as \[P=\frac{5\cdot10^5}{210.15}\approx2379.251.\]

1.5 Problem 2.3.5

A person want to have $500000 at the end of 22 years. To obtain that aim, this person deposit $P at the end of each year during 22 years, with a constant interest rate 16% p.a. compounded annually. What is P?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{22}P\cdot(1+16\%)^{22-i}\approx157.415P,\] yielding the annual deposit as \[P=\frac{5\cdot10^5}{157.415}\approx3176.318.\]

1.6 Problem 2.6.1

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 \(17^{\textrm{th}}\) birthday. Assume that the first deposit is $P and thereafter the parents increase the deposited amount by 8% annually. Suppose that the bank fixed interest rate at 6% p.a. compounded monthly. What should the minimum initial deposits be in order for the amount in the fund to reach at least $300,000 after her \(17^{\textrm{th}}\) deposit?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{17}P\cdot(1+8\%)^{i-1}\cdot\left(1+\frac{6\%}{12}\right)^{204-12i}\approx50.969P,\] yielding the initial deposit as \[P=\frac{3\cdot10^5}{50.969}\approx5885.938.\]

1.7 Problem 2.6.2

Suppose that you open a retirement fund and deposit to the fund $2000 at the end of each month. If the fund pays 5% p.a. compounded monthly, how much would you accumulate at the end of 15 years?

Solution. The future value of all deposits are \[\textrm{FV}=\sum_{i=1}^{180}2000\cdot\left(1+\frac{5\%}{12}\right)^{180-i}\approx534577.888.\]

1.8 Problem 2.6.3

At the beginning of this year you made a 20-year loan of $100,000 with interest rate 16% p.a. compounded annually. You are going to pay off the loan by paying a constant amount of $P at the end of each year, starting from the end of this year. What is P?

Solution. The present value of all payments are \[\textrm{PV}=\sum_{i=1}^{20}\frac{P}{(1+16\%)^i}\approx5.929P,\] yielding the annual deposit as \[P=\frac{10^5}{5.929}\approx16866.703.\]

1.9 Problem 2.9.2

The grand prize for a lottery is $1,000 per year for 10 years and then $500 per year in perpetuity (i.e., the first $500 payment is at the end of year 11). If the relevant interest rate is 10%, what is the grand prize worth today?

Solution. The present value of the first 10 payments is \[\textrm{PV}_1=\sum_{i=1}^{10}\frac{1000}{(1+10\%)^i}\approx6144.567\] while the present value of the perpetuity is \[\textrm{PV}_2=\frac{500:10\%}{(1+10\%)^{10}}\approx1927.716\] yielding the grand prize worth as \[\textrm{PV}=\textrm{PV}_1+\textrm{PV}_2=8072.283.\]

1.10 Problem 2.9.3

You are saving to retire with $1 million 30 years from today. You can start the savings plan with a $5,000 deposit today. Additionally, you can deposit $7,500 10 years from today, $10,000 20 years from today, and $15,000 upon retirement. You need to set up an ordinary annuity plan to reach your goal. What annual payment $K must you make in the plan to have $1 million upon retirement if you can invest at 12%?

Solution. The future value of the remaining payments is \[\begin{align*}&10^6-5000\cdot(1+12\%)^{30}-7500\cdot(1+12\%)^{20}-10^4\cdot(1+12\%)^{10}-15000\\ &\approx731794.709\end{align*}\] while the same quantity, expressing in terms of annual payments, is \[\textrm{FV}=\sum_{i=1}^{30}K\cdot(1+12\%)^{30-i}\approx241.333K,\] yielding the annual payment of \[K=\frac{731794.709}{241.333}\approx3032.303.\]

2.1 Problem 3.1

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% p.a. compounded annually?

Solution. The present value of all payments is \[\textrm{PV}=\sum_{i=1}^{360}\frac{K}{(1+8\%)^{i/12}}\approx139.978K,\] yielding the monthly payment of \[K=\frac{8\cdot10^8}{139.978}\approx5715183.81.\]

2.2 Problem 3.2

A person receives a loan of 500 million VND from a local bank with loan tenor 13 years and interest rate 9% p.a. compounded annually. Due to Covid-19, the customer only pays the interest in the first 2 years and then pay fixed monthly payments K over the next 11 years. Compute the payment K.

Solution. The future value at 2 years from now of the remaining payments is \[\textrm{FV}_2=\sum_{i=1}^{132}\frac{K}{(1+9\%)^{i/12}}\approx84.978K,\] yielding the monthly payment of \[K=\frac{5\cdot10^8}{84.978}\approx5883844.807.\]

3.1 Problem 4.1

A person receives a loan of 25 million VND on 20/05/2021 from a finance consumer company with the interest rate 34% p.a. 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.

monthly_payment(25 * 10^6, 0.34, '2021-05-20', '2021-06-16', 17)

## [1] 1868062

3.2 Problem 4.2

A person receives a loan of 25 million VND on 20/05/2021 from a finance consumer company with the interest rate is 34% p.a. with daily simple interest. Due to Covid-19, the customer has to pay interest 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 payment K. The contract closed date is on 16/11/2022. Compute the payment K.

monthly_payment(25 * 10^6, 0.34, '2021-11-16', '2021-12-16', 12)

## [1] 2485231