Financial Math FM/Loans

Learning objectives

The Candidate will understand key concepts concerning loans and how to perform related calculations.

Learning outcomes

The Candidate will be able to:

• Define and recognize the definitions of the following terms: principal, interest, term of loan, outstanding balance, final payment (drop payment, balloon payment), amortization.
• Calculate:

• The missing item, given any four of: term of loan, interest rate, payment amount, payment period, principal.
• The outstanding balance at any point in time.
• The amount of interest and principal repayment in a given payment.
• Similar calculations to the above when refinancing is involved.

Introduction

In this chapter, two methods of repaying a loan will be discussed, namely amortization method and sinking fund method. In particular, for each of these two methods, we will discuss how to determine he outstanding loan balance at any point in time, and the amount of interest and principal repayment in each payment made by borrower.

Amortization method

Definition. (Amortization method)

• For amortization method, the borrower repays the lender by a series of payments at regular intervals.
• Each payment is applied first to interest due on the outstanding balance at the time just before the payment is made to pay the interest, and
• after deducting the amount of interest from each payment, the amount left in each payment is going as the principal repayment to reduce the loan balance (i.e. how much the borrower owes).
• Payments are made to reduce the loan balance to exactly zero.

Amortization of level payment

The series of payments made by borrower is level in this subsection, and payments form annuity-immediate in our discussion . To illustrate this, consider the following diagrams.

Borrower's perspective:

   L     R     R  ...  R  ...     R
   ↑     ↓     ↓       ↓          ↓
---|-----|-----|-------|----------|---
   0     1     2  ...  k  ...     n     

Lender's perspective:

   L     R     R  ...  R  ...     R
   ↓     ↑     ↑       ↑          ↑  
---|-----|-----|-------|----------|---
   0     1     2  ...  k  ...     n     

in which

• ↑ means the amount is received, ↓ means the amount is paid;
• $L$ is the amount borrowed (i.e. the amount of loan);
• $n$ is the number of payments;
• $R$ is the level payment made by the borrower (this is return from the lender's perspective).

• Let $B_{k}$ be the outstanding balance at time $k$, just after the $k$th payment ($B_{0}=L$, which is the initial balance).
• Let $i$ be the effective interest rate during each interval for payments.

Proposition. (Recursive method to determine outstanding balance (level payment)) $B_{k+1}=(1+i)B_{k}-R$.

Proof.

• First, $B_{k}$ will accumulate to $B_{k}(1+i)$ from time $k$ to $k+1$.
• The interest due on $B_{k}$ is $iB_{k}$.
• So, the reduction of outstanding balance from the payment of $R$ at $t=k+1$ is $R-iB_{k}$.
• It follows that the outstanding balance at time $k+1$ is $B_{k}-(R-iB_{k})=B_{k}(1+i)-R$.

$\Box$

Proposition. (Fundamental relationship between amount of loan and payments) $L=Ra_{{\overline {n}}|i}$.

Proof.

• Using the above recursive method, $B_{1}=B_{0}(1+i)-R=L(1+i)-R$;
• $B_{2}=B_{1}(1+i)-R=L(1+i)^{2}-R(1+i)-R$;
• $B_{3}=L(1+i)^{3}-R(1+i)^{2}-R(1+i)-R$;
• ...
• $B_{n}=L(1+i)^{n}-R(1+i)^{n-1}-R(1+i)^{n-2}-\dotsb -R$.
• Since $B_{n}=0$ for amortization method (the loan balance is reduceed to zero at the end by definition), we have

$${\begin{aligned}&&L(1+i)^{n}-R(1+i)^{n-1}-R(1+i)^{n-2}-\dotsb -R&=0\\&\Rightarrow &L(1+i)^{n}&=R(1+i)^{n-1}+R(1+i)^{n-2}+\dotsb +R\\&\Rightarrow &L(1+i)^{n}v^{n}&=R(1+i)^{n-1}v^{n}+R(1+i)^{n-2}v^{n}+\dotsb +Rv^{n}\\&\Rightarrow &L&=Rv+Rv^{2}+\dotsb +Rv^{n}{\overset {\text{ def }}{=}}Ra_{{\overline {n}}|i}.\\\end{aligned}}$$

$\Box$

Remark.

• Because of this relationship, we can calculate $L$ by pressing $-R$ PMT $n$ N $100i$ I/Y CPT PV to get $L$.

• In particular, $-R$ is inputted since payments are cash outflow (we use borrower's perspective here).

Proposition. (Prospective method to determine outstanding balance (level payment)) $B_{k}=Ra_{{\overline {n-k}}|i}$.

Proof.

• From the proof of fundamental relationship between $L$ and $R$, we have

$${\begin{aligned}B_{k}&=L(1+i)^{k}-R(1+i)^{k-1}-R(1+i)^{k-2}-\dotsb -R\\&=(1+i)^{k}(Rv+Rv^{2}+\dotsb +Rv^{n})-R(1+i)^{k-1}-R(1+i)^{k-2}-\dotsb -R\\&={\cancel {R(1+i)^{k-1}+R(1+i)^{k-2}+\dotsb +R(1+i)^{k-(k-1)}+R(1+i)^{k-k}}}+R(1+i)^{k-(k+1)}\dotsb +R(1+i)^{k-n}{\cancel {-R(1+i)^{k-1}-R(1+i)^{k-2}-\dotsb -R}}\\&=R(1+i)^{-1}+R(1+i)^{-2}+\dotsb +R(1+i)^{-(n-k)}\\&=Rv+Rv^{2}+\dotsb +Rv^{n-k}\\&=Ra_{{\overline {n-k}}|i}.\end{aligned}}$$

.

$\Box$

Proposition. (Retrospective method to determine outstanding balance (level payment))

$$B_{k}=L(1+i)^{k}-Rs_{{\overline {k}}|i}.$$

Proof.

• From the proof of fundamental relationship between $L$ and $R$, we have

$${\begin{aligned}B_{k}&=L(1+i)^{k}-R(1+i)^{k-1}-R(1+i)^{k-2}-\dotsb -R\\&=L(1+i)^{k}-(1+i)^{k}(Rv+Rv^{2}+\dotsb +Rv^{k})\\&=L(1+i)^{k}-Rs_{{\overline {k}}|i}.\end{aligned}}$$

$\Box$

• Another method to determine outstanding balance (and also principal and interest paid in different payments) is using BA II Plus.
• Procedure:

1. Input $-R$ into PMT (if $R$ is unknown, it should be determined first).
2. Input $L$ into PV

• We can also compute PMT or PV given sufficient information.

1. Press 2ND PV
2. Press the starting payment number (for $k$th payment, press $k$) and press ENTER ↓.
3. Press the ending payment number (for $k$th payment, press $k$) and press ENTER ↓ (press the same number as the starting payment number for selecting exactly one payment ).
4. Then, outstanding balance just after the selected payment(s) is displayed (BAL=... is displayed).
5. Press ↓ and loan (or "principal") paid in the selected payment(s) is displayed (PRN=... is displayed).
6. Press ↓ and interest paid in the selected payment(s) is displayed (INT=... is displayed).

Example.

• Suppose a loan of 2000 is repaid by 15 annual payments of $R$ in arrears.
• With annual interest rate of 10%, $R={\frac {2000}{a_{{\overline {15}}|0.1}}}\approx 262.95$.

Exercise.

Calculate $R$ with annual interest rate of 10% payable semiannually.

	192.68
	262.95
	266.71
	804.23
	964.35

Example.

• Suppose a loan of $L$ is repaid by 20 monthly payments of $0.1L$ in arrears.
• Calculate the annual effective interest rate $i$.

Solution:

• The monthly effective interest rate $j$ is calculated by $L=0.1La_{{\overline {20}}|j}\Rightarrow a_{{\overline {20}}|j}=10\Rightarrow j\approx 7.75\%$ (using BA II Plus).
• So, the annual effective interest rate $i=(1+j)^{12}-1\approx 145.04\%$.

Exercise.

Calculate $i$ if there are 30 such monthly payments.

	190.93%
	217.56%
	274.23%
	290.93%
	317.56%

Example.

• A loan of 1000 is repaid by 12 annual level payments in arrears.
• The annual interest rate is 8%.
• Then, the outstanding balance at the end of 5th year is 690.86 by pressing 1000 PV 12 N 8 I/Y CPT PMT 2ND PV 5 ENTER ↓ 5 ENTER ↓, and outstanding balance is displayed.

Exercise.

Calculate the outstanding balance at the end of 8th year.

	42.38
	90.31
	341.97
	439.50
	529.81

Exercise.

A loan is repaid by 10 annual payments at the end of each year. In the first 3 years, payments of 1000 are made at the end of each year. In the remaining 7 years, payments of 2000 are made at the end of each year.

1 Choose the correct expression for the amount of loan.

	$1000a_{{\overline {10}}|}+1000v^{3}a_{{\overline {7}}|}$
	$1000a_{{\overline {3}}|}+1000v^{3}a_{{\overline {7}}|}$
	$1000a_{{\overline {10}}|}+2000v^{3}a_{{\overline {7}}|}$
	$3000a_{{\overline {10}}|}$
	$2000a_{{\overline {10}}|}-1000v^{3}a_{{\overline {7}}|}$

2 Choose the correct expression for the amount of outstanding balance at year 3.

	$1000s_{{\overline {3}}|}$
	$1000s_{{\overline {3}}|}+2000a_{{\overline {7}}|}$
	$2000a_{{\overline {7}}|}$
	$1000a_{{\overline {3}}|}+2000v^{3}a_{{\overline {7}}|}$
	None of the above

3 Choose the correct expression for the amount of outstanding balance at year 7.

	$\left(1000a_{{\overline {3}}|}+2000v^{3}a_{{\overline {7}}|}\right)(1+i)^{7}-1000\left(s_{{\overline {7}}|}+s_{{\overline {4}}|}\right)$
	$\left(1000a_{{\overline {3}}|}+2000v^{3}a_{{\overline {7}}|}\right)(1+i)^{7}-1000\left(s_{{\overline {7}}|}+s_{{\overline {3}}|}\right)$
	$\left(1000a_{{\overline {10}}|}+1000v^{3}a_{{\overline {7}}|}\right)(1+i)^{7}-1000\left(s_{{\overline {7}}|}+s_{{\overline {3}}|}\right)$
	$1000a_{{\overline {3}}|}$
	None of the above

Now, we consider the amount of interest and principal repayment in each payment made by borrower.

Proposition. (Splitting an installment into principal and interest repayments (level payment)) Let $P_{k}$ be the principal repaid in the $k$th installment (i.e. $k$th payment made by borrower), and $I_{k}$ be the amount of interest paid in the $k$th installment. (This notation has different meaning in the context of Chapter 1.) Suppose the installments made by borrower is level, and each of them equals $R$. Then,

$$P_{k}=Rv^{n-k+1},\quad I_{k}=R-P_{k}.$$

Proof.

• First, by definitions, $R=P_{k}+I_{k}$ because the installment is first deducted by the interest due ($I_{k}$), and the remaining amount ($R-I_{k}$) is used to repay principal. Therefore, $I_{k}=R-P_{k}$.
• It remains to prove the formula for $P_{k}$. By definition, $I_{k}=iB_{k-1}$ because the interest is due on the outstanding balance (before the $k$th installment).

$$P_{k}=R-I_{k}=R-iB_{k-1}=R-i\underbrace {Ra_{{\overline {n-k+1}}|}} _{\text{prospective}}=R\left({\cancel {1}}-{\cancel {i}}\left({\frac {{\cancel {1}}-v^{n-k+1}}{\cancel {i}}}\right)\right)=Rv^{n-k+1}$$

$\Box$

Remark.

• We can use BA II Plus to determine $P_{k}$ and $I_{k}$, which is discussed before.

After splitting each installment, we can make an amortization schedule which illustrates the splitting of each repayment in a tabular form. An example of amortization schedule is as follows:

Amortization schedule for a loan of $a_{{\overline {n}}|}$ repaid over $n$ periods at rate $i$

Period	Payment	Interest paid	Principal repaid	Outstanding loan balance
0	0	0	0	$a_{{\overline {n}}|}$ (prospective)
1	1	$i\underbrace {a_{{\overline {n}}|}} _{B_{0}}=\underbrace {1} _{R}-\underbrace {v^{n}} _{P_{1}}$	$\underbrace {v^{n}} _{1(v^{n-1+1})}$	$\underbrace {a_{{\overline {n}}|}} _{B_{0}}-\underbrace {v^{n}} _{P_{1}}=\underbrace {a_{{\overline {n-1}}|}} _{\text{prospective}}$
2	1	$ia_{{\overline {n-1}}|}=1-v^{n-1}$	$v^{n-1}$	$a_{{\overline {n-1}}|}-v^{n-1}=a_{{\overline {n-2}}|}$
...	...	...	...	...
$k$	1	$ia_{{\overline {n-k+1}}|}=1-v^{n-k+1}$	$v^{n-k+1}$	$a_{{\overline {n-k+1}}|}-v^{n-k+1}=a_{{\overline {n-k}}|}$
...	...	...	...	...
$n-1$	1	$ia_{{\overline {2}}|}=1-v^{2}$	$v^{2}$	$a_{{\overline {2}}|}-v^{2}=a_{{\overline {1}}|}$
$n$	1	$ia_{{\overline {1}}|}=1-v$	$v$	$a_{{\overline {1}}|}-v=0$
Total	$n$	$n-a_{{\overline {n}}|}$	$a_{{\overline {n}}|}$	not important

(You may verify the recursive method to determine outstanding balance using this table, e.g. $a_{{\overline {n}}|}(1+i)-1={\ddot {a}}_{{\overline {n}}|}-1=a_{{\overline {n-1}}|}$)

It can be seen that total payment ($n$) equals total interest paid ($n-a_{{\overline {n}}|}$) plus total principal repaid ($a_{{\overline {n}}|}$), and each payment equals the interest paid plus principal repaid in the corresponding period (read horizontally), as expected, because the payment is either used for paying interest, or used for repaying principal.

It can also be seen that the total principal repaid equals the amount of loan (i.e. outstanding loan balance in period 0) ($a_{{\overline {n}}|}$), as expected, because the whole loan is repaid by the payments in $n$ periods.

Example.

• A loan of 1000 is repaid by seven annual payments of $R$ in arrears.
• The annual interest rate is 5%.
• The principal repaid in 3rd payment is approximately 135.41 (press 1000 PV 7 N 5 I/Y CPT PMT 2ND PV 3 ENTER ↓ 3 ENTER ↓ ↓);
• the interest paid in 3rd to 6th payment is approximately 107.65 (press ↑ ↑ 6 ENTER ↓ ↓ ↓, continuing from the above pressing sequence).

Exercise.

A loan is being repaid with a series of payments at the end of each month for 5 years, and amount of each payment is 100. The annual interest rate is 10%.

1 Choose the correct expression for the amount of principal repaid in the 2nd payment.

	$100(1.1)^{-4}$
	$100(1.1)^{-3}$
	$100(1.1)^{-59/12}$
	$100(1.1)^{-29/6}$
	None of the above

2 Calculate the sum of amount of principal repaid in 3rd payment and payments after this payment.

	63.09
	124.68
	126.68
	4629.17
	4691.76

3 Choose the correct expression for the outstanding balance at the end of year 4.

	$100a_{{\overline {13}}|1.1^{1/12}-1}$
	$100a_{{\overline {2}}|0.1}^{(12)}$
	$100a_{{\overline {1}}|1.1^{1/12}-1}^{(12)}$
	$100a_{{\overline {60}}|1.1^{1/12}-1}(1.1)^{4}-100s_{{\overline {48}}|1.1^{1/12}-1}$
	$100a_{{\overline {5}}|1.1}^{(12)}(1.1)^{48}-100s_{{\overline {4}}|1.1}^{(12)}$

Amortization of non-level payment

In this subsection, we will consider amortization of non-level payment. The ideas and concepts involved are quite similar to the amortization of non-level payment. Borrower's perspective:

   L    R_1   R_2 ... R_k  ...   R_n
   ↑     ↓     ↓       ↓          ↓
---|-----|-----|-------|----------|---
   0     1     2  ...  k  ...     n     

Lender's perspective:

   L    R_1   R_2 ... R_k  ...   R_n
   ↓     ↑     ↑       ↑          ↑  
---|-----|-----|-------|----------|---
   0     1     2  ...  k  ...     n     

in which $R_{1},R_{2},\ldots ,R_{n}$ are non-level payments, and the other relevant notations used in amortization of level payment have the same meaning.

Because the payments are now non-level, we need formulas different from that for the amortization of level payment to determine amount of loan and outstanding balance at different time, and to split the payment into interest payment and principal repayment. They are listed in the following.

Proposition. (Relationship between amount of loan and payments (non-level payment))

$$L=R_{1}v+R_{2}v^{2}+\cdots +R_{n}v^{n}.$$

Proof. Omitted since the main idea is identical to the proof for the level payment version.

$\Box$

Proposition. (Prospective method to determine outstanding balance (non-level payment))

$$B_{k}=R_{k+1}v+R_{k+2}v^{2}+\cdots +R_{n}v^{n-k}.$$

Proof. Omitted since the main idea is identical to the proof for the level payment version.

$\Box$

Proposition. (Retrospective method to determine outstanding balance (non-level payment))

$$B_{k}=L(1+i)^{k}-R_{1}(1+i)^{k-1}-R_{2}(1+i)^{k-2}-\cdots -R_{k}.$$

Proof. Omitted since the main idea is identical to the proof for the level payment version.

$\Box$

Proposition. (Recursive method to determine outstanding balance (non-level payment))

$$B_{k}=B_{k-1}(1+i)-R_{k}.$$

Proof. Omitted since the main idea is identical to the proof for the level payment version.

$\Box$

Remark. Recursive method is more useful for non-level payment than level payment.

Proposition. (Splitting an installment into principal and interest repayments (non-level payment))

$$I_{k}=iB_{k-1},\quad P_{k}=R_{k}-I_{k}.$$

Proof.

• $I_{k}=iB_{k-1}$: It follows from definition of $I_{k}$.
• $P_{k}=R_{k}-I_{k}$: It follows from $R_{k}=P_{k}+I_{k}$ which is true by definition.

$\Box$

Exercise.

A loan is repaid with repayments that start at 100 at the 1st year, and increase by 100 per year until a payment of 1000 is made, and thereafter no payments will be made. The annual effective interest rate is 5%.

1 Choose a correct expression for outstanding balance at the end of 4th year.

	$400a_{{\overline {5}}|}+100(Ia)_{{\overline {6}}|}$
	$400a_{{\overline {6}}|}+100(Ia)_{{\overline {5}}|}$
	$500a_{{\overline {6}}|}+100(Ia)_{{\overline {6}}|}$
	$400a_{{\overline {6}}|}+100(Ia)_{{\overline {6}}|}$
	$500a_{{\overline {5}}|}+100(Ia)_{{\overline {6}}|}$

2 Choose a correct expression for principal repaid at the end of 7th year.

	$700-i(700a_{{\overline {2}}|}+100(Ia)_{{\overline {2}}|})$
	$700-i(700a_{{\overline {2}}|}+100(Ia)_{{\overline {3}}|})$
	$700-i(700a_{{\overline {3}}|}+100(Ia)_{{\overline {3}}|})$
	$700-i(800a_{{\overline {2}}|}+100(Ia)_{{\overline {3}}|})$
	$700-i(800a_{{\overline {3}}|}+100(Ia)_{{\overline {3}}|})$

Amortization of payments that are made at a different frequency than interest is convertible

In this situation, we can obtain the amount of loan, outstanding balance, and principal repaid and interest paid in a payment, by calculating the equivalent interest rate that is convertible at the same frequency at which payments are made. Then, the previous formulas can be used directly, at this equivalent interest rate. This method is analogous to the method for calculating the annuity with payments made at a different frequency than interest is convertible.

Exercise.

A loan is repaid with level payments of 2000 each at the end of each quarter for 10 years. The semiannual effective interest rate is 5%. Calculate the interest paid in the 3rd payment. Correct your answer to two decimal places.

Sinking fund method

After discussing amortization method, we discuss another way to repay a loan, namely sinking fund method.

Definition. (Sinking fund method) For sinking fund method, all principal (i.e. amount of loan) is repaid by the borrower in a single payment at maturity. Interest due on the principal is paid at the end of each period and a deposit is made into a sinking fund at the end of each period (same amount of deposit for each of the time points), so that the accumulated value of sinking fund equals the amount of principal at maturity.

Borrower's perspective:

Loan repayment:
   L     Li    Li     ...         Li    L
   ↑     ↓     ↓                  ↓     ↓
---|-----|-----|------------------|-----|---
   0     1     2      ...        n-1    n
   \    / \   /                   \    /
    \  /   \ /       ...           \  / 
     i      i                        i       rate

Sinking fund:
         D     D      ...         D     D  L
         ↓     ↓                  ↓     ↓ ↗
---|-----|-----|------------------|-----|---
   0     1     2      ...        n-1    n
   \    / \   /                   \    /
    \  /   \ /       ...           \  / 
     j      j                        j       rate

Lender's perspective: (Lender do not know how the borrower repays the loan, so sinking fund is not shown)

Loan repayment:
   L     Li    Li     ...         Li    L
   ↓     ↑     ↑                  ↑     ↑ 
---|-----|-----|------------------|-----|---
   0     1     2      ...        n-1    n
   \    / \   /                   \    /
    \  /   \ /       ...           \  / 
     i      i                        i       rate

in which

• $L$ is the amount borrowed
• $n$ is the number of payment periods
• $i$ is the effective interest rate paid by borrower to lender
• $j$ is the effective interest rate earned on the sinking fund (which is usually strictly less than $i$ in practice)
• $D$ is the level sinking fund deposit

Let $R$ is the level payment made by borrower at the end of each period, which equals $D+$ interest paid to lender, i.e. $R=Li+D$.

By definition of sinking fund method, $L=Ds_{{\overline {n}}|j}$ because the accumulated value of sinking fund equals amount of loan at maturity.

Using these two equation, we can have the following theorem.

Proposition. (Relationship between each payment made by borrower and amount of loan in sinking fund method)

$$R=L\left(i+{\frac {1}{s_{{\overline {n}}|j}}}\right).$$

Proof. Because $L=Ds_{{\overline {n}}|j}\Rightarrow D={\frac {L}{s_{{\overline {n}}|j}}}$

$$R=Li+D=Li+{\frac {L}{s_{{\overline {n}}|j}}}=L\left(i+{\frac {1}{s_{{\overline {n}}|j}}}\right).$$

$\Box$

Recall that ${\frac {1}{a_{{\overline {n}}|i}}}=i+{\frac {1}{s_{{\overline {n}}|i}}}$. We can observe that a similar expression compared with the right hand side appears in above equation ($i+{\frac {1}{s_{{\overline {n}}|j}}}$). In view of this, we define

$${\frac {1}{a_{{\overline {n}}|i\&j}}}=i+{\frac {1}{s_{{\overline {n}}|j}}}.$$

(we use '$i\&j$' because the right hand side involves both $i$ and $j$.) Then, if the amount of loan is 1, then the payment made by borrower at the end of each period is ${\frac {1}{a_{{\overline {n}}|i\&j}}}$.

Naturally, we would like to know what $a_{{\overline {n}}|i\&j}$ equals. We can determine this as follows:

$${\begin{aligned}{\frac {1}{a_{{\overline {n}}|i\&j}}}&=i+{\frac {1}{s_{{\overline {n}}|j}}}\\&=\left({\frac {1}{a_{{\overline {n}}|j}}}-j\right)+i\qquad {\text{because }}{\frac {1}{a_{{\overline {n}}|j}}}={\frac {1}{s_{{\overline {n}}|j}}}+j\\&={\frac {1}{a_{{\overline {n}}|j}}}+(i-j)\\&={\frac {1+(i-j)a_{{\overline {n}}|j}}{a_{{\overline {n}}|j}}}\\\Rightarrow a_{{\overline {n}}|i\&j}&={\frac {a_{{\overline {n}}|j}}{1+(i-j)a_{{\overline {n}}|j}}}.\end{aligned}}$$

(The right hand side also involve $i$ and $j$, as expected, because the reciprocal of an expression involving $i$ and $j$ should also involve $i$ and $j$) In particular, if $i=j$, $a_{{\overline {n}}|i\&j}=a_{{\overline {n}}|i}=a_{{\overline {n}}|j}$ as expected, and

$$R=Li+D=L\left(i+{\frac {1}{s_{{\overline {n}}|i}}}\right)={\frac {L}{a_{{\overline {n}}|i}}}.$$

Therefore, each level payment made by borrower in the sinking fund method is the same as the levelpayment in the amortization method, because $L=Ra_{{\overline {n}}|i}$ in amortization method of level payment.

Using this notation, we can express the relationship between $R$ and $L$ as follows:

$$R={\frac {L}{a_{{\overline {n}}|i\&j}}}={\frac {L(1+(i-j)a_{{\overline {n}}|j})}{a_{{\overline {n}}|j}}}$$

Exercise.

A borrower borrows a loan of 1000 for ten years, and repays it using the sinking fund method. The annual effective interest rate charged for the loan is 5%, and the annual effective interest rate earned for the sinking fund is 3%.

1 Choose correct expression(s) for the amount of each payment made at the end of year.

	$1000\left(3\%+{\frac {1}{s_{{\overline {10}}|5\%}}}\right)$
	${\frac {1000}{a_{{\overline {10}}|3\%\&5\%}}}$
	$1000\left(5\%+{\frac {1}{a_{{\overline {10}}|5\%\&3\%}}}\right)$
	$1000(5\%)+{\frac {1000(3\%)}{1.03^{10}-1}}$
	$1000a_{{\overline {10}}|5\%\&3\%}$

If we assume that the balance in the sinking fund could be used to reduce the amount of loan, then net amount of loan after $k$th payment is

$$L-Ds_{{\overline {k}}|j},$$

net amount of interest paid in the $k$th period is

$$Li-j(Ds_{{\overline {k-1}}|j}),$$

and principal repaid in the $k$th period is

$$Ds_{{\overline {k}}|j}-Ds_{{\overline {k-1}}|j}=D(1+j)^{k-1}.$$

Exercise.

TODO