15 Amortized Loans
15.1 General Case
Consider a loan of \(L\) that is to be repaid with a series of payments over \(n\) units of time. The term of the loan is \(n\) units of time. Let’s denote a payment at time \(t\) as \(P_t\). The balance of the loan at the outset is equal to the present value of the payments to be made in the future: \(L= PV_0(\text{Payments})\)
Each payment can be broken down into 2 components of interest and principal:
- \(I_t\) = Interest portion of the time-t payment
- \(Prin_t\) = Principal portion of the time-t payment
The sum of the two components is equal to the payment amount: \(P_t = Prin_t + I_t\)
The interest component is the product of the outstanding balance after the previous payment \(B_{t-1}\) and the periodic interest rate appropriate for the length of the period:
- \(I_t = iB_{t-1}\)
- \(Prin_t = P_t - I_t\)
The outstanding balance at time \(t\) is denoted by \(B_t\), and it is equal to the previous loan balance minus the principal payment made at time \(t\):
- \(B_t = B_{t-1} - Prin_t\)
Example: A loan is to be repaid with payments of \(\$250\) in 1 year, \(\$300\) in 2 years, \(\$100\) in 3 years, and \(\$490.35\) in 4 years. The annual effective interest rate is 5%.
- What is the initial amount of hte loan?
- What is the interest portion of the first payment?
- What is the principal portion of the first payment?
- What is the loan balance after the first payment?
Solution:
- The initial amount of the loan is: \[L = B_0 = \frac{250}{1.05} + \frac{300}{1.05^2} + \frac{100}{1.05^3} + \frac{490.35}{1.05^4} = 1,000\]
- The interest portion of the first payment is: \[I_1 = iB_0 = 0.05 \times 1000 = 50\]
- The principal portion of the first payment is: \[Prin_1 = P_1 - I_1 = 250 - 50 = 200\]
- The loan balance after the first payment is: \[B_1 = B_0 - Prin_1 = 1000 - 200 = 800\]
For a loan, the outstanding balance at time \(t\) can be calculated two ways, and is illustrated through the following equation of value: \[(1+i)^tB_0 - AV_t(\text{Payments made on or before time $t$}) = PV_t (\text{Payments made after time $t$})\]
The left side represents the retrospective method, and it finds the loan balance at time \(t\) as the accumulated value of the loan minus the accumulated value of the payments made on or before time \(t\). The right side represents the prospective method, and it finds the loan balance at time \(t\) as the present value of the remaining payments to be made after time \(t\):
- Retrospective Method: \[B_t = (1+i)^tB_0 - AV_t(\text{Payments made on or before time $t$})\]
- Prospective Method: \[B_t = PV_t (\text{Payments made after time $t$})\]
15.2 Level Payment Amortized Loans
A loan of \(L\) can be repaid in \(n\) equal payments, each of which occurs at the end of each unit of time. The amount of each payment is denoted by \(P\), and this amount can be obtained by solving the equation of value: \[L = B_0 = P\times a_{\overline{n|}} \iff P = \frac{L}{a_\overline{n|}}\] Each payment can be broken down into its components of interest and principal: \[P = I_t + Prin_t\] where \[\begin{align}I_t &= B_{t-1}\times i \\ Prin_t &= P_t - I_t\end{align}\] The loan balance at time \(t\) is the previous loan balance minus the principal payment made at time \(t\): \[B_t = B_{t-1} - Prin_t\]
Example: A loan of \(\$10,000\) is to be repaid with level annual payments over \(5\) years. The annual effective interest rate is \(5\%\).
- What is the amount of each annual payment?
- Create an amortization table showing the principal and interest components of each payment. Also show the outstanding principal at the end of each year.
Solution:
- The payment amount is: \[Pmt = \frac{B_0}{a_\overline{n|}} = \frac{10000}{a_\overline{5|}} = 2,309.75\]
- We can construct the amortization table:
| Year \(t\) | \[P\] | \[Prin_t = P - I_t\] | \[I_t = iB_{t-1}\] | \[B_t\] |
|---|---|---|---|---|
| 0 | 10,000.00 | |||
| 1 | 2,309.75 | 1,809.75 | 500.00 | 8,190.25 |
| 2 | 2,309.75 | 1,900.24 | 409.51 | 6,290.02 |
| 3 | 2,309.75 | 1,995.25 | 314.50 | 4,294.77 |
| 4 | 2,309.75 | 2,095.01 | 214.74 | 2,199.76 |
| 5 | 2,309.75 | 2,199.76 | 109.99 | 0.00 |
We can use the calculator’s amortization worksheet to find the interest and principal portions of each payment. For example, to obtain the interest and principal portions of the third payment we use the calculator as follows:
5 [N] 5 [I/Y] 10,000 [PV] [CPT] [PMT] -> PMT = -2,309.75
[2nd] [AMORT] (P1 = ) 3 [ENTER] ↓ (P2 = ) 3 [ENTER] ↓
Continuing to hit the down arrow key, we observe:
BAL =4,294.77, PRN = -1,995.25, INT = -314.50
The settings for P1 and P2 tell the calculator to determine the principal and interest included in payments P1 through P2. When we want to find the principal and interest components of a single payment, we set P1 equal to P2.
15.2.1 Formulae for Loans Repaid with Level Payments
By the prospective method, we see that the outstanding balance at time \(t\) equals to: \[B_t = P \times a_{\overline{n-t|}}\] and the retrospective method yields: \[B_t = B_0\times (1+i)^t - P\times s_{\overline{t|}}\] We can also get the formulae for \(I_t\) and \(Prin_t\): \[I_t = (1-v^{n-t+1})\times P\] \[Prin_t = v^{n-t+1} \times P\] Notice that the values of \(Prin_t\) form a geometric series with ratio \(v\).