# Loan Repayment Analysis Given data- Principal \( = \$12000 \) - Original term: 29 years - Payments: 15 level payments, payable at the end of every **alternate year** (i.e., every 2 years) - First payment: at end of 1 - Effective annual interest rate \( i = 4\% \) - Extension: 6 additional years, after the 4th payment --- ## Part (a): Find the original installment amount ### Step 1: Determine the number of payments and timing - Payments are made every 2 years. - Total original term: 29 years - Number of payments: \( n = 15 \) ### Step 2: Find the present value of the payments at time - Payments are level, paid at the end of years 1, 3, 5, ..., 29. - The timing of payments: at years \( 1, 3, 5, ..., 29 \) ### Step 3: Calculate the present value of the payment stream - Let \( R \) be the amount of each payment. The present value at time of the payments: \[ PV = R \times \sum_{k=1}^{15} v^{2k - 1} \] where \( v = \frac{1}{1 + i} = \frac{1}{1.04} \). The sum: \[ \sum_{k=1}^{15} v^{2k - 1} = v \times \frac{1 - (v^2)^{15}}{1 - v^2} \] - \( v = \frac{1}{1.04} \approx .961538 \) - \( v^2 = (.961538)^2 \approx .924556 \) Calculate numerator: \[ 1 - (v^2)^{15} = 1 - (.924556)^{15} \] Calculate \( (.924556)^{15} \): \[ (.924556)^{15} \approx e^{15 \times \ln .924556} \approx e^{15 \times (-.0784)} = e^{-1.176} \approx .308 \] Thus, \[ \text{Sum} = .961538 \times \frac{1 - .308}{1 - .924556} = .961538 \times \frac{.692}{.075444} \] Calculate denominator: \[ .075444 \] Calculate numerator: \[ .692 / .075444 \approx 9.169 \] Finally, \[ \text{Sum} \approx .961538 \times 9.169 \approx 8.810 \] ### Step 4: Find the installment \( R \) The initial loan value: \[ PV = 12,000 \] \[ 12,000 = R \times 8.810 \] \[ R = \frac{12,000}{8.810} \approx \$1,362.42 \] --- ## **Answer (a):** **Original installment \( R \approx \$1,362.42 \)** --- ## Part (b): Outstanding balance just before the 5th payment ### Step 1: Identify timing - The 5th payment is due at the end of year 9. - Just before the 5th payment: at the end of year 8. ### Step 2: Compute the present value at time 8 of remaining payments Remaining payments after year 8 are at: - years 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 (since total payments are 15, every 2 years). Number of remaining payments at the moment just before the 5th payment: - Payments remaining: after the 4th payment (which occurs at year 7), the 5th payment is at year 9. Remaining payments: \[ \text{Payments at years } 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 \] Number of remaining payments: \[ 11 \] ### Step 3: Calculate the present value at year 8 The present value of these remaining payments at year 8: \[ PV_{8} = R \times \sum_{k=1}^{11} v^{2k - 1} \] where: \[ \text{payments at years } 9, 11, \dots, 29 \] which correspond to: \[ k=1: v^{1} \quad (year 9) \] \[ k=2: v^{3} \quad (year 11) \] \[ \vdots \] \[ k=11: v^{21} \quad (year 29) \] Sum: \[ \sum_{k=1}^{11} v^{2k - 1} = v \times \frac{1 - (v^{2})^{11}}{1 - v^{2}} \] Calculate: \[ (v^{2})^{11} = (.924556)^{11} \approx e^{11 \times -.0784} = e^{-.862} \approx .422 \] Then, \[ \text{Sum} = .961538 \times \frac{1 - .422}{1 - .924556} = .961538 \times \frac{.578}{.075444} \approx .961538 \times 7.666 \approx 7.377 \] Now, compute the outstanding balance: \[ \text{Outstanding balance} = R \times 7.377 \approx 1362.42 \times 7.377 \approx \$10,065 \] --- ## **Answer (b):** **Outstanding balance just before the 5th payment: \(\boxed{\$10,065}\)** --- ## Part (c): Extra interest due to the extension ### Step 1: Original remaining term at extension point - Remaining term after 4 payments: 10 payments remaining (up to year 29). ### Step 2: Extended term - The loan is extended by 6 more years. - Additional payments: 3 more payments (every 2 years), totaling 3 extra payments at years 31, 33, 35. ### Step 3: Calculate the new total interest paid - **Original plan:** 15 payments, each \$1,362.42. - **Extended plan:** original 4 payments made, remaining balance \$10,065 at year 8. **New schedule:** - Payments: at years 9, 11, ..., 29, plus 3 additional payments at years 31, 33, 35. ### Step 4: Calculate the future value of the original payments and compare **Interest paid originally:** \[ \text{Total original payments} = 15 \times \$1,362.42 \approx \$20,436 \] **Interest accrued:** - Total paid minus principal: \( \$20,436 - \$12,000 = \$8,436 \) **Interest with extension:** - Additional payments at years 31, 33, 35. - The outstanding balance at year 8: approximately \$10,065. - These are paid off over the extended period, adding extra interest. ### Step 5: Calculate the extra interest - The extension adds **extra interest** because the principal is paid later, accruing more interest. - The **additional interest** is roughly: \[ \text{Extra interest} \approx R \times \text{present value of extended payments} - \text{additional interest from original plan} \] **Alternatively**, since the principal is paid later, the difference in total interest paid can be approximated by the interest accrued on the remaining balance during the extension period. - The additional 6 years of interest on the remaining balance: \[ \text{Interest on } \$10,065 \text{ over 6 years} \approx 6 \times (4\%) \times \$10,065 \approx 6 \times .04 \times 10,065 \approx \$2,415 \] --- ## **Final answer (c):** **Additional interest due to extension \(\boxed{\approx \$2,415}\)** --- # Summary | Part | Result | | --- | --- | | (a) | Original installment \( R \approx \$1,362.42 \) | | (b) | Outstanding balance before 5th payment \(\approx \$10,065\) | | (c) | Extra interest due to extension \(\approx \$2,415\) | --- **Note:** Precise calculations may vary slightly depending on rounding during intermediate steps.