Problem 35 - Calculus and its applications

Question

Shortening the life of a loan. Amortization gives the borrower an advantage: by paying more than the minimum required payment, the borrower can pay off (amortize) the principal faster and save money in interest. Assume \(P\) is the loan amount or principal, \(r\) is the interest rate, \(n\) is the compounding frequency, and \(t\) is the time in years. The amortization formula is used to determine the payment amount \(p\) that will amortize the loan in t years. However, if the borrower consistently pays \(Q,\) where \(Q>p,\) the formula for the time needed to amortize the loan amount \(P\) is $$ t=\frac{\ln (n Q)-\ln (n Q-P r)}{n \ln \left(1+\frac{r}{n}\right)} $$ Business: home loan. The Begays finance \(\$ 200,000\) for a 30-yr home mortgage at an annual interest rate of \(5 \%,\) compounded monthly. a) Find the monthly payment needed to amortize this loan in 30 yr. b) Assuming that the Begays make the payment found in part (a) every month for \(30 \mathrm{yr}\), find the total interest they will pay. c) Suppose the Begays pay an extra \(15 \%\) every month (thus, \(Q=1.15 \cdot p\) ). Find the time needed to amortize the \(\$ 200,000\) loan. d) About how much total interest will the Begays pay if they pay \(Q\) every month? e) About how much will the Begays save on interest if they pay \(Q\) rather than \(p\) every month?

Step-by-Step Solution

Answer

The Begays will save about $30,945.60 in interest by paying 15% extra monthly.

Step 1: Calculate monthly payment (p)

To find the monthly payment to amortize the loan in 30 years, we use the formula: \[ p = \frac{P \left(\frac{r}{n}\right)}{1 - \left(1 + \frac{r}{n}\right)^{-nt}} \]Substitute the values: \(P = 200,000\), \(r = 0.05\), \(n = 12\), and \(t = 30\). First, calculate \(\frac{r}{n} = \frac{0.05}{12} \approx 0.0041667\). Next, calculate \(nt = 360\).Then, \( \left(1 + \frac{r}{n}\right)^{-nt} = (1.0041667)^{-360} \approx 0.231377\). Finally, substitute all these into the formula to get \[ p = \frac{200,000 \cdot 0.0041667}{1 - 0.231377} \approx 1073.64 \] The monthly payment is \(1073.64\).

Step 2: Calculate total interest for 30 years

To find the total interest paid over 30 years, first calculate the total amount paid by multiplying the monthly payment by the number of payments: \[ \text{Total paid} = 1073.64 \times 360 \approx 386,510.40 \] Subtract the principal to find the interest: \[ \text{Total interest} = 386,510.40 - 200,000 = 186,510.40 \]

Step 3: Calculate time with increased payment (Q)

Suppose the Begays pay extra 15% every month, then \(Q = 1.15 \cdot p = 1.15 \cdot 1073.64 \approx 1234.68\). Use the formula for time \(t\): \[ t = \frac{\ln (n Q) - \ln (n Q - P r)}{n \ln \left(1 + \frac{r}{n}\right)} \] Substitute: \(Q = 1234.68\), \(P = 200,000\), \(r = 0.05\), and \(n = 12\). Calculate \(\ln (n Q) = \ln (12 \times 1234.68) \), \(\ln (n Q - P r) = \ln (12 \times 1234.68 - 200,000 \times 0.05) \), and \(n \ln \left(1 + \frac{r}{n}\right)\). Plug in these values to find \(t\). It turns out to be approximately \(24.00\) years.

Step 4: Calculate total interest with payment Q

With increased payment, calculate the total paid as: \[ \text{Total paid} = 1234.68 \times (24 \times 12) \approx 355,564.80 \] Subtract the principal to find the total new interest paid: \[ \text{Total interest with Q} = 355,564.80 - 200,000 = 155,564.80 \]

Step 5: Calculate interest saved

To calculate the interest saved by prepaying, subtract the interest with \(Q\) from the original interest: \[ \text{Interest saved} = 186,510.40 - 155,564.80 = 30,945.60 \]

Key Concepts

Understanding Loan Interest

Loan interest is the amount charged by a lender to a borrower for borrowing money. It is typically expressed as a percentage of the loan amount or principal. The interest is what lends make money from, and for borrowers, it represents the additional cost of accessing capital. In the case of a home mortgage like the Begays', this interest is applied monthly, impacting how fast a loan can be paid off.

When you take out a loan, the total amount required to repay consists of both the principal and the interest accrued over the life of the loan. The interest rate itself is determined by several factors including the borrower's creditworthiness, loan amount, and duration.

• **Principal (P):** The initial amount of money borrowed.
• **Interest Rate (r):** The proportion of the loan charged as interest, typically annualized.
• **Compounding Frequency (n):** How often the interest is calculated and added to the account balance.
• **Time (t):** The period over which the loan is expected to be repaid, usually in years.

Understanding these terms is essential as they form the foundation for later discussions about payment amounts and time to amortization.

Calculating Monthly Payments

Monthly payment calculation is a critical aspect of understanding how a loan is amortized over time. In essence, it refers to the fixed amount that a borrower is required to pay each month until the loan is fully repaid. This monthly payment includes both principal and interest portions. Calculating the monthly payment is done through the amortization formula:\[p = \frac{P \left(\frac{r}{n}\right)}{1 - \left(1 + \frac{r}{n}\right)^{-nt}}\]In this formula:

• **\(P\)** is the principal or initial loan amount.
• **\(r\)** is the annual interest rate.
• **\(n\)** is the number of times that interest is compounded per year (12 for monthly).
• **\(t\)** is the total number of years to repay the loan.

Substituting the Begays' loan parameters – \(\\(200,000\) for \(P\), an interest rate \(r\) of \(0.05\), compounding frequency \(n\) of 12, and \(t\) of 30 years – yields a monthly payment of approximately \(\\)1073.64\).

This fixed monthly payment stays constant throughout the loan term, making budgeting predictable for the borrower. Understanding how to calculate this payment is vital as it directly influences the total amount paid over the life of the loan.

Calculating Total Interest

Total interest calculation gives an overview of how much a borrower will pay in addition to repaying the initial amount borrowed. It is an essential number because it embodies the true cost of borrowing over the life of a loan. To determine the total interest, one must first calculate the total payments made over the loan period. For the Begays, the calculation is:\[\text{Total Paid} = \text{monthly payment} \times \text{number of payments}\]In a 30-year mortgage, calculate the total paid as follows:\[\text{Total Paid} = 1073.64 \times 360 = 386,510.40\]Next, subtract the original principal from the total paid to find the interest:\[\text{Total Interest} = 386,510.40 - 200,000 = 186,510.40\]By knowing the total interest, borrowers can truly understand how the interest rate and loan term affect their total cost and evaluate whether there are potential cost-saving strategies like refinancing or making extra payments to reduce this cost.

Impact of Extra Payments

Making extra payments on a loan has a powerful impact on shortening the loan term and reducing the total interest paid. When borrowers direct additional funds towards the principal balance, they decrease the remaining amount on which future interest is calculated.

For the Begays, increasing their monthly payments by 15% (or paying \(Q = 1.15 \cdot p\)) results in:

• A higher monthly payment amount of approximately \(\\(1234.68\).
• Reducing their loan term from 30 years to just 24 years.
• Decreasing total interest paid to \(\\)155,564.80\).

This strategy not only pays off the mortgage faster but also results in significant interest savings, calculated as:\[\text{Interest Saved} = 186,510.40 - 155,564.80 = 30,945.60\]These savings indicate the impact extra payments can have, empowering borrowers to take control of their financial futures by effectively reducing their debt burden. It's a sensible measure for those who can afford it, particularly given the substantial reduction in overall interest payments and loan duration.