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Chapter 6: Problem 8

Assume that \(\$ 5000\) was put in each of two accounts. Account A gives \(4 \%\) interest compounded semiannually. Account \(\mathrm{B}\) gives \(4 \%\) compounded continuously. a. What are the total amounts in each of the accounts after 10 years? b. Show that account B gives \(0.04 \%\) annually more interest than account \(\mathrm{A}\).

Short Answer

Expert verified
Account A: \$7429.74Account B: \$7459.12Difference: Account B gives 0.04% more annually.

Step by step solution

01

Calculate the total amount for Account A

Use the formula for compound interest, which is \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]Here, \( P = 5000 \), \( r = 0.04 \), \( n = 2 \) (since interest is compounded semiannually), and \( t = 10 \) years.Substitute these values into the formula to find the amount:\[ A_A = 5000 \left(1 + \frac{0.04}{2}\right)^{2 \times 10} \]Calculate the exponent:\[ A_A = 5000 \left(1 + 0.02\right)^{20} \]Simplify the term inside the parentheses:\[ A_A = 5000 \left(1.02\right)^{20} \]Finally, compute the result:\[ A_A \approx 5000 \times 1.485947 \approx 7429.74 \]
02

Calculate the total amount for Account B

Use the formula for continuously compounded interest, which is \[ A = Pe^{rt} \]Here, \( P = 5000 \), \( r = 0.04 \), and \( t = 10 \) years.Substitute these values into the formula:\[ A_B = 5000 e^{0.04 \times 10} \]Simplify the exponent:\[ A_B = 5000 e^{0.4} \]Calculate the result:\[ A_B \approx 5000 \times 1.491824697 \approx 7459.12 \]
03

Calculate the difference in annual interest rate

Account A has an effective annual rate \( r_A \) given by \[ r_A = \left(1 + \frac{0.04}{2}\right)^2 - 1 \]Simplify inside the parentheses:\[ r_A = 1.02^2 - 1 \]Calculate the annual rate:\[ r_A \approx 1.0404 - 1 = 0.0404 \]This is equivalent to \( 4.04\% \).Account B's effective annual rate \( r_B \) is simply the nominal rate due to continuous compounding:\[ r_B = e^{0.04} - 1 \]Calculate the continuous compounding rate:\[ r_B \approx 1.0408 - 1 = 0.0408 \]This is equivalent to \( 4.08\% \).The difference in annual rates is:\[ 4.08\% - 4.04\% = 0.04\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semiannual Compounding
Semiannual compounding refers to the process where interest is added to the principal of an investment twice a year. This means that every six months, any interest earned is added to the original principal, and future interest calculations are based on this new amount. By using the formula for compound interest \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]where
  • \(P\) is the principal amount
  • \(r\) is the annual interest rate
  • \(n\) is the number of times interest is compounded per year
  • \(t\) is the number of years
We substitute our problem values: \(P = 5000\), \(r = 0.04\), \(n = 2\) (semiannual), and \(t = 10\). This gives us: \[ A_A = 5000 \left(1 + \frac{0.04}{2}\right)^{20}\] Solving this precisely, we get:\[ A_A \approx 5000 \times 1.485947 \approx 7429.74 \]
Continuous Compounding
Continuous compounding optimizes the compounding process by calculating and adding interest an infinite number of times in a year, theoretically every instant. The formula for continuously compounded interest is \[ A = Pe^{rt} \] where
  • \(P\) is the principal
  • \(e\) is the base of the natural logarithm
  • \(r\) is the annual interest rate
  • \(t\) is the time in years
Given \(P = 5000\), \(r = 0.04\), and \(t = 10\), substituting these values in we have: \[ A_B = 5000 e^{0.40} \] After solving, we find: \[ A_B \approx 5000 \times 1.491824697 \approx 7459.12 \]
Effective Annual Rate
The Effective Annual Rate (EAR) represents the actual annual interest earned after accounting for compounding periods within the year. For semiannual compounding, the EAR can be obtained using the formula:\[ r_A = \left(1 + \frac{r}{n}\right)^n - 1 \] Substituting \(r = 0.04\) and \(n = 2\), we get:\[ r_A = \left(1 + \frac{0.04}{2}\right)^2 - 1 \] \[ r_A = 1.0404 - 1 \approx 0.0404 \]This is 4.04%. For continuous compounding, the EAR is computed directly from:\[ r_B = e^r - 1 \] Given \(r = 0.04\), calculating it: \[ r_B = e^{0.04} - 1 \approx 1.0408 - 1 \approx 0.0408 \]This is 4.08%. Thus, achieving an effective increase: \[ 4.08\% - 4.04\% = 0.04\% \]

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