Comprehension

Find the effective rate which is equivalent to a normal rate of 10% p.a. compounded:

Given:

\[ (1.05)^2 = 1.1025, \quad (1.025)^4 = 1.1038. \]

Question: 1

Semi-annually.

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The effective annual rate (EAR) accounts for compounding and is given by \( r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1 \). It is always greater than or equal to the nominal rate.
Updated On: Feb 11, 2025
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Solution and Explanation

Step 1: For compounding semi-annually, the effective annual rate \( r_{{eff}} \) is calculated using the formula: \[ r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1, \] where: - \( r \) is the nominal annual interest rate, - \( n \) is the number of compounding periods per year. 
Step 2: Substitute the given values \( r = 0.10 \) (10%) and \( n = 2 \): \[ r_{{eff}} = \left(1 + \frac{0.10}{2}\right)^2 - 1 = (1.05)^2 - 1. \] 
Step 3: Compute the value: \[ r_{{eff}} = 1.1025 - 1 = 0.1025 \quad {or} \quad 10.25\%. \]

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Question: 2

Quarterly.

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The more frequently interest is compounded, the higher the effective annual rate (EAR). The formula to calculate EAR is: \[ r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1. \]
Updated On: Feb 11, 2025
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Solution and Explanation

Step 1: The effective annual rate \( r_{{eff}} \) for a given nominal rate \( r \) compounded \( n \) times per year is given by: \[ r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1. \] where: - \( r = 0.10 \) (10% nominal rate), - \( n \) is the number of compounding periods per year. 
Step 2: First, compute for semi-annual compounding (\( n = 2 \)): \[ r_{{eff}} = \left(1 + \frac{0.10}{2}\right)^2 - 1 = (1.05)^2 - 1. \] 
Step 3: Compute the value: \[ r_{{eff}} = 1.1025 - 1 = 0.1025 \quad {or} \quad 10.25\%. \] 
Step 4: Now, compute for quarterly compounding (\( n = 4 \)): \[ r_{{eff}} = \left(1 + \frac{0.10}{4}\right)^4 - 1 = (1.025)^4 - 1. \] 
Step 5: Compute the value: \[ r_{{eff}} = 1.1038 - 1 = 0.1038 \quad {or} \quad 10.38\%. \]

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