A Solved problem for compound interest.

6b-Introduction to a solved problem for compound interest

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A Solved problem for compound interest.

Detailed expression for compounded continuously.

This is the equation used to estimate the future value of continuously compounded payments, where (e) is Euler’s number. We will need to use the equation for the third option in the next solved problem as we will see later on.

The equation for the compound continuously.

A Solved problem, the reference is the Time value of money part-1.

It is required to select the correct option of the four given options, what is the least present value to be deposited as of today to get $10000 after given years from today, The interest rate options are varying for both the value of interest and the frequency of interest.

Option-1 for the solved problem for compound interest.

This is the first option, how to find the value of the present value P0 to get $10,000 after 4 years, with the interest of 8 % compounded quarterly?

The relation of P0=F*(P/F,i%,n) will be used provided that the new interest rate is adjusted to be (0.08/4). The term (1+i/n)^nt=(1+(0.08/4))^(4*4), since the number of years=4, the present value=F/(1+i/n)^nt=10,000/(1+0.02)^(4*4)=$7284.45.

Solved example 9-select between options

Option-2 for the solved problem for compound interest.

This is the second option, how to find the value of the present value P0 to get $10,000 after 5 years, with an interest of 7 % compounded yearly?

The solution of the problem for option b.

The relation of P0=F*(P/F,i%,n) will be used provided that the new interest rate is, The term (1+i/n)^nt=(1+(0.07/1))^(1*5), since the number of years=5, the present value=F/(1+i/n)^nt=10,000/(1+0.07)^(5)=$7192.86.

Option-3 for the solved problem for compound interest.

This is the third option, how to get the present value P0 to get $10,000 after 10 years, with an interest of 4 % compounded continuously?

The relation of P0=F*(P/F,i%,n) will be used provided that the new interest rate is, e the Euler’s value is taken as 2.7183 and is used for the case of compounded continuously.The term (e)^it=(2.718))^(0.04*10), since the number of years=10, the present value=F/(e^)it=10,000/(2.718))^(0.04*10)=$6703.1825.

The solution of the problem for the case of option c.

Option-4 for the solved problem for compound interest.

This is the fourth option, how to find the value of the present value P0 to get $10,000 after 8 years, with an interest of 4 % compounded semi-annually?

The relation of P0=F*(P/F,i%,n) will be used provided that the new interest rate is 4%, compounded semi-annual.The term (1+i/n)^nt=(1+(0.04/2))^(2*8), since the number of years=8, the present value=F/(1+i/n)^nt=10,000/(1+0.02)^(16)=$7284.458.

From the above different options, the least P0 value obtained is $ 6703.2, which is option C.

The final answer for the solved problem.

This is the pdf file used in the illustration of this post.

For a useful external resource, Engineering Economy. Applying Theory to Practice.

The next post is Introduction to Uniform series of compound interest.

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