Brief description of post 6b economy

6b-Introduction to a solved problem for compound interest.

A Solved problem for compound interest.

In the next video, a solved problem was given for the selection of the correct option of the four given options, what is the least present value to be deposited as of today to get $10000 after a certain given year from today? how to select the least deposit as of today. The video has a closed caption in English.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

Detailed expression for compounded continuously.

The equation for the compound continuously.

This is the equation used to estimate the future value of continuously compounded payments, where (e) is the 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.

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.

Solved example 9-select between options

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.

Option-2 for the solved problem for compound interest.

The solution of the problem for option b.

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 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.

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

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’ value is taken as 2.7183 and is use 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.

Option-4 for the solved problem for compound interest.

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

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.

The final answer for the solved problem.

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

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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