Last Updated on September 1, 2024 by Maged kamel
Uniform series of compound interest-2/2 in the Economy.
Solved example 4-1, how to determine the F value in terms of P, A, and I%?
In compound interest-2/2, we start to have a look at the Solved example 4-1 from Prof. Donald G Newman’s book, Engineering Economic Analysis.
A man deposits $500 in a credit union at the end of each year for 5 years. Our timeline starts from 0 to 5. The deposits were done for 5 years at the end of years 1&2&3&4&5. The final value, or F the future value to be estimated which is cash from the person’s point of view.
Solution: this example is a direct application of the uniform series.
The symbolic form of the future value can be written as F value=F*(A/ F, i%,n), where f is the required future value,n is the time in years, while A is the amount of uniform series. I is the interest rate yearly.
![Compound interest-2/2-Solved problem 4-1-How to find The final value F for known uniform series A?](https://magedkamel.com/wp-content/uploads/2022/01/Page-1-post-7a-economy.jpg)
We have the interest i=5%, the uniform series value A=$500, and n=5. Then the future value can be estimated as F=A*((1+i)^n-1)/i. F=500*((1+0.05)^5-1)/0.05=(500)*0.27625)/0.05. F=$2763, this is the future money that will be received after 5 years.
![Part B of the Solved example 4-1](https://magedkamel.com/wp-content/uploads/2022/01/Page-2-post-7a-economy.jpg)
Solved example 2.5, how to determine the F value in terms of P, A, n, and I%?
The Solved example is from Profs. Leland Blank, Anthony Tarquin’s book of Engineering economy.
Example 2.5 The president of Ford Motor Company wants to know the equivalent future worth of a $1 million capital investment each year for 8 years, starting 1 year from now.
Ford Capital earns at a rate of 14% per year. The author uses $1000 units, for F/a the author has estimated the value as (1+0.14)^8-1/0.14=1.852586/0.14= 13.2328 by the author. F=1000*13.2328 =$13,232.8 in units, estimated when i%= a compound interest %14.
![Solved example 2.5-what is F value If uniform series A=$1 million, i=14%, and n=8?](https://magedkamel.com/wp-content/uploads/2022/01/Page-3-post-7a-economy.jpg)
If we wish to estimate the future value F value by relations, The present value p=1000, we have interest rate i=14%, F=1000* (1+0.14)^8-1 divided by I. The future value of the uniform series F=1000* (1/0.14) * (1+0.14)^8-1=$13,232.0, which is the same value as estimated by the author.
Solved example 4.3, how to determine the A value in terms of P, n, and I%?
The Solved example is from Prof. Donald G Newman’s book, Engineering Economic Analysis. Solved example 4.3 explains how to estimate the uniform series of compound interest.
Example 4.3 Consider a situation in which you borrow $5000. You will repay the loan in five ends of the year payments. The first payment is due one year after you receive the loan. Interest on the loan is 8%. What is the size of each of the five payments?
Solution:1- We draw the timeline from 0 to 5, the cash in at time t=0, where P= $5000, but the loan will be repaid in series with an equal amount.
![Solved example 4.3-Uniform series of compound interest.](https://magedkamel.com/wp-content/uploads/2022/01/Page-4-post-7a-economy.jpg)
2-The first payment is due after one year, for 5 years with an interest rate of 8%. We have 5 payments in years 1 &2& 3&4&5.
3-This relation is not a relation between A&F but between P and A.
4-A value= F(i/(1+i)^n-1), since F=P*(1+i)^n. We adjust the relation to be A=P*(1+i)^(i/((1+i)^n-1)).
5-We have (1+i)^n= (1+0.08)^4.
The value of A can be found in the following equation. A=5000*(1+0.0.08)^5*(0.0.08/((1.08)^5-1)=0.11754/0.4693* A=5000* 0.25044=$1252.
The next post title will be post 8- Easy Illustration of the Arithmetic Gradient. The post includes an easy illustration of a new type of cash flow which is the arithmetic gradient series cash flow involves an increase or decrease of a constant amount in the cash flow of each analysis period.
For a useful external resource, Engineering Economy. Applying Theory to Practice.