# 7a- Uniform series of compound interest-2/2 in the Economy.

## 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.

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.

### 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.

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.

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.

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

The next post title will be 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.

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