Brief content of post-8 -economy.

8- Easy Illustration of the Arithmetic Gradient.

Illustration of the Arithmetic Gradient.

Brief description of the video content.

In this video, I made a revision for all of the known relations between P & F& A, the present value, the future value, and the arithmetic series. In the second part, a new subject will be explained which is the arithmetic gradient what is the idea of the arithmetic gradient and how we can find the value of P/G? or the present value with the given gradient in terms of i% the compound interest and n the number of years.


A solved example 4-8 from Prof. Donald G Newman’s book,  Engineering economic analysis-9ths. The video has a subtitle and 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.

The discussion will be about the Arithmetic gradient.

Content of Engineering economy post.

Our new subject for discussion will be the arithmetic gradient, but we will review the relationship between the P and F, P is the present- value, F is the future value. 

Review of the different types of investments.

Deposit money to get the retirement amount at the end of your investment (P/F) relation.

If you deposit money in a bank for the investment of amount=P, for which, you will receive a compound interest i% to get a future  F-value for your investment=F.

We will find that we have different signs, P is represented by a downward arrow, or minus sign since it is a cash-out but, when you receive F value it will be cash-in and the value is drawn by an upward arrow. The P-value in terms of F, i%, n, and F in terms of P, i%, and n are shown in the slide.

  For P-value,  it is P=F*(P/F, i%, n). For F- value it is=F=P*(1+i)^n. P- value= F* (1+i) ^-n.

The relation between Future value F with present value P with known I, n.

The relation is based on annual compound interest i%, but, if we have changed the annual interest to become a quarterly compounded interest. The new interest rate i2=(i1/4%) and raised to the power of 4*(number of years), or 4n. That was the relation of F with a given P, i, and n.

Deposit money to buy a car at the end of your investment (P/F) relation

There is another relation, for instance, if somebody wishes to buy a car, he or she took a loan from a bank with interest i% and payment F will be at the end of time n, this is from the bank’s point of view.

Cash in will be with a plus sign  + and the money received F is will be cash-out, with a negative sign (-).

There is another type of relationship when someone receives money to buy goods and pays through installment or uniform series of compound interest after one year of receiving the cash-in for the agreed period. The P-value is cash in or with a positive sign.

The relation between uniform series A with present value P with known I, n.

The cash out with equal amounts will be cash-out with negative signs, its value =A. The value for A in terms of P, i, n, A=i*P*(1+i)^n/((1+i)^n-1)), where i is the interest rate, and n is the time in years. The values of A are equal, same as uniformly distributed load with A value as cash in installments in n the bank account, with an interest I and for some time.

Deposit money as uniform series to get insured money at the end of an investment (A/F) relation

For instance, if somebody wants to receive insured money after some time, he will pay equal installments of value  A with an interest rate i% and for n time, start after one year, the diagram shows that type, the A values are shown as upward cash in and at the end of time, for instance, time=8 years.

How to find F with given A, i, n?

The F- value received at the end of year 8, can be estimated based on a compound interest i%, the  F- value= A*(F/A, i%, n), also F=A/i*((1+i)^n-1).

The arithmetic gradient factors.

We will discuss a new subject, which is the arithmetic gradient factors.
While paying installments, these installments are not of equal value, but the difference between each consecutive payment is constant and equals  G, which is the same difference between the second and third installments also between the third and fourth installments as well as between the fourth and fifth installments.

After one year of the agreement, somebody pays an installment with the value A. The cash-in diagram has a shape of a trapezium shape, for which the smaller value is A and the larger base=A+4G  for the sketch shown where n=5 years.

Introduction to Arithmetic gradient.

But if we have n of years, the larger base value of the trapezium will be A+(n-1)G. But if we have n of years, the larger base value will be  A+(n-1)G.

The previous shape consists of a trapezium and can be solved by converting it into two shapes.

The first shape is in the form of an equal installment diagram with a value of A for 5 years as cash-in.
The second part is the shape of a triangle with increasing hypotenuse from 0 to 4G in the case of 5 years as shown in the sketch with a slope of 4G/4 =G.
If we want to estimate the future- value in terms of A, i%, n. The relation is F=(A/i)*((1+i)^n-1)). If we want to estimate A in terms of P,  i, n, The relation is F=(i*P)*(1+i)^n/((1+i)^n-1)).

How to find F with given arithmetic gradient G, i, n?

While for the triangular shape, the value for F is in terms of i%, n, G. F=(G/I)*(1+i)^n-1)/I)-n. P/G Find P in terms of G. The( P/G, i,n) has a relation, which is shown in the slide.

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

External resource in Engineering Economy. This is a great link that illustrated different types of economies, and how to make economic decisions. the time value of money, Applying Theory to Practice.

The next post: Solved problem 4-8-Arithmetic Gradient

-part 2. how to find P, with given I%,n, G?

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