Last Updated on September 7, 2026 by Maged kamel
Arithmetic Gradient-part 2 in the Economy.
Arithmetic Gradient-part 2-Solved problem 4-8: how to find P, with given I%,n, G?
In Arithmetic Gradient Part 2, we will discuss solved problem 4-8.
A man has purchased a new automobile. He wishes to set aside enough Money in a bank account to cover the car’s maintenance for the first 5 years.
It has been estimated that the maintenance is as follows: For year 1, the maintenance Cost = is $120; for year 2, the maintenance Cost is $150; for year 3, the maintenance Cost is $180; for year 4, the maintenance Cost is $210; and for year 5, the maintenance Cost is $ 240.
Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest.
How much should the car owner deposit in the bank now?
Solution: In the Arithmetic Gradient-part 2. A-The man deposits P at year 0 to repay unequal maintenance costs at the end of each year. We draw the timeline diagram: at t = 1, we have a value of $120; at t = 22, we have a value of $150; at t = 33.
We have a value of $180 at t=4, $210 at t=5, and $240 at t=6. The shape of a Trapezium is created.
B-We check whether G has a constant value by subtracting the value at t=1 from t=2, which is 150-120=$30, which is the same value obtained by subtracting the value at t=2 from t =3, which is 180-150=$30.
Arithmetic Gradient-part 2- G value for the deposit.
This is the same value obtained by subtracting the value at t3 from t4: 210-180=$30, and the same value is obtained by subtracting the value at t3 from t4: 240-210=$30. This is called the G value.
We will consider the Trapezium as consisting of two shapes: the first is rectangular and runs from t=1 to t=5, with i=5% and n=5 years; the second is triangular and runs from t=1 to t=5.
The value at t=1 is 0, and the value at t=5 is $120; $120 represents the difference between 240 and 120.
For the first shape, the value of A is $120, a constant. Our G value is $30.
The present value P is estimated as the sum of P1, due to a uniform series with known values of A, I%, and n, and P2, due to an arithmetic gradient with known values of G, I%, and n.
Arithmetic Gradient-part 2-Estimate P/A and P/G values by using formulae.
A/P=(I%)*(1+i%)^n/(1+i%)^n -1)(0.05)*(1+0.05)^5/(1.05)^5-1, we have i5=5%,n=5 years, substitute in the formula we can get the value of A/P which is equal to 0.2309.
For the value of P/A, it is the reciprocal of A/P, which will be =1/0.2309=4.329. For the P/G value, we have the relation, (P/G)=((1+i)^n-i*n-1))/(i^2*(1+i)^n)). If we want to estimate the value of P/G using the relations, we substitute P/G into the equation shown in the slide image.
We can estimate the final P-value as the sum of P1 and P2; these values are shown in the slide image.
We get the P/A value, and in the present worth column, find P given G, which is 8.237. P1=120*4.329. For the value of P2, we have P2=30*(8.237).
To get the present value of the Money to be deposited in the bank, we will sum P1 and P2 as follows: P = P1 + P2 = 519 + 247.11 = $766.
Arithmetic Gradient-part 2-Estimate P/A and P/G values by using a table.
We use the table for I = 5%. For the compound amount factors, check the known data: n = 5. We need to estimate P/A for the first shape and P/G for the second shape. We move horizontally to the left, under the column of present worth factors, to find P for the given A.
P/G is shown by using the table to be =8.236. We will obtain the final value of P by summing P1 and P2. The final result matches the previous estimates obtained by substitution.
Can we use an Excel built-in function to model the relationship between P and G? Please refer to the next slide for details.
The PDF for this post can be viewed or downloaded from the following button.
External resource in Engineering Economy. This is a great link that illustrates different types of economies and how to make economic decisions.
The Time value of Money is a good reference
The next post will be about Types of assets.