Last Updated on September 6, 2026 by Maged kamel
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Solved problems for P-F and F-P.
There are three problems for the P-F Value.
A Problem.5 estimates the future value for a given present value, Time, and interest.
We have solved problem 3.5 from Newnan’s book, Engineering Economic Analysis: If $500 were deposited in a bank savings account, how much would be in the account for three years hence if the bank paid 6% interest compounded annually?
We draw a diagram: the x-axis is for Time. We divide Time into three spaces starting from 0, or Time now. The cash deposited is $500, and we draw a downward arrow.
An upward arrow represents the Future value at the end of 3 years.
The relation between the future value and the present value states that F=P*(1+i)^n.
The P-value is the present value=$500, i=6% n=3 years, we will substitute F=500*(1+0.06)^3=$595.51. The future value F, which we have estimated, is shown with a downward arrow because the bank will pay it.
The diagram has 3 spaces labeled 0, 1, 2, 3 in years; i = 6%.
By symbols, the same value is obtained as F=P(F/P,i=0.06,n=3)=5001.06^3=$595.508.
The second solved problem is 3.6, one of the three Present/Future Value problems.
How to determine the p-value from a given F-value, n, and I%?
This is the second of the three solved problems for P-f, quoted from Prof. Donald G. Newnan’s book, Engineering Economic Analysis. If you wish to have $800 in a savings account at the end of 4 years, we draw in Time scale, 4 equal spaces as 0,1,2,3,4, the F value is shown as an upward arrow=$800 at the end of year 3, and the 5% interest was paid annually; i = 5%, n=4.
How much should you put in the savings account now? P at Time 0 is unknown, shown as a downward arrow.
If F=P(1+i)^n, readjust the formula to be P=F(1+i)^(-n), For F=800, i=0.05, n=4. We substitute to get P=800*(1+0.05)^-4, or P=800/1.05^4=$658.16. This is the amount to be deposited in the bank to get $800. Notice that the P-value is < F value.
There are tables from which we can estimate the compound interest factors.
For the case of interest rate I=5%, to find the present worth factor P/F for a single payment, given F, we use the table with I=5%, n=4, highlighted in green.
We draw a line from n=4, proceed to the left, and find that the value of P/F is 0.8227.
That is the direct method for using tables to obtain the P-value for a given F.
Problem 2.2 concerns finding the expected investment at Time 0, given a specific investment’s future value and the cost of Money.
The third solved problem 2.2 of the three solved problems for Present/Future Value.
How to determine F-value from a given P-value, n, I%?
This is the third solved problem of the three solved problems for P-F. Solved problem 2.2- As discussed in the introduction to this chapter, the Houston American Cement factory will require an investment of $200 million to construct. Delays beyond the anticipated 2012 implementation year will require additional funds to construct the factory.
Assuming a 10% annual interest rate, use both tabulated factor values and spreadsheet functions to determine the following for the board of directors of the Brazilian company that plans to develop the plant.
(a) The equivalent investment is needed if the plant is built in 2015.
(b) The equivalent investment needed had the plant been constructed in the year 2008.
We have a Time interval from 2012 to 2015, which spans three years.
The i = 10% table is used, with n = 3.
We will estimate the F from the relation, F=P(F/P, i,n) F=200(F/P,10%, n=3) from the table, F/P=1.3310, then multiply by $200 millions& F=200*1.3310=$266.20.
Using an Excel sheet to solve for the P-value and F-value.
Solved problem 2.2: How to find the expected investment at Time 0 for a given investment future value when the cost of Money is provided, but using an Excel sheet? If we want to use the Excel sheet, create a new table with the following specifications: P = $200 in millions, i = 10%, n = 3. This is for part a) of the solved problem.
The result obtained is the same.
The function used is FV(10%, 3, 200),, with two commas.
For the second part of the solved problem, we have Fe F 00,200 million 1million 4n, = 4nn million 4nn million the presentown.
The related excel function is PV(10%,4,,200)=$136.60.
We will use the table for n=4, which is the difference between 2012& 2008, I=10%,P=F(P/F,i,n)=P/F=0.6830, P=200*0.6830=$136.60 million.
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