Brief content of post 5-Engineering economy.

5- Step-by-step illustration of solved problems for P-F value.

Solved Problems For P-F Value Relation.

Brief description of the video content.

The video includes three solved problems for p-f. The first solved problem 3.5, For P-F Value Relation for deposited money in the bank it is required to estimate the compounded value after three years, the interest rate of 6%, the solution is made by using the equivalence relation between the value of time now and the value after three years, another form of solution is given by symbols.



Second problem 3.60 for which it is required to estimate the value of money to be deposited now to get a value of $800 after 4 years with a compound interest of 5%.

The third solved problem is 2.2, where the money is invested to construct a cement factory, but due to some problems, the factory is to be constructed after some time, given the interest cost, it is required to estimate the cost of investment. 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.

Solved problems for P-F, F-P.

There are Three solved problems for P-F Value.

Solved problem 3.5 to estimate the future value for a given present value, time, and interest.

We have a 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 the time, we divide time into, three spaces starting from 0, or time now, spaces starting from 0, or time now,  the cash is deposited is $500, and we draw a downward arrow.

Solved problems for P-F Value.

For the Future value at the end of 3 years, it is represented by an upward arrow.
 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 as a downward arrow since the bank will pay this value.

A solved problem 3.5 to estimate future value by symbols.

The diagram has 3 spaces as 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)=500*1.06^3=$595.508.

The second solved problem 3.6 of the solved problems for Present/Future Value.

How to determine p-value from a given F-value, n, I%?

This is the second solved problem 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 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 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.

A solved problem 3.6.

We substitute, to get P=800*(1+0.05)^-4, or P=800/1.05^4=$658.16, this is the money to be deposited in the bank to get$800, notice that 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 get the present worth factor P/F under a single payment, find P with given F, we use a table, with  I=5%, n=4, which is highlighted by a yellow color.

A solved problem 3.6 how to find the present value?

We draw a line from n=4, proceed to the left, then the value of P/F can be found, as  0.8227.
That is the direct method by using tables to get the P-value with given F.

A solved problem 2.2 how to find the expected investment at time 0, for given investment future value, the cost of money is given?

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 the implementation year of 2012 will require additional money to construct the factory.

Assuming that the cost of money is 10% per year, compound interest, 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.

A solved problem 2.2 how to find the expected investment at time 0, for given investment future value, the cost of money is given?

We have a time interval from 2012 to 2015, these are three years.
The table of i=10% is used, n=3.

A solved problem 2.2 how to find the expected investment at time 0?

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.

A solved problem 2.2 how to find the expected investment at time 0, for given investment future value, the cost of money is given, but using an excel sheet? If we want to use the excel sheet, make a new table P=$200 in millions, i=10%, n=3, this is for part a) of the solved problem. The result obtained is the same.

A solved problem 2.2 how to find the expected investment at time 0?

The function used is FV (10%,3,,200) with two commas.
For the second part of the solved problem, we have F=$200 million, i=10%, n=4, the present value is unknown.

The related excel function is  PV(10%,4,,200)=$136.60.

The second part of the solved example 2.2.

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.

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

For a useful external resource, Engineering Economy. Applying Theory to Practice.

This is a link to the next post: Easy approach to compounding technique.

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