### Single-Amount Factors ( F/P and P/F )

of money F accumulated after n years (or periods) from a single present worth P, with interest

compounded one time per year (or period). Recall that compound interest refers to interest paid

on top of interest. Therefore, if an amount P is invested at time t = 0, the amount F1 accumulated

1 year hence at an interest rate of i percent per year will be

where the interest rate is expressed in decimal form. At the end of the second year, the amount

accumulated F2 is the amount after year 1 plus the interest from the end of year 1 to the end of

year 2 on the entire F1

The amount F2 can be expressed as

Similarly, the amount of money accumulated at the end of year 3, using Equation [2.1], will be

Figure 2–1 Cash flow diagrams for single-payment factors: (a) ﬁnd F, given P, and (b) ﬁ nd P, given F. |

A standard notation has been adopted for all factors. The notation includes two cash ﬂ ow sym- bols, the interest rate, and the number of periods. It is always in the general form ( X Y , i , n ). The letter X represents what is sought, while the letter Y represents what is given. For example, F /P means ﬁ nd F when given P. The i is the interest rate in percent, and n represents the number of periods involved.

Using this notation, ( F P ,6%,20) represents the factor that is used to calculate the future amount F accumulated in 20 periods if the interest rate is 6% per period. The P is given. The standard notation, simpler to use than formulas and factor names, will be used hereafter. Table 2–1 summarizes the standard notation and equations for the F/P and P/F factors. This information is also included inside the front cover.

To simplify routine engineering economy calculations, tables of factor values have been pre- pared for interest rates from 0.25% to 50% and time periods from 1 to large n values, depending on the i value. These tables, found at the rear of the book, have a colored edge for easy identiﬁ cation. They are arranged with factors across the top and the number of periods n down the left side. The word discrete in the title of each table emphasizes that these tables utilize the end-of-period convention and that interest is compounded once each interest period. For a given factor, interest rate, and time, the correct factor value is found at the intersection of the factor name and n . For example, the value of the factor ( P/F ,5%,10) is found in the P/F column of Table 10 at period 10 as 0.6139. This value is determined by using Equation [2.3].

**EXAMPLE 2.1**

Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be invested immediately. With the expectation of earning at the rate of 8% per year, Sandy hopes to take the entire amount out in exactly 20 years to pay for a family vacation when the oldest daughter is due to graduate from college. Find the amount of funds that will be available in 20 years by using

**(a)**hand solution by applying the factor formula and tabulated value and

**(b)**a spreadsheet function.

SOLUTION

The cash ﬂow diagram is the same as Figure 2–1a. The symbols and values are

P = $10,000 F = ? i = 8% per year n = 20 years

**(a)**Factor formula: Apply Equation [2.2] to ﬁ nd the future value F. Rounding to four decimals, we have

Standard notation and tabulated value: Notation for the F/P factor is (F/ P,i%,n).

Table 13 provides the tabulated value. Round-off errors can cause a slight difference in the final answer between these two methods.

**(b)**Spreadsheet: Use the FV function to ﬁ nd the amount 20 years in the future. The format is that shown in Equation [2.4]; the numerical entry is FV(8%,20,,10000). The spread- sheet will appear similar to that in the right side of Figure 1–13, with the answer ($46,609.57) displayed. (You should try it on your own computer now.) The FV function has performed the computation in part (a) and displayed the result. The equivalency statement is: If Sandy invests $10,000 now and earns 8% per year every year for 20 years, $46,610 will be available for the family vacation.

**EXAMPLE 2.2 The Cement Factory Case**

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 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 spread- sheet functions to determine the following for the board of directors of the Brazilian company that plans to develop the plant.

**(a)**The equivalent investment needed if the plant is built in 2015.

**(b)**The equivalent investment needed had the plant been constructed in the year 2008.

Solution

Figure 2–2 is a cash ﬂ ow diagram showing the expected investment of $200 million ($200 M) in 2012, which we will identify as time t = 0. The required investments 3 years in the future and 4 years in the past are indicated by F3 = ? and P -4 = ?, respectively.

**(a)**To ﬁnd the equivalent investment required in 3 years, apply the F P factor. Use $1 million units and the tabulated value for 10% interest (Table 15).

Now, use the FV function on a spreadsheet to ﬁnd the same answer, F3 = $266.20 million. (Refer to Figure 2–3, left side.)

**(b)**The year 2008 is 4 years prior to the planned construction date of 2012. To determine the equivalent cost 4 years earlier, consider the $200 M in 2012 (t 0) as the future value F and apply the P/ F factor for n = 4 to ﬁnd P -4. (Refer to Figure 2–2.) Table 15 supplies the tabulated value.

The PV function PV(10%,4,,200) will display the same amount as shown in Figure 2–3, right side.

This equivalence analysis indicates that at $136.6 M in 2008, the plant would have cost about 68% as much as in 2012, and that waiting until 2015 will cause the price tag to increase about 33% to $266 M.

## 0 comments:

## Post a Comment