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

  The  most fundamental factor in engineering economy  is the one that determines the amount
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.