## 3

P (1+i)2

P (1+i)2 i

P (1+ i)2 + P (1 + i)2 i = P (1 + i)2 (1 + i) = P (1+i)3

n

P (1+i)n-1

P (1+i)n-1 i

P (1+ i)n-1 + P (1 + i)n-1 i = P (1 + i)n-1 (1 + i) = P (1+i)n

nipulation, this can be written conveniently mathematically as P*(1+i)n dollars ($100*1.082 =$116.64).

Table 4.5 extends the above logic to year 3 and then generalizes the approach for year n. If we return our attention to our original goal of developing a formula for Fn which is expressed only in terms of the present amount P, the annual interest rate i, and the number of years n, the above development and Table 4.5 results can be summarized as follows:

For Compound Interest Fn = P (1+i)n

### Example 4

Repeat Example 3 using compound interest rather than simple interest. Fn = P * (1 + i)n F4 = 500 * (1 + 0.10)4 F4 = 500 * (1.10)4 F4 = 500 * (1.4641) F4 = $732.05

Notice that the balance available for withdrawal is higher under compound interest ($732.05 > $700.00). This is due to earning interest on principal plus interest rather than earning interest on just original principal. Since compound interest is by far more common in prac tice than simple interest, the remainder of this chapter is based on compound interest unless explicitly stated otherwise.

### 4.6.5 Single Sum Cash Flows

Time value of money problems involving compound interest are common. Because of this frequent need, tables of compound interest time value of money factors can be found in most books and reference manuals that deal with economic analysis. The factor (1+i)n is known as the single sum, future worth factor or the single payment, compound amount factor. This factor is denoted (FI P,i,n) where F denotes a future amount, P denotes a present amount, i is an interest rate (expressed as a percentage amount), and n denotes a number of years. The factor (FIP,i,n) is read "to find F given P at i% for n years." Tables of values of (FIP,i,n) for selected values of i and n are provided in Appendix 4A. The tables of values in Appendix 4A are organized such that the annual interest rate (i) determines the appropriate page, the time value of money factor (F IP) determines the appropriate column, and the number of years (n) determines the appropriate row.

Example 5

Repeat Example 4 using the single sum, future worth factor.

Fn = P * (1 + i)n Fn = P * (FIP,i,n) F4 = 500 * (FIP,10%,4) F4 = 500 * (1.4641) F4 = 732.05

The above formulas for compound interest allow us to solve for an unknown F given P, i, and n. What if we want to determine P with known values of F, i, and n? We can derive this relationship from the compound interest formula above:

The factor (1+i)-n is known as the single sum, present worth factor or the single payment, present worth factor. This factor is denoted (PIF,i,n) and is read "to find P given F at i% for n years." Tables of (PIF,i,n) are provided in Appendix 4A.

Example 6

To accumulate $1000 five years from today in an account earning 8%/yr compound interest, how much must be deposited today?

P = Fn * (1 + i)-n P = F5 * (PIF,i,n) P = 1000 * (PIF,8%,5) P = 1000 * (0.6806) P = 680.60

To verify your solution, try multiplying 680.60 * (FI P,8%,5). What would expect for a result? (Answer: $1000) If you're still not convinced, try building a table like Table 4.5 to calculate the year end balances each year for five years.

### 4.6.6 Series Cash Flows

Having considered the transformation of a single sum to a future worth when given a present amount and vice versa, let us generalize to a series of cash flows. The future worth of a series of cash flows is simply the sum of the future worths of each individual cash flow. Similarly, the present worth of a series of cash flows is the sum of the present worths of the individual cash flows. Example 7

Determine the future worth (accumulated total) at the end of seven years in an account that earns 5%/yr if a $600 deposit is made today and a $1000 deposit is made at the end of year two?

for the $600 deposit, n=7 (years between today and end of year 7)

for the $1000 deposit, n=5 (years between end of year 2 and end of year 7)

F7 = 600 * (FIP,5%,7) + 1000 * (FIP,5%,5) F7 = 600 * (1.4071) + 1000 * (1.2763) F7 = 844.26 + 1276.30 = $2120.56

### Example 8

Determine the amount that would have to be deposited today (present worth) in an account paying 6%/ yr interest if you want to withdraw $500 four years from today and $600 eight years from today (leaving zero in the account after the $600 withdrawal).

for the $500 deposit n=4, for the $600 deposit n=8 P = 500 * (PIF,6%,4) + 600 * (PIF,6%,8) P = 500 * (0.7921) + 600 * (0.6274) P = 396.05 + 376.44 = $772.49

### 4.6.7 Uniform Series Cash Flows

A uniform series of cash flows exists when the cash flows in a series occur every year and are all equal in value. Figure 4.3 shows the cash flow diagram of a uniform series of withdrawals. The uniform series has length 4 and amount 2000. If we want to determine the amount of money that would have to be deposited today to support this series of withdrawals starting one year from today, we could use the approach illustrated in Example 8 above to determine a present worth component for each individual cash flow. This approach would require us to sum the following series of factors (assuming the interest rate is 9%/yr):

P = 2000*(PIF,9%,1) + 2000*(PIF,9%,2) + 2000*(PIF,9%,3) + 2000*(PIF,9%,4)

After some algebraic manipulation, this expression can be restated as:

(PIF,9%,3) + (PIF,9%,4)] P = 2000*[(0.9174) + (0.8417) + (0.7722) + (0.7084)] P = 2000*[3.2397] = $6479.40

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