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Future Value
Below are links to the following topics:
- Power of Interest Compounding
- Future Value
- Calculating Future Values
- Future Value Of Annuities
- Growing Perpetuities (Dividend Discount Model)
- Compounding Interest- Single sums of money
- Future Value Examples
- Series of Cash Flows
- Uniform Series of Cash Flows
- FV of an Annuity
- Gradient Series of Cash Flows
- Geometric Series of Cash Flows
- Multiple Compounding Periods in a Year
- Loan Amortization
- Computing Car Payments
- Effective Annual Interest Rate
- Effective Annual Interest Rate II
- Credit Card Interest Rate
- Discrete Flows
- Continuous Flows
- Equivalence and Indifference
- Introduction to Measures of Worth
- Inflation Considerations
- Net Present Value Buying Pressure Washer
- Annuities
- Perpetuities and Capitalized Worth Method
- Break-even Analysis
- Company Marginal Growth Rate
- Reinvesting at a Different Rate
- Yield-to-Maturity or Internal Rate of Return
The Power of Interest Compounding [viewable here in Excel]
Interest compounding is a powerful tool, the more, the better if you're in control of the money. The opposite if you're borrowing the money. Why? If you're earning 10% compounded annually on $10,000, every year you will make $1,000. If you're earning 10% compounded daily on $10,000, every year you will make $1,051.72 with no added risk.
Keep in mind interest compounding is a double edge sword. Take your credit card for example, it will most likely state an annual interest rate of 18%. Now read the fine print; most likely the 18% is compounded monthly, and when computed will equal 19.54% annually.
| Effective Interest Rate = (1 + (Stated Annual Interest Rate/n)n)-1 | |||||
| Compounding Periods | Deposit | 1 year with interest | |||
| 1 | 10.000% | $10,000 | $11,000 | 10% | =((1 + (.10/1))1)-1 |
| 6 | 10.426% | $10,000 | $11,043 | 10% | =((1 + (.10/6))6)-1 |
| 12 | 10.471% | $10,000 | $11,047 | 10% | =((1 + (.10/12))12)-1 |
| 365 | 10.516% | $10,000 | $11,052 | 10% | =((1 + (.10/365))365)-1 |
| Continuous compounding | 10.517% | $10,000 | $11,052 | 10% | =(EXP(2.7183)-1) |
| exponent (exp) = | 2.7183 |
Future Value
Future value is the amount of money an investment will grow to at some date in the future by earning interest at some compound rate. For example, suppose you put $1,000 (the present value (PV)) into an account earning an interest rate of 10% per year. The amount you will have in five years, assuming you take nothing out of the account before then, is called the future value of $1,000 at an interest rate of 10% per year for five years.
Let us define our terms more precisely:
PV = Present value or beginning amount in your account. Here it is: $1,000.
i= Interest rate, usually expressed in percent per year. Here it is 10% (or .10 as a decimal).
n = Number of years the account will earn interest.
FV = future value at the end of n years.
Now let's calculate the future value in this example one step at a time. First, how much will you have after the first year? You will have your original $1,000 plus interest of $100 (10% if $1,000 or .1 x $1,000). Your future value at the end of year 1 will therefore by $1,100.
FV = $1,000 x 1.10 = $1,100
If you redeposit this entire sum of $1,100 for another year, how much will you have at the end of year 2? During year 2 you will earn 10% interest on the entire $1,100. The interest earned is thus .10 x $1,100 = $110. You will therefore have $1,210 at the end of year 2.
To gain further understanding of the nature of compound interest, we can break this future value of $1,210 into three components. First, there is the original principle of $1,000. Next, there is the interest on this principal - $100 in the first year and another $100 in the second year. The interest on the original principle is called simple interest ($200 in our example). Finally, there is $10 of interest earned in the second year on the $100 of interested earned in the first year. Interest earned on interest already earned is compound interest. The total interest earned ($210) is the sum of the simple interest ($200) plus the compound interest ($10).
Practically speaking, you do not care how much of your total interest of $210 is simple interest and how much is compound interest. All you really care about is how much you will have in your account in the future, that is, the future value. The most direct way to calculate the future value at the end of year 2 is to recognize that it is the original principal multiplied by 1.10 and then multiplied by 1.10 again.
FV = $1,000 x 1.10 x 1.10 = $1,000 x 1.102 = $1,210
After three years you will have:
FV = $1,000 x 1.10 x 1.10 x 1.10 = $1,000 x 1.103 = $1,311
By this chain or reasoning, we can find future value after five years by repeated multiplication:
FV = $1,000 x 1.10 x 1.10 x 1.10 x 1.10 x 1.10 = $1,000 x 1.105 = $1,610.51
Thus, we have our answer to the original question. The future value of $1,000 at an interest rate of 10% per year for five years is $1,610.51. The total interest earned over the five years is $610.51, of which $500 is simple interest and $110.51 compound interest.
More generally, if ¡ is the interest rate and n is the number of years, the future value is the $1,000 is given by the formula:
FV = 1,000 (1 +¡)n
The expression multiplying the PV of $1,000 is the future value of $1 and is known as the future-value factor. In our example, it is 1.61051 (easily found using Excel). The formula for the future-value factor is:
Future value factor = (1 +¡)n
The future-value of any amount invested at 10% per year for five years is just the amount time the same future value factor of 1.61501. Thus, the future value of $500 invested at 10% per year for 5 years is $500 x 1.6501 = $804.254. The future-value factor is greater the higher the interest rate and the longer the time the investment is held.
Calculating Future Values
In practice, there are a variety of ways to calculate future values, which we can illustrate with the example of calculating the future value of $1,000 at an interest rate of 10% per year for five years.
1. We could multiply $1,000 by 1.10 five times:
FV = $1,000 x 1.10 x 1.10 x 1.10 x 1.10 x 1.10 = $1,000 x 1.105 = $1,610.51
This method is fine if the holding period is not too long. But when the number of periods, n, gets large, this method becomes tedious. Using Excel we can compute =FV(0.1,5,,-1000) = $1,610.51
2. We can use A table of future value factors to compute future values.
3. Finally, there is a handy rule-of-thumb that can help you estimate future values when you do not have your calculator or a table available. It is called the Rule of 72. This rule says that the number of years it takes for a sum of money to double in value (the "doubling time") is approximately equal to the number 72 divided by the interest rate expressed in percent per year:
Doubling time = 72/interest rate
So at an interest rate of 10% per year, it should take approximately 7.2 years to double your money. If you start with $1,000 you will have $2,000 after 7.2 years, $4,000 after 14.4 years, $8,000 after 21.6 years, and so on.
Future Value Of Annuities (End-Of-The-Period) [viewable here in Excel]
In some cases, an individual may plan to set aside a fixed annuity each period for a number of periods and will want to know how much he or she will have at the end of the period. The future value of an end-of-the-period annuity can be calculated as follows:
FV of an Annuity = FVA (A,r,n) = A [(1+r)n - 1 / r]
Thus, the notation we will use throughout this book for the future value of an annuity will be FV(A,r,n).
Illustration : Individual Retirement Accounts (IRA)
Individual retirement accounts (IRAs) allow some tax payers to set aside $2,000 a year for retirement and exempts the income earned on these accounts from taxation. If an individual starts setting aside money in an IRA early in her working life, the value at retirement can be substantially higher than the nominal amount actually put in. For instance, assume that this individual sets aside $2,000 at the end of every year, starting when she is 25 years old, for an expected retirement at the age of 65, and that she expects to make 8% a year on her investments. The expected value of the account on her retirement date can be estimated as follows:
Expected Value of IRA set - aside at 65 = $2,000 [(1.08)40 - 1 / .08] = $518,113
The tax exemption adds substantially to the value because it allows the investor to keep the pre-tax return of 8% made on the IRA investment. If the income had been taxed at say 40%, the after-tax return would have dropped to 4.8%, resulting in a much lower expected value:
Expected Value of IRA set - aside at 65 = $2,000 [(1.048)40 - 1 / .048] = $230,127
As you can see, the available funds at retirement drops by more than 55% as a consequence of the loss of the tax exemption.
What's the value of an IRA if you contribute $2,000 annually at 8% for 40 years (tax free)?
| FV | =A(((1+r)n)-1)/r) | |
| A= | $2,000 | |
| r= | 8% | |
| n= | 40 | |
| FV= | $518,113 | =2000*(((1+0.08)40)-1)/0.08 |
| Calculated (end-of-period) | $518,113 | =FV(0.08,40,-2000,,0) |
| Calculated (beginning-of-period) | $559,562 | =FV(0.08,40,-2000,,1) |
What's the value if you contribute $2,000 annually at 8% into the S&P 500 for 40 years (notice NOT tax free)?
| FV | =A(((1+r)n)-1)/r) | |
| A= | $2,000 | |
| r= | 4.800% | =.08*(1-.4) |
| n= | 40 | |
| FV= | $230,127 | =2000*(((1+0.048)40)-1)/0.048 |
| Calculated (end-of-period) | $230,127 | =FV(0.048,40,-2000) |
| Calculated (beginning-of-period) | $241,174 | =FV(0.048,40,-2000,,1) |
Future Value of an Annuity: Sinking Fund
A company received $10 million bond due in 10 years, company earns 8% on money deposited in sinking fund. How much must be deposited annually to payoff $10 million bond.
| Amount | $10,000,000 | |
| Rate | 8% | |
| Number of deposits | 10 | |
| Sinking fund payment | $690,295 | =10000000*(0.08/(((1+0.08)10)-1)) |
| Using "Payment" function (not PV or FV) (end-of-period) | $690,295 | =PMT(0.08,10,,-10000000) |
| Using "Payment" function (not PV or FV) (beginning-of-period) | $639,162 | =PMT(0.08,10,,-10000000,1) |
Growing Perpetuities (Dividend Discount Model)
What's the value of a company stock paying $2.73 annual dividend with a 6% growth rate?
| PV = CF/(r-g) | ||
| Dividend (CF) | $2.73 | |
| Growth (g) | 6% | |
| Rate of return (r) | 12.23% | |
| Value of stock | $46.50 | =(2.73*(1+0.06)/(0.12223-0.06)) |
Compounding Interest - Single Sums of Money
To illustrate the mathematical operations involved in modeling cash flow profiles using compounded interest, first consider the investment of a single sum of money, P, in a savings account for n interest periods. Let the interest rate per interest period be denoted by ¡, and let the accumulated total in the fund n periods in the future be denoted by F. Assuming no monies are withdrawn during the interim, the amount in the fund after n periods equals p x (1+i)n. As a convenience in computing values of F (the future worth) when given values of P (the present worth). The quantity (1+i)n is referred to as the single sum, future worth factor and is denoted by (F|P ¡,n). The expression (F|P ¡,n) is read as "the F, given P factor at i% for n periods." The information is summarized as follows:
Let P = the equivalent value of an amount of money at time zero, or present worth
F = the equivalent value of an amount of money at time n, or future worth
¡ = the interest rate per interest periods
n = the number of interest periods
Thus, the future worth is related to the present worth as follows:
Equation (2.1) F = P(1+i)n
where ¡ is expressed as a decimal amount. Equivalently,
F = (F|P ¡,n)
where ¡ is expressed as a percentage amount. A cash flow diagram depicting the relationship between F and P for the savings account example is given in Figure 2.4. Remember that F occurs n periods after P.
As individual borrows $1,000 at 12% compounded annually. The loan is paid back after 5 years.
How much should be repaid?F = P(1+i)n
= $1,000(1 + 0.12)5= $1,000(1.7723)
=$1,762.30
The amount to be repaid equals $1,762.30
Since we are able to determine conveniently values of F when given values of P, ¡, and n, it is a simple matter to determine values of P when given values of F, ¡ and n. In particular, since F = P(1+i)n on dividing both sides by (1+i)n, we find that the present worth and future worth have the relation:
Equation (2.3) P = F(1+i)-n
or
Equation (2.4) P = F(P|F ¡,n)
where (1+i)-n and (P|F ¡,n) are referred to as the single sum, present worth factor.
To illustrate the computation of P given F,¡, and n, suppose you wish to accumulate $10,000 in a savings account 4 years from now and the account pays interest at a rate of 9% compounded annually. How much must be deposited today?
=10000*((1+0.09)-4) = $7,084
You want to accumulate $10,000 in a money market account 4 years from now, the account pays 9% compounded annually. How much do you have to deposit today?
| =10000*((1+0.09)-4) = $7,084 | =PV(0.09,4,,-10000) = $7,084.25 | =10000*((1+0.09)-4) =$7084.25 |
| You want to accumulate $10,000 in a money market account 4 years from now, the account pays 9% compounded annually. How much do you have to deposit today? | $7,084 | $7,084.25 |
Future Value Examples [viewable here in Excel]
| Future Value Examples | Using Excel | Manually | ||
| How much does $1,000 accumulate after three years at 7% (compounded annually)? | $1,225 | =FV(0.07,3,0,-1000) | $1,225.04 | =1000/(1/(1.073)) = $1,225.04 |
| What is the present value of $1,225.04 at 7% for 3 years (compounded annually)? | $1,000 | =PV(0.07,3,0,-1225) | $1,000.00 | =1225.04/(1.073) = $1,000 |
| What is the interest rate (compounded annually) if you deposit $1,000 and after 3 years it's worth $1,225.04? | 7.00% | =RATE(3,0,-1000,1225,0) | 7.00% | =(1.225043(1/3))-1 = 7.00% |
| You deposit $1,000 at 7% (compounded annually) per year, how long will it take to earn $1,225.04? | 3 | =NPER(.07,0,-1000,1225) | 3 | =LN(1225/1000)/LN(1+7.0%) = 3 |
| $1,000 has accumulated to $2,000 in 8 yrs, what is the annual growth rate (compounded annually)? | 9.051% | =RATE(8,0,-1000,2000,0) |
| What is the present value of $2,000 at 9.051% (compounded annually) for 2 years? | ($1,000) | =PV(.09051,8,,2000) |
| What is the interest rate (compounded annually) if you deposit $1,000 and after 8 years it's worth $2,000? | $2,000 | =FV(.09051,8,,-1,000) |
| You deposit $1,000 at 9.051% per year (compounded annually), how long will it take to earn $2,000? | 8 | =NPER(.09051,,-1,000,2,000) |
| Rule of 72 | 7.96 | =72/(.0951*100) |
| A $100k investment earns 14% per year (compounded annually), how many years before it reaches $1 million? | 17.57 | =NPER(0.14,0,-100000,1000000,0) | 17.573194 | =LN(1000000/100000)/LN(1+0.14) = 17.573194 |
| What is the interest rate (compounded annually) if you deposit $100,000 and after 17.57 years it's worth $1 million? | 14.000% | =RATE(17.57,0,-100000,1000000,0) | 14.00% | =(1000000/100000)(1/17.573194)-1 = 14.00% |
| What is the present value of $1,000,000 at 14.00% (compounded annually) for 17.57 years? | ($100,000) | =PV(.14,17.57,,1000000) | ($100,000) | =-1000000/(1.1417.573194) = -$100,000 |
| What is the future value of $100,000 at 14.00% (compounded annually) for 17.57 years? | $1,000,000 | =FV(.14,17.57,,-100,000) | $1,000,000 | =100000*(1.1417.573194) = $$1,000,000 |
| What is the future value of $100,000 at 14.00% (compounded annually) for 17.57 years? | $1,000,000 | =100,000*(1+.14)17.57 |
Series of Cash Flows
Having considered the transformation of a single sum of money to a future worth equivalent when given a present worth amount and vice versa, we generalize that discussion to conversion of a series of cash flows to present worth and future equivalents. In particular, let At denote the magnitude of a cash flow (receipt or disbursement) at the end of time period t. Using discrete compounding, the present worth equivalent for the cash flow series is equal to the sum of the present worth equivalents for the individual cash flows. Consequently,
Equation (2.5) P = A1(1+i)-1 + A2(1+i)-2 + …An-1(1+i)-(n-1) + An(1+i)-n
or using the summation notation
Equation (2.6) P = nΣt=1At(1+i)-t
or, equivalently,
Equation (2.7) P = nΣt=1At(P|F ¡,t)
Consider the series of cash flows depicted by the cash flow diagram given in Figure 2.5. Using an interest rate of 6% / interest period, the present value equivalent is given by:
P = $300(P|F 6,1) - $300(P|F 6,3) + $200(P|F 6,4) + $400(P|F 6,6) + $200(P|F 6,8)
P = $300(1+.06)1 - $300(1+.06)3 + $200(1+.06)4 + $400(1+.06)6+ $200(1+.06)8
The future worth equivalent is equal to the sum of the future worth equivalents for the individual cash flows. Thus,
Equation (2.8) F = A1(1+i)-1 + A2(1+i)-2 + ….An-1(1+i)-(n-1) + An(1+i)-n
or, using the summation notation
Equation (2.9) F = nΣt=1At(1+i)n-t
Notice in Equation 2.8 and 2.9 that the exponent of the interest factor counts the number of periods between the cash flow and the time period where F is located.
Equation (2.10) F = nΣt=1At(F|P ¡,n-t)
Alternatively, since we know the value of future worth is given by
Equation (2.11) F = P(1+i)n
Equation (2.12) F = nΣt=1At(1+i)n-t
See the example below with a slight modification: allow initial deposit to earn interest for two years before the initial withdrawal (earning 12% interest), how much must be deposited?
| $2,000 | |
| 3.604776202 | =(((1.125)-1)/(0.12*((1+0.12)5))) |
| 0.797193878 | =1/(1.125) |
| $5,747.41 | =2000*3.604776202*0.79719388 |
Uniform Series of Cash Flows
A uniform series of cash flows exists when all of the cash flows in a series are equal. In the case of a uniform series the present worth equivalent is given by:
Equation (2.13) P = nt=1ΣAt(1+i)-t
where A is the magnitude of an individual cash flow in the series.
Letting X = (1+i)-1 and bringing A outside the summation yields
P = A nΣt=1 Xt-1
Letting h = t-1
Equation (2.14) P = AX n-1Σh=o Xh
Since the summation in Equation 2.14 represent the first n terms of a geometric series, the closed form value for the summation is give by:
Equation (2.15) X n-1Σh=o = 1-xn/1-X
P = AC(1-Xn/1)
Replacing X with (1+i)-1 yields the following relationship between P and A
Equation (2.16) P = A[(1+i)n-1]]/[i(1+i)n]
more commonly expressed as
Equation (2.17) P = A(P|A ¡,n)
where P = A(P|A ¡,n) is referred to as the uniform series, present worth factor.
An individual wishes to deposit a single sum of money in a saving account so that five equal annual withdrawals of $2,000 can be made before depleting the fund. If the first withdrawal is to occur 1 year after the deposit and the fund pays interest at 12% compounded annually, how much should be deposited?
Future Value of an Annuity
How much must be deposited to withdraw $2,000 annually @ 12% for 5 yrs (PMT), this is an annuity.
| Deposit | $2,000 |
| =PV(0.12,5,-2000) | $7,210 |
| =2000*((((1+0.12)5)-1))/(0.12*((1+0.12)5)) | $7,210 |
As depicted in Figure 2.7, the value of P to be determined occurs at t = 0, whereas a straight-forward application of the P = A[(1+i)n-1/i(1+i)n] will yield a single sum equivalent to t =2.
Consequently, the value obtained at t = 2 must be moved backward in time to t = 0. The latter operating is easily performed using the P = A[(1+i)n-1/i(1+i)n]*F (1+i)-n factor.
Therefore:
| P=((((1+0.12)5)-1))/(0.12*((1+0.12)5)) | 3.6048 |
| P=(1+.12)-2 | 0.797 |
| P = (3.6048)*(0.7972) | 2.87 |
| P = $2,000 * (3.6048)*(0.7972) | $5,747 |
Deferring the first withdrawal for 2 years reduces the amount of the deposit by:
$7,209.60 - $5,747.49 = $1,462.11
The reciprocal relationship between P and A can be expressed as:
Equation (2.18) A = P[i(1+i)n/(1+i)n-1]
or as
Equation (2.19) P = P(A|P ¡,n)
The expression P(A|P ¡,n) is called the capital recovery factor for reasons that will become clear later. The P(A|P ¡,n) factor is used frequently in both personal financing and in comparing economic investment alternatives.
Suppose $10,000 is deposited into an account that pays interest at a rate of 15% annually. If 10 equal, annual withdrawals are made from the account, with the first withdrawal occurring 1 year after the deposit, how much can be withdrawn each year in order to deplete the fund with the last withdrawal?
Since we know that A and P are related by
A = P[i(1+i)n/(1+i)n-1]
then
=PMT(0.15,10,10000) = $1,993
=10000*(0.15*(1.15)10)/(((1.15)10)-1) = $1,993
Suppose that in the above example, the first withdrawal is delayed for 2 years, as shown in Figure 2.8. How much can be withdrawn each of the 10 years?
P=(1+.15)2 =10000*(1.15)2= $13,325
P=((((1+0.15)10)))/(((1+0.15)10)-1) =$13,225*((1.15)10)/(((1.15)10)-1) = $2,635
=PMT(0.15,10,-10000,13225) = $1,341
The future worth of a uniform series is obtained by recalling that:
Equation (2.20) F = P(1 + i)N
Substituting Equation 2.16 into Equation 2.20 for P and reducing yields
Equation (2.21) F = A[((1+i)N-1)/i]
or Equivalently
Equation (2.22) F = A(F|A ¡,n)
where (F|A ¡,n) is referred to as the uniform series, future worth factor.
If annual deposit of $1,000 are made into a saving account for 30 years, how much will be in the fund immediately after the last deposit if the fund pays interest at a rate of 8% compounded annually?
| =FV(0.08,30,-1000) | $113,283 |
| =1000*(((1+0.08)30)-1)/0.08 | $113,283 |
| A = F((1+i)n-1)/(i)) =FV(0.08,30,-1000) = $113,283 | $113,283 |
| =A((1+i)n-1)/(i)) =1000*(((1+0.08)30 )-1)/0.08 = $113,283 | $113,283 |
The reciprocal relationship between A and F is easily obtained from Equation 2.21. Specifically, we find that:
Equation (2.23) A = F[(i/(1+i)n-1)]
or
Equation (2.24) A = F(A|F ¡,n)
the expression (A|F ¡,n) is referred to as the sinking fund factor, since the factor is used to determine the size of a deposit one should place (sink) in a fund in order to accumulate a desired future amount. As depicted in Figure 2.9, F occurs at the same time as the last A. Thus, the last A or deposit earns no interest.
If $150,000 is to be accumulated in 35 years, how much must be deposited annually in a fund paying 8% compounded annually in order to accumulate the desired amount immediately after the last deposit?
A =PMT(0.08,35,0,-150000) = $870
A = F[(i/(1+i)n-1)/] = 150000*(0.08/(((1+0.08)35)-1)) = $870
Gradient Series of Cash Flows [viewable here in Excel]
A gradient series of cash flows occurs when the value of a given cash flow is greater than the value of the previous cash flow by a constant amount, G. Consider the series of cash flows depicted in Figure 2.10. The series can be represented by the sum of a uniform series and a gradient series. By convention, the gradient series is defined to have the first positive cash flow occur at the end of the second time period. The size of the cash flow in the gradient series occurring at the end of period t given by:
Equation (2.25) At = (t - 1)G t = 1, …, n)
The gradient series arises when the value of an individual cash flow differs from the preceding cash flow by a constant, G. As an illustration, if an individual receives an annual bonus and the size of the bonus increases by $100 each year, then the series is a gradient series. Also, operating and maintenance cost tend to increase over time because of both inflation effects and a gradual deterioration of equipment; such costs are often approximated by a gradient series.
The present worth equivalent of a gradient series is obtained by recalling
Equation (2.26) P = nΣt=1At(1+i)-t
Substituting Equation 2.25 into Equation 2.26 gives
Equation (2.27) P = nΣt=1(t-1)G(1+i)-t
or, equivalently,
Equation (2.28) P = G nΣt=1(t-1)(1+i)-t
As an exercise, you may wish to show that the summation reduces to:
Equation (2.29) P = G [1-(1 + ni)(1+i)-n / i2)
Equation (2.30) P = G [(P|G ¡,n) - n(P|F ¡,n)/ i)
or equivalently,
Equation (2.31) P = G(P|G ¡,n)
where (P|G ¡,n) is the gradient series, present worth factor.
A uniform series equivalent to the gradient series is obtained by multiplying the value of the gradient series present worth factor by the value of the (A|P ¡,n) factor to obtain:
Equation (2.32) P = G [((1 - n)/¡)] x F[(i/(1+i)n-1)]
or equivalently,
A = G (A|G ¡,n)
where the factor (A|G ¡,n) is referred to as the gradient-to-uniform series conversion factor.
To obtain the future worth equivalent of a gradient series at time n, multiply the value of the (A|G ¡,n) factor by the value of the (F|A ¡,n) factor.
It is not uncommon to encounter a cash flow series that is the sum of difference of a uniform series and a gradient series. To determine present worth and future worth equivalents of such a composite, one can deal with each special type of series separately.
Maintenance costs for a particular machine increases by $1,000/year over the 5-year life of the equipment. The initial maintenance cost is $3,000. Using an interest rate of 8% compounded annually, determine the present worth equivalent for the maintenance costs.
A cash flow diagram for this example is given in Figure 2.11. Note that the cash flow series consists of the sum of a uniform series of $3,000 and a gradient series of $1,000. Converting the uniform series to a present worth amount gives:
P = A[(1+i)n-1/i(1+i)n] = =PV(0.08,5,-3000) = $11,978
Converting the gradient series to a present worth amount gives:
| =1000*(1-(1+(0.08*5))*((1+0.08)-5))/(0.082) = $7,372 | $7,372 |
| =PV(0.08,5,-3000) = $11,978 | $11,978 |
| Hence, the present worth equivalent of the maintenance cost will be: $19,351 | $19,351 |
(Notice that n equals 5 even though only four positive cash flows are present in the gradient series.)
| Initial maintenance and annul cost | Maintenance cost discounted |
| $3,000 | $2,778 |
| $4,000 | $3,429 |
| $5,000 | $3,969 |
| $6,000 | $4,410 |
| $7,000 | $4,764 |
| $25,000 | $19,351 |
The uniform series equivalent of the cash flow series given in Figure 2.11 can be determined by summing the base series of $3,000 and the uniform series equivalent of the gradient series.
P = G [((1 - n)/¡)] x F[(i/(1+i)n-1)]
=(1/0.08)-(5/0.08)*(0.08/(((1+0.08)5)-1)) = 1.8465
= $3,000 x $1,000 x (1.8465) = $4847
Converting the maintenance costs to a future value amount can be performed either by converting the present worth amount to a future worth amount,
P = F(1+i)-n + $19,350.50 = $28,432
or by observing the uniform series amount to a future worth amount,
= G [((1 - n)/¡)] x [((1+i)n-1)/i]
=(1/0.08)-(5/0.08)*(0.08/(((1+0.08)5)-1)) = 1.8465
= $3,000 x $1,000(1.8465) = $4,847
F =(((1+0.08)5)-1)/0.08 = 5.8660
= $3,000 x $1,000(1.8465)*(((1+0.08)5)-1)/0.08 = $28,432
Five annual deposit are made into a money market account earning 8% compounded annually. The deposits are made sequentially in the following order: $800, $700, $600, $500 and $400. What is the account value after the fifth deposit (beginning of period)?
P = G [((1 - n)/¡)] x F[(i/(1+i)n-1)]
=(1/0.08)-(5/0.08)*(0.08/(((1+0.08)5)-1)) = 1.8465
= $100 x (1.8465) = $184.65
= $800 - $184.65 = $615.35
Therefore, a uniform series having cash flows equal to $800 - $184.65 = $615.35, is equivalent to the original cash flow series. The future worth equivalent is found to be:
| Fx[((1+i)n-1)/i] | ||
| =F(((1+0.08)5)-1)/0.08 = 5.86660 | 5.86660 | |
| = 5.8666 x 184.65 = $3,610.01 | $3,610.01 | |
| Deposits | Deposit at the end of the year | Deposit at the beginning of the year |
| $800 | $1,088 | $1,175 |
| $700 | $882 | $952 |
| $600 | $700 | $756 |
| $500 | $540 | $583 |
| $400 | $400 | $432 |
| $3,000 | $3,610 | $3,899 |
Geometric Series of Cash Flows
The geometric cash flow series, as depicted in Figure 2.13, occurs when the size of the cash flow increases (decreases) by a fixed percent from one time period to the next. If j denotes the percent change in the size of a cash flow from one period to the next, the size of the tth flow can be given by:
At = At-1(1 + j) t = 2, …..,n
or more conveniently,
Equation (2.33) At = A1(1 + j)t-1 t = 1, …..,n
The geometric series is used to represent the growth (positive j) or decay (negative j) of costs and revenues undergoing annual percentage changes. As an illustration, if labor costs increase by 10% a year, then the resulting series representation of labor costs will be geometric series.
The present worth equivalent of the cash flow series is obtained by substituting Equation 2.33 into Equation 2.6 to obtain:
Equation (2.34) P = nΣt=1A1(1+j)t-1 x (1 + i)-t
or
Equation (2.35) P = A1(1+j)-1 nΣt=1 [(1+j)/(1 + i)]t)
or
Equation (2.37) P = A1(P|A ¡,j,n)
where A1(P|A ¡,j,n) is the geometric series, present worth factor.
For the case of j ≥ 0 and ¡ ≠ j, the relationship between P and A can be conveniently expressed in terms of compound interest factors previously considered:
Equation (2.38) P = A1[1 - (F|P j,n)(P|F ¡,n)/ i-j] j ≥ 0 and ¡ ≠ j
Your company is looking to buy a new air conditioning system. The air conditioning maintenance cost is $1,000 per year (end of year) increasing 8% annually for 15 yrs. You need to setup a reserve fund to cover maintenance for 15 years, the reserve account will earn 10% per year compounded annually. How much must be deposited today to cover the 15 years of maintenance costs?
=A*(1-((1+in))*((1+j-n)))/(j-i)
j = .10, l = .08, n = 15, a = 1,000
a=1000
=1000*(1-((1.0815))*((1.1-15)))/(0.1-0.08) = $12,030
The future worth equivalent of the geometric series is obtained by multiplying the value of the geometric series present worth factor and the (F\P ¡,n) factor:
Equation (2.39) F =A*(((1+i)n))-((1+j)n)))/(i-j) ¡ ≠ j
Equation (2.40) nA1(1+i)n-1 j ≥ 0
where (F|A ¡,n) is the geometric series, future worth factor).
An individual receives an annual bonus and deposits it in a money market account that pays 8% compounding annually. The size of the bonus increases 10% per year; the initial deposit was $500. What is the value after the tenth deposit?
=A*(((1+in))-((1+jn)))/(i-j)
j = .1, l = .08, n = 10, a = 500
=500*(((1.0810))-((1.110)))/(0.08-0.1) = $10,870
=500*(((EXP(0.08)10))-(EXP((0.1)10)))/EXP(0.1-0.08) = $517
=10870+517 = $11,387
Multiple Compounding Periods in a Year [viewable here in Excel]
Not all interest rates are stipulated as annul compounding rates. In the business worlds it is common practice to express a nonannual interest rate as follows:
12% per year compounded monthly or 12% per year per month
When expressed in this form, 12% per year per month is known as the nominal annual interest rate, r1. Unfortunately, the techniques we have used up to now cannot be used directly to solve an economic analysis problem of this type because the interest period (per year) and com-pounding period (monthly) are not the same. To solve this type of problem, one of two approaches must be employed; the period interest rate approach or the effective interest rate approach.
To utilize the period interest rate approach, we must define a new term, the period interest rate:
Period interest rate = Nominal annual interest rate / Number of interest periods per year
Using the pervious example:
Period interest rate = 12% per year per month / 12 month per year = 1% per month/month
Since the interest period and the compounding period are now the same (monthly).
$2,000 is invested in an account which pays 12% per year compounded monthly. What is the balance in the account after 3 years?
Nominal annual interest rate = 12%/year/month
Period interest rate - (12%/year/month)/(12 month/year = 1%/month/month
Number of interest periods = 3 years x 12 month/year = 36 interest periods (month)
Deposit: $2,000
F = P(F|P, ¡,n) = $2,000(F|P,1,36) = $2,000 * (1 + .12/36))36 = $2254.54
=(1 + 0.12/12)36 = 1.431
=$2,000 x (1 + 0.12/12)36 = $2,862
In a similar manner:
$2,000 deposited into a money market account paying 12% per year compounded semi-annually.
Nominal annual interest rate = 12%/year/6-month
Period interest rate - (12%/year/6-month)/(2 6-month/year) = 6%/6-month/6-month
Number of interest periods = 3 years x 2 6-month/year = 6 interest periods (month)
What is the balance after 3 years?
Deposit: $2,000
F = P(F|P, ¡,n) = $2,000(F|P,6,6) = $2,000 * (1 + .12/6))6 = $2,837
=FV(0.12/2,6,,-2000) = $2,837
=((0.12/2+1)3) = 1.42
Balance after 3 years =1.42*$2,000 = $2,837
$2,000 deposited into a money market account paying 12% per year compounded quarterly.
What is the balance after 3 years?
Nominal annual interest rate = 12%/year/quarter
Period interest rate - (12%/year/quarter)/(4quarter/year = 3%/quarter/quarter
Number of interest periods = 3 years x 4 quarter/year = 12 interest periods (month)
What is the balance after 3 years?
F = P(F|P, ¡,n) = $2,000(F|P,3,12) = $2,000 * (1 + .12/3))12 = $2,852
=(((0.12/4)+1)4)3 = 1.43
=1.43*$2,000 = $2,852
=FV(0.12/4,12,,-2000) = $2,852
Balance after 3 years =(((0.12/4)+1)4)3 * 2000 = $2,852
$2,000 deposited into a money market account paying 12% per year compounded daily.
What is the balance after 3 years?
Nominal annual interest rate = 12%/year/daily
Period interest rate - (12%/year/day)/(365 days/year = 0.03293%/day/day
Number of interest periods = 3 years x 365 days/year = 1095 interest periods (month)
What is the balance after 3 years?
Deposit: $2,000
F = P(F|P, ¡,n) = $2,000(F|P,.0329,1095) = $2,000 * (1 + .12/3))12 = $2,866
=(((0.12/365)+1)365)3 = 1.43
=2,000*1.43 = $2,866
=FV(0.12/365,365*3,,-2000) = $2,866
Balance after 3 years =(((0.12/365)+1)365)3 * 2000 = $2,866
The Cost of a Cup of Coffee
You have a habit of drinking a cup of Starbuck's coffee ($2.00 a cup) on the way to work every morning for 30 years. If you put the money in the bank for the same period, how much would you have, assuming your accounts earns 5% interest compounded daily.
Rate (daily) =0.05/365 = .00127
Periods (number of days) =365*30 = $10,950
Daily accrual = -$2.00
Amount saved =FV(0.000137,10950,-2) = $50,831
Mutual Fund Growth
Suppose you make equal quarterly deposits of $1,000 into a fund that pays interest at 12% compounded monthly. What is the balance at the end of year 3?
Deposits = $1,000
Number of periods = 12
Monthly compounded interest rate paid quarterly (keep in mind, 3 = 3 months = 1 quarter =((1+0.01)3)-1 = 3.03%
Balance =FV(0.0303,12,1000)*-1 = $14,216
Calculate the effective annual interest rate for the following case: 12% compounded monthly (i = ([1 + r/CK]C)-1):
Effective interest rate per payment period i = [1 + r/CK]C-1
C = number of interest periods per payment period
K = number of payment periods per year
CK = total number of interest periods per year, or M
r / K = nominal interest rate per payment period
Effective Interest Rate Per Quarter
| i=(1+0.01)3-1 = 3.03% [Why is this cubed? Because 12% annually is 3% quarterly.] | |
| Effective annual interest rate | |
| i=((1+0.01)12)-1 = 12.68% | 12.68% |
| i=((1+0.03030)4)-1 = 12.68% | 12.68% |
Suppose your savings account pays 9% interest compounded quarterly. If you deposit $10,000 for one year, how much would you have at the end of the year?
Excel "Effective Rate" calculation =EFFECT(0.09,4) = 9.31%
Excel "Nominal Rate" calculation
a) Interest rate per quarter: 9%/4=2.25
b) Annual effective interest rate: =NOMINAL(.0931,4) = 9.0%
i(annual)=(1+0.0225)4-1=9.31%
c) Balance at the end of one year (after 4 quarters)
F=$10,000(F/P,2.25%,4) =FV(0.0225,4,,-10000) = $10,931
You need $1,800 per month for twenty years on a $250,000 investment, what rate do you need to earn this? 6.06%
| $1,800 | =PMT(0.060618/12,20*12,-250000) | $7,000 |
| 6.0618% | =RATE(12*20,-1800,250000)*12 | 33.5551% |
| $1,165,239 | =FV(0.08,20,,-250000) | $371,487 |
| $462,041 | =FV(0.06/12,20*12,-1000) | $462,041 |
Loan Amortization
Many loans, such as home mortgage loans and car loans, are repaid in equal periodic installments. Part of each payment is interest on the outstanding balance of the loan and part is repayment of principal. After each payment, the outstanding balance is reduced by the amount of principal repaid. Therefore, the portion of the payment that goes toward the payment of interest is lower than the previous period's interest payment, and the portion going toward repayment of principal is greater than the previous period's.
For example, let's assume you take out a $12,500 car loan at an interest rate of 6% per year to be repaid in 60 monthly installments. First we calculate the monthly payments by finding the PMT that has a PV of $12,500 when discounted at 6% for 5 years or 60 payments.
Computing Car Payments [viewable here in Excel]
What is the monthly car payment on a 5-year loan of $12,500 at 6% compounded monthly? Answer: $241.66
| Nominal annual interest rate = 6%/year/month | |
| Period interest rate - (6%/year/month)/(12 months/year = 0.50%/month/month | |
| Number of interest periods = 5 years x 365 days/year = 1,825 interest periods (month) | |
| A = P(A|P, ¡,n) = $12,500(A|P,.50,60) = $12,500 * (1 + .06/2)30 | |
| =(12500*(((0.06/12)*((1+(0.06/12))60))/(((1+(0.06/12))60)-1))) = $241.66 | $241.66 |
| =PMT(0.06/12,60,-12500,0) = $241.66 | $241.66 |
| Total Interest paid =14499.6-12500 = $1,999.60 | $1,999.60 |
| Total amount repaid =241.66*60 = $14,499.60 | $14,499.60 |
So the monthly payment is $241.66. In the first year, how much of the 12,500 - 10,290 = $2,210.04 is interest. You can compute this using Excel or by building your own amortization schedule. The interest payment is $698.88 with an outstanding balance of $10,289.96 (the amortization).
| Monthly car payment | -$689.88 | =CUMIPMT(0.08/12,36,20000,6,12,0) |
| Monthly car payment | 689.88 | See Chart Below |
| Interest Payment (month 12) | -$52.40 | =IPMT(0.06/12,12,60,12500) = -$52.40 |
| Principal Payment | -$189.26 | =PPMT(0.08/12,36,36,20000) = -$189.26 |
| Monthly car payment | -$241.66 | =-$189.26 -$52.40 = $241.66 |
| Cumulative Interest Payments (months 1-12) = 12 months of payments | -$689.88 | =CUMIPMT(0.08/12,36,20000,6,12,0) |
| Cumulative Principal Payments (months 1-12) = 12 months of payments | -$2,210.04 | =CUMPRINC(0.08/12,36,20000,6,12,0) |
| Cumulative car payment after year 1 | -$2,899.92 |
Buying a New Car
You buy a new car and borrow $35,000 for 5 years at 6% per year compounded monthly. What is the monthly payment?
=PMT(0.06/12,60,-35000,0) = $676.65
P=(35000*(((0.06/12)*((1+(0.06/12))60))/(((1+(0.06/12))60)-1))) = $676.65
You have a 3-year, $20,000 car loan with monthly payments at 8%, what is the interest and principal payment for the final payment (compounded monthly)?
| Monthly Payment | $626.73 | =(20000*(((0.08/12)*((1+(0.08/12))36))/(((1+(0.08/12))36)-1))) |
| Interest Payment | -$4.15 | =IPMT(0.08/12,36,36,20000) |
| Principal Payment | -$622.58 | =PPMT(0.08/12,36,36,20000) |
| Cumulative Interest Payments (months 6-12) = 7 months of payments | -$744.46 | =CUMIPMT(0.08/12,36,20000,6,12,0) |
| Cumulative Principal Payments (months 6-12) = 7 months of payments | -$3,642.64 | =CUMPRINC(0.08/12,36,20000,6,12,0) |
| $626.73 | =PMT(0.08/12,36,-20000) |
| Car Payment Amortization Schedule |
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| Time (in months) | Loan Amount | Monthly Payment | Monthly interest Factor | Interest | Principal | Outstanding Balance |
| 1 | 12500 | $241.66 | 0.005 | 62.50 | 179.16 | 12,320.84 |
| 2 | 12,321 | $241.66 | 0.005 | 61.60 | 180.06 | 12,140.78 |
| 3 | 12,141 | $241.66 | 0.005 | 60.70 | 180.96 | 11,959.83 |
| 4 | 11,960 | $241.66 | 0.005 | 59.80 | 181.86 | 11,777.97 |
| 5 | 11,778 | $241.66 | 0.005 | 58.89 | 182.77 | 11,595.20 |
| 6 | 11,595 | $241.66 | 0.005 | 57.98 | 183.68 | 11,411.51 |
| 7 | 11,412 | $241.66 | 0.005 | 57.06 | 184.60 | 11,226.91 |
| 8 | 11,227 | $241.66 | 0.005 | 56.13 | 185.53 | 11,041.39 |
| 9 | 11,041 | $241.66 | 0.005 | 55.21 | 186.45 | 10,854.93 |
| 10 | 10,855 | $241.66 | 0.005 | 54.27 | 187.39 | 10,667.55 |
| 11 | 10,668 | $241.66 | 0.005 | 53.34 | 188.32 | 10,479.22 |
| 12 | 10,479 | $241.66 | 0.005 | 52.40 | 189.26 | 10,289.96 |
| 13 | 10,290 | $241.66 | 0.005 | 51.45 | 190.21 | 10,099.75 |
| 14 | 10,100 | $241.66 | 0.005 | 50.50 | 191.16 | 9,908.59 |
| 15 | 9,909 | $241.66 | 0.005 | 49.54 | 192.12 | 9,716.47 |
| 16 | 9,716 | $241.66 | 0.005 | 48.58 | 193.08 | 9,523.39 |
| 17 | 9,523 | $241.66 | 0.005 | 47.62 | 194.04 | 9,329.35 |
| 18 | 9,329 | $241.66 | 0.005 | 46.65 | 195.01 | 9,134.34 |
| 19 | 9,134 | $241.66 | 0.005 | 45.67 | 195.99 | 8,938.35 |
| 20 | 8,938 | $241.66 | 0.005 | 44.69 | 196.97 | 8,741.38 |
| 21 | 8,741 | $241.66 | 0.005 | 43.71 | 197.95 | 8,543.43 |
| 22 | 8,543 | $241.66 | 0.005 | 42.72 | 198.94 | 8,344.49 |
| 23 | 8,344 | $241.66 | 0.005 | 41.72 | 199.94 | 8,144.55 |
| 24 | 8,145 | $241.66 | 0.005 | 40.72 | 200.94 | 7,943.61 |
| 25 | 7,944 | $241.66 | 0.005 | 39.72 | 201.94 | 7,741.67 |
| 26 | 7,742 | $241.66 | 0.005 | 38.71 | 202.95 | 7,538.72 |
| 27 | 7,539 | $241.66 | 0.005 | 37.69 | 203.97 | 7,334.75 |
| 28 | 7,335 | $241.66 | 0.005 | 36.67 | 204.99 | 7,129.76 |
| 29 | 7,130 | $241.66 | 0.005 | 35.65 | 206.01 | 6,923.75 |
| 30 | 6,924 | $241.66 | 0.005 | 34.62 | 207.04 | 6,716.71 |
| 31 | 6,717 | $241.66 | 0.005 | 33.58 | 208.08 | 6,508.64 |
| 32 | 6,509 | $241.66 | 0.005 | 32.54 | 209.12 | 6,299.52 |
| 33 | 6,300 | $241.66 | 0.005 | 31.50 | 210.16 | 6,089.36 |
| 34 | 6,089 | $241.66 | 0.005 | 30.45 | 211.21 | 5,878.14 |
| 35 | 5,878 | $241.66 | 0.005 | 29.39 | 212.27 | 5,665.87 |
| 36 | 5,666 | $241.66 | 0.005 | 28.33 | 213.33 | 5,452.54 |
| 37 | 5,453 | $241.66 | 0.005 | 27.26 | 214.40 | 5,238.15 |
| 38 | 5,238 | $241.66 | 0.005 | 26.19 | 215.47 | 5,022.68 |
| 39 | 5,023 | $241.66 | 0.005 | 25.11 | 216.55 | 4,806.13 |
| 40 | 4,806 | $241.66 | 0.005 | 24.03 | 217.63 | 4,588.50 |
| 41 | 4,589 | $241.66 | 0.005 | 22.94 | 218.72 | 4,369.78 |
| 42 | 4,370 | $241.66 | 0.005 | 21.85 | 219.81 | 4,149.97 |
| 43 | 4,150 | $241.66 | 0.005 | 20.75 | 220.91 | 3,929.06 |
| 44 | 3,929 | $241.66 | 0.005 | 19.65 | 222.01 | 3,707.05 |
| 45 | 3,707 | $241.66 | 0.005 | 18.54 | 223.12 | 3,483.92 |
| 46 | 3,484 | $241.66 | 0.005 | 17.42 | 224.24 | 3,259.68 |
| 47 | 3,260 | $241.66 | 0.005 | 16.30 | 225.36 | 3,034.32 |
| 48 | 3,034 | $241.66 | 0.005 | 15.17 | 226.49 | 2,807.83 |
| 49 | 2,808 | $241.66 | 0.005 | 14.04 | 227.62 | 2,580.21 |
| 50 | 2,580 | $241.66 | 0.005 | 12.90 | 228.76 | 2,351.45 |
| 51 | 2,351 | $241.66 | 0.005 | 11.76 | 229.90 | 2,121.55 |
| 52 | 2,122 | $241.66 | 0.005 | 10.61 | 231.05 | 1,890.50 |
| 53 | 1,890 | $241.66 | 0.005 | 9.45 | 232.21 | 1,658.29 |
| 54 | 1,658 | $241.66 | 0.005 | 8.29 | 233.37 | 1,424.92 |
| 55 | 1,425 | $241.66 | 0.005 | 7.12 | 234.54 | 1,190.38 |
| 56 | 1,190 | $241.66 | 0.005 | 5.95 | 235.71 | 954.68 |
| 57 | 955 | $241.66 | 0.005 | 4.77 | 236.89 | 717.79 |
| 58 | 718 | $241.66 | 0.005 | 3.59 | 238.07 | 479.72 |
| 59 | 480 | $241.66 | 0.005 | 2.40 | 239.26 | 240.46 |
| 60 | 240 | $241.66 | 0.005 | 1.20 | 240.46 | (0.00) |
| $14,499.60 | $1,999.60 | $12,500.00 | ||||
Effective Annual Interest Rate [viewable here in Excel]
The second approach to solving problems where the compounding is not annual is the effective interest rate approach. The effective annual interest rate is the annual interest rate that would be equivalent to the period interest rate as previously calculated.
For example: If the nominal interest rate is 12%/year/quarter, then period interest rate is 3%/quarter/quarter. One dollar invested for 1 year at 3%/quarter/quarter would have a future worth of:
F = P(F|P, ¡,n) = $1(F|P, 3,4) = $1 (1.03)4 = $1 (1.1255) = $1.1255
To get this same value in one year would require an annual interest rate of 12.55%/year/year.
This value is called the effective annual interest rate. The effective annual interest rate is give by (1.03)4 -1 = 0.1255 or 12.55%.
Equation (2.42) Effective annual interest rate = ieff = (1+r/m)m -1
where r = nominal annual interest rate
m = number of interest periods per year
Calculate the effective annual interest rate for each of the following cases:
12% effective interest compounded semi-annually [=EFFECT(0.12,2)] = 12.36%
nominal annual interest rate = 12%/year/6-month [=EFFECT(0.12,2)] = 12.35%
period interest rate = 6%/6-month/6-month
effective annual interest rate = (1 + .012/2)2 - 1 = .1236 or 12.36%
12% effective interest compounded monthly [=EFFECT(0.12,12)] = 12.68%
nominal annual interest rate = 12%/year/month [ =(0.12/12+1)12 -1] = 12.68%
period interest rate = 1%/month/month
effective annual interest rate = (1 + .012/12)12 - 1 = .1268 or 12.68%
12% effective interest compounded daily
nominal annual interest rate = 12%/year/day [ =EFFECT(0.12,365)] = 12.75%
period interest rate = 1%/month/day [=(0.12/365+1)365-1] = 12.75%
effective annual interest rate = (1 + .012/12)365 -1 = .1275 or 12.75%
18% effective interest compounded daily
=EFFECT(0.18,365) = 19.72%
=(0.18/365+1)365-1 = 19.72%
You borrow $1,000 and paid off the loan with interest after 4.5 years. The amount paid was $1,500. What was the effective annual interest rate for this transaction? Letting the interest period be a 6-month period, it is seen that the payment of $1,500 and the debt of $1,000 are related by the expression:
F = P(1+i)n
F = (F|P ¡,n)
Number of 6-month periods = 4.5*2 = 9
Thus,
$1,500 = $1,000 (F|P ¡,n)
or
$1,500 = $1,000 (1 + i)9
IRR: $1000 = $1500/(1+i)9
IRR: $1000 = $1500/(1+i)4.5
I = ((1500/1000)1/4.5)-1 = 9.43%
Dividing both sides by $1,000 gives
1.50 = (1 + i)9
Taking the logarithm of both sides
log 1.50 = 9 log(1 + i)
=log (1.5) = 17.61%
=0.1761/9 = 1.96%
=101.96%-1 = 4.62%
=(1+4.62%)2-1 = 9.45%
Dividing both sides by 9 and taking the antilog (10^) of the result provides the relation annual interest rate yields:
ieff = (1 + i)2 - 1
= (1 + 0.046)2 -1 = .0943
The effective annual interest rate for the loan transaction was approximately 9.43%.
Modern Approach
Loan =$1,000
Periods = 9
Total principal and interest paid = $1,500
6 month interest rate =RATE(F120,,-1000,1500) = 4.61%
Annual effective interest rate =((1+0.0461)2)-1 = 9.43%
6 month interest rate =RATE(9,,-1000,1500) = 4.61%
6 month interest rate =(1500/1000)(1/9)-1 = 4.61%
Annual effective interest rate =(1500/1000)(1/4.5)-1 = 9.43%
The effective annual interest rate for the loan transaction was approximately 9.43%.
Effective Annual Interest Rate II [viewable here in Excel]
Credit Card Interest Rate
In the United States, the federal government requires that anyone borrowing money be informed of the total amount of interest paid and the annual percentage rate, the percentage quoted can be either the nominal annual interest rate, the effective annual interest rate, or something else entirely. To illustrate, many revolving charge accounts charge an interest rate of 1.5%/month, which is equivalent to 18% compounded monthly or 19.56% compounded annually. In many cases, the annual percentage rate is stated to be 18%.
Since, in personal financing, the annual percentage rate might be computed differently among the alternative sources of funds, the effective annual interest rate is a very good basis for comparing alternative financing plans.
Continuous Compounding
You should note that as the frequency of compounding in a year increases, the effective interest rate increases. Since business transactions occur daily or hourly in most businesses, and money is normally "put to work" for the business as soon as it is received, compounding is occurring quite frequently. If one wishes to account explicitly for such rapid compounding, then continuous compounding relations should be used. Continuous compounding means that each year is divided into an infinite number of interest periods. Mathematically, the single payment compound amount factor under continuous compounding is given by:
limm-∞(1 + r/m)mn = exprn
where n is the number of years, m is the number of interest periods per year, and r is the nominal annual interest rate. Given P, r, and n, the value of F can be computed using continuous compounding as follows:
Equation (2.43) F = P exprn
or
Equation (2.44) F = P(F|P r,n)∞
where (F|P r,n)∞ denotes the continuous compounding, single sum, future growth factor. The subscript ∞ is provided to denote that continuous compounding is being used.
Credit Card Interest: Monthly Compounding
| Months | Balance | 18% compounded monthly | FV |
| 1 | $1.000 | 1.015 | $1.015 |
| 2 | $1.015 | 1.015 | $1.030 |
| 3 | $1.030 | 1.015 | $1.046 |
| 4 | $1.046 | 1.015 | $1.061 |
| 5 | $1.061 | 1.015 | $1.077 |
| 6 | $1.077 | 1.015 | $1.093 |
| 7 | $1.093 | 1.015 | $1.110 |
| 8 | $1.110 | 1.015 | $1.126 |
| 9 | $1.126 | 1.015 | $1.143 |
| 10 | $1.143 | 1.015 | $1.161 |
| 11 | $1.161 | 1.015 | $1.178 |
| 12 | $1.178 | 1.015 | $1.196 |
If $2,000 is invested in a mutual fund earning 12% with continuous compounding, what is the total valve after year 5∞
Investment = $2,000
Continuous Compounding = (EXP(0.12)5)*2000 = $3,644
Continuous Compounding =2000*(EXP(1)(5*0.12)) = $3,644
Growth factor =(EXP(0.12)5) = 1.82
Continuous Compounding =1.82*2000 = $3,644
Thus, a withdrawal of $3,644.20 will deplete the fund after 5 years
Discrete Cash Flows [viewable here in Excel]
If it is assumed that cash flows are discretely spaced over time, then the continuous compounding relations for the uniform, gradient, and geometric series can be obtained. Substituting exp-rn for (1 + i)-n expr -1 for ¡, and exprn for (1+i)n in the remaining discrete compounding formula yields the continuous compounding interest factors summarized in Table 2.4.
Table 2.4 Summary of Continuous Compounding and Discrete Interest Factors
| To Find | Given | Factor (Continuous) | Symbol (Continuous) | Factor (Discrete) | Symbol (Discrete) | |
| P | F | e-rn | (P|F r,n)8 | (1+ i)-n | (P|F i,n)8 | Single payment, present value |
| F | P | ern | (F|P r,n)8 | (1+i)n | (F|P i,n)8 | Single payment, future value |
| F | A | e-rn-1/er-1 | (F|A r,n)8 | (1+i)n-1))/i | (F|A i,n)8 | Uniform series, compound amount |
| A | F | er-1/ern-1 | (A|F r,n)8 | i/((1+i)n-1) | (A|F i,n)8 | Uniform series, sinking fund |
| P | A | ern-1/ern(er-1) | (P|A r,n)8 | ((1+i)n-1)/(i(1+i)n) | (P|A i,n)8 | Uniform series, present value |
| A | P | ern(er-1)/ern-1 | (A|P r,n)8 | (i(1+i)n)/((1+i)n-1) | (A|P i,n)8 | Uniform series, capital recovery factor |
| P | G | ern-1-n(er-1)/ern(e r-1)2 | (P|G r,n)8 | ((1-(1+ni)(1+i)-n))/i2 | (P|G i,n)8 | Gradient series, present value |
| A | G | 1/er-1-(n/(ern-1)) | (A|G r,n)8 | ((1+i)n-(1+ni))/(i((1+i)n-1)) | (A|G i,n)8 | Gradient series, uniform series |
| P | A1, c | 1-e(c-r)n/er-ec | (P|A1 r,c,n)8 | (1-(1+j)n(1+i)-n)/(i-j) | (P|A1 i,j,n)8 | Geometric series, present value factor |
| F | A1, c | ern-ecn/er-ec | (F|A1 r,c,n)8 | ((1+i)n-(1+j)n)/(i-j) | (F|A1 i,j,n)8 | Geometric series, future value factor |
Table 2.5 Summary of Continuous Compounding Interest Factors for Continuous Flows
| P | A | ern-1/rern | (P|A r,n)8 |
| A | P | rern/ern-1 | (A|P r,n)8 |
| F | A | ern-1/r | (F|A r,n)8 |
| A | F | r/ern-1 | (A|F r,n)8 |
To illustrate the use of the continuous compounding interest factors, suppose $1,000 is deposited each year into an account that pays interest at a rate of 12% compounded continuously. Determine both the amount in the account immediately after the tenth deposit and the present worth equivalent for 10 deposits.
| $1,000 deposited each month for 10 years at 12% continuous compounding | 3.32 | =(EXP(0.12)10) |
| PV of the $1,000 | $1.00 | =PV(D113,10,,-D132) |
| 20% effective interest compounded daily | 22.13% | =EFFECT(0.2,365) |
| 20% nominal interest compounded daily | 18.24% | =NOMINAL(0.2,365) |
You invest $1,000 every year for the next 10 years at a 12% continuous rate. What is the end value?
| Deposit | Exp Annual 12% | Value | Discounted | Exp Annual 12% |
| $1,000 | 1.0000 | $1,000 | $893 | 1.0000 |
| $1,000 | 1.1275 | $1,127 | $887 | 1.0101 |
| $1,000 | 1.2712 | $1,271 | $787 | 1.2712 |
| $1,000 | 1.4333 | $1,433 | $698 | 1.4333 |
| $1,000 | 1.6161 | $1,616 | $619 | 1.6161 |
| $1,000 | 1.8221 | $1,822 | $549 | 1.8221 |
| $1,000 | 2.0544 | $2,054 | $487 | 2.0544 |
| $1,000 | 2.3164 | $2,316 | $432 | 2.3164 |
| $1,000 | 2.6117 | $2,612 | $383 | 2.6117 |
| $1,000 | 2.9447 | $2,945 | $340 | 2.9447 |
| $18,197 | $6,073 | |||
| $1,000 | 18.1974 | $18,197 | =((EXP(1)(0.12*10))-1)/(((EXP(1)0.12))-1) | |
| $1,000 | 5.4810 | $5,481 | =((EXP(1)(0.12*10))-1)/(((EXP(1)(0.12*10)))*((EXP(1).12-1)) |
An individual receives an annual bonus and deposits it in a savings account that pays 8% compounded continuously. The size of the bonus increases each year at a rate of 10% compounded continuously; the initial deposit was $500. Determine how much will be in the fund immediately after the tenth deposit.
In this case, A1 = $500, r = 8%, c = 10%, and n = 10.| Yr | Exp | Bonus Increase | Interest Rate | Value |
| 0 | 1.00 | $500 | 1.0000 | $500 |
| 1 | 1.11 | $553 | 1.0833 | $599 |
| 2 | 1.22 | $611 | 1.1735 | $717 |
| 3 | 1.35 | $675 | 1.2712 | $858 |
| 4 | 1.49 | $746 | 1.3771 | $1,027 |
| 5 | 1.65 | $824 | 1.4918 | $1,230 |
| 6 | 1.82 | $911 | 1.6161 | $1,472 |
| 7 | 2.01 | $1,007 | 1.7507 | $1,763 |
| 8 | 2.23 | $1,113 | 1.8965 | $2,110 |
| 9 | 2.46 | $1,230 | 2.0544 | $2,527 |
| $11,505 |
Continuous Cash Flows [viewable here in Excel]
Thus far only discrete cash flows have been assumed. It was assumed that cash flows occurred at, say, the end of the year. In some cases money is expected throughout the year on a somewhat uniform basis. Costs of labor, carrying inventory, and operating and maintaining equipment are typical examples. Others included capital improvement projects that conserve energy water or process steam.
Consequently, as a mathematical convenience, instead of assuming that money flows in discrete increments at the end of monthly, weekly, daily, or hourly time periods, it is assumed that money flows continuously during the time period at a uniform rate. Instead of having a uniform series of discrete cash flows of magnitude A, it is assumed that a total of ? dollars flows uniformly and continuously throughout a given time period. Such an approach to modeling cash flows is referred to as the continuous flow approach.
To illustrate the continuous flow concept, suppose you are able to divide $1,000 into k equal amounts to be deposited at equally spaced points in time during a year. The interest rate per period is defined to be r/k, where r is the nominal rate. Thus, the present worth of the series of k equal amounts is:
P = $1,000/k(P|A r/k,k)
or
P = $1,000/k[(1 + (r/k))k-1/(r/k)(1 + (r/k))k
which reduces to
P = $1,000k[1/r - 1/(r(1 + (r/k))k)
taking the limit of P as k approaches infinity gives
k-1lim P = $1,000(1/r - 1/rer)
or
P = $1,000(expr - 1/rer)
In general, for n years
Equation (2.51) P = Ā(exprn - 1 /rern)
or
Equation (2.52) P = Ā(P|Ā r,n)
where (P|Ā r,n) is referred to as the continuous compounding uniform series present worth factor.
With continuous flow and continuous compounding, the continuous annual cash flow, Ā, is equivalent to Ar/(expr - 1), when discrete flow and continuous compounding is used. Hence, the discrete flow equivalent of Ā is given by:
A = Ā(expn - 1/r)
or
A = Ā(F|Ā r,1)
What is the present value of $10,000 with annual contributions per year for 10 years earning 20% continuous compounding?
$10,000
=((EXP(1)(0.2*10))-1)/(0.2*(EXP(1)(0.2*10))) = 4.32
= $10,000 X 4.32 = $43,233
What is the future value of $10,000 with annual contributions for 10 years earning 20% continuous compounding?
| What is the future value of $10,000 with annual contributions for 10 years earning 20% continuous compounding? | $10,000 | 31.95 | $319,453 | =((EXP(1)(10*0.2))-1)/0.2 = 31.95 |
| FV Exp() | FV | Year | FV Exp() | |
| $10,000 | 1.11 | $11,070 | 1 | =((EXP(1*0.2))-1)/0.2 |
| $10,000 | 2.46 | $24,591 | 2 | =((EXP(2*0.2))-1)/0.2 |
| $10,000 | 4.11 | $41,106 | 3 | =((EXP(3*0.2))-1)/0.2 |
| $10,000 | 6.13 | $61,277 | 4 | =((EXP(4*0.2))-1)/0.2 |
| $10,000 | 8.59 | $85,914 | 5 | =((EXP(5*0.2))-1)/0.2 |
| $10,000 | 11.60 | $116,006 | 6 | =((EXP(6*0.2))-1)/0.2 |
| $10,000 | 15.28 | $152,760 | 7 | =((EXP(7*0.2))-1)/0.2 |
| $10,000 | 19.77 | $197,652 | 8 | =((EXP(8*0.2))-1)/0.2 |
| $10,000 | 25.25 | $252,482 | 9 | =((EXP(9*0.2))-1)/0.2 |
| $10,000 | 31.95 | $319,453 | 10 | =((EXP(10*0.2))-1)/0.2 |
| 31.95 | $319,453 | =31.95 |
Equivalence and Indifference [viewable here in Excel]
Throughout the preceeding sections, we have used the term equivalence without defining precisely what the term meant. This was done intentionally in order to introduce subtly the notion that two cash flow series of profiles are equivalent at some specific interest rate k%, if their present values are equal using an interest rate of k%. We will now define equivalence (and a related concept, indifference) more formally and discuss their importance in finance.
Equivalence
In finance, "equivalence" means "the state of being equal in value." The concept is primarily applied in the comparison of two or more cash flow profiles. Specifically, two (or more) cash flow profiles are equivalent if their time-value-of-money (present values) at a common point in time are equal.
Question: Are the following two cash flows equivalent at 15%/yr?
Cash flow 1: Receive $1,322.50 two years from today
Cash flow 2: Receive $1,00.00 today
Analysis Approach 1: Compare worth's at t = 0 (present value)
Present Value (1) = $1,322.50 x (1.15)-2 = $1,000.00
Present Value (2) = $1,000.00
Answer: Cash flow 1 and cash flow 2 are equivalent
Analysis Approach 1: Compare worth's at t = 0 (future value)
Future Value (1) = $1,000.00
Future Value (2) = $1,000.00 x (1.15)2 = $1,322.50
Answer: Cash flow 1 and cash flow 2 are equivalent
Analysis Approach 1: Compare worth's at t = 1 (future and present value)
Present Value (1) = $1,322.50 x (1.15)-1 = $1,150.00
Future Value (2) = $1,000.00 x (1.15)1 = $1,150.00
Answer: Cash flow 1 and cash flow 2 are equivalent
Analysis Approach 1: Compare worth's at t = 4 (future and present value)
Present Value6 (1) = $1,322.50 x (1.15)-4 = $2,313.00
Future Value6 (2) = $1,000.00 x (1.15)6 = $2,313.00
Answer: Cash flow 1 and cash flow 2 are equivalent
Note that the selection of the point in time, t, at which to make the comparison is completely arbitrary. Clearly, however, some choices are more intuitively appealing than others (t = 0 and t = 2 in the above example).
What sum of money at t = 6 is equivalent to the cash flow profile shown in Figure 2.14 if ¡ = 10%?
The present value of the cash flow profile is given by:
Using the Single payment present value formula (1+i)-n, (F|P, ¡,n) =1.1-5 = .6209
Using the Uniform series formula (1+i)n-1/(i(1+i)n)
PV = -$400(1+.10)-1 + $100(1.10)-1 x (((1+0.1)3)-1)/(0.1*(1+0.1)3)) + 100 * (1.10)-5 x (((1+0.1)3)-1)/(0.1*(1+0.1)3)) = 16.86
Using the Single payment future value formula (1+i)n, (P|F, ¡,n) =1.106 = 1.77
Thus, at 10% a positive cash flow of $29.86 at t = 6 =16.86*1.77 = $29.84 (This is equivalent to the cash flow profile shown in Figure 2.14).
Figure 2.14 Cash flow diagram for the equivalence example.
Introduction to Measures of Worth [viewable here in Excel]
Most frequently the time value of money concepts and factors presented in this section are used to calculate measures of worth of investments alternatives. Several important measures of worth will now be introduced by way of the following example. These measures of worth, as well as several others, will be formally treated in later sections.
A firm purchases a machine for $30,000, keeps it for 5 years, and sells if for $6,000. During the time the machine was owned by the company, operating maintenance costs totaled $8,000 the first year, $9,000 the second year, $10,000 the third year, $11,000 the fourth year, and $12,000 the fifth year. The firm uses a 15% interest rate in performing economic analysis. Determine the single sum of money occurring at (1) t = 0 and (2) t = 5, which is equivalent to the cash flow history for the machine. Also, determine the uniform series occurring over the interval [1,5] that is equivalent to the cash flow profile for the machine.
In the economic analysis literature, the single sum equivalent at time zero for a cash flow profile is called the present worth or present value for the cash flow profile. We will use the term present value and denote it as PV. For the example:
PV = -$30,000 - $8,000/(P|A 15,5) - $1,000(P|G 15,5) + $6,000(P|F 15,5) = $59,609.57
Hence, a single expenditure of $59,609.57 at time zero would have been equivalent to the cash flows experienced during the ownership of the machine.
The single sum equivalence at the end of the life of a project is termed the future worth or future value for the project. We denote the future value by FV and compute its value as follows:
FV = -$30,000 (F|P 15,5) - [$8,000 + $1,000(A|G 15,5)] x (F|A 15,5) + $6,000 = -$119,897
Alternatively, the future value can be obtained from the present value:
FV = PV(F|P ¡,n)
FV = -$59,608.57(F|P 15,5)
FV = $119,898.69
Hence, a single expenditure of $119,897 at time 5 would have been equivalent to the cash flow associated with the machine.
A uniform series equivalent for a series of yearly cash flows is referred to as the annual worth or equivalent uniform annual cost for the project. The latter expression is most appropriate for the type of example under consideration, since the resulting uniform series is a cost, not an income series. The annual worth designation is AW; EUAC denotes the equivalent uniform annual cost.
For the example problem, the EUAC determination is performed as follows:
EUAC = -$30,000 (A|P 15,5) - [$8,000 + $1,000(A|G 15,5)] x (F|A 15,5) + $6,000(A|F 15,5) = $17,782
Hence, an annual expenditure of $17,782/year for 5 years is equivalent to the cash flow profile associated with the machine investment. Alternatively, the equivalent uniform cost can be obtained from either the present value or future value.
EUAC = -PW(A|P ¡,n)
EUAC = -FW(A|F ¡,n)
Present worth, future worth and annual worth computations are used often in comparing economic investment alternatives having different cash flow profiles. Consequently, we will have need for PV, FV and AW calculations throughout this section.
Alternative and Modern Day Approach
| Interest Rate | 15% | PV | FV |
| Project Purchase | -$30,000 | -$30,000 | -$34,500 |
| Maintenance | -$8,000 | -$6,957 | -$10,580 |
| Maintenance | -$9,000 | -$6,805 | -$13,688 |
| Maintenance | -$10,000 | -$6,575 | -$17,490 |
| Maintenance | -$11,000 | -$6,289 | -$22,125 |
| Maintenance + 6000 residual | -$6,000 | -$2,983 | -$13,878 |
| NPV | -$59,609 | -$59,609 | -$112,261 |
| 5 year cost is w/out interest is $74,000, this is the cash flow of the machine | -$119,896 | ||
| Annual worth (Annual cost each year of the machine) | $17,839 |
Changing Interest Rates
The preceding example assumed that the interest rate did not change during the time period of concern. Past experience indicates that such a situation is not likely if the time period of interest extends over several years (i.e., more than one interest rate may be applicable). Considering a single sum of money and discrete compounding, if ¡, denotes the interest rate appropriate during time period t, the future value for a single sum of money can be expressed as:
Equation (3.1) F = P(1+i1)(1+i2)…(1+in-1)(1+in)
and the inverse relation
Equation (3.2) P = F(1+in)-1(1+in-1)-1…(1+i2)-1(1+i1)-1
Consider the example depicted in Figure 3.1 in which an individual deposited $1,000 in a savings account that paid interest at an annual compounding rate of 8% for the first 3 years, 10% for the next four years, and 12% for the next two years. Note how the cash flow diagram has been modified to incorporate the notations of changing interest rates. How much was in the fund at the end of the ninth year?
Old School Confusing Engineering Approach
Letting Vt denote the value of the account at the end of time period t, we see that
V3 = $1,000(F|P 8,3) =$1,000(1.2597) = $1,259.70
Likewise,
V7 = $1,259.70(F|P 10,4) =$1,259.70(1.4641) = $1,844.33
Likewise,
V9 = $1,844.33(F|P 12,2) =$1,844.33(1.2544) = $2,313.53
| Year | New Simple Finance Approach | Compounding Rate | FV | |
| 1 | $1,000 | 108.00% | $1,080.00 | 80.00 |
| 2 | $1,080 | 108.00% | $1,166.40 | 86.40 |
| 3 | $1,166 | 108.00% | $1,259.71 | 93.31 |
| 4 | $1,260 | 110.00% | $1,385.68 | 125.97 |
| 5 | $1,386 | 110.00% | $1,524.25 | 138.57 |
| 6 | $1,524 | 110.00% | $1,676.68 | 152.43 |
| 7 | $1,677 | 110.00% | $1,844.34 | 167.67 |
| 8 | $1,844 | 112.00% | $2,065.67 | 221.32 |
| 9 | $2,066 | 112.00% | $2,313.55 | 247.88 |
| 1,313.55 |
You have $10,573.45 earning 12% annually (1% per month) for 12 months, what was the original deposit? =PV(0.01,12,0,-10573,0) = $9,383
What is the future value of $9,383 at 12% compounded monthly for one year? =FV(0.01,12,,-9983) = $10,573
What is the compounded monthly interest rate? =RATE(12,0,-9983,10573,0) = 1.00%
What is the interest rate (compounded monthly) if you deposit $9,383 and after 12 months it's worth $10,573? =NPER(.01,0,-9383,10573,0) = 12
What is the present value of $10,573 at 12.00% (compounded monthly) for one year? =10,573/(1.01)12 = $9,383
You deposit $300.00 per month (today) earning 12% annually (1% per month [compounded monthly]) for 12 months, what is the balance after 2 years (total principal deposits is $300 x 12 months = $3,600)?
| Period | Deposit | Future Value | |
| 1 | $300 | $303 | =$300*(1.01)1 = $303 |
| 2 | $300 | $306 | =$300*(1.01)2 = $306 |
| 3 | $300 | $309 | =$300*(1.01)3 = $309 |
| 4 | $300 | $312 | =$300*(1.01)4 = $312 |
| 5 | $300 | $315 | =$300*(1.01)5 = $315 |
| 6 | $300 | $318 | =$300*(1.01)6 = $318 |
| 7 | $300 | $322 | =$300*(1.01)7= $322 |
| 8 | $300 | $325 | =$300*(1.01)8 = $325 |
| 9 | $300 | $328 | =$300*(1.01)9 = $328 |
| 10 | $300 | $331 | =$300*(1.01)10 = $331 |
| 11 | $300 | $335 | =$300*(1.01)11 = $335 |
| 12 | $300 | $338 | =$300*(1.01)12 = $338 |
| $3,843 | After yr 1 | ||
| $4,329 | After yr 2 |
Extending the consideration of changing interest rates to series of cash flows, the PV of a series of cash flows can be represented as:
(Equation 3.3) P = A1(1+i1)-1+ A2(1+i1)-1…An(1+i1)-1(1+i2)-1
and the inverse relation
(Equation 3.4) F = An + An-1(1+in) + An-2(1+in-1)…A1(1+i2)(1+i3)
Consider the cash flow diagram given in Figure 3.2 with the appropriate interest rates indicated.
Determine the present value, future value and uniform series equivalents for the cash flow series.
Computing the present value gives (F = P(1+i)-n):
P = $200 *(1+.10)-1 - $200(1+.10)-1(1.10)-1+$300(1+.08)-1(1.10)-2+$200(1+.12)-1(1.08)-2(1.10)-2 = $372.62
Computing the future value gives (F = P(1+i)n):
F=(200-(200*((1.08)2)*(1.12)1)+(300*(1.08)1*((1.12)1))+(200*(1.12)1*((1.08)2)*((1.1)))) = $589.01
| Cash Flow | Interest Rate | Cumulative Discount Factor | PV | Discounted Cash Flow | FV | Cumulative Discount Factor | FV |
| $200.00 | 10% | 90.91% | $182 | 181.8 | 1.10 | 1.10 | $200.00 |
| -$200.00 | 10% | 82.64% | -$165 | -165.3 | 1.21 | 1.21 | $(261.27) |
| $300.00 | 8% | 76.52% | $230 | 229.6 | 1.31 | 1.31 | $362.88 |
| $200.00 | 8% | 70.85% | $0 | 0.0 | 1.41 | 1.41 | |
| $200.00 | 12% | 63.26% | $126.5 | 126.5 | 1.58 | 1.58 | $287.40 |
| $372.6 | 372.6 | $589.01 |
Inflation Considerations
Inflation is characterized by a decrease in the purchasing power of money caused by an increase in general price levels of goods and services without an accompanying increase in the value of the goods and services. Inflationary pressure is created when more dollars are put into an economy without an accompanying increase in goods and services. In other words, printing more money without an increase in economic output generates inflation. A complete treatment of inflation is beyond the scope of this section.
Any consideration of cash flows in today's economy must include a consideration of inflation. Although it is important to consider the impact of inflation on investments made within our country, it is especially important to do so in multinational investment situations. The inflation rates in Argentina, Brazil, Canada, France, Great Britain, Italy, Japan, Mexico and the United States, for example, can be dramatically different. Hence, a firm that is faced with making decisions concerning the investment of capital in various nations must give strong consideration to the inflationary economics in the countries in question.
Inflation affects adversely the purchasing power of money. To illustrate what we mean by the term, purchasing power of money, suppose a firm purchases one-million pounds of material each year, and the price of the material increases by 10% per year. Obviously, the quantity of material the firm can purchase with a fixed amount of money, that is, purchasing power of the firm's money, decreases over time. The only way the firm can afford to continue purchasing the material is decrease its usage rate or to increase its own source of funds. In the latter case, the purchasing power of its customers' money will be decreased. In these situations, the continuing spiral of prices increases does not contribute real increases in the firm's profits; instead, it results in an inflated representation of the firm's profits. The overall process is referred to as inflation.
Let's say over the past five decades we have experienced price level increases at both single-digit and double-digit rates. In some countries, triple-digit rates describe the price increases they have experienced. It has become widely accepted to refer to price level increases as inflation and to refer to price level decreases as deflation. (Under deflationary conditions, the deflation rate can be incorporated in economic analyses as a negative inflation rate.)
Although we have become accustomed to living in inflationary times, we are often very uncertain about how inflation should be treated, if at all, in an economic analysis. This section will present some background material on inflation, followed by two equivalent approaches for performing economic analysis under inflation.
Background on Inflation
One of the most difficult aspects of inflation is measuring it. To illustrate our claim, consider the following types of inflation: inflation rates, real interest rates, combined interest rates, and commodity escalation rates.
As noted previously, in recent years it has been apparent that the cost of goods and services are often affected by inflation. As a result, it is appropriate for interest rates and rates of return to reflect both the time value of money and the effects of inflation. For example, your decision to invest in a fund paying 12% interest would be quite different under negligible inflation conditions and 10% inflation conditions. We call the time value of money (or desired return) in the absence of inflation the real interest rate and denote it by d. The combined interest rate is denoted by ¡ and the inflation rate is denoted by j. The three rates are related as follows:
1 + ¡ = (1 + d)(1 +j)
or
Equation (3.5) ¡ = d + j + dj
If the inflation rate (j) is 3%/year and the real time value of money (d) is 15%/year, what is the combined interest rate (¡)?
1 + ¡ = (1 + j) x (1 /d)
1 + ¡ = (1 + .03) x (1 /0.15)
1 + ¡ = 1.1845
¡ = 18.45%
If the combined interest rate (real + nominal) is 26% per year and the nominal rate is 20%, what is the real interest rate?
solve for j = 1.26/1.20-1 = 5.0%
Inflation, the real rate of interest, and future value
The real interest rate is the nominal interest minus inflation. So the real rate of return of a 8% bond with 5% inflation is not 3%.
The real rate = nominal interest rate - rate of inflation / 1 + rate of inflation: .08 - .05 / 1 + .05 = 2.875%
To complicate matters, it is not uncommon to find that prices and expenses for various goods and services increase (and decrease) at different rates, due to different external factors. For example, the price paid for a dozen eggs has scarcely changed over the past decade or more; whereas, the cost of new housing has increased dramatically in some parts of the Unites States. Oil prices change at different rates than the price of bread; changes in aluminum and steel prices are not closely correlated with changes in textile prices or the prices for computers. In this case, we would consider the rate at which unit prices changed for each commodity to be an individual commodity escalation rate. (It is important to note that the commodity escalation rate for an individual commodity can be greater or less than the general inflation rate).
In practice, when it comes time to measure the inflation rate, what we observe are the values of commodity escalation rates and the combined interest rates. The value of the inflation rate is not as evident. As a result, in estimating the value of the underlying inflation rate, a surrogate measure is generally used. One popular surrogate measure of the inflation rate in the Unites States is the yearly change in the Consumer Price Index (CPI); another is the Producer Price Index (PPI), as know previously as the Wholesale Price Index (WPI).A number of composite indexes are used by firms to forecast the inflation rate; for example both Wall Stet Journal and Forbes publish their own indexes.
The challenge is not in obtaining values of indexes. Rather, it is in knowing how to convert the values into forecasts of future inflation rates. As an example, the federal government regularly publishes monthly updates on hundreds of indexes; it also provides monthly updates on hundreds of "business barometers," including aluminum output (in thousands of metric tons), electric power output (in billions of kilowatt-hours), machine tool orders (in millions of dollars), new construction (in billions of dollars), prime interest rates, and retail sales (in billions of dollars).
Table 3.1 present CPI values for the years since the base year 1993. From annual changes in the CPI, a value can be assigned to the overall annual inflation rate.
| Date | % Change in CPI |
| 1/1/1993 | 2.95% |
| 1/1/1994 | 2.61% |
| 1/1/1995 | 2.81% |
| 1/1/1996 | 2.93% |
| 1/1/1997 | 2.34% |
| 1/1/1998 | 1.55% |
| 1/1/1999 | 2.19% |
| 1/1/2000 | 3.38% |
| 1/1/2001 | 2.83% |
| 1/1/2002 | 1.59% |
| 1/1/2003 | 2.27% |
| 1/1/2004 | 2.68% |
| 1/1/2005 | 3.39% |
| 1/1/2006 | 3.23% |
| 1/1/2007 | 2.85% |
| 1/1/2008 | 3.84% |
| 1/1/2009 | -0.36% |
| 1/1/2010 | 1.64% |
| 1/1/2011 | 3.16% |
| 1/1/2012 | 2.07% |
| 1/1/2013 | 1.46% |
| 1/1/2014 | 1.62% |
| 1/1/2015 | 0.12% |
| 1/1/2016 | 1.26% |
| 1/1/2017 | 2.13% |
| 1/1/2018 | 2.44% |
Some do not agree with the use of changes in the CPI as estimates of inflation. They argue that the inflation rate cannot be an exact representation of the purchasing power of your money without your purchases being accurately reflects by the makeup of commodities included in the computation of the CPI.
As an example, if you never intend to buy or sell a house, then it would seem that the change of prices in houses will not have an impact on the purchasing power of your money. Likewise, if you are a vegetarian, then the change in price for beef, pork and chicken would not seem to affect you directly, they will affect others - and those people, in turn, will increase the prices of their goods and services in order to recover the increased amount of money spent on items you do not buy.
Because the overall economy is so interdependent, it is difficult to determine the inflation rate for an individual firm. Hence, changes in composite indexes, such as the CPI and PPI (or WPI), are used by individuals and organizations to estimate the impact of inflation on their economic investments. Since the rates of change in CP and PPI (or WPI) reflect the changes in prices of a mixture of commodities, then any estimate of inflation derived from these rates will be a weighted average of the annual rates of change in the costs of a wide mixture of commodities, then any estimate of inflation derived from these rates will be a weighted average of the annual rates of change in the costs of a wide range of commodities. For this reason, the resulting estimate of the inflation rate is referred to as a "market basket" rate.
In addition to consumer prices, producer prices, and wholesale prices, the Commerce Department and the Labor Department, among others, publish a wide range of economic data for a variety of industry groups. Depending on the industry, inflation estimates might be more accurately developed using one or more such sources.
When consideration of inflation is introduced into economic analysis, future cash flows can be stated in terms of either constant-worth dollars or then-current dollars. Then-current cash flows are expressed in terms of the face amount of dollars (actual number of dollars) that will change hands when the cash flow occurs. Alternatively, constant-worth cash flows are expressed in terms of the purchasing power of dollars relative to a fixed point in time known as the base period. Then-current cash flows explicitly incorporate inflation into the cash flows; constant-worth cash flows do not.
For the next four years, a family anticipates buying $1,000 worth of groceries each year. If inflation is expected to be 3%/year what are the then-current cash flows required to purchase the groceries.
To buy the groceries, the family will need to take the following face amount of dollars to the store. We will somewhat artificially assume that the family only shops once per year, buys the same set of items each year, and that the first trip to the store will be one year from today:
Year 1: dollars required $1,000.00 x (1.03) = $1,030.00
Year 2: dollars required $1,030.00 x (1.03) = $1,060.90
Year 3: dollars required $1,060.90 x (1.03) = $1,092.73
Year 4: dollars required $1,092.73 x (1.03) = $1,125.51
Inflation and Maintenance Costs
The need for maintenance occurs at an increasing rate of 8%/year; the maintenance cost this year totals $1,000. Furthermore, the labor and parts-required to maintain the equipment increase due to inflation at a rate of 10%/year. The real interest rate for the firm is 12%/year. It is desired to determine the present value equivalent of the then-current and constant-worth maintenance cost in the fifth year. In this case, v = 0.08, j = 0.10, T0 = $1,000 and d = 0.12.
The commodity escalation rate for maintenance cost is made up of an 8% real escalation rate and a 10% inflation rate. Hence, the commodity escalation rate will be equal to the sum of 8%, 10% and their product, or 18.8% =(1+0.08)*(1.1)-1.
The then-current maintenance cost will total $1,000(1.188)5 or $2,366.31 in the fifth year; the constant-worth maintenance costs in the fifth year will be $1,000(1.08)5 =$1,469.33. To obtain the present value equivalent of the maintenance cost, the real interest rate of 12% is applied to the constant-worth amount, whereas a combined interest rate of 0.12 + 0.10 + 0.12(.10) or 0.232 (23.2%) is applied to the then-current amount. Thus:
P = $1,469.30 (1.12)-5 = $833.68
or
P = 2366.31(1.232)-5 = $833.72
Net Present Value Method [viewable here in Excel]
The present value of investment alternative j can be represented as:
PVj(i) = NΣt=0 Ajt(1+i)-t
with PVj(i) = present value of Alternative j using MARR of i%
N = planning horizon
Ajt = net cash flow for Alternative j at the end of the period t
¡ = Minimum Attractive Rate of Retuen (MARR)
The present value method is the most popular measure of merit available. Where it is used to compare alternatives, the one having the greatest present value is the alternative recommended.
Buying a Pressure Washer - NPV
A pressure washer was purchased for $16k, with a life of 5 years and sold for $3k.
Maintenance is $4k year with 12% annual interest cost. What is the NPV?
| Interest Rate | 12% | PV | PV |
| Project Purchase | -$16,000 | -$16,000 | =-16000/((1+0.12)0) |
| Maintenance | -$4,000 | -$3,571 | =-4000/(1+0.12)1 |
| Maintenance | -$4,000 | -$3,189 | =-4000/(1+0.12)2 |
| Maintenance | -$4,000 | -$2,847 | =-4000/(1+0.12)3 |
| Maintenance | -$4,000 | -$2,542 | =-4000/(1+0.12)4 |
| Maintenance + $3,000 residual | -$1,000 | -$567 | =-1000/((1+0.12)5) |
| Net Present Value | -$28,717 | -$28,717 |
We can see that a single expenditure of $28,717 at time zero is equivalent to the cash flow profile for the pressure washer.
=FV(0.12,5,,-28717) = $50,609
=PMT(0.12,5,-16000,3000)+4000 $7966
Upgrading a Tooling Machines - NPV
Upgrading a tooling machine will cost $10,000 with an extended life of 6 years and no salvage value. Income will increase $2,525/yr with a hurdle rate of 15%. What is the NPV?
| Interest Rate | 15% | PV | PV |
| Project Purchase | -$10,000 | -$10,000 | =-10000/((1+0.15)0) |
| Income | $2,525 | $2,196 | =2525/(1+0.15)1 |
| Income | $2,525 | $1,909 | =2525/(1+0.15)2 |
| Income | $2,525 | $1,660 | =2525/(1+0.15)3 |
| Income | $3,840 | $2,196 | =3840/(1+0.15)4 |
| Income | $3,840 | $1,909 | =3840/((1+0.15)5) |
| Income | $3,840 | $1,660 | =3840/((1+0.15)6) |
| NPV | $1,530 | $1,530 | |
| Future Value | $3,539 | ||
| IRR | 20.00% |
Annuities [viewable here in Excel]
Often the future cash flows in a savings plan, an investment project, or a loan repayment schedule are the same each year. We call such a level stream of cash flows or payments an annuity. The term comes from the life-insurance business, where an annuity contract is one that promises a stream of payments to the purchaser for some period of time. In finance, it is applied more generally and applies to any level stream of cash flows; thus, the stream of payments on an installment loan or mortgage are called annuities too.
If the cash flows start immediately, as in a savings plan or a lease, it is called an immediate annuity. If the cash flows start at the end of the current period rather than immediately, it is called an ordinary annuity. A mortgage is an example of an ordinary annuity. There are some convenient formulas, tables and calculator functions for computing the present and future values of annuities which come in handy when the stream of cash flows last for many periods.
Future Value of Annuities
For example, suppose you intend to save $100 each year for the next three years, how much will you have accumulated at the end of that time if the interest rate is 10% per year? If you start saving immediately, you will have:
FV = $100 x 1.101 x $100 x 1.102 x $100 x 1.103 = $364.19
FV = $100 x (1.1011 x 1.103 x 1.103) = $364.19
The result if a future value of $364.10. The factor multiplying the $100 is the future value of a $1 payment per year for 3 years.
In computing the future value of an annuity it matters whether it is an immediate-annuity, as in our example, or whether it is an ordinary annuity. In the case of an ordinary annuity, the first $100 contribution is made at the end of the first period. Figure 4.7 shows a time line that contrasts the two situations.
| Time | Contribution | Rate | Compounding | Immediate Annuity | Ordinary Annuity |
| 1 | $100 | 10% | 1.10 | $110.00 | $110.00 |
| 2 | $100 | 10% | 1.21 | $121.00 | $121.00 |
| 3 | $100 | 10% | 1.33 | $133.10 | $100 |
| $300 | $364 | $331 |
Although in both cases there are the same number of payments, under the immediate annuity pattern, the entire amount earns interest for an additional year. Thus, an immediate annuity would have a FV equal to that of the ordinary annuity multiplied by 1 + i. For an ordinary annuity of $1 per year the formula for future value is:
FV = (1 + i)n - 1 / I or (=100*(((1.1)3-1)/0.1)) = $331.00
=100*(((1.1)3-1)/0.1) = $331.00
=FV(0.1,3,,-100) = $331.00
We find that the future value of our savings plan of $100 per year for three years is $364.10 if the first deposit is made immediately (an immediate annuity), and $331 if delayed until the end of the first year (an ordinary annuity).
Perpetuities and Capitalized Worth Method
A specialized type of cash flow series is a perpetuity, a uniform Series of cash flows which continues indefinitely. This is a special case, since an infinite series of cash flows would rarely be encountered in the business world; rather, a finite series of cash flows is the general rule. However, for such very long-term investment projects as bridges, highways, forest harvesting, or the establishment of endowment funds where the estimated life is 50 years or more, an infinite cash flow series may be appropriate.
If a present value P is deposited into a fund at interest rate ¡ per period so that a payment of size A may be withdrawn each and every period forever, then the following relation holds between P, A and ¡.
Pi = A
Thus as depicted in Figure 4.3, P is a present value that will pay out equal payments of size A indefinitely if the interest rate per period is ¡. The present value P is termed the capitalized worth of A, the size of each of the perpetual payments.
Capital Recovery Formula
You purchase a computer for $82k, with a 7 yr life, a $5k salvage value and interest cost of 15%. What is the PV? (=PMT(0.15,7,-82000,5000) = $19,258)
Computed Surplus (=FV(0.15,7,19258,-82000) = $4,997)
Computed Cost (=PV(0.15,7,-19258,-4997) = $82,000)
Computed term (=NPER(0.15,19258,-4997,82000)*-1 = 7.00)
Computed interest cost (=RATE(7,19258,,-82000-5000)*-1 = 15%)
Comparison of Alternatives
Defining Investment Alternatives
Consider the two cash flows in Comparison A and Comparison B below. The two alternatives are one-shot investments. We are unable to predict what investment alternatives will be available in the future, but we do anticipate that cash flows can be reinvested and earn a 15% return.
| Comparison A (4 year investment) | ||
| Interest Rate | 15% | PV |
| Project Purchase | -$4,000 | -$4,000 |
| Income | $3,500 | $3,043 |
| Income | $3,500 | $2,647 |
| Income | $3,500 | $2,301 |
| Income | $4,500 | $2,573 |
| NPV | $6,564 | $6,564 |
| Future Value | $15,183 | $15,183 |
| IRR | 81.279% | 57.634% |
| Standard Deviation of income | $500 | $307 |
|
|
|
|
| Comparison B (6 year investment) | ||
| Interest Rate | 15% | PV |
| Project Purchase | -$5,000 | -$5,000 |
| Income | $1,000 | $870 |
| Income | $2,000 | $1,512 |
| Income | $3,000 | $1,973 |
| Income | $4,000 | $2,287 |
| Income | $5,000 | $2,486 |
| Income | $6,000 | $2,594 |
| NPV | $6,721 | $6,721 |
| Future Value | $15,547 | $15,547 |
| IRR | 44.62% | 25.76% |
| Standard Deviation of income | $1,871 | $660 |
Break-even Analysis [viewable here in Excel]
Suppose a firm is considering manufacturing a new product and the following data is provided.
What is the unit break-even number?
| Sales Price per unit | $12.50 |
| Equipment Cost | $200,000 |
| Overhead cost per year | $50,000 |
| Operating and maintenance per operating hour | $25 |
| Production time/1000 units (hours) | 100 |
| Machine Life | 5 |
| Required return | 15% |
| Salvage value | 0 |
| Break-Even Units =(PMT(0.15,5,-200000)+50000)/((-100/1000)*25+12.5) = 10,966 | 10,966 |
| 137,078.89 |
Should You Rent or Buy a Compressor?
Consider a contractor who experiences a seasonal pattern of activity for air compressors. The company owns eight air compressors and feels that this number will not be adequate to meet future peak demand. The contractor realizes that there will be times when more than eight air compressors will be needed, and is considering purchasing an additional compressor for use during peak times.
A local equipment rental firm will rent air compressors at a cost of $50/day. Air compressors can be purchased for $6,000. The difference in operating and maintenance cost between owned and rented air compressors is estimated to be $3,000/year.
Letting X denote the number of days a year that more than eight compressors are required, the following break-even analysis is performed. A planning horizon of 5 years, zero salvage value, and 20% minimum attractive rate of return are assumed.
| Cost or Rate of Return | 20% |
| Compressor life with zero salvage value | 5 |
| Return factor =(0.2*((1+0.2)5))/(((1+0.2)5)-1) | 33.4% |
| Daily Rental | $50 |
| Purchase Compressor | $6,000 |
| Required return | $2,006 |
| Maintenance (per year) | $3,000 |
| Cost to purchase compressor vs. renting | $5,006 |
| Break-Even in hours (per year) | 100 |
Based on past history, if you plan to rent a compressor for more then 100 hours per year over the next 5 years, it makes sense to purchase a compressor vs. renting. Taxes or deductions are not considered in this example.
Determining a Company's Growth Rate
Motorola's current return on capital is 12.18% and its reinvestment rate is 52.99%.
We expect Motorola's return on capital to rise to 17.22% over the next 5 years (which is half way towards the industry average).
Expected Growth Rate
= ROC New on new investments*Reinvestment Rate Current+ {[1+(ROC In 5 years-ROC Current)/ROC Current]1/5-1}
= .1722 *.5299 +{ [1+(.1722-.1218)/.1218](1/5)-1}
= .1629 or 16.29%
- One way to think about this is to decompose Motorola's expected growth into:
- Growth from new investments: .1722 * .5299 = 9.12%
- Growth from more efficiently using existing investments: 16.29% - 9.12% = 7.17%
Note that we are assuming that the new investments start making 17.22% immediately, while allowing for existing assets to improve returns gradually.
Reinvesting at a Different Rate
You have $10,000 to invest for two years. You have a choice, invest in a bank certificate of deposit (CD) paying 7% or one-year CD paying 6% today and hopefully 8% in year two. Hopefully, because unless you have a working crystal ball, you are hoping or guessing CD rates will increase to 8% when the 6% CD matures. Without getting into future contracts or options, let's stick with the CD example for now. What is the best investment, the 2 yr CD or the one year CD?
FV = 10,000 x 1.072 = $11,449
FV = 10,000 x 1.06 x 1.08 = $11,448
Yield to Maturity or Internal Rate of Return [viewable here in Excel]
The internal rate of return is the discount rate that makes the present value of future cash flows equal to the outlay required. Or, it is the interest rate where NPV equals zero.
If the NPV is zero and the IRR is greater than zero, we know that the return on capital is greater than the cost of capital. This is the fundamental rule of finance, successful projects have returns on capital greater than the cost of capital and for equity investors the return on equity should be greater than the cost of equity.
IRR = $75 = $100/(1+i)5
I = (100/76)(1/5)
IRR or Yield to Maturity is: =(100/75)(1/5)-1 = 5.922%
You're thinking about buying a new car, the lender offers you a 4 year, $5,000 loan where you pay back $9,000. A friend offers you a $5,000 loan at 12% payable over the next 4 years.
What's the better deal?
Lender = IRR = $5000 = $9000/(1+i)(4) =(9000/5000)(1/4)-1 = 15.83%
Friend's loan = 12%
Clearly you friends deal is better then the lenders.
| Payment at | Payment at | |
| 12% | 15.83% | |
| Interest year 1 | 600 | 792 |
| Interest year 2 | 672 | 886 |
| Interest year 3 | 753 | 1,057 |
| Interest year 4 | 861 | 1,267 |
| Total Interest | 2,886 | 4,002 |
| Loan Amount | 5,000 | 5,000 |
| Total P & I | 7,886 | 9,002 |
| Annual Payment | 1,971 | 2,250 |
Sources: Damodaran, Hull, Basu, Kolb, Brueggeman, Fisher, Bodie, Merton, White, Case, Pratt, Agee.
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