Overview

In this second lesson, we will enhance our knowledge of calculating present, annual, and future values, and then the rate of return analysis and break-even method will be explored. The calculation of present, annual, and future values is essential to project evaluation. And the rate of return and break-even methods are a critical framework to make investment decisions.Proper application of these different approaches to analyzing the relative economic merit of alternative projects depends on the type of projects being analyzed. As noted in Lesson 1, two basic classifications of investments are:

1. revenue-producing investment alternatives
2. service-producing investment alternatives

The application of these methods differs for revenue and service-producing projects. This lesson concentrates on the application of present worth, annual worth, future worth, and rate of return techniques and their examples. These methods are illustrated here on a before-tax analysis basis.

Learning Objectives

At the successful completion of this lesson, students should be able to:

• enhance their understanding of present, annual, and future values;
• understand the framework of break-even and rate of return analysis;
• use present, annual, and future values to make investment decisions; and
• use break-even and rate of return analysis to make investment decisions.

What is due for Lesson 2?

This lesson will take us one week to complete. Please refer to the Course Syllabus for specific timeframes and due dates. Specific directions for the assignment below can be found within this lesson.

Questions?

If you have any questions, please post them to the discussion forum, located under the Modules tab in Canvas. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.

Nominal, Period and Effective Interest Rates Based on Discrete Compounding of Interest

Usually, financial agencies report the interest rate on a nominal annual basis with a specified compounding period that shows the number of times interest is compounded per year. This is called simple interest, nominal interest, or annual interest rate. If the interest rate is compounded annually, it means interest is compounded once per year and you receive the interest at the end of the year. For example, if you deposit 100 dollars in a bank account with an annual interest rate of 6% compounded annually, you will receive $100\ast \left( 1+0.06 \right)=106$ dollars at the end of the year.

But, the compounding period can be smaller than a year (it can be quarterly, monthly, or daily). In that case, the interest rate would be compounded more than once a year. For example, if the financial agency reports quarterly compounding interest, it means interest will be compounded four times per year and you would receive the interest at the end of each quarter. If the interest is compounding monthly, then the interest is compounded 12 times per year and you would receive the interest at the end of the month.

For example: assume you deposit 100 dollars in a bank account and the bank pays you 6% interest compounded monthly. This means the nominal annual interest rate is 6%, interest is compounded each month (12 times per year) with the rate of 6/12 = 0.005 per month, and you receive the interest at the end of each month. In this case, at the end of the year, you will receive $100\ast {\left( 1+0.005 \right)}^{12}=106.17$ dollars, which is larger than if it is compounded once per year: $100\ast {\left( 1+0.06 \right)}^1=106$ dollars. Consequently, the more compounding periods per year, the greater total amount of interest paid.

Please watch the following video, Nominal and Period Interest Rates (Time 3:52).

Period interest rate i = r/m
Where m = number of compounding periods per year
r = nominal interest rate = mi

"An effective interest rate is the interest rate that when applied once per year to a principal sum will give the same amount of interest equal to a nominal rate of r percent per year compounded m times per year. Annual Percentage Yield (APY) is the standard term used by the banking industry to identify an effective interest rate."

The future value, F1, of investing P at i% per period for m period after one year:

And if the effective interest rate, E, is applied once a year, then future value, F2, of investing P at E% per year:

Then:

$\begin{array}{c}F1=F2 \\ P{\left( 1+i \right)}^m=P{\left( 1+E \right)}^1\end{array}$

Since P the same in both sides: ${\left( 1+i \right)}^m=E+1$

Then:

If the effective Annual Interest, E, is known and equivalent period interest rate i is unknown, the equation 2-1 can be written as:

Going back to the previous example, $$\begin{array}{c}i=6/12=0.005 \\ \text{so, }E={\left( 1+0.005 \right)}^{12}-1=1.0617-1=0.0617\text{ or }6.17\%\end{array}$$

Please watch the following video, Effective Interest Rate (Time 4:02).

Example 2-1:

Assume an investment that pays you 2000 dollars in the end of the first, second, and third year for an annual interest rate of 12% compounded quarterly. Calculate the time zero present value and future value of these payments after three years.

Quarterly period interest rate i = 12/4 = 3%

$$\begin{array}{c}P =2,000*\left( P/F_{3\%,4}\right)+2,000*\left( P/F_{3\%,8}\right)+2,000*\left( P/F_{3\%,12}\right) \\ =2000\left[ 1/{\left( 1+0.03 \right)}^4\right]+2000\left[ {\left( 1/(1+0.03 \right)}^8\right]+2000\left[ 1/{\left( 1+0.03 \right)}^{12}\right] \\ =\$4,758.55\end{array}$$

$$\begin{array}{c}F =2,000*\left( F/P_{3\%,(12-4)}\right)+2,000*(F/P_{3\%,(12-8)})+2,000*\left( F/P_{3\%,(12-12)}\right)\\ =2,000*\left( F/P_{3\%,8}\right)+2,000*\left( F/P_{3\%,4}\right)+2,000 \\ =2000*{\left( 1+0.03 \right)}^8+2000*{\left( 1+0.03 \right)}^4+2000\\ =\text{ \$}6,784.56\end{array}$$

Please note that since the interest rate is compounded quarterly, we have to structure the calculations in a quarterly base. So there will be 12 quarters (three years and 4 quarters per each year) on the time line.
The 2000 dollars interest is paid at the end of the first, second, and third year, which are going to be the last quarters of each year (4^th quarter, 8^th quarter, and 12^th quarter).

Please watch the following video, Nominal and Period Interest Rates Example (Time 3:45).

Continuous Compounding of Interest

If an annual interest rate compounds annually, then it should be compounded once a year.
If an annual interest rate compounds semi-annual, then it should be compounded twice a year.
If an annual interest rate compounds quarterly, then it should be compounded 4 times per year.
If an annual interest rate compounds monthly, then it should be compounded 12 times per year.
If an annual interest rate compounds daily, then it should be compounded 365 times per year.
And if the compounding period becomes smaller, then the number of compoundings per year, m, becomes larger. In the limit as m goes to infinity, period interest, i, approaches zero. This case is called Continues Compounding of Interest. Using differential calculus, Continues Interest Single Discrete Payment Compound-Amount Factor (F/P_r,n) can be calculated as:

And, Continues Interest Single Discrete Payment Present Worth Factor (P/Fr,n)

r is nominal interest rate compounded continuously
n is number of discrete valuation periods
e is base of natural log (ln) = 2.7183

Example 2-2:

Lets recalculate example 2-1 considering continues compound interest rate of 12%:

$$\begin{array}{c}P =2,000*\left( P/F_{12\%,1}\right)+2,000*\left( P/F_{12\%,2}\right)+2,000*\left( P/F_{12\%,3}\right)=2000\left[ 1/e^{0.12*1}\right]+2000\left[ 1/e^{0.12*2}\right]+2000\left[ 1/e^{0.12*3}\right] \\ =\text{ \$}4,742.45 \\ F =2,000*\left( F/P_{12\%,2}\right)+2,000*\left( F/P_{12\%,1}\right)+2,000=2000*e^{0.12*2} +2000*e^{0.12*1}+2000=\text{ \$}6,797.49\end{array}$$

Note: The following links explains how to use the excel function (EXP) to calculate e raised to the power of number:

Link 1: EXP Function in Excel
Link 2: Excel Functions

Please watch the following video, Continuous Compounding of Interest (Time 4:54).

“Flat” or “Add-on” Interest Rate

A flat or add-on interest rate is applied to the initial investment principal each interest compounding period. This means total interest received for the investment on a flat interest is calculated linearly and simply is the summation of interest on all periods. For example, if you invest 1000 dollars at the present time in a project with flat interest rate of 12% per annum for 100 days, you will receive 32.88 dollars after 100 days:
$$1000*0.12*\left( 100/365 \right)=32.88\text{ dollars interest}$$
The flat interest rate is usually applied when interest is calculated for a portion of a year or period.

Note: In engineering economics, the term “simple interest” is usually used as “add-on” or “flat” interest rate as defined here.

Example 2-3:

If an investment gives you 8% interest compounded annually, how long will it take to double your money, invested in present time?

$$F=P*F/P_{i,n}$$$$F/P =2\text{ or} F/P_{i,n} =2$$$${\left( 1+i \right)}^n =2$$$${\left( 1+0.08 \right)}^n =2$$

By taking ln (natural log) or log from each side, we will have:
$$\begin{array}{c}ln{\left( 1.08 \right)}^n =Ln\left( 2 \right) \\ n*ln\left( 1.08 \right)=Ln\left( 2 \right) \\ n=Ln\left( 2 \right)/ln\left( 1.08 \right)=9\text{ years}\end{array}$$

It takes 9 years to double your money for an investment with 8% interest compounded annually.

The following links show how to calculate natural log using Excel:

Link 1: LN Function
Link 2: How to Return the Natural Logarithm of a Number using Formulas

Example 2-4:

Calculate the present value of following payments assuming the interest rate of 10% (compounded per period)

$$P=1000\left( P/A_{10\%,5}\right)\left( P/F_{10\%,1}\right)$$$$\begin{array}{c}\text{Using Table 1-12: }P/A_{i,n}=\left[ {\left( 1+i \right)}^n-1 \right]/\left[ i{\left( 1+i \right)}^n\right] \\ P=1000\left[ {\left( 1+0.1 \right)}^5-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^5\right]*1/\left( 1+0.1 \right) \\ P=1000*3.7908*0.9090=\text{ \$}3,446.17\end{array}$$

Note that here, uniform series of $1000 start from year 2. However, factor $\left( P/A_{10\%,5}\right)$ returns the P in the year before beginning of the first payment, which is year 1 here. Therefore, to calculate the present value of these uniform series of payments, we need to discount that for one year by multiplying it by $\left( P/F_{10\%,1}\right)$.

Example 2-5:

What is the present value and equivalent series of annual end-of-period values for payments occurred in the following timeline, assuming the interest rate of 10% (compounded per period)?

$$\begin{array}{c}P=1000\left( P/A_{10\%,3}\right)+2000\left( P/A_{10\%,3}\right)\left( P/F_{10\%,3}\right)+3000\left( P/A_{10\%,3}\right)\left( P/F_{10\%,6}\right) \\ \text{Using Table 1-12: }P/A_{i,n}=\left[ {\left( 1+i \right)}^n-1 \right]/\left[ i{\left( 1+i \right)}^n\right] \\ P=1000\left[ {\left( 1+0.1 \right)}^3-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^3\right]+2000\left[ {\left( 1+0.1 \right)}^3-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^3\right]*1/{\left( 1+0.1 \right)}^3 +3000\\ \left[ {\left( 1+0.1 \right)}^3-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^3\right]*1/{\left( 1+0.1 \right)}^6\\ P=1000*2.4869+2000*2.4869*0.7513+3000*2.4869*0.5645=\text{ \$}10434.96\end{array}$$

$$\begin{array}{c}A=P*\left( A/P_{10\%,9}\right) \\ \text{Using Table 1-12}:A/P_{i,n} =\left[ i{\left( 1+i \right)}^n\right]/\left[ {\left( 1+i \right)}^n-1 \right] \\ A=10435.28*\left[ 0.1{\left( 1+0.1 \right)}^9\right]/\left[ {\left( 1+0.1 \right)}^9-1 \right]=10435.28*0.1736=\text{ \$}1,811.99\end{array}$$

Note that: 
There are three equal series of 1000 dollars from year 1 to year 3 so the present value (at time 0) of those can be calculated as: $1000\left( P/A_{10\%,3}\right)$.

There are three equal series of 2000 dollars from year 4 to year 6: Because $2000\left( P/A_{10\%,3}\right)$ gives the P of these three payments at the year 3 (one year before the first one) so we need to discount the value for three years to have the present value for time 0 so present value of three equal series of 2000 dollars from year 4 to year 6 equals:

$$2000\left( P/A_{10\%,3}\right)\left( P/F_{10\%,3}\right)$$

There are three equal series of 3000 dollars from year 7 to year 9: and $3000\left( P/A_{10\%,3}\right)$ gives the P at the year 6 (one year before the first one) so we need to discount the value for six years to have the present value for time 0 so present value of three equal series of 3000 dollars from year 7 to year 9 equals:

$$3000\left( P/A_{10\%,3}\right)\left( P/F_{10\%,6}\right)$$

Please watch the following video, Applications of Compound Interest Formulas (Time 4:56).

Note: As displayed in Figure 2-1, using Microsoft Excel, you can calculate all the present values and then add them together much more conveniently.

Figure 2-1: Calculating the total present value of all payments occurring in the future by determining each payment's present values and then adding them together using Microsoft Excel

Credit: Farid Tayari

Example 2-6:

Assume you can invest in a machine that can yield the income after all expense of 1000 dollars twice in the first and second years, 2000 dollars twice in the third and fourth years, and 3000 dollars twice in the fifth and sixth years. At the end of the sixth year, the machine has a resale value of $10,000. How much can be paid for this machine at the present time with the interest rate of 10% compounded annually?

$$\begin{array}{c}P=1000\left( P/A_{10\%,2}\right)+2000\left( P/A_{10\%,2}\right)\left( P/F_{10\%,2}\right)+3000\left( P/A_{10\%,2}\right)\left( P/F_{10\%,4}\right)+10000\left( P/F_{10\%,6}\right) \\ UsingTable1-12:P/A_{i,n}=\left[ {\left( 1+i \right)}^n-1 \right]/\left[ i{\left( 1+i \right)}^n\right] \\ P=1000\left[ {\left( 1+0.1 \right)}^2-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^2\right]+2000\left[ {\left( 1+0.1 \right)}^2-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^2\right]*1/{\left( 1+0.1 \right)}^2+3000 \\ \left[ {\left( 1+0.1 \right)}^2-1 \right]/\left[ 0.1{\left( 1+0.1 \right)}^2\right]*1/{\left( 1+0.1 \right)}^4+10000*1/{\left( 1+0.1 \right)}^6\\ P=1000*1.7355+2000*1.7355*0.8265+3000*1.7355*0.6830+10000*0.5645=\text{\$}13,805.12\end{array}$$

Here we have:

Two 1000 dollars at year 1 and 2, so the present value can be calculated as $1000\left( P/A_{10\%,2}\right)$

Two 2000 dollars at year 3 and 4, so the present value can be calculated as $2000\left( P/A_{10\%,2}\right)( P/F_{10\%,2})$. Because, similar to explanation in example 2-4 and 2-5, $2000\left( P/A_{10\%,2}\right)$ gives the present value of these two payments at the year 2 (one year before the first one) it needs to be discounted for two years to have the present value for time 0 and present value of two 2000 dollars at year 3 to year 4 equals 2000 $2000\left( P/A_{10\%,2}\right)\left( P/F_{10\%,2}\right)$.

Two 3000 dollars at year 5 and 6: similarly, PV of these two payments will be $3000\left( P/A_{10\%,2}\right)\left( P/F_{10\%,4}\right)$. Because $3000\left( P/A_{10\%,2}\right)$ returns the present value at year 4 and it is required to be discounted for 4 years to give the present value of these payments at time zero.

Figure 2-2 displays how you can calculate the present value in Microsoft Excel by adding up all the present values of payments occurring in different time periods.

Figure 2-2: Calculating the total present value of all payments occurring in the future by determining each payment's present values and then adding them together using Microsoft Excel

Credit: Farid Tayari

Example 2-7:

In order to pay off a 100,000 dollars mortgage in 20 years with interest rate of 8% per year (compounded annually), how much will the annual end-of-year mortgage payments be?

$$\begin{array}{c}A=100,000*\left( A/P_{8\%,20}\right) \\ \text{Using Table 1-12}:A/Pi,n=\left[ i{\left( 1+i \right)}^n\right]/\left[ {\left( 1+i \right)}^n-1 \right] \\ A=100,000*\left[ 0.08{\left( 1+0.08 \right)}^{20}\right]/\left[ {\left( 1+0.08 \right)}^{20}-1 \right]=100,000*0.101852=10,185.22\text{ dollars per year}\end{array}$$

Similar to what we had in previous sections (such as Example 2-6), there are problems that require you to calculate present value (as an unknown variable) for payments occurring in the future as revenue, with interest rate or rate of return (as known variable). These types of calculations are called break-even and enable you to determine the initial investment cost that can break-even the future payments considering a specified interest rate. It gives you the equivalent amount of money that needs to be invested at present time for receiving the given payments in the future with the desired interest rate.

As explained in Lesson 1, time value of money affects present value calculations. Consequently, the size of the payments, interest rate, and also payment schedule are influential factors in determining present value and break-even calculations.

Example 2-8:

Assume two investments of A and B with the payment schedule as shown in Figure 2-3. Calculate the present value of these investments considering minimum rates of return of 10% and 20%. The calculation will give the initial cost that can be invested to break-even with 10% and 20% rate of return.

Please notice that cumulative payments for investment A and B are equal and the difference between two investments is in the payment schedule.

Figure 2-3a: In investment A, the payment (revenue) schedule will be 100, 200, 300, and 400 dollars at the end of the first, second, third and fourth year. In investment B, the payment (revenue) schedule will be 400, 300, 200, and 100 dollars at the end of the first, second, third and fourth year.

Assuming rate of return 10%:

Assuming rate of return 20%:

This example shows the effect of time on future payments.Cumulative payments for investment A and B are equal, and the difference between two investments is in the payment schedule. In investment B, the investor receives a larger amount of revenue in the closer future, which amortizes the investor’s principal more rapidly than “A."

Example 2-9:

Investing on an asset is expected to yield 2,000 dollars per year in income after all expenses for each of the next ten years. It is also expected to have a resale value of $25,000 in ten years. How much can be paid for this asset now if a 12% annual compound interest rate of return before taxes is desired? Note that the wording of this example can be changed to describe a mineral reserve, petroleum, chemical plant, pipeline, or other general investment, and the solution will be identical.

Figure 2-3b: Cash flow: 2,000 dollars per year in income after all expenses for 10 years and resale value of $25,000 in the tenth year.
C: Cost
I: Income
L: Salvage Value

Present Value Equation:
Let’s equate costs and income at the present time.
Present value of all costs =present value of all incomes plus present value of salvage

The result will be similar, if costs and revenue plus salvage is equated in any time.

Future Value Equation
If we equate costs and income by the end of the 10^thyear, then:
future value of cost = future value of income + future value of salvage

Annual Value Equation
Let’s equate the annual value costs and incomes,
annual value of cost = annual value of income + annual value of return

Please note that an equation can be written to equate costs and incomes at any point in time and the same break-even initial cost of $19,350 can be obtained.

So far, we have learned how to determine the unknown variables including present value, future value, uniform series of equal investments, and so on. In these question types, the interest rate was a given parameter. But, there are situations where the interest rate, i, is the unknown variable. In such cases, we know (or expect) the amount of money to be invested and the revenue that will occur in each time period, and we are interested in determining the period interest rate that matches these numbers. This category of problems is called rate of return (ROR) calculation type. In these problems, we are interested to find the interest rate that yields a Net Present Value of zero (the break-even interest rate). This break-even rate is sometimes called the Internal Rate of Return.

For example, assume for an investment of 8000 dollars at present time, you will receive 2000 dollars annually in each of year one to year five. What would be the interest rate (compounded annually) for which this project would break even?

The problem can be written as:
$8000=2000\left( P/A_{i,5}\right)$ or
$\left( P/A_{i,5}\right)=4$

With a trial and error procedure, we can find the interest rate that fits into this equation (i= 7.93%). Therefore, the rate of return on this investment (or Internal Rate of Return) is i= 7.93% per year.

Again, assume all the parameters are known and specified except the rate of return i. In order to determine i, usually, a trial and error method is used that will be explained in Example 2-10 and the following video.

Example 2-10:

In Example 2-9, assume 20,000 dollars is paid for the asset in present time (C = 20,000 dollars), a yield of 2,000 dollars per year in income after all expenses is expected for each of the next ten years and also the resale value in the tenth year will be 25,000 dollars. What annual compound interest rate, or return on investment dollars, will be received for this cash flow?

Figure 2-4: Cash flow: 20,000 dollars investment at present time, 2,000 dollars per year in income after all expenses for 10 years and resale value of $25,000 in the tenth year.

An equation can be written setting costs equal to income at any point in time and the project rate of return i can be calculated, i.e., the beginning or end of any period. Here, we will use the present value method to determine internal rate of return, i.
In order to solve this problem, an equation that equates costs to income at any point in time (for example beginning or end of any period) should be written with the project rate of return i as an unknown variable.

present value equation at present time to calculate i:
$\text{present value of cost }=\text{ present value of income }+\text{ present value of salvage}$

$\begin{array}{c}\text{present value of all costs }\left( C \right)=20,000 \\ \text{present value of all incomes }=2,000*\left( P/A_{i,10}\right)=2000*\left[ {\left( 1+i \right)}^{10}-1 \right]/\left[ i{\left( 1+i \right)}^{10}\right] \\ \text{present value of salvage }=25,000*\left( P/F_{i,10}\right)=25,000*\left[ 1/{\left( 1+i \right)}^{10}\right]\end{array}$

$20,000=2000*\left[ {\left( 1+i \right)}^{10}-1 \right]/\left[ i{\left( 1+i \right)}^{10}\right]+25,000*\left[ 1/{\left( 1+i \right)}^{10}\right]$

It is very difficult to solve this explicitly for i. By trial and error, we can easily find the i that makes the right side of the equation equal to the left side.

For the initial guess of i=10% , the left side is:
$2,000*6.1446+25,000*0.3855=\$21,930$

And for i=12% , the left side is:
$2,000*5.6502+25,000*0.3220=\$19,350$

Then, we can try i=11% (the middle point) and i=11.5% to find 11.5% is the rate of return to make the left side to equal to the right side.

In Excel specifically, another way to calculate the break-even rate of return is to use the IRR function. As long as the project has an investment cost in the present year and subsequent cash flows, you can use the IRR function to calculate the Internal Rate of Return. (If the project has a different cost and cash flow structure, then it's harder to use the Excel function here.) This video has a short example (without any narration) of the Excel IRR function. The Excel help file for IRR is also very useful.

For an illustration of the trial and error method, see the following video, Trial and error problem in Excel (6:52).

(Please use 1080p HD resolution to view it).

Summary

In Lesson 2 we have learned:

• nominal, period, and effective interest rates based on a discrete compounding of interest;
• present, annual, and future values and the conversions between the values as one of the most important fundamentals of the class; and
• how to conduct rate of return and break-even analysis, which are very critical frameworks for investment decision-making.

Reminder - Complete all of the Lesson 2 tasks!

You have reached the end of Lesson 2! Double-check the to-do list on the Lesson 2 Overview page to make sure you have completed all of the activities listed there before you begin Lesson 3.