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BASIC ENGINEERING ECONOMICS

Keywords: engineering economic analysis, cash flow tables and diagram, present worth, future worth, uniform series, and uniform gradient series, interest factors, and effective and nominal interest rates.

Introduction

Economic analysis is very important to engineers. An important goal of engineers, and their clients, is to implement projects that are sustainable. Sustainable projects heed the triple-bottom-line: meeting environmental, social, AND economic criteria. Because Civil Engineering projects are often very expensive, millions and even billions of dollars, multiple alternatives are almost always considered. One very important criterion in comparing alternatives is the economic benefit.

The economic analysis methods we learn in this class are primarily concerned with comparing two or more alternative engineering projects. This entails finding a way to fairly compare money spent or earned at different times over the life of a project, from construction to operation to dismantling. The value of money varies with time (time-value), thus a fair comparison of projects will take this into account.

Here is a simple example that illustrates one aspect of the time-value of money, that it can be invested and return additional money. Your rich aunt offers to give you $100 now (Alternative 1) or $150 five years from now (Alternative 2). If you don’t understand the time-value of money, you might ask for the $150 five years from now, as $150 is more than $100. Or if you need money right away, you might take the $100, regardless of long-term benefit. But which alternative is best in the long run, i.e., five years from now? One way to compare your alternatives is to determine their value to you at some common point in time. For example, you could compare their worth five years from now. How much will the $100 be worth at the end of 5 years? More or less than the $150 you could wait for? You will need to predict how much you could make by investing the money. Let us assume you can make 10 % interest, 0.1 as a fraction, each year, paid at the end of each year. In five years, the $100 will be worth

$100 x (1 + 0.10) x (1 + 0.10) x (1 + 0.10) x (1 + 0.10) x (1 + 0.10) = $100 (1.1)5 = $161.

Because $161 > $150, Alternative 1 is preferred.

Another way to compare your options is to evaluate their worth to you right now. In this case, Alternative 1 is worth $100. But, what is Alternative 2 worth? The important question is how much money do you need to invest right now to have $150 at the end of 5 years? If we again assume that we can make 10 % interest each year, the mathematical equation is, where X = the amount invested now in order to have $150 after 5 years,

X (1 + 0.1)5 = 150 or X = 150 / (1.1)5 = 150 (1.1)-5 = $93

This is the present worth of the $150. What would you rather have right now, $100 or $93? As expected, Alternative 1 is preferred.

If you really want to get fancy, you could convert the $100 now and the $150 at the end of five years into a uniform series of payments, one at the end of each of the next five years. Having $100 now is equivalent (at 10% interest) to getting $26.4 at the end of each of the next five years (a uniform series of payments). Getting $150 five years from now is equivalent to getting $24.6 at the end of each of the next five years. As expected, Alternative 1 is still best.

These three examples are essentially equivalent. In each analysis, the two alternatives are compared using payments at the same time: the present, five years from now, or a Uniform series of payments over the next five years. While more complicated, the economic evaluation of engineering projects is essentially the same. The costs and benefits must be compared taking into account the time value of money. An engineer typically prefers to compare projects based on their present worth, not a future worth. The uniform series is also used. An actual engineering project will have monetary amounts occurring in many years, both receipts and payments, but the comparison process is fundamentally the same as our simple example.

These examples illustrate the time-value of money based on interest. Because money can “work” for us, providing us interest, we prefer to have a given amount of money now rather than later. We’d rather have $100 now, compared to having $100 five years from now (or even $150!). If we are going to lose the use of the $100 for five years, we will want $161 at the end of the five-year period, if a 10% interest rate is reasonable. Similarly, if an engineering project costs $100M to build, benefits should have a present worth greater than $100M, or a future worth greater than $161M at the end of five years, or provide more than $26.4M at the end of each of the next five years, or provide some other cash flow that exceeds the $100M initial investment.

We will learn five different ways to evaluate Engineering Projects: Present Worth Analysis, Annual Cash Flow Analysis, Rate of Return Analysis, Benefit-Cost Ratio Analysis, and Payback Period Analysis. The second example above is a simple illustration of the first method. The third example above is a simple illustration of the second method. All of these methods are based on an economic model of a project, either a cash flow diagram or table. Mathematical models are then generated based on the cash flow model. The models are descriptive and deterministic. Most can be solved analytically. The Rate of Return Analysis method is solved numerically.

Economic Project Evaluation

Five economic parameters, generated by their respective analysis methods, are used to compare alternatives: Present Worth Net Benefit (PWNB), Annual Net Benefit (ANB), Rate of Return (i*), Benefit Cost Ratio (BC Ratio), and Payback Period (PP). The respective methods are Present Worth Analysis, Annual Cash Flow Analysis, Rate of Return Analysis, Benefit-Cost Ratio Analysis, and Payback Period Analysis. The first two can be used to directly identify projects that maximize net benefits. The higher the present worth of a project, the more net benefit it provides. Similarly, the higher the uniform annual cash flow of a project, the more net benefit it provides. The last three parameters do not identify the project with greatest net benefit, but facilitate other useful comparisons.

The rate of return of a project is the Interest Rate it provides, e.g., a project with a rate of return of 10% is equivalent to investing money in a bank at 10%. All else equal, a larger rate of return is better than a smaller one. The benefit-cost ratio is simply the ratio of benefits to costs for a project. Projects with ratios greater than 1 have benefits that outweigh costs. All else equal, a larger benefit-cost ratio is better than a smaller one. Finally, the payback period is the time it takes a project to return its initial investment. All else equal, shorter payback periods are better than longer ones. Both the benefit-cost ratio and payback period are for a given interest rate.

Engineers design products and facilities for private and public clients. Clients want the benefits associated with a project to outweigh the costs; in fact, they generally want to maximize net benefits. Net Benefits are Total Benefits minus Total Costs, where benefits and costs are measured in equivalent units and time. For example, early in a project to build a large bridge, cost estimates will be generated for a number of alternative designs. All else equal, the alternative with the largest net benefit will be chosen.

Calculating the Present Worth Net Benefit (benefits minus costs, in present dollars) of each alternative involves converting all the costs and benefits associated with each alternative to their present worth. Alternatively, costs and benefits can be converted into a Uniform Gradient Series of annual payments (annual cash flow method/uniform series). In this case, the project with the highest annual net benefit is best.

Other clients may want to compare a project to alternative investments. A client that is used to getting 10 % return on investments may not be willing to consider a project with a Rate of Return of only 8 %. Large government projects are sometimes justified using the Benefit-Cost Ratio. Only projects with ratios greater than one will be considered. Finally, many companies use the Payback Period to justify investments such as new equipment. Many companies only invest in projects with fairly short payback periods, as short as 1 or 2 years.

All of the parameters discussed here compare projects fairly, because monetary cash flows are compared for equivalent time frames; however, only the present worth and annual cash flow parameters always give the same results (i.e., the project with maximum PWNB will always be the project with maximum ANB, but may not be the project with highest i* or BC Ratio or the

project with lowest PP). It’s like evaluating athletes based on their ability to bench press and their percent body weight as fat. Both values may be valid measures of athleticism, but ranking athletes by one or the other may not give identical results.

Before we can learn more about using these Engineering Economic Evaluation methods, we must learn some basic concepts: cash flow tables and diagram, present worth, future worth, uniform series, and uniform gradient series, interest factors, and effective and nominal interest rates.

Cash Flow Diagram or Table

A Cash Flow Diagram or Table is a schematic model of a project. Either can be used to identify the proper equations (called interest factors) to use to convert a project to an equivalent present worth or uniform series. A cash flow diagram (see below) consists of a horizontal time line with vertical arrows at regular intervals. The regular intervals represent the Compounding Period, usually years. Arrows projecting upward represent benefits, while arrows projecting downward represent costs. The length of the arrow is proportional to its amount; however, diagrams are not always drawn to scale. While not included in the diagram below, sufficient information is usually given on the diagram to identify the amount each arrow represents. When a single arrow is used at each time interval, it usually represents the net benefit (or cost).

A cash flow table ( e.g., Table 1) provides the same information as a cash flow diagram, but in a tabular form. Each row represents the end of a Compounding Period, from the end of period zero (i.e., the beginning of period 1) to the end of period n. Positive monetary amounts represent net benefits, while negative amounts (or in red, parentheses, or both) represent net costs.

The amount represented by a single arrow (Diagram) or given in a row (Table) is the total amount for that period, i.e., the sum of all benefits and costs for that period. The diagram above could be used to represent Alternative 1 in Table 1.

Table 1: Cash Flow Table

End of Period, yr

Alternative 1, $

Alternative 2a, $

Alternative 2b, $

0

-$20,000

-$40,000

-$40,000

1

$10,000

$20,000

$20,000

2

$10,000

$40,000

$0

3

$10,000

NA

$20,000

4

$10,000

NA

$0

5

$10,000

NA

$20,000

6

$20,000

NA

$40,000

NA = Not Applicable

Typical amounts used to create a cash flow table or diagram are Capital Costs (CC), Revenue (R), Operation and Maintenance Costs (O&M) costs, and Salvage Value (SV). Capital Costs and O&M costs are indeed costs. Capital Costs are, e.g., the cost of land, equipment and buildings. They always occur at the beginning of an alternative, but may also occur periodically, e.g., when equipment is periodically replaced over an alternative’s lifetime. O&M costs are typically recorded at the end of each period. They are the cost of operating and maintaining a given project, e.g., wages, supplies, energy costs, and repairs. Revenues and Salvage Value are benefits. Like O&M costs, Revenues are typically recorded at the end of each period. They are the gross profits, realized from the sale of products or services. Salvage Values occur at the end of a lifetime, e.g., the end of equipment or facility lifetimes. They consist of the remaining value of any facilities, land, and/or equipment that are sold. They always occur at the end of the alternative’s lifetime, but can also occur periodically, e.g., if used equipment is periodically sold over an alternative’s lifetime.

The project lifetime is represented by the length of the timeline (Diagram) or the number of rows (Table). The sum of all costs and benefits for a given period is shown in the associated arrow or row. For example, the sum of revenues and O&M costs during period 1 for Alternative 1 is $10,000 (Table 1).

Cash flow diagrams or tables are often created from project summaries. Project summaries for Alternatives 1 & 2 are given in Table 2.

Table 2: Project Summaries

Description

Alternative 1, $

Alternative 2, $

Capital Costs (facilities &

equipment only), $

$20,000

$40,000

Revenue, $/yr

$20,000

$30,000

O&M, $/yr

$10,000

$10,000

Salvage Value, $

$10,000

$20,000

Lifetime

6

2

Capital Costs are always placed at the beginning of the time line of a Cash Flow Diagram and in the first row of a Cash Flow Table. The difference between Revenue and O&M (operating and maintenance) costs are placed in the arrows or rows representing periods 1 through n. Salvage Values are added to the amount for the last period. Following this procedure, the project summaries given in Table 2 above can be used to create the Alternative 1 and Alternative 2a cash flow tables given in Table 1.

In some cases, a fair comparison of projects requires that each project operate for the same time span. For alternatives 1 & 2 in Tables 1 & 2, this can be accomplished by repeating Alternative 2 three times to give a total Project Lifetime of 6 years. When a project is repeated, the equipment Capital Costs are repeated at the beginning of each new lifetime, while Salvage Values are applied at the end of each project lifetime. Thus, for Alternative 2, the total amount at the end of year 2 is calculated from the Capital Costs of new equipment, the Salvage Value of the old equipment, the Revenue, and the O&M. The same is done for year 4. The amount for year 6 does not include the purchase of new equipment as the alternative is over. The cash flow table resulting from repeating Alternative 2 for 6 years is given in the third column of Table 1 (Alternative 2b).

It can be helpful to view alternatives as cash flow diagrams. Alternative 2b is shown in Cash Flow Diagrams 2 and 3, below. The first diagram shows the cash flow with a single arrow for each period. The second diagram uses a separate arrow for different cash flow types (Capital Costs, Revenue minus Operation and Maintenance Costs, and Salvage Value), which can be useful for certain types of economic evaluations.

Present Worth, P

Present worth occurs now. Shown on a cash flow diagram, it appears at the beginning of the first period.

Various equations employing Interest Factors can be used to determine the Present Worth of various cash flows.

Present worth of a Future Worth: ( )

Present worth of a Uniform Series: ( ) ( )

Present worth of a Uniform Gradient Series: (( ) ( ) ( ) )

Where P = Present Worth, F = Future Worth, A = Uniform Series, G = Uniform Gradient Series, i = interest rate applied per compounding period, and n = number of compounding periods. A uniform series consists of a series of uniform amounts, A, each occurring at the end of periods 1 to n. A uniform gradient series consists of an amount 0 at the end of the first period, an amount G the end of the second period, than an amount of 2G and so on until an amount of (n-1)G occurs at the end of period n.

Note: In the equations above, the Present Worth must occur at the beginning of the cash flow diagram. Similarly, the Future Worth must occur at the end of period n. The Uniform Gradient Series and Uniform Series must each start at the end of period 1 and end at the end of period n.

Future Worth

Future worth is the value of an amount of money at some future time. Shown on a cash flow diagram, it appears anywhere but at the beginning.

Various equations employing Interest Factors can be used to determine the Future Worth of various cash flows.

Future worth of a Present Worth: ( )

Future worth of a Uniform Series: ( )

Future worth of a Uniform Gradient Series: (( ) )

Where P = Present Worth, F = Future Worth, A = Uniform Series, G = Uniform Gradient Series, i = interest rate applied per compounding period, and n = number of compounding periods.

Note 1: In the equations above, the Future Worth occurs at the end of period n. This means that the equation that converts a Uniform Series to its equivalent Future Worth does so only for a Future Value that occurs at the end of the Uniform Series. Similarly, the equation that converts a Uniform Gradient Series to its equivalent Future Worth does so only for a Future Value that occurs at end of the Uniform Series.

Note 2: Two steps are required to convert a Uniform Series to an equivalent future amount that does not occur at the end of the Uniform Series. First convert the Uniform Series to its equivalent Present Worth, second convert the Present Worth to the appropriate Future Worth. The same applies to a Uniform Gradient Series.

Uniform Series

A Uniform Series consists of uniform amounts repeating over the length of a cash flow diagram, each period, from the end of period 1 to period n.

For a Uniform Series, the cash amount at the end of periods 1 through n is A.

Various equations employing Interest Factors can be used to determine the equivalent Uniform Series of various cash flows.

Uniform Series equivalent to a Present Worth: ( ) ( )

Uniform Series equivalent to a Future Worth: ( )

Uniform Series equivalent to a Uniform Gradient Series: ( ( ) )

Where P = Present Worth, F = Future Worth, A = Uniform Series, G = Uniform Gradient Series, i = interest rate applied per compounding period, and n = number of compounding periods.

Note: In the equations above, the Present Worth must occur at the beginning of the cash flow diagram. Similarly, the Future Worth must occur at the end of period n. The Uniform Gradient Series must have the same start and end point as the Uniform Series.

Uniform Gradient Series

A Uniform Gradient Series is zero at the end of period 1, then increases an amount G each period to period n.

Where P = Present Worth, F = Future Worth, A = Uniform Series, G = Uniform Gradient Series, i = interest rate applied per compounding period, and n = number of compounding periods.

For a Uniform Gradient Series, the cash amount at the end of period x is G(x – 1), for x ≥ 1.

Interest Factors

In engineering economic evaluations one often converts one cash flow to an equivalent one using Interest Factors. The basic equation used is X = Y·(X/Y,i,n). X and Y are cash flows with equivalent worth, but that occur at different times. (X/Y,i,n) is the Interest Factor, which represents the equation used to convert Y to an equivalent cash flow X. i is the interest rate applied to each period. n is the number of compounding periods, often years, but can take other lengths, including quarters, months or days. Interest factors for a given Y, I and n can be obtained from Factor Tables, formulas or software.

At their simplest, X and Y are single amounts, i.e., benefits or costs that occur once. Y might be a future value occurring after n periods that one wants to convert to an equivalent present amount, X, using an interest rate of i. At their most complex, X and/or Y are series of amounts, which may be uniform or varying. A number of Interest Factors are described below. Let P = Present Amount, F = Future Amount, A = an amount repeated in a uniform series, G = the increment in a Uniform Gradient Series (a series that is zero at the end of the first period, then increases by an amount G each period thereafter), i = the effective interest rate, n = the number of compounding periods.

Single Payment Compound Amount: (F/P,i,n) = ( ) (Converts PW to FW)

Single Payment Present Worth: (P/F,i,n) = ( ) (Converts FW to PW)

Uniform Series Sinking Fund: (A/F,i,n) = ( ) (Converts FW to Uniform Series)

Capital Recovery: (A/P,i,n) = ( ) ( ) (Converts PW to Uniform Series)

Uniform Series Compound Amount: (F/A,i,n) = ( ) (Converts Uniform Series to FW)

Uniform Series Present Worth: (P/A,i,n) = ( ) ( ) (Converts Uniform Series to PW)

Uniform Gradient Present Worth: (P/G,i,n) = ( ) ( ) ( ) (Converts Uniform Gradient Series to PW)

Uniform Gradient Future Worth: (F/G,i,n) = ( ) (Converts Uniform Gradient Series to FW)

Uniform Gradient Uniform Series: (A/G,i,n) = ( ) (Converts Uniform Gradient Series to Uniform Series)

A uniform series consists of a series of uniform amounts, A, each occurring at the end of periods 1 to n. A uniform gradient series consists of an amount 0 at the end of the first period, an amount G the end of the second period, than an amount of 2G and so on until an amount of (n-1)G occurs at the end of period n. Of course, the use of spreadsheets and other computer software makes it possible to avoid using Interest Factors. However, problem formulation, especially on tests, is best accomplished by first writing down simple economic problems in terms of interest factors.

When working with a cash flow table or diagram one can take each amount and convert it to a present worth using the Single Payment Present Worth factor; however, many cash flows encountered in practice can be converted more easily using a combination of the factors above. For example, a cash flow consisting of an amount A at the end of the first period that increases an amount G every period thereafter can be separated into a Uniform Series and a Uniform Gradient Series.

Effective Interest Rate

A key part of using any Interest Factor is the interest rate. An effective interest rate (sometimes called simply an interest rate) will increase or decrease an amount by a fraction each compounding period. It is often given as a percent. For example, money invested at an annual interest rate of 10% increases by an amount 0.1 X in a year, where X is the amount at the beginning of the year. In this case, the compounding period is one year. For example, $500 will grow by $500 x 0.1 = $50 in a year. In the second year the $550 would grow an additional $550 x 0.1 = $55.

Any compounding period can be used, but days, months, quarters and years are most common. Interest on mortgages is often compounded monthly. Credit card interest is generally compounded daily or monthly. Interest rates are applied to money.

Growth or decline rates are similar to interest rates, but can be applied to any type of value. For example, a town of 100,000 growing at 1% per year will have 100,000 x (1 + 0.01) = 101,000 people after one year. A single bacterium that doubles every 12 hours has a growth rate of 100% in 12 hours. In a week (fourteen 12-hour periods) it will become 1 x (1+1)14 = 16,384 bacteria.

The compounding period determines how often the interest is applied. Unless specifically indicated otherwise, it is best to assume that a given interest rate is compounded yearly, called an effective annual interest rate. A Subperiod Effective Interest Rate is one that is compounded over a smaller period than the effective interest rate of interest. The Subcompounding Periods can be repeated an integer number of times in the period of interest. For example, if the Subcompounding Periods is months, it can be repeated 12 times in a year. The following equation can be used to convert a subperiod effective interest rate to an effective interest rate of some period larger than the subperiod.

( )

Where: i = Effective Interest Rate, (unitless). j = Subperiod Effective Interest Rate, (unitless). m = Number of Subcompounding Periods in the Period of Interest, (unitless).

If the Subperiod Effective Interest Rate is 5% compounded quarterly, the effective annual interest rate is i = (1 + 0.05)4 – 1 = 0.215, or 21.5 %.

Interest rates are sometimes reported as a Nominal Interest Rate. A Nominal Interest Rate can be used to determine the effective interest rate given sufficient information. For example, a mortgage rate described as “8 % compounded monthly” is most likely a nominal rate; in this case the effective interest rate is actually 0.67 % interest applied every month. The 8 % (nominal) rate is divided by the number of compounding periods in a year (in this case 12) to

obtain the actual interest rate applied each compounding period (in this case, each month). The effective interest rate is calculated from a nominal interest as

( ⁄)

Where: i = Effective Interest Rate, (unitless). s = Nominal Interest Rate, (unitless). m = Number of Subcompounding Periods in the Period of Interest, (unitless).

For this example, the effective subperiod interest rate is 8/12 = 0.67% each month and the effective annual interest rate is [1 + (0.08/12)]12 – 1 = 0.083 = 8.3%.

Effective interest rates are used to calculate Present Worth Net Benefit or Future Worth. What is the Future Worth of $100 after 7 years at a Nominal Interest Rate of 8% compounded monthly? Using the effective subperiod interest rate, one has a 0.67% interest rate compounded over 84 months (12 compounding periods each year for 7 years): F = 100(1 + 0.0067)84 = $175. Using the effective annual interest rate, one has an 8.3% interest rate compounded over 7 years (1 compounding period each year for 7 years): F = 100(1 + 0.083)7 = $175. As expected, the same Future Worth is calculated.

Mortgages are compounded monthly, but mortgage interest rates are usually advertised as Nominal Interest Rate. For example, a 6 % mortgage interest rate (nominal) is compounded monthly, so you will pay 6/12 = 0.5 % interest on your principal each month. The effective annual interest rate is 6.17%. Credit card interest rates are compounded monthly or even daily, but credit card interest rates are given as nominal values, called the APR or annual percentage rate. For example, a 12 % APR compounded daily (nominal), means you will pay 12/365 = 0.033 % interest per day on any purchases not paid off in time. The effective interest rate is 12.75%. A 12 % APR compounded monthly, means you will pay 12/12 = 1 % interest per month on any purchases not paid off in time. The effective interest rate is 12.68%.