Engineering Economics Blog

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The  most fundamental factor in engineering economy  is the one that determines the amount of money   F  accumulated after   n  years (or periods) from a   single  present worth P, with interest compounded one time per year (or period). Recall that compound interest refers to interest paid on top of interest. Therefore, if an amount   P  is invested at time t = 0, the amount F1 accumulated 1 year hence at an interest rate of i percent per year will be


where the interest rate is expressed in decimal form. At the end of the second year, the amount accumulated   F2 is the amount after year 1 plus the interest from the end of year 1 to the end of year 2 on the entire F1

The amount  F2 can be expressed as

Similarly, the amount of money accumulated at the end of year 3, using Equation [2.1], will be

Figure 2–1 Cash fl ow diagrams for single-payment factors: (a) fi  nd F, given P, and (b) fi  nd P, given F.

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The cash fl ow is fundamental to every economic study. Cash fl ows occur in many confi  gurations and amounts - isolated single values, series that are uniform, and series that increase or decrease by constant amounts or constant percentages.

This chapter develops derivations for all the commonly used engineering economy factors that take the time value of money into account.
  
The application of factors is illustrated using their mathematical forms and a standard notation format. Spreadsheet functions are used in order to rapidly work with cash fl  ow series and to perform sensitivity analysis.
  

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A Japan-based architectural fi rm has asked a United States–based software engineering group to infuse GPS sensing capability via satellite into monitoring software for high-rise structures in order to detect greater than expected horizontal movements. This software could be very benefi  cial as an advance warning of serious tremors in earthquake-prone areas in Japan and the United States. The inclusion of accurate GPS data is estimated to increase annual revenue over that for the current software system by $200,000 for each of the next 2 years, and by $300,000 for each of years 3 and 4. The planning horizon is only 4 years due to the rapid advances made internationally in building-monitoring software. Develop spreadsheets to answer the questions below.  
 
(a) Determine the total interest and total revenue after 4 years, using a compound rate of r eturn of 8% per year.   
(b) Repeat  part  (  a ) if estimated revenue increases from $300,000 to $600,000 in years 3 and 4.  
(c) Repeat  part  (  a ) if infl ation is estimated to be 4% per year. This will decrease the   real rate of return  from 8% to 3.85% per year (Chapter 14 shows why).  
 
Solution by Spreadsheet

Refer to  Figure 1–14   a  to   d  for the solutions. All the spreadsheets contain the same information, but some cell values are altered as required by the question. (Actually, all the questions can be answered on one spreadsheet by changing the numbers. Separate spreadsheets are shown here for explanation purposes only.)
 
The Excel functions are constructed with reference to the cells, not the values themselves, so that sensitivity analysis can be performed without function changes. This approach treats the value in a cell as a   global variable  for the spreadsheet. For example, the 8% rate in cell B2 will be referenced in all functions as B2, not 8%. Thus, a change in the rate requires only one alteration in the cell B2 entry, not in every relation where 8% is used.

See Appendix A for additional information about using cell referencing and building spreadsheet relations. 
 
(a) Figure 1–14 a shows the results, and  Figure 1–14 b presents all spreadsheet relations for estimated interest and revenue (yearly in columns C and E, cumulative in columns D and F). As an illustration, for year 3 the interest I3  and revenue plus interest R3 are


 The detailed relations shown in  Figure 1–14   b  calculate these values in cells C8 and E8.


The equivalent amount after 4 years is $1,109,022, which is comprised of $1,000,000 in total revenue and $109,022 in interest compounded at 8% per year. The shaded cells in Figure  1–14   a  and   b  indicate that the sum of the annual values and the last entry in the cumulative columns must be equal. 
 
(b) To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000, use the same spreadsheet and change the entries in cells B8 and B9 as shown in Figure  1–14   c . Total interest increases 22%, or $24,000, from $109,222 to $133,222. 
 
(c) Figure  1–14   d  shows the effect of changing the original   i  value from 8% to an infl  ationadjusted rate of 3.85% in cell B2 on the fi rst spreadsheet. [Remember to return to the $300,000 revenue estimates for years 3 and 4 after working part (  b ).]  Infl  ation has now reduced total interest by 53% from $109,222 to $51,247, as shown in cell C10.  

Figure 1–14  Spreadsheet solutions with sensitivity analysis, Example 1.17  a  to   c .

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The functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency computations involving   compound interest  and the terms   P, F, A , i, and n . The use of a calculator to solve most simple problems is preferred by many students and professors as described in Appendix D. However, as cash fl  ow series become more complex, the spreadsheet offers a good alternative.  Microsoft Excel   is used throughout this book because it is readily available and easy to use. Appendix A is a primer on using spreadsheets and Excel. The functions used in engineering economy are described there in detail, with explanations of all the parameters. Appendix A also includes a section on spreadsheet layout that is useful when the economic analysis is presented to someone else - a coworker, a boss, or a professor.
 
A total of seven Excel functions can perform most of the fundamental engineering economy calculations. The functions are great supplemental tools, but they do not replace the understanding of engineering economy relations, assumptions, and techniques. Using the symbols   P, F, A, i , and n defi  ned in the previous section, the functions most used in engineering economic analysis are formulated as follows.



If some of the parameters don’t apply to a particular problem, they can be omitted and zero is assumed. For readability, spaces can be inserted between parameters within parentheses. If the parameter omitted is an interior one, the comma must be entered. The last two functions require that a series of numbers be entered into contiguous spreadsheet cells, but the fi  rst fi  ve can be used with no supporting data. In all cases, the function must be preceded by an equals sign ( ) in the cell where the answer is to be displayed.
 
To understand how the spreadsheet functions work, look back at Example 1.6a , where the equivalent annual amount   A  is unknown, as indicated by   A =?. (In Chapter 2, we learn how engineering economy factors calculate   A , given   P, i, and n .)  To  fi  nd  A using a spreadsheet function, simply enter the PMT function PMT(5%,5,5000).  Figure 1–13  is a screen image of a spreadsheet with the PMT function entered into cell B4. The answer ($1154.87) is dis-
played. The answer may appear in red and in parentheses, or with a minus sign on your screen
to indicate a negative amount from the perspective of a reduction in the account balance. The
right side of  Figure 1–13  presents the solution to Example 1.6  b.  The future value   F  is determined by using the FV function. The FV function appears in the formula bar; and many examples throughout this text will include cell tags, as shown here, to indicate the format of important entries. 
 
The following example demonstrates the use of a spreadsheet to develop relations (not built-in functions) to calculate interest and cash fl ows. Once set up, the spreadsheet can be used to perform sensitivity analysis for estimates that are subject to change. We will illustrate the use of spreadsheets throughout the chapters. (  Note:  The spreadsheet examples may be omitted, if spreadsheets are not used in the course. A solution by hand is included in virtually all examples.)

Figure 1–13 Use of spreadsheet functions PMT and FV, Example 1.6.

EXAMPLE 1.17 - Introduction to Spreadsheet Use

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For any investment to be profi  table, the investor (corporate or individual) expects to receive more money than the amount of capital invested. In other words, a fair   rate of return,  or   return on investment,  must be realizable. The defi nition of ROR in Equation [1.4] is used in this discussion, that is, amount earned divided by the principal.
  
Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be expected.

Therefore, some reasonable rate must be established for the selection criteria (step 4) of the engineering economy study ( Figure 1–1 ).

The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established for the evaluation and selection of alternatives. A project is not economically viable unless it is   expected to return at least the MARR.  MARR is also referred to as the hurdle rate, cutoff rate, benchmark rate, and minimum acceptable rate of return.
    
Figure 1–12  indicates the relations between different rate of return values. In the United States, the current U.S. Treasury Bill return is sometimes used as the benchmark safe rate. The MARR will always be higher than this, or a similar, safe rate. The MARR is not a rate that is calculated as a ROR. The MARR is established by (fi nancial) managers and is used as a criterion against which an alternative’s ROR is measured, when making the accept/reject investment  decision.     
  
To develop a foundation-level understanding of how a MARR value is established and used to make investment decisions, we return to the term   capital  introduced in Section 1.1.   Although the MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally connected to how much it costs to obtain the needed capital funds.   It always costs money in the form of interest to raise capital. The interest, expressed as a percentage rate per year, is called the   cost of capital.  As an example on a personal level, if you want to purchase a new widescreen HDTV, but do not have suffi cient money (capital), you could obtain a bank loan for, say, a cost of capital of 9% per year and pay for the TV in cash now. Alternatively, you might choose to use your credit card and pay off the balance on a monthly basis. This approach will probably cost you at least 15% per year. Or, you could use funds from your savings account that earns 5% per year and pay cash. This approach means that you also forgo future returns from these funds. The 9%, 15%, and 5% rates are your cost of capital estimates to raise the capital for the system by different methods of capital fi nancing. In analogous ways, corporations estimate the   cost of capital  from different sources to raise funds for engineering projects and other types of projects.

Figure 1–12 Size of MAAR relative to other rate of return  values.

In general, capital is developed in two ways—equity fi nancing and debt fi  nancing. A combination of these two is very common for most projects. Chapter 10 covers these in greater detail, but a snapshot description follows.
  
Equity fi  nancing  The corporation uses its own funds from cash on hand, stock sales, or retained earnings. Individuals can use their own cash, savings, or investments. In the example above, using money from the 5% savings account is equity fi  nancing.  
   
Debt fi  nancing  The corporation borrows from outside sources and repays the principal and interest according to some schedule, much like the plans in Table 1–1. Sources of debt capital may be bonds, loans, mortgages, venture capital pools, and many others. Individuals, too, can utilize debt sources, such as the credit card (15% rate) and bank options (9% rate) described above.  
  
Combinations of debt-equity fi nancing mean that a   weighted average cost of capital (WACC)  results. If the HDTV is purchased with 40% credit card money at 15% per year and 60% savings account funds earning 5% per year, the weighted average cost of capital is 0.4(15) + 0.6(5) =9% per year.
  
For a corporation, the   established MARR  used as a criterion to accept or reject an investment alternative will usually be equal to or   higher than the WACC  that the corporation must bear to obtain the necessary capital funds. So the inequality


 must be correct for an accepted project. Exceptions may be government-regulated requirements (safety, security, environmental, legal, etc.), economically lucrative ventures expected to lead to other opportunities, etc.
  
Often there are many alternatives that are expected to yield a ROR that exceeds the MARR as indicated in  Figure 1–12 , but there may not be suffi cient capital available for all, or the project’s risk may be estimated as too high to take the investment chance. Therefore, new projects that are undertaken usually have an expected return at least as great as the return on another alternative that is not funded. The expected rate of return on the unfunded project is called the opportunity cost.            
  
The opportunity cost is the rate of return of a forgone opportunity caused by the inability to pursue a project. Numerically, it is the   largest rate of return of all the projects not accepted (forgone) due to the lack of capital funds or other resources.  When no specifi  c MARR is established, the de facto MARR is the opportunity cost, i.e., the ROR of the fi rst project not undertaken due to unavailability of capital funds.          
  
As an illustration of opportunity cost, refer to Figure 1–12 and assume a MARR of 12% per year. Further, assume that a proposal, call it A, with an expected ROR = 13% is not funded due to a lack of capital. Meanwhile, proposal B has a ROR = 14.5% and is funded from available capital. Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13% is the   opportunity cost;  that is, the opportunity to make an additional 13% return is forgone.

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Table 1–1 details four different loan repayment plans described below. Each plan repays a $5000 loan in 5 years at 8% per year compound interest. 
 
• Plan 1: Pay all at end.  No interest or principal is paid until the end of year 5. Interest accumulates each year on the total of principal and all accrued interest. 
 
• Plan 2: Pay interest annually, principal repaid at end.  The accrued interest is paid each year, and the entire principal is repaid at the end of year 5. 

• Plan 3: Pay interest and portion of principal annually.  The accrued interest and one-fi  fth of the principal (or $1000) are repaid each year. The outstanding loan balance decreases each year, so the interest (column 2) for each year decreases. 
  
• Plan 4: Pay equal amount of interest and principal.  Equal payments are made each year with a portion going toward principal repayment and the remainder covering the accrued interest. Since the loan balance decreases at a rate slower than that in plan 3 due to the equal end-of-year payments, the interest decreases, but at a slower rate.   



(a)Make a statement about the   equivalence  of each plan at 8% compound interest. 
(b)Develop an 8% per year   simple  interest repayment plan for this loan using the same approach as plan 2. Comment on the total amounts repaid for the two plans. 
 
Solution 
 
(a) The amounts of the annual payments are different for each repayment schedule, and the total amounts repaid for most plans are different, even though each repayment plan  requires exactly 5 years. The difference in the total amounts repaid can be explained by the time value of money and by the partial repayment of principal prior to year 5.
    
A loan of $5000 at time 0 made at 8% per year compound interest is equivalent to each of the following:
  
    Plan 1   $7346.64 at the end of year 5 
    Plan 2   $400 per year for 4 years and $5400 at the end of year 5 
    Plan 3    Decreasing payments of interest and partial principal in years 1 ($1400) through 5 ($1080) 
    Plan 4   $1252.28 per year for 5 years   
  
An engineering economy study typically uses plan 4; interest is compounded, and a constant amount is paid each period. This amount covers accrued interest and a partial amount of principal repayment. 
 
(b)The repayment schedule for 8% per year simple interest is detailed in Table 1–2. Since the annual accrued interest of $400 is paid each year and the principal of $5000 is repaid in year 5, the schedule is exactly the same as that for 8% per year compound interest, and the total amount repaid is the same at $7000. In this unusual case, simple and compound interest result in the same total repayment amount. Any deviation from this schedule will cause the two plans and amounts to differ.

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Assume an engineering company borrows $100,000 at 10% per year compound interest and will pay the principal and all the interest after 3 years. Compute the annual interest and total amount due after 3 years. Graph the interest and total owed for each year, and compare with the previous example that involved simple interest.
 
Solution

To include compounding of interest, the annual interest and total owed each year are calculated
by Equation [1.8]. 

Figure 1–11  Interest    I  owed and total amount owed for (  a ) simple interest (Example 1.14) and (  b ) compound interest (Example 1.15).

A  more  effi  cient way to calculate the total amount due after a number of years in Example 1.15 is to utilize the fact that compound interest increases geometrically. This allows us to skip the year-by-year computation of interest. In this case, the   total amount due at the end of each year   is


This allows future totals owed to be calculated directly without intermediate steps. The general form of the equation is


where    i  is expressed in decimal form. Equation [1.10] was applied above to obtain the $133,100 due after 3 years. This fundamental relation will be used many times in the upcoming chapters.
  
We can combine the concepts of interest rate, compound interest, and equivalence to demonstrate that different loan repayment plans may be equivalent, but differ substantially in amounts paid from one year to another and in the total repayment amount. This also shows that there are many ways to take into account the time value of money.

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GreenTree Financing lent an engineering company $100,000 to retrofi t an environmentally unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money will the fi rm repay at the end of 3 years?

Solution

The interest for each of the 3 years is

Total interest for 3 years from Equation [1.7] is


The amount due after 3 years is

                                             Total  due = $100,000 + 30,000 = $130,000  

The interest accrued in the fi rst year and in the second year does not earn interest. The interest due each year is $10,000 calculated only on the $100,000 loan principal.


In most fi nancial and economic analyses, we use   compound interest   calculations.

For compound interest,  the interest accrued for each interest period is calculated on the principal plus the total amount of interest accumulated in all previous periods.   Thus, compound interest means interest on top of interest.
  
Compound interest refl ects the effect of the time value of money on the interest also. Now the
interest for one period is calculated as


In mathematical terms, the interest I t for time period t may be calculated using the relation.

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The  terms    interest,     interest period,  and   interest rate  (introduced in Section 1.4) are useful in calculating equivalent sums of money for one interest period in the past and one period in the future.

However, for more than one interest period, the terms   simple interest  and   compound interest   become important.
  
Simple interest  is calculated using the principal only, ignoring any interest accrued in preceding interest periods. The total simple interest over several periods is computed as where I is the amount of interest earned or paid and the interest rate I is expressed in decimal form.

EXAMPLE 1.14 - Simple and Compound Interest

EXAMPLE 1.15 - Simple and Compound Interest

EXAMPLE 1.16 - Simple and Compound Interest

 

 

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Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to work on a wider variety of electronic items and increase his annual revenue. Howard received 2-year repayment options from banks A and B.


After reviewing these plans, Howard decided that he wants to repay the $10,000 after only 1 year based on the expected increased revenue. During a family conversation, Howard’s brother-in-law offered to lend him the $10,000 now and take $10,600 after exactly 1 year.

Now Howard has three options and wonders which one to take. Which one is economically
the best?
 
Solution

The repayment plans for both banks are economically equivalent at the interest rate of 5% per year. (This is determined by using computations that you will learn in Chapter 2.) Therefore, Howard can choose either plan even though the bank B plan requires a slightly larger sum of money over the 2 years.

The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the principal of $10,000, which makes the interest rate 6% per year. Given the two 5% per year options from the banks, this 6% plan should not be chosen as it is not economically better than the other two. Even though the sum of money repaid is smaller, the timing of the cash fl  ows and the interest rate make it less desirable.   The point here is that cash fl  ows themselves, or their sums, cannot be relied upon as the primary basis for an economic decision. The interest rate, timing, and economic equivalence must be considered. 

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Manufacturers make backup batteries for computer systems available to Batteries + dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries + outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs.
 
  (a)    The amount of $98 now is equivalent to a cost of $105.60 one year from now. 
  (b)    A truck battery cost of $200 one year ago is equivalent to $205 now. 
   (c)    A $38 cost now is equivalent to $39.90 one year from now. 
  (d)    A $3000 cost now is equivalent to $2887.14 one year earlier. 
  (e)    The carrying charge accumulated in 1 year on an investment of $20,000 worth of  batteries  is  $1000.  

Solution 

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Economic equivalence is a fundamental concept upon which engineering economy computations are based. Before we delve into the economic aspects, think of the many types of equivalency we may utilize daily by transferring from one scale to another. Some example transfers between scales are as follows:

Often equivalency involves two or more scales. Consider the equivalency of a   speed  of 110 kilometers per hour (kph) into miles per minute using conversions between distance and time scales with three-decimal accuracy.



Four scales—time in minutes, time in hours, length in miles, and length in kilometers are combined to develop these equivalent statements on speed. Note that throughout these statements, the fundamental relations of 1 mile 1.609 kilometers and 1 hour 60 minutes are applied. If a fundamental relation changes, the entire equivalency is in error.

Now  we  consider  economic  equivalency.  
  

Economic equivalence  is a combination of   interest rate  and   time value of money  to determine the different amounts of money at different points in time that are equal in economic value.  

As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to $106 one year from today.     


If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective. In either case you have $106 one year from today. However, the two sums of money are equivalent to each other   only  when the interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106
one year from today.

In addition to future equivalence, we can apply the same logic to determine equivalence for previous years. A total of $100 now is equivalent to $100 /1.06 = $94.34 one year ago at an interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year.

The fact that these sums are equivalent can be verifi ed by computing the two interest rates for 1-year interest periods.

and


The cash fl ow diagram in  Figure 1–10  indicates the amount of interest needed each year to make these three different amounts equivalent at 6% per year.

Figure 1–10 Equivalence of money at 6% per year interest.

EXAMPLE 1.12 - Economic Equivalence

EXAMPLE 1.13 - Economic Equivalence

 

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A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. The $100 spent on maintenance the fi rst year has increased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash fl ow diagram from the company’s perspective and indicate where the present worth now is located.


Solution

Let now be time t = 0. The incomes and costs for years 7 through 1 (next year) are tabulated below with net cash fl ow computed using Equation [1.5]. The net cash fl ows (one negative, eight positive) are diagrammed in  Figure 1–9 . Present worth   P  is located at year 0.

Figure 1–9 Cash fl ow diagram, Example 1.11.

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An electrical engineer wants to deposit an amount   P  now such that she can withdraw an equal annual amount of  A1 = $2000 per year for the fi rst 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash fl ow diagram appear if i = 8.5% per year?
 
Solution

The cash fl  ows are shown in Figure 1–8. The negative cash outfl  ow   P  occurs now. The withdrawals (positive cash infl ow) for the A1 series occur at the end of years 1 through 5, and A2 occurs in years 6 through 8.

Figure 1–8  Cash  fl  ow diagram with two different   A  series, Example 1.10.