Engineering Economy

Engineering economy, the analysis of the economic consequences of engineering decisions, was originated by A. M. Wellington in his The Economic Theory of Railway Location, published in 1887. For more information on Wellington, see A. M. Wellington. Engineering economy is now considered a part of the education of every engineer.

There are many bad ways to make decisions. Because of the uncertainties of the future, even a rational method of decision-making can sometimes result in bad choices. However, a bad result is almost guaranteed by a poor decision-making process. A fundamental principle is that choices can be made only among alternatives, and that only the differences between these alternatives are material. A course of action cannot usefully be compared only with itself. We hear every day enthusiasts for one course of action or another recommending their hobby on the basis of its peculiar beauty, with alternatives deprecated if considered at all. In any engineering decision, all reasonable alternatives should be discovered and fairly considered.

If the consequences of a course of action can be reduced to monetary terms, it is easy to compare alternatives on the basis of maximum return. Such considerations expressible in money may be called reducible. In terms of reducible considerations, the decision criterion is that of greatest return. Of course, this may not produce the best decision, since not all considerations can be reduced to money. These non-monetary considerations are called irreducible, and may often be of compelling importance. For example, a course of action may be illegal though profitable. While recognizing irreducible considerations, engineering economy usually ignores them in its objective recommendations, which concern monetary matters only. The expression of truly irreducible considerations in terms of money is usually inappropriate and misleading. The engineering economic analysis is only one consideration when making a decision.

Most decisions involve money flows at different times. The comparison of these flows can only be usefully made if they are reduced to some sort of equivalent amount at the same time, usually the present. The model for this comparison is furnished by the familiar concept of interest. If I have $1000, and do nothing with it, I will still have $1000 at any later time if nobody steals it. If I lend it at interest, I give the money in return for the promise to be repaid at a later date, with the addition of a certain amount of interest. This amount is generally expressed as a percentage, say 5% per annum. Then, the interest will be 5% of $1000, or $50, so I will receive $1050 a year from now. The person borrowing the money intends to use it in some way that will return more than $50 worth, perhaps only in personal satisfaction. This is particularly the case with consumer borrowing, which turns out to be a very profitable way for a financial institution to bleed a sucker dry.

Let P be the present value of the money, $1000, and F the future value. Then, after 1 year, F = P(1 + i), where i is the interest rate as a decimal fraction. In this case, F = P(1 + 0.05). Our deal may be for the borrower to retain the money, and to add the interest each year to the principal. After n years, then F = P(1 + i)n. The amount is said to be compounded annually, as interest is charged on the interest. Instead of years, any time period may be specified, say months. Then, after one year, F = P(1 + i')12, where i' is the interest per month.

It is usual to quote a nominal interest rate per year. If we compound monthly, then the annual nominal rate is i = 12i', where i' is the interest per month. For a 5% nominal rate, the monthly interest rate is i' = 5/12 = 0.42%, approximately. Now, (1 + .05/12)12 = 1.05117, so the interest for a year would be $51.17 instead of $50.00. The effective interest rate is 5.117% for 5% per annum compounded monthly. The nominal rate is only used to find the rate for other periods. It is so easy to turn a low nominal rate that attracts suckers into a high effective rate that lenders must now usually quote both. The effective rate is usually called the APR, the annual percentage rate.

Financial people actually do compound at finite intervals. Technical people might be uncomfortable with this, since it seems to them as if continuous compounding would be more logical. For continuous compounding, F = Peni. If i << 1, then this is approximately F = P(1 + i)n, the result of periodic compounding. The differential equation is dF/dn = iPeni = iF, showing that the growth (interest) is proportional to the amount (principal). In one time unit, F = Pei, so the effective interest rate is ei/100. For i = 0.05, this is 5.127%, which can be compared with 5.117% for monthly compounding. For the usual interest rates, the difference between monthly compounding and continuous compounding is immaterial.

A very common pattern of cash flow is the annuity, a payment or recipt of equal amounts of money A at equal intervals of time. This is actually a level annuity, meaning that the amounts are the same. We can find the final value of an annuity by multiplying each payment by the interest factor taking it to the final time (or, equivalently, to the initial time) and adding the results. It is conventional to consider the payment A to be made at the end of each interval. There is no payment at t = 0, and a payment of A at time t = n. The series can be summed explicitly, with the result F = (A/i)[(1 + i)n - 1], or A = F{i/[(1 + i)n - 1]}. When i << 1, the factor in braces becomes i/ni = 1/n. Then, F = (A/i)[ni] = nA, the sum of the payments, and A = F/n. For a nonzero interest rate, the payments are smaller, since the fund also earns interest. Dividing by (1 + i)n, the value F can be converted to a present value P. This is the value of receipts of equal amounts A for the life n of the annuity. One pays an amount P now for its equivalent in future annuity receipts.

If the interest factor (1 + i)n is much larger than 1, which can occur for either a high interest rate i or a large number of compounding periods n, then we see that P = A/i, approximately. If A is a regular cost that continues indefinitely, then A/i may be called the capitalized cost. It is the amount of money that will yield the cost A regularly when it attracts an interest i. It is easily calculated, and can be used to compare different investments on the basis of first cost (present value). It need not represent an actual fund. On the other hand, it may do so. If a fund is established to endow an annual $1000 scholarship, and the interest rate is 5%, then the capitalized cost is $1000/.05 = $20,000.

The formulas for a level annuity are so convenient that the end of year convention is often used. Costs may occur at any point in an interval, but little error is introduced when the interest rate is not too high if they are assumed to occur at the end of the year. Similarly, if the costs vary by small amounts, there will be little error if they are replaced by an average.

In a graded annuity, the payments or receipts differ by a fixed increment G. That is, we have amounts of A, A + G, A + 2G, ..., A + (n - 1)G at the ends of n periods. G may be negative as well as positive. We can consider A and G separately, adding the final results. The graded annuity then has the amounts 0, G, 2G, ..., (n - 1)G. Note that the payment at the end of the first period is 0. Since 1 + 2 + 3 + ... + n = n(n + 1)/2, the sum of these payments is n(n - 1)G/2, and the average payment is A = (n - 1)G/2. If the interest rate is not zero, the later payments have a smaller equivalent. The simplest way to handle this is to convert the graded annuity to a constant annuity, which depends on the interest rate. We find A = (G/i){1 - ni/[(1 + i)n - 1]}, which converts the graded annuity G to an equal series of payments A, including a payment at t = 1.

The expression for A seems to give the limit A = 0 when i becomes small, but this is incorrect. In this case we must expand to second order: (1 + i)n - 1 = ni + n(n - 1)i2/2 + ... . Then, A = (G/i){1 - 1/[1 + (n - 1)i/2]} = (G/i)[(n - 1)i/2] = (n - 1)G/2, which is the correct result. The factor {i/[(1 + i)n - 1]} may be recognized as the ratio A/F for an annuity of A for n periods. It can be evaluated by taking F = 1 and finding A.

The classic text of Grant and Ireson in the References assumes that interest tables will be used to perform calculations. However, since then much more convenient means of calculation have become available. The HP-48G pocket calculator, for example, does interest calculations with ease and accuracy. The Time Value of Money screen accepts values for n, i, P, A and F. It also allows the choice of payments at the end or beginning of a period, and for multiple payments per period. If you specify 12 payments per year, then n = 12 for compounding monthly. The HP-48 will solve for any one of the five quantities in terms of the other four, not merely evaluate a formula. In particular, it will solve for an interest rate i, which is very convenient. I certainly don't recommend doing engineering economy without a calculator.

To see how to use the HP-48, let us find the present value of an annuity of $1000 per year for 20 years, if the interest rate is 5%. Press → SOLVE, then ↑ (or ↓ 4 times) and OK (soft key F). Highlight a quantity to be changed, key in the value, and press ENTER. Set N = 20, %/yr = 5, PMT: 1000, P/YR = 1, FV = 0, END (the +/- key toggles this). Now highlight PV and press SOLVE (soft key F). The value -12,462.21 will appear in PV. The present value of this annuity is $12,462.21. The signs of the values distinguish payments and receipts. A negative A will reduce a positive P to a smaller F, perhaps zero, while a positive A will increase a zero P to a counterbalancing negative F in the future. The signs that appear may be disconcerting, but the HP-48 considers them rational. Always check a result for reasonableness.

If A = 0, we are using only F = P(1 + i)n. To find a future value, set P equal to the present value, and solve for F, which will appear as negative. To find a present value, set F equal to the future value, and solve for P, which will be negative. The difference in signs may be interpreted to show that one amount is a payment, the other a receipt. Which is which depends on whether one takes the viewpoint of the lender or the borrower.

I made a user-defined function on the HP-48 for finding the level annuity equivalent to a graded annuity. I called the function AG(i,n). It expects i and n on the stack, and is called by pressing VAR (unless this menu is already displayed) and then the soft key AG. After this function is evaluated, it is multiplied by the particular value of G. To create this function, enter 'AG(I,N)=1/i-1/((1+i)^N-1)', press ENTER, and then ←DEF. To remove a function like this, press →MEMORY and highlight the desired function. Then press NXT and then soft key PURGE to remove it. Or recall its name to the stack with 'AG and press ←PURGE.

It sometimes helps to construct a cash flow diagram for a problem. Time periods are marked off on the horizontal axis, while cash receipts are represented by upward arrows of scaled length, and cash disbursements by downward arrows. These diagrams can be related to the values entered into the HP-48, and the timing of the cash flows can be clearly presented.

Textbooks usually present a formidable array of six compound interest factors. If you examine them for a while, it becomes evident that they are only simple combinations of two factors, the compound amount factor (caf) F/P = (1 + i)n and the sinking-fund factor (sff) A/F = i/[(1 + i)n - 1]. All the rest are just reciprocals of these, or products of the two to convert between P and F. The values in the tables given by Grant and Ireson are easily computed by the HP-48, which can, therefore, completely replace them. On a four-function calculator, the caf can be found by repeated multiplication, using the memory (press * and MR repeatedly), and then the sff is easily calculated. The caf is conveniently stored in the memory for multiple uses when solving a problem. As friendly advice, it is not a good idea to perform any calculation on the HP-48 that you do not completely understand and you could not carry out longhand.

The general problem of engineering economy is rather simple. For each alternative we predict the expenses and receipts, then reduce them to the same epoch using P = F/(1 + i)n, where n can be positive or negative. The alternative that offers the most return, or is of lowest cost, for the same investment, is then selected. In this case, we may presume that the interest rate i is known. Alternatively, for each alternative we can determine a rate of return by solving for i, and choose the alternative offering the greatest rate of return, irreducible considerations being equal. We can also reduce all costs and benefits to their values at one time, and then take their ratio, the Benefit-Cost ratio. The alternative with the greatest benefit-cost ratio is then chosen. This method is often used for governmental decisions, since it is very easy to inflate the benefits by assigning high values to irreducibles. Let's now look at the various methods for deciding between alternatives on an economic basis.

Making an economic decision when all the cash flow is current is straightforward. If the irreducible considerations are equivalent, then the alternative with the lowest cost is selected. We may also confront differences in irreducibles with differences in cost and make a more subjective choice. This is done all the time in making purchases in a shop. Engineering decisions are usually more difficult, since the cash flows may occur at different times. Cash flows in the past are common to all alternatives, and so will not affect the choice between alternatives. These are often called sunk costs. Often, it is thought that having already spent a great deal of money on a certain bad idea means that it should be continued, no matter how disadvantageous it may be, to "avoid throwing away the investment." Actually, the investment is already thrown away. This is called "staying the course" in today's politics, which means pursuing a bad idea in spite of negative evidence, or "throwing good money after bad." The desirability of alternatives depends only on the differences between the alternatives in the future. The past is common to all alternatives. Cash flows in the future may occur at different times, and this will affect the relative attractiveness of alternatives.

Many economic alternatives begin with an investment at the present that will result in future receipts. Suppose that two alternatives each require a current investment of $1000. One investment will result in the receipt of $1276 in 5 years, while the other will result in the receipt of $1500 at the same time. Other things equal, we will, of course, choose the alternative yielding $1500. However, suppose the second alternative returns $1500 the end of 6 years instead of 5. Now, which is better? It is better to receive money sooner than later, and this recommends the $1276. It is better to receive more money than less, and this recommends the $1500. To make an objective decision, we may ask what interest rate on our $1000 would produce the stated outcomes. Using the HP-48G, we find that the $1276 is equivalent to 5.00%, while the $1500 is equivalent to 6.99%. The $1500 alternative is better on this basis, since it corresponds to a greater percentage return. No borrowing or lending of money is involved in this decision, though we talk of interest rates.

An alternative investment would be to lend the $1000 at the prevailing interest rate. If that rate were 5%, then this alternative would be just as good as the one offering $1276 in 5 years. It would also involve very little effort, and if the money were placed with a reputable bank, would be secure. Any attractive investment of mony would have to have a rate of return greater than the prevailing interest rate. The prevailing interest rate serves as a floor for acceptable rates of return. The interest rate at which an investment becomes attractive is, therefore, considerably higher than the prevailing interest rate. If the rate of return of an investment is greater than the prevailing interest rate, then we can borrow money to make the investment, and enjoy profit in addition to being able to pay off the loan.

Another way to analyze our choices is to compare the present worth of the alternatives. To do this, we reduce the future receipts to the present by choosing a discount rate. It is called discount because the present worth is always less than the future value, and the smaller as the time of the receipt is farther in the future. It is a measure of how much we prefer a quick return to a later return. If we choose 8%, then the receipt of $1276 after 5 years has a present worth of $868.42, while the $1500 after 6 years has a present value of $945.25. The second alternative is preferable, by $76.83, as we concluded from the rate of return.

If the discount rate is 12%, then the present values become $724.04 and $759.95. The difference is now only $35.91, but the second alternative is still preferred, though by a smaller amount. This demonstrates that selection on the basis of present worth gives the same result as rate of return unless the discount rate is unusually high. Note that in all these cases, the present worth is less than $1000, so it would not be economical to invest this amount to receive the proposed benefits.

A 5%, 10-year, $10,000 bond pays interest of $500 each year and its face value of $10,000 at the end of the 10th year. If the prevailing interest rate is 5%, then the present worth of the interest payments is $3,860.87 and of the principal amount $6,139.13. The sum of these is $10,000, which will be a fair price for the bond. If the prevailing interest rate were 3%, then the present worth of the interest payments would be $4,625, and of the principal $7,440.94, total $12,066.04. This would now be the price of the bond. On the other hand, at 7%, the value of the bond would be only $3,511.79 + $5,083.49 = $8,595.28. This shows how the value of a bond depends on the prevailing interest rate.

If the payments or receipts are irregular, each one may be converted to its equivalent value at a chosen time. With the HP-48G, this is no longer as arduous as it once was, and becomes a practical solution to many problems. A present value is easily converted to a finite or perpetual annuity, and vice versa.

We have illustrated how to compare alternatives on the basis of rate of return or on present worth. Alternatives of the same life can also be compared by equivalent uniform annual costs. To do this, simply convert present worth to a level annuity for the life of the investment. This is convenient for investments with long life, which can often be compared on the basis of the capitalized cost. The benefit to cost ratio can also be used. The "benefit" is the present worth of the receipts, and the "cost" is the present worth of the investment. At 5%, an investment of $1000 that returns $1500 in 5 years has a present worth of $1,175.29, and so the B/C ratio is 1.175. However, at 10%, the benefit is only $931.38, so the B/C ratio has fallen to 0.931. Whether the alternative is desirable or not depends on the interest rate.

Let's give another example of what we mean by rate of return on an investment. Consider the repayment of a debt of $20,000 that will become due in 10 years. One way to do this is to establish a sinking fund that will allow us to repay, or extinguish the debt when it becomes due. This is often called amortization. If there were no interest, then putting aside $20,000/10 = $2000 per year would suffice. However, if the interest rate is 10% per annum, we must set aside more than this. First of all, we must pay the interest on the principal of $20,000, which is $2000. We must also pay into the sinking fund. The annuity necessary to receive $20,000 in 10 years at 10% is $1255. The difference between 10 x $1255 = $12,550 and the future value of the fund is provided by the interest. The total payment will be $3255 per year, the sum of the annuity and the interest on the principal.

The payment in the preceding paragraph may be called the capital recovery annuity amount. There are two ways to calculate it on the HP-48. We may calculate the annuity part by setting P = 0, F = 20,000 and solving for A, which is $1255. Then we add Pi = $2000 to this amount. Alternatively, we may set P = 20,000, F = 0 and solve for A. This gives the $3255 directly. Incidentally, this shows that A/P = A/F + i.

If a loan is repaid by an annuity of equal amounts, one may consider that the payment is first applied to the interest due on the present value at the start of the period. The remainder of the payment reduces the balance owed. The "amortization" function of the HP-48 calculates the part of a payment that is applied to the interest, and the remaining part that is applied to retiring the principal. First, the annuity payment is calculated from the present value, interest rate and number of periods, with future value zero, just as we have done above. The AMOR function then calculates the interest, the amount applied to the principal, and the balance remaining to be paid. It can do this for a group of N payments. If N = 1, it is done payment by payment. One goes on to the next payment by pressing B→PV, which adjusts the present value, and then AMOR again. For example, if $100,000 is borrowed for 20 years at 5%, the annual payment to retire the debt is $8024.26. Of the first payment, $5000 is applied to the interest, leaving $3084.26 to reduce the principal. The balance is now $96,915.74. The interest for the next year will be $4845.79. Over the life of the loan, the total interest paid will be $60,484.20.

If the $20,000 in the paragraph before last is not a debt but an investment, suppose it yields an annual profit A. If the payback varies, we convert it to an equivalent annual amount A. If A = $3255, then this is the same amount that will pay back a $20,000 debt at an interest rate of 10%. This is the rate of return of the investment. If we make an investment, we forgo the income we would receive from competing investments. One of these alternatives is simply to loan the money at the current interest rate. Therefore, the current interest rate is a lower bound on acceptable rates of return. This is the reason why interest plays a part in engineering economy, and why acceptable rates of return are usually a good deal higher than the prevailing rate of interest. Indeed, any concern usually has places to put money that will always pay off better than some larger interest rate, such as 10% or 15%.

The HP-48 makes it very easy to find rate of return. If an investment of $P returns an equivalent annual profit of $A for N years, set the future value F = 0 and solve for the interest rate, which will be the rate of return. If the future value is given instead, set A = 0 and solve for the interest rate. These calculations used to be done by assuming interest rates and interpolating.

An example of a graded series is given in Examples 5-22 and 5-23 in Grant and Ireson. A piece of equipment costs $6000 and has a life of 6 years, with no salvage value. Its operating costs are $1500 the first year, increasing by $200 in each following year. The interest rate is assumed to be 12%. What is the present cost of buying and using the equipment for its 6-year life? The purchase cost is $6000, a present value. The operating costs are to be referred to the present. These are a level annuity of $1500 for six years, and a graded annuity with G = $200. Note that the graded annuity is $0 at the end of the first year, $200 at the end of the second year, and so on, so it agrees with the conventions we adopted above.

The present value of the level annuity is easily found with the HP-48. We have n = 6, i = 12, A = 1500 and F = $0. The annuity represents a charge of $6167.11 at the present time. For n = 6 and i = 12, the ratio A/F is found to be 0.123225. To find this, set F = 1 and solve for A. The equivalent flat annuity is given by A = (200/0.12)[1 - (6)(0.123225)] = $434.42. The present value of a flat annuity of this value is then $1786.08. The total present value of purchasing and using the equipment for 6 years is $6000 + $6167.11 + $1786.08 = $13,953.19. This value may now be compared with the cost of alternatives.

If the equipment had a salvage value of, say, $1000 after 6 years, we can find the present value of this benefit at 12%, which will be $506.69, reducing the cost to $13,446.50. This cost can be annuitized at $3270.53 per annum, perhaps for accounting purposes. Practically any problem in engineering economy can be solved by the methods illustrated here, after it has been properly analyzed and understood. It is dangerous to solve such problems by a rote procedure without full understanding of the value of money at different times. This especially includes computer programs where the assumptions made by the programmer may not be clear. Programmers have far too much time on their hands. Grant and Ireson give 25 examples in Chapter 5 that are easily and quickly solved with the HP-48G in an hour or little more. These examples will give you confidence in solving nearly any problem in engineering economy.

Care should be taken in considering depreciation "costs". These are not actual costs, but the accountant's way of allocating the first cost of a capital asset to different time periods when the asset has a long life. The book value of an asset is the difference between its cost and the accumulated depreciation. It has no relation to any real value of the asset. The depreciation account is a contra-asset account, an expense account that is used to determine profit or loss. No cash flow is involved, and it is not a fund for replacement of equipment. Since depreciation may affect income taxes, it can have an effect on investment decisions, but usually only a small one.

Grant and Ireson explain that many methods used to determine rate of return on an investment by considering book values are erroneous and misleading. These methods are popular among the financial crowd, however, since accounts are more familiar to them than engineering economy. By book values we mean the accounts that are used to determine profit and loss for an accounting period. Usually, a ratio between the profits for a year to the book value of the assets is taken as a rate of return. The result depends, of course, on the method of depreciation, and may not reflect the actual timing of payments. Only a proper consideration of the actual cash flows related to an investment by the methods of compound interest can provide a meaningful rate of return.


E. L. Grant and W. G. Ireson, Principles of Engineering Economy, 5th ed. (New York: Ronald Press, 1970).

Hewlett-Packard, HP48G Series User's Guide (Corvallis, OR: H-P Company, 1993). pp. 18-13 to 18-20.

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Composed by J. B. Calvert
Created 6 February 2005
Last revised 13 February 2005