Engineering Economics

Eng ineeri ng Economy Third Edition Leland T. Blank, P. E. Department of Industrial Engineering Assistant Dean of Engineering Texas A & M University Anthony J. Tarquin, P. E. Department of Civil Engineering Assistant Dean of Engineering The University of Texas at EI Paso McGraw-Hill Book Company New York S1. Louis San Francisco Auckland Bogota Caracas Colorado Springs Hamburg Lisbon London Madrid Mexico Milan Montreal New Delhi Oklahoma City Panama Paris San Juan Silo Paulo Singapore Sydney Tokyo Toronto 4 Level One 1. Define and recognize in a problem statement the economy symbols P, F, A, n, and i. 1. 6 Define cash flow, state what is meant by end-of-period convention, and construct a cash-flow diagram, given a statement describing the amount and times of the cash flows. Study Guide 1. 1 Basic Terminology Before we begin to develop the terminology and fundamental concepts upon which engineering economy is based, it would be appropriate to define what is meant by engineering economy. In the simplest terms, engineering economy is a collection of mathematical techniques which simplify economic comparisons.

With these techniques, a rational, meaningful approach to evaluating the economic aspects of different methods of accomplishing a given objective can be developed. Engineering economy is, therefore, a decision assistance tool by which one method will be chosen as the most economical one. In order for you to be able to apply the techniques, however, it is necessary for you to understand the basic terminology and fundamental concepts that form the foundation for engineering-economy studies.

Some of these terms and concepts are described below. An alternative is a stand-alone solution for a give situation. We are faced with alternatives in virtually everything we do, from selecting the method of transportation we use to get to work every day to deciding between buying a house or renting one. Similarly, in engineering practice, there are always seveffl ways of accomplishing a given task, and it is necessary to be able to compare them in a rational manner so that the most economical alternative can be selected.

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The alternatives in engineering considerations usually involve such items as purchase cost (first cost), the anticipated life of the asset, the yearly costs of maintaining the asset (annual maintenance and operating cost), the anticipated resale value (salvage value), and the interest rate (rate of return). After the facts and all the relevant estimates have been collected, an engineering-economy analysis can be conducted to determine which is best from an economic point of view.

However, it should be pointed out that the procedures developed in this book will enable you to make accurate economic decisions only about those alternatives which have been recognized as alternatives; these procedures will not help you identify what the alternatives are. That is, if alternatives ,4, B, C, D, and E have been identified as the only possible methods to solve a Particular problem when method F, which was never recognized as an alternative, is really the most attractive method, the wrong decision is certain to be made because alternative F could never be chosen, no matter what analytical techniques are used.

Thus, the importance of alternative identification in the decision-making process cannot be overemphasized, because it is only when this aspect of the process has been thoroughly completed that the analysis techniques presented in this book can be of greatest value. In order to be able to compare different methods for accomplishing a given objective, it is necessary to have an evaluation criterion that can be used as a basis Terminology and Cash-Flow Diagrams 5 for judging the alternatives. That is, the evaluation criterion is that which is used to answer the question "How will I know which one is best? Whether we are aware of it or not, this question is asked of us many times each day. For example, when we drive to work, we subconsciously think that we are taking the "best" route. But how did we define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, depending upon which criterion is used to identify the best, a different route might be selected each time! (Many arguments could have been avoided if the decision makers had simply stated the criteria they were using in determining the best). In economic analysis, dollars are generally used as the basis for comparison.

Thus, when there are several ways of accomplishing a given objective, the method that has the lowest overall cost is usually selected. However, in most cases the alternatives involve intangible factors, such as the effect of a process change on employee morale, which cannot readily be expressed in terms of dollars. When the alternatives available have approximately the same equivalent cost, the nonquantifiable, or intangible, factors may be used as the basis for selecting the best alternative, For items of an alternative which can be quantified in terms of dollars, it is important to recognize the concept of the time value of money.

It is often said that money makes money. The statement is indeed true, for if we elect to invest money today (for example, in a bank or savings and loan association), by tomorrow we will have accumulated more money than we had originally invested. This change in the amount of money over a given time period is called the time value of money; it is the most important concept in engineering economy. You should also realize that if a person or company finds it necessary to borrow money today, by tomorrow more money than the original loan will be owed. This fact is also explained by the time value of money.

The manifestation of the time value of money is termed interest, which is a measure of the increase between the original sum borrowed or invested and the final amount owed or accrued. Thus, if you invested money at some time in the past, the interest would be Interest = total amount accumulated - original investment (1. 1) On the other hand, if you borrowed would be Interest money at some time in the past, the interest (1. 2) = present amount owed - original loan In either case, there is an increase in the amount of money that was originally invested or borrowed, and the increase over the original amount is the interest.

The original investment or loan is referred to as principal. Probs. Interest Calculations When interest is expressed as a percentage of the original amount per unit time, the result is an interest rate. This rate is calculated as follows: . Percent interest rate = interest accrued per unit time 00% .. I x 1 0 origma amount (1. 3) 6 Level One By far the most common time period used for expressing interest rates is 1 year. However, since interest rates are often expressed over periods of time shorter than 1 year (i. e. 1% per month), the time unit used in expressing an interest rate must also be identified and is termed an interest period. The following two examples illustrate the computation of interest rate. Example 1. 1 The Get-Rich-Quick (GRQ) Company invested $100,000 on May 1 and withdrew a total of $106,000 exactly one year later. Compute (a) the interest gained from the original investment and (b) the interest rate from the investment. Solution (a) Using Eq. (1. 1), Interest = 106,000 - 100,000 = $6000 (b) Equation (1. 3) is used to obtain Percent interest rate = 6000 per year 100,000 x 100% = 6% per year.

The two sums of money are therefore equivalent to each other when the interest rate is 12% per year. At either a higher or a lower interest rate, however, $100 today is not equivalent to $112 one year from today. In addition to considering future equivalence, one can apply the same concepts for determining equivalence in previous years. Thus, $100 now would be equivalent to 100/1. 12 = $89. 29 one year ago if the interest rate is 12% per year. From these examples, it should be clear that $89. 29 last year, $100 now, and 112 one year from now are equivalent when the interest rate is 12% per year.

The fact that these sums are equivalent can be established by computing the interest rate as follows: 112 100 = 1. 12, or 12% per year and 8~~~9 = 1. 12, or 12% per year The concept of equivalence can be further illustrated by considering different loan-repayment schemes. Each scheme represents repayment of a $5000 loan in 5 years at 15%-per-year interest. Table 1. 1 presents the details for the four repayment methods described below. (The methods for determining the amount of the payments are presented in Chaps. 2 and 3. )

• Plan 1 a interest or principal is recovered until the fifth year.

Interest accumulates each year on the total of principal and all accumulated interest.

• Plan 2 The accrued interest is paid each year and the principal is recovered at the end of 5 years.

• Plan 3 The accrued interest and 20% of the principal, that is, $1000, is paid each year. Since the remaining loan balance decreases each year, the accrued interest decreases each year.

• Plan 4 Equal payments are made each year with a portion going toward princi- pal recovery and the remainder covering the accrued interest. Since the loan balance decreases at a rate which is slower than in plan 3 because of the equal end-of-year payments, the interest decreases, but at a rate slower than in plan.

That the total amount repaid in each case would be different, even though each repayment scheme would require exactly 5 years to repay the loan. The difference in the total amounts repaid can of course be explained by the time value of money, since the amount of the payments is different for each plan. With respect to equivalence, the table shows that when the interest rate is 15% per year, $5000 at time 0 is equivalent to $10,056. 80 at the end of year 5 (plan 1), or $750 per year for 4 years and $5750 at the end of year 5 (plan 2), or the decreasing amounts shown in years 1 through 5 (plan 3), or $1,491. 8 per year for 5 years (plan 4). Using the formulas developed in Chaps. 2 and 3, we could easily show that if the payments in Terminology and Cash-Flow Diagrams 9 each plan (column 4) were reinvested at 15% per year when received, the total amount of money available at the end of year 5 would be $10,056. 80 from each repayment plan. Additional Examples 1. 14 and 1. 15 Probs. 1. 8 and 1. 9 1. 4 Simple and Compound Interest The concepts of interest and interest rate were introduced in Sees. 1. 1 and 1. 2 and ed in Sec. 1. 3 to calculate for one interest period past and future sums of money equivalent to a present sum (principal).

Vote that the total investment in plan 1 is $15,660 while the total required in plan _ to accumulate the same amount of money is nearly six times greater at $88,400. Both the power of compounding and the wisdom of planning for your retirement at he earliest possible time should be quite evident from this example. An interesting observation pertaining to compound-interest calculations in-olves the estimation of the length of time required for a single initial investment to double in value. The so-called rule of 72 can be used to estimate this time.

It should be understood that a present sum P represents a single sum of money at some time prior to a future sum or uniform series amount and therefore does not necessarily have to be located at time t = O. Example 1. 11 shows a P value at a time other than t = O. The units of the symbols aid in clarifying their meaning. The present sum P and future sum F are expressed in dollars; A is referred to in dollars per interest period. It is important to note here that in order for a series to be represented by the symbol A, it must be uniform (i. e. the dollar value must be the same for each period) and the uniform dollar amounts must extend through consecutive interest periods. Both conditions must exist before the dollar value can be represented by A. Since n is commonly expressed in years or months, A is usually expressed in units of dollars per year or dollars per month, respectively. The compound-interest rate i is expressed in percent per interest period, for example, 5% per year. Except where noted otherwise, this rate applies throughout the entire n years or n interest periods. The i value is often the minimum attractive rate of return (MARR).

At any point in time, the net cash flow would be represented as Net cash flow = receipts - disbursements (1. 5) Since cash flow normally takes place at frequent and varying time intervals within an interest period, a simplifying assumption is made that all cash flow occurs at the end of the interest period. This is known as the end-of-period convention. Thus, when several receipts and disbursements occur within a given interest period, the net cash flow is assumed to occur at the end of the interest period.

However, it should be understood that although the dollar amounts of F or A are always considered to occur at the end of the interest period, this does not mean that the end of the period is December 31. In the situation of Example 1. 7, since investment took place on May 1, 1988, the withdrawals will take place on May 1, 1989 and each succeeding May 1 for 10 years (the last withdrawal will be on May 1, 1998, not 1999). Thus, end of the period means one time period from the date of the transaction (whether it be receipt or disbursement).

Construct the cash-flow diagram for this case, assuming an interest rate of 12% per year. Solution Figure 1. 3 presents the cash-flow diagram. Comment While it is not necessary to use an exact scale on the cash-flow axes, you will probably avoid errors later on if you make a neat diagram. Note also that the present sum P is a receipt at year 0 and the future sum F is a disbursement at the end of year 5. Example 1. 10 If you start now and make five deposits of $1000 per year (A) in a 17%-per-year account, how much money will be accumulated (and can be withdrawn) immediately after you have made the last deposit?

Assume that you want to deposit an amount P into an account 2 years from now in order to be able to withdraw $400 per year for 5 years starting 3 years from now. Assume that the interest rate is 151% per year. Construct the cash-flow diagram. Figure 1. 5 presents the cash flows, where P is to be found. Note that the diagram shows what was given and what is to be found and that a P value is not necessarily located at time t = O. Solution Additional Examples 1. 18 to 1. 20 Probs. 1. 30 to 1. 46 Additional Examples Example 1. 12 Calculate the interest and total amount accrued after 1 year if $2000 is invested at an interest rate of 15% per year.

Assume an interest rate of 12% per year. Solution Comment Figure 1. 8 presents the cash-flow diagram. The two $50 negative cash flows form a series of two equal end-of-year values. As long as the dollar values are equal and in two or more consecutive periods, they can be represented by A, regardless of where they begin or end. However, the $150 positive cash flow in 1999 is a single-occurrence value in the future and is therefore labeled an F value. It is possible, however, to view all of the individual cash flows as F values. The diagram could be drawn as shown in Fig. . 9. In general, however, if two or more equal end-of-period amounts occur consecutively, by the definition in Sec. 105 they should be labeled.

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