Chapter 44

An Overview of Engineering Economics

44.1. Introduction

Understanding the basics of economics is important both in the business world and in one’s personal life. It is an understatement to say “Mistakes can be costly!” This chapter is intended to provide a short overview of engineering economics that can be used both in your future work activities as well as your personal life.

44.2. Simple and Compound Interest

Assume that you decide that it is necessary to borrow $10,000 for four years from your parents, who agree if you will pay them 8% interest. The $10,000 is referred to as the principal. The 8% is defined as the interest rate and it is not compounded. This means that at the end of each of the first three years that you will pay $800 (8% of $10,000) for that year’s interest back to your parents. At the end of the four years you will still owe them the $10,000 plus 8% of the $10,000 or $800 for interest. This is referred as simple interest. In other words, at the end of the four-year period, you will have paid a total of $13,200 for the loan of the $10,000. Table 44-1 illustrates this case as plan number one. The table should be self-explanatory.

To illustrate compound interest, a similar example will be used. Suppose your parents are unable or unwilling (if you are a bad credit risk) to loan you the money and you are forced to go to a bank. If you have the necessary collateral (something whose value is worth $10,000), the bank might be willing to loan the $10,000 for the same four-year period at 8% interest rate compounded annually (once per year) on the unpaid balance. This means that you will pay 8% on the unpaid balance each year. Many choices exist on how you repaid the loan. Three different choices will be presented and are included in Table 44-1.

Plan number two illustrates what would happen if you made no payments until the end of the fourth year. At that time, you would be required to make a single payment of $13,605. Plan number three represents making a payment each year equaling one-fourth of the principal and the required interest payment. Its total cost is $12,000. Finally, plan number four indicates the sequence of events that would occur if you were to make four equal payments. How the equal payment value was obtained is not important at this time. Later it should become obvious. The total cost for this plan is $12,076. It should also become apparent that the sooner one starts repaying the principal and interest, the lowest the total cost of the loan. From these examples, you should recognize that time is money (see Figure 44-1). All of the multi-year payments neglect the time value of money. What does this mean? For example, it means that if you could have deposited the money and earned more than 8% compound annually, then plans three and four might not be the best alternatives. In these four examples, all values are rounded to the nearest dollar value.

Table 44-1 Simple and compound interest examples
Plan	End of Year	Interest Due (8% of the Money owed at start of year)	Total Money Owed Before Year-End Payment	Year-End Payment	Money Owed After Year-End Payment
1	0				$10,000
	1	$800	$10,800	$800	$10,000
	2	$800	$10,800	$800	$10,000
	3	$800	$10,800	$800	$10,000
	4	$800	$10,800	$10,800	$0
2	0				$10,000
	1	$800	$10,800	$0	$10,800
	2	$864	$11,664	$0	$11,664
	3	$933	$12,597	$0	$12,597
	4	$1,008	$13,605	$13,605	$0
3	0				$10,000
	1	$800	$10,800	$3,300	$7,500
	2	$600	$8,100	$3,100	$5,000
	3	$400	$5,400	$2,900	$2,500
	4	$200	$2,700	$2,700	$0
4	0				$10,000
	1	$800	$10,800	$3,019	$7,781
	2	$622	$8,403	$3,019	$5,384
	3	$431	$5,815	$3,019	$2,796
	4	$224	$3,019	$3,019	$0

44.3. Terminology

It is helpful to define the terms and associated symbols that will be used in the remainder of this chapter. The necessary terms and their symbols are shown in Table 44-2. As will be explained in the next several sections of this chapter, the financial value of an alternative can be computed at any point in time, which allows the comparison of multiple alternatives.

44.4. Financial–Time diagrams

Financial-time diagrams are a visual method of representing a financial alternative that spans one or more interest periods. Time and, thus, interest periods are represented on the independent (horizontal) axis while the financial values are represented on the dependent (vertical) axis. Financial values above the horizontal axis (positive values) represent receipts or incomes, while values below the horizontal axis (negative values) represent expenses or payments. All transactions are always assumed to occur at the end of an interest period.

Figure 44-2 is provided solely as a means of illustrating all of the terms except i in Table 44-2. P is a present sum of money at time (interest period) zero. The U values represent a uniform series that provides income at periods 1, 2, 3, and 4. The right triangle and its arrows represent a gradient income series that increases from zero to 3G between periods 1 and 4 (G per period). Finally, S in period 4 represents a one-time future income. This example probably is unrealistic in the sense that all of the values are positive. In a practical problem, the sum of the positive values must equal the sum of the negative values when the time value of money is considered.

A more realistic example is given in Figure 44-3. This might represent plan number one Table 44-1 where P equals the $10,000 loan from your parents, U equals the $800 interest payments made at the end of each year, and S equals the $10,000 repayment of the principal at the end of four years.

Suppose that you were given the following situation, could you draw the financial-time diagram to represent it? The result of an at-home production activity produces an income of $9,000 in the first year and decreases by $1000 each year thereafter. The activity is abandoned after 8 years. What is the present worth at time zero of this activity if the interest rate is 10% compounded annually? Analyzing the situation, it should be recognized that there is a decreasing gradient of $1,000 each year. However, since gradients must be positive, a decreasing gradient can be represented as the sum of a positive uniform series and a negative increasing gradient series. The resultant financial diagram is given in Figure 44-4.

Table 44-2 Terms and symbols
Term	Symbol
Interest rate per interest period	i
Number of interest periods involved	n
Present sum of money	P
Future sum of money (see Note 1)	S
Uniform series of money (see Note 2)	U
Uniform gradient of money (see Note 3)	G
Note 1: This term represents a sum of money at the end of
n interest periods from the present date that is equivalent
to P with an interest rate i.
Note 2: This term represents an end-of-period payment or
receipt in a uniform series that continues for the coming n
periods, the entire series is equivalent to P at an interest
rate i.
Note 3: This term represents an end-of-period payment or
receipt in a uniformly increasing series is equivalent to P
at an interest rate of i. The series starts at zero and
increases either positively or negatively by G in each
interest period

It is suggested that you practice by drawing the financial-time diagrams for plan numbers two, three, and four in Table 44-1. The correct answers are shown in Figures 42-5 through Figure 44-7.

44.5. Financial Relationships

This section provides the formulas relating P, U, G, and S in terms of i and n. The most commonly used formulas and their names are as follows:

Given P, to find S: S = P*(1+i)ⁿ

= P*(single payment compound amount factor)

= P*(caf’-i%-n)

Given S, to find P: P = S*[1/(1+i)ⁿ]

= S*(single payment present worth factor)

= S*(pwf’-i%-n)

Given S, to find U: U = S*[i/{(1+i)ⁿ – 1}]

= S*(uniform series sinking fund factor)

= S*(sff-i%-n)

Given P, to find U: U = P*[{i*(1+i)ⁿ}/{(1+i)ⁿ – 1}]

= P*(uniform series capital recovery factor)

= P*(crf-i%-n)

Given U, to find S: S = U*[{(1+i)ⁿ – 1}/i]

= U*(uniform series compound amount factor)

= U*(caf-i%-n)

Given U, to find P: P = U*[{(1+I)ⁿ – 1}/{i*(1+i)ⁿ}]

= U*(uniform series present worth factor)

= U*(pwf-i%-n)

Given G, to find U: U = G*[{(1+i)ⁿ – n*I - 1}/{i*(1+i)ⁿ – i}]

= G*(gradient series uniform factor)

= G*(gsuf-i%-n)

Given U, to find G: G = U*[{i*(1+i)ⁿ – i}/{(1+i)ⁿ – n*i – 1}]

= U*(uniform series gradient factor)

= U*(gsrf-i%-n)

The symbols, such as (pwf’-i%-n), are generally used in writing the equations to be solved. The actual numerical interest rate is substituted for the symbol “i” and the actual number of interest periods is substituted for the symbol “n.” Note that single payment factor names are followed by an appositive while the series factors are not.

The names for the various uniform series factors were not plucked out of the air, but have a practical basis. A fund established to produce a specific financial amount at the end of a specific time period by means of a uniform series of payments throughout the period is referred to a sinking fund. The capital recovery factor relates how a present sum can be recouped through a series of uniform payments. The compound amount factor indicates how much money will result at the completion of a uniform series of deposits. The present worth factor indicates the present worth of a uniform series of payments.

44.6. Compounding Period and Interest Rate

Thus far, each time an interest rate has been mention, it has been stated that it was compounded annually or yearly. Interest rates are almost always specified on an annual basis with an indication of how often the interest is compounded. For example, if the interest rate is specified as twelve percent compounded annually for five years, then the interest rate used would be twelve percent and the number of periods would be five. If, however, this same twelve percent interest rate were compounded semiannually, then the interest rate used would be six percent and the number of periods would be ten. See Table 44-3 for more examples. After reviewing this table, you should recognize that you should divide the annual interest rate by the number of times per year that the interest is compounded and multiply the number of periods by that same number. Today, interest is compounded as often as daily.

This can be expressed mathematically in the following manner:

Let the interest rate “r” be compounded “m” times per year.

This produces an interest rate of “r/m” per compounding period.

The nominal interest rate per annum is m*(r/m) = r.

The effective interest rate per annum is (1 + r/m)^m - 1.

What is the big deal about how often interest is compounded? The effective interest rate per annum is the interest that will produce the same bottom line result by compounding annually. This can be best illustrated by looking at Table 44-4. As can be seen from this table, the more often that one compounds interest the larger the effective interest rate is. This is another way that companies can make additional money from less knowledgeable people.

44.7. Methods of Comparison

Comparison of alternatives must involve the same period of time. For example, suppose alternative number one has a period of six years and alternative number two has a period of eight years. You should use twenty-four years as analysis period, since that is the smallest number that both six- and eight-years will both divide into without a remainder. The original financial-time diagram for alternative number one should be repeated four times with starts in years one, seven, thirteen, and nineteen. Similarly, alternative number two should be repeated three times with starts beginning in years one, nine, and seventeen. It may be desirable to increase the costs associated the repeats of an alternative by some reasonable rate of inflation.

Table 44-3. Compounding effects
Compounding Period	Nominal Interest Rate	Number Of Periods
Annually	12%	5
Semiannually	6%	10
Quarterly	3%	20
Monthly	1%	60

Table 44-4. Impact of compounding
Compounding Period	Nominal Interest Rate	Effective Interest Rate
Annually	12%	12%
Semiannually	6%	12.36%
Quarterly	3%	12.55%
Monthly	1%	12.68%

Several different methods of comparison exist and are commonly used. The present worth method brings everything back to year zero (a P value at time zero) where the present worth values of the alternatives can be compared. The annual cost method uses a uniform series over the entire life of the alternatives (a U value over n periods) to compare alternatives. The rate of return method only can be used to compare two alternatives at a time. In this method, one determines what interest rate (i) will make the two alternatives equal. That interest rate is then used to make a selection between the two alternatives. If more than two alternatives exist, the process is repeated until only one alternative is left. One can also compare alternatives at the end of their life (an S value at the end of n periods) as another method.

44.8. Example Problems

For the following example problems, two types of situations have been selected. The first group involves situations that you might encounter as an individual or family, while the second group illustrates a few business examples. It is hoped that these examples will prove to be enlightening.

42.8.1. How Much Are Grades Worth?

Problem statement:

Jack and Jill started college at least four years ago and both majored in the same field of engineering. Jack partied almost every night, but worked hard enough to maintain a 2.5/4.0 cumulative grade point. Jill, on the other hand, studied hard Sunday through Thursday nights. She still found time to party sensibly on Friday night and Saturday. The result was that Jill amassed a 3.5/4.0 cumulative grade point. During their college careers, both did a co-op with the same company. During their last semester, they both interviewed a number of companies, including several of the same ones, looking for a job. Jack only received two permanent job offers while Jill received five. Both received a permanent job offer from Amalgamated Engineering Products (AEP). Jack’s offer was for $45,000 per year, while Jill’s offer was for $55,000 per year.

Some other information is also needed; the following conservative estimates are necessary:

· Both are currently twenty-two years of age.

· Both will accept their job offers with AEP.

· Both will work for AEP for the next 38 years, retiring when they reach sixty years of age.

· Both will receive a five percent cost-of-living salary increase each year.

· Jack will receive a promotion at the beginning of his eleventh and twenty-first years of service.

· Jill will work hard during her professional career and receive a promotion at the beginning of her ninth, seventeenth, and twenty-fifth years of service.

· Each time a promotion is received the recipient will receive a one-time five percent salary increase that includes their cost-of-living increase for that year.

· All other things are considered equal.

Required activities and questions:

a) Setup a table showing the income each will receive on a yearly basis during their professional careers. Number the years by their years of employment.

b) How much pre-tax difference will occur over their professional careers? Neglect the time values of money for this part (just add the amounts ignoring the points in time where they were received.)

c) Draw the financial financial-time diagram for each.

d) Write and solve the equation for present worth in year zero for each person if the interest rate is assumed to be six percent compounded annually.

e) What is the difference in the two present worth values?

f) Was Jack’s partying in college really worth it?

Solution:

a) See Table 44-5.

b) $6,959,719 - $5,499,187 = $1,460,532

c) See Figures 42-8 and 42-9.

d) P_JACK = 47,250*(pwf’-6%-1) + 49,613*(pwf’-6%-2) + … + 316,799*(pwf’-6%-38)

= 47,250*(0.94340) + 49,613*(0.89000) + … + 316,799*(0.10924)

= 44,575 + 44,155 + … + 34,607

= $1,511,395

P_JILL = 57,750*(pwf’-6%-1) + 60,638*(pwf’-6%-2) + … + 406,559*(pwf’-6%-38)

= 57,750*(0.94340) + 60,638*(0.89000) + … + 406,559*(0.10924)

= 54,481 + 53,968 + … + 44,412

= $1,893,491

See Table 44-6 for complete calculations

e) $1,893,491 - $1,511,395 = $382,096

f) Jack probably felt it was when he was in school, but not a few years later, especially if Jill was his supervisor.

Table 44-5 Jack and Jill's salary record
Year	Jack				Jill
	Beginning of the Year Salary	Cost of Living Raise	Promotion Increment	End of the Year Salary	Beginning of the Year Salary	Cost of Living Raise	Promotion Increment	End of the Year Salary
0				$45,000				$55,000
1	$45,000	$2,250		$47,250	$55,000	$2,750		$57,750
2	$47,250	$2,363		$49,613	$57,750	$2,888		$60,638
3	$49,613	$2,481		$52,093	$60,638	$3,032		$63,669
4	$52,093	$2,605		$54,698	$63,669	$3,183		$66,853
5	$54,698	$2,735		$57,433	$66,853	$3,343		$70,195
6	$57,433	$2,872		$60,304	$70,195	$3,510		$73,705
7	$60,304	$3,015		$63,320	$73,705	$3,685		$77,391
8	$63,320	$3,166		$66,485	$77,391	$3,870		$81,260
9	$66,485	$3,324		$69,810	$81,260	$4,063	$4,266	$89,589
10	$69,810	$3,490		$73,300	$89,589	$4,479		$94,069
11	$73,300	$3,665	$3,848	$80,814	$94,069	$4,703		$98,772
12	$80,814	$4,041		$84,854	$98,772	$4,939		$103,711
13	$84,854	$4,243		$89,097	$103,711	$5,186		$108,896
14	$89,097	$4,455		$93,552	$108,896	$5,445		$114,341
15	$93,552	$4,678		$98,229	$114,341	$5,717		$120,058
16	$98,229	$4,911		$103,141	$120,058	$6,003		$126,061
17	$103,141	$5,157		$108,298	$126,061	$6,303	$6,618	$138,982
18	$108,298	$5,415		$113,713	$138,982	$6,949		$145,931
19	$113,713	$5,686		$119,398	$145,931	$7,297		$153,228
20	$119,398	$5,970		$125,368	$153,228	$7,661		$160,889
21	$125,368	$6,268	$6,582	$138,219	$160,889	$8,044		$168,934
22	$138,219	$6,911		$145,129	$168,934	$8,447		$177,380
23	$145,129	$7,256		$152,386	$177,380	$8,869		$186,250
24	$152,386	$7,619		$160,005	$186,250	$9,312		$195,562
25	$160,005	$8,000		$168,006	$195,562	$9,778	$10,267	$215,607
26	$168,006	$8,400		$176,406	$215,607	$10,780		$226,387
27	$176,406	$8,820		$185,226	$226,387	$11,319		$237,707
28	$185,226	$9,261		$194,487	$237,707	$11,885		$249,592
29	$194,487	$9,724		$204,212	$249,592	$12,480		$262,072
30	$204,212	$10,211		$214,422	$262,072	$13,104		$275,175
31	$214,422	$10,721		$225,143	$275,175	$13,759		$288,934
32	$225,143	$11,257		$236,401	$288,934	$14,447		$303,381
33	$236,401	$11,820		$248,221	$303,381	$15,169		$318,550
34	$248,221	$12,411		$260,632	$318,550	$15,927		$334,477
35	$260,632	$13,032		$273,663	$334,477	$16,724		$351,201
36	$273,663	$13,683		$287,346	$351,201	$17,560		$368,761
37	$287,346	$14,367		$301,714	$368,761	$18,438		$387,199
38	$301,714	$15,086		$316,799	$387,199	$19,360		$406,559
Total				$5,499,187				$6,959,719

Self-check learning activities:

a) Change the annual cost-of-living increases from five percent to six percent.

b) Change the promotion increases from five percent to ten percent.

c) Change the interest rate from six percent compounded annually to eight percent compounded annually.

d) Repeat parts a) through e) above.

Answers:

a) Jack: $7,588,097 Jill: $9,964,991

b) $2,376,894

c) Similar

d) Jack: $1,344,973 Jill: $1,722,051

e) $377,078

42.8.2. Paying Off College Loans

Problem statement:

Both Jack’s parents and Jill’s parents were not wealthy, so Jack and Jill had to establish student loans in order to complete college. Assume that both Jack and Jill graduated from college owing $50,000 in college loans. Immediately after graduation, twelve percent interest compounded monthly is assessed on each loan. Jack plans to continue his partying ways and wants a new sports car immediately. In order to make his new car payments, Jack decides he must stretch his loan repayment over an eight-year period. Jill, on the other hand, decides that she can drive her “old clunker’ for another year. She decides to repay her loan over a four-year period. Both Jack and Jill chose to make equal monthly payments.

Required activities and questions:

a) Draw the financial-time diagram for each person.

b) Write and solve the necessary equations to determine each person’s monthly payment values?

c) Neglecting the time value of money, how much more did it cost Jack to spread his payments out over an additional four years?

d) What is the present worth of each of their repayment plans?

Solution:

Since the interest rate is twelve percent compounded monthly, the interest rate that will be used is one percent and the number of periods is equal to the number of years multiplied by twelve.

a) See Figures 42-10 and 42-11.

b) U_JACK = P*(crf-1%-96)

= 50,000*(0.01625)

Table 44-6 Jack’s and Jill's present worth of salary
Year	Jack			Jill
	End of the Year Salary	Present Worth Factor	Present Worth of Salary	End of the Year Salary	Present Worth Factor	Present Worth Of Salary
1	$47,250	0.94340	$44,575	$57,750	0.94340	$54,481
2	$49,613	0.89000	$44,155	$60,638	0.89000	$53,968
3	$52,093	0.83962	$43,738	$63,669	0.83962	$53,458
4	$54,698	0.79209	$43,326	$66,853	0.79209	$52,954
5	$57,433	0.74726	$42,917	$70,195	0.74726	$52,454
6	$60,304	0.70496	$42,512	$73,705	0.70496	$51,959
7	$63,320	0.66506	$42,111	$77,391	0.66506	$51,469
8	$66,485	0.62741	$41,714	$81,260	0.62741	$50,984
9	$69,810	0.59190	$41,320	$89,589	0.59190	$53,028
10	$73,300	0.55839	$40,930	$94,069	0.55839	$52,528
11	$80,814	0.52679	$42,572	$98,772	0.52679	$52,032
12	$84,854	0.49697	$42,170	$103,711	0.49697	$51,541
13	$89,097	0.46884	$41,772	$108,896	0.46884	$51,055
14	$93,552	0.44230	$41,378	$114,341	0.44230	$50,573
15	$98,229	0.41727	$40,988	$120,058	0.41727	$50,096
16	$103,141	0.39365	$40,601	$126,061	0.39365	$49,623
17	$108,298	0.37136	$40,218	$138,982	0.37136	$51,613
18	$113,713	0.35034	$39,839	$145,931	0.35034	$51,126
19	$119,398	0.33051	$39,463	$153,228	0.33051	$50,644
20	$125,368	0.31180	$39,090	$160,889	0.31180	$50,166
21	$138,219	0.29416	$40,658	$168,934	0.29416	$49,693
22	$145,129	0.27751	$40,274	$177,380	0.27751	$49,224
23	$152,386	0.26180	$39,894	$186,250	0.26180	$48,760
24	$160,005	0.24698	$39,518	$195,562	0.24698	$48,300
25	$168,006	0.23300	$39,145	$215,607	0.23300	$50,236
26	$176,406	0.21981	$38,776	$226,387	0.21981	$49,762
27	$185,226	0.20737	$38,410	$237,707	0.20737	$49,293
28	$194,487	0.19563	$38,048	$249,592	0.19563	$48,828
29	$204,212	0.18456	$37,689	$262,072	0.18456	$48,367
30	$214,422	0.17411	$37,333	$275,175	0.17411	$47,911
31	$225,143	0.16425	$36,981	$288,934	0.16425	$47,459
32	$236,401	0.15496	$36,632	$303,381	0.15496	$47,011
33	$248,221	0.14619	$36,286	$318,550	0.14619	$46,568
34	$260,632	0.13791	$35,944	$334,477	0.13791	$46,128
35	$273,663	0.13011	$35,605	$351,201	0.13011	$45,693
36	$287,346	0.12274	$35,269	$368,761	0.12274	$45,262
37	$301,714	0.11579	$34,936	$387,199	0.11579	$44,835
38	$316,799	0.10924	$34,607	$406,559	0.10924	$44,412
Total			$1,511,395			$1,893,491

= $812.64

U_JILL = P*(crf-1%-48)

= 50,000*(0.02633)

= $1,316.69

c) 96*812.64 – 48*1,316.69 = 78,013.64 – 63,201.21 = $14,812.43

d) P_JACK = U*(pwf-1%-96) = 812.64*(61.52770) = $50,000

P_JILL = U*(pwf-1%-48) =1,316.69*(37.97396) = $50,000

This is what should be expected.

Self-check learning activities:

a) Assume Jack paid off $500 of his loan each month for the first two years, what equal monthly payments would he have to make to still pay the loan of in eight years? (Answer: $977.51)

b) What increasing gradient value starting at zero over the same four-year period would be equal to Jill’s uniform monthly payment of $1316.69? (Answer: $60.96)

42.8.3. Establishing a Supplemental Retirement Nest Egg

Problem statement:

After going to work, both Jack and Jill decide to create their own retirement nest eggs to supplement their company-matched retirement accounts. Jack faithfully deposits $100 per month into an account that pays six percent compounded monthly. Jack feels that he can not lower his living standard because he feels the need to purchase a new sports car every three years. He also buys the latest electronic equipment including a 48” HDTV ($15,000) and upgrades his expensive audio system. Jill, on the other hand, deposits ten percent of her pre-tax salary into a bank account that also pays six percent compounded monthly. She also drives more conservative automobiles until they have amassed 100,000+ miles. Her audio/video system is not state-of-the- art, but most people would be more than satisfied to have what she has.

Required activities and questions:

Both Jack and Jill’s estimated salary data is available in Table 44-5. The following activities are to be performed and the following questions are to be answered:

a) Draw the financial-time diagrams for both Jack and Jill.

b) Determine how much money is available in each of their supplemental retirement accounts when they retire.

Solution:

Since the interest is compounded on a monthly basis, the interest rate will be 0.5% (6% / 12) and 456 (38 * 12) interest periods.

a) The financial-time diagrams are shown in Figures 42-12 and 42-13. Jill’s data will be based on the data in Table 44-7.

b) Jack’s supplemental retirement account final amount is relatively simple to determine; it is

S_JACK = U*(caf-0.5%-456)

= 100*(1744.259173)

= $174,426

Table 44-7 Jill's supplement retirement account
	End of the Year Salary	10% of Annual Salary	Monthly Deposit	Sum of caf' factors for Year	Future Worth of Fund
1	$57,750	$5,775	$481	112.95108	$54,358
2	$60,638	$6,064	$505	106.38922	$53,760
3	$63,669	$6,367	$531	100.20858	$53,168
4	$66,853	$6,685	$557	94.38699	$52,584
5	$70,195	$7,020	$585	88.90361	$52,005
6	$73,705	$7,371	$614	83.73879	$51,433
7	$77,391	$7,739	$645	78.87401	$50,868
8	$81,260	$8,126	$677	74.29185	$50,308
9	$89,589	$8,959	$747	69.97589	$52,242
10	$94,069	$9,407	$784	65.91067	$51,668
11	$98,772	$9,877	$823	62.08161	$51,099
12	$103,711	$10,371	$864	58.47500	$50,538
13	$108,896	$10,890	$907	55.07791	$49,981
14	$114,341	$11,434	$953	51.87818	$49,432
15	$120,058	$12,006	$1,000	48.86434	$48,888
16	$126,061	$12,606	$1,051	46.02558	$48,350
17	$138,982	$13,898	$1,158	43.35174	$50,209
18	$145,931	$14,593	$1,216	40.83323	$49,657
19	$153,228	$15,323	$1,277	38.46104	$49,111
20	$160,889	$16,089	$1,341	36.22666	$48,571
21	$168,934	$16,893	$1,408	34.12208	$48,036
22	$177,380	$17,738	$1,478	32.13977	$47,508
23	$186,250	$18,625	$1,552	30.27262	$46,986
24	$195,562	$19,556	$1,630	28.51395	$46,469
25	$215,607	$21,561	$1,797	26.85744	$48,255
26	$226,387	$22,639	$1,887	25.29716	$47,725
27	$237,707	$23,771	$1,981	23.82753	$47,200
28	$249,592	$24,959	$2,080	22.44328	$46,681
29	$262,072	$26,207	$2,184	21.13945	$46,167
30	$275,175	$27,518	$2,293	19.91136	$45,659
31	$288,934	$28,893	$2,408	18.75461	$45,157
32	$303,381	$30,338	$2,528	17.66507	$44,660
33	$318,550	$31,855	$2,655	16.63883	$44,169
34	$334,477	$33,448	$2,787	15.67220	$43,683
35	$351,201	$35,120	$2,927	14.76173	$43,203
36	$368,761	$36,876	$3,073	13.90415	$42,728
37	$387,199	$38,720	$3,227	13.09639	$42,258
38	$406,559	$40,656	$3,388	12.33556	$41,793
Total	$6,959,716	$695,972			$1,836,566

Because Jill’s deposits are uniform within a year, but not from year to year, her computation is more complex. At least two approaches may be used in this case. The effort required is approximately the same. The first approach is to convert a uniform series (U) for each year into a future amount (S) at the end of that year. Each of those S values is then considered to be a P value, which is moved forward to obtain the desired S value at the end of the 38 years.

The second approach is to first compute the single payment compound amount factor (caf’) for each month in a year. For example, in year one, the first deposit has to be moved forward 455 interest periods; the second deposit is moved forward 454 interest periods, etc. Since each deposit is the same for that year, the twelve factors can be added together before multiplying by the monthly deposit value. Finally, all of the resultant values are summed. The second approach was used in Table 44-7. Jill’s final amount is $1,836,566 as shown in Table 44-7.

Some interesting observations can be made by analyzing the results. Jack’s total deposits of $45,600 resulted in $174,426, which is a ratio of 3.83 to 1.00. Jill’s total deposits of $695,972 produced a balance of $1,836,566, which is a ratio of 2.64 to 1.00. Note that Jill’s annual contribution produce approximately a uniform distribution to her balance. This is because of her significant increase in earning throughout her professional career. Likewise, Jill’s ratio is less than Jack’s because her deposits increase throughout her career.

Self-check learning activities:

a) What balance would have resulted for Jack if he had deposited $100 each month for the first year and increased the amount of his deposits by $10 per month of each subsequent year ($110 per month the second year, $120 per month the third year, etc.) (Answer: $359,080)

b) How would Jill’s final balance have changed if the six percent interest rate were compounded annually? (Answer: $1,733,357)

42.8.4. The Value of Delaying a Purchase

Problem statement:

Both Jack and Jill rented unfurnished apartments after graduation. Jack immediately went to Super Sam’s Electronics and spent $2,500 on a new audio-visual system for his bedroom. He followed that up with a visit Shyster Harry’s Bargain Furniture Barn where he spent $5,000 on living room, dining room, and bedroom furniture. He charged the entire $7,500 on his credit card, which has an interest rate of eighteen percent compounded monthly.

Jill, on the other hand, decided to use her old bedroom furniture, a used sofa and matching chair that her mom and dad wanted to replace, and a card table and chairs for the dining room. She plans to use this furniture for the first year while she saves $3,000 for her new furniture purchases. At the end of the first year, she managed to save $3,200 towards new furniture. At that time, Jill purchased a new audio-visual system on sale for $800, which she paid cash for from her $3,200 furniture savings. She then went to Fashion Interiors, a high-quality local furniture store, where she purchased new living room and dining room for $9,900. Jill also had to borrow $7,500 to pay for the remainder of her furniture purchases. Rather than using her credit card, Jill went to her company’s credit union, where she borrowed the remaining $7,500 for three years with an interest rate of nine percent compounded monthly.

Required activities and questions:

a) Draw the financial-time diagrams for both Jack and Jill.

b) If Jack takes three years to pay off his credit card debt on these two purchases, what will his out-of-pocket payments be each month? Assume that all payments are equal.

c) Jill’s monthly payments, which also will be equal, are how much?

d) How much difference is there in Jack’s and Jill’s annual payments?

Solution:

a) See Figures 42-14 and 42-15.

b) U_JACK = P*(crf-1.5%-36)

= 7500*(0.03615)

= $271.14 per month

c) U_JILL = P*(crf-0.75%-36)

= 7500*(0.03180)

= $238.50

d) ($271.14 - $238.50)*12 = $32.64*12 = $391.74

Self-check learning activities:

a) How much did each pay for borrowing the money?

(Answer: Jack: $2,261.15 Jill: $1,085.93)

b) If the both had gotten a 15% rate compounded quarterly, what would their monthly payments have been?

(Answer: $262.53)

42.8.5. Obtaining a New Car

Problem statement:

Jack is considering a three-year lease on a new $35,000 automobile. In order to obtain the lease, Jack must put $5,000 down in the beginning. If he returns the car in good shape and within the allotted mileage, he will receive credit for thirty percent of the original price at the time of the return. Assume the interest rate is ten percent compounded quarterly.

Required activities and questions:

a) Draw the financial-time diagram.

b) What is his monthly payment for the three-year lease?

c) Suppose Jack bought the car with $5,000 down and a three-year loan of ten percent compounded quarterly, what would his monthly payment be?

Solution:

a) See Figure 44-16.

b) $35,000 = $5,000 +U*(pwf-2.5%-12) + $10,500*(pwf’-2.5%-12)

$30,000 = U*(10.257765) + $10,500*(0.743556)

$30,000 = U*(10.257765) + $7,807

U = ($30,000 - $7,807)/10.257765

= $2163

Monthly payment = $2163/3 = $721 per month

c) $35,000 = $5,000 + U*(pwf-2.5%-12)

$30,000 = U*(10.257765)

U = $30,000/10.257765

= $2925

Monthly payment = $2925/3 = $975 per month

Self-check learning activities:

a) How much more would just buying the car cost Jack per month and in total? (Answer: $254 per month or $9,144 in total)

b) Why should he consider buying the car?

(Answer: If he plans to keep the car beyond three years or to put high mileage on it, he should consider buying it.)

42.8.6. Obtaining a First Home

Problem statement:

Jack finally decided it was time to settle done and get married. Jack and Jane had been living together for three years and were paying $1,000 per month for rent. Since they got no return on the money they paid for rent, they decided to buy a home. Since they live and work in a major metropolitan area, cheap housing does not exist. They finally found a house they both liked for $250,000. Home loans were going for seven percent compounded monthly if one makes a twenty percent down payment, which Jack and Jane don’t have. However, Jack’s parents are willing to loan them $30,000 interest free for five years and Jane’s parents agreed to loan them $18,000 interest free for six years. Jack and Jane sell her old car for $2,000 to complete the twenty percent down payment.

Required activities and questions:

a) Draw the financial-time diagram for their home purchase assuming a thirty-year loan.

b) For a thirty-year loan, what will their monthly payments be, including repaying the both sets of parents? How much interest will they pay?

c) Repeat part a) for a twenty-year loan.

d) Repeat part a) for a fifteen-year loan.

Solution:

a) See Figure 46-17.

b) 30-tear loan:

U₃ = $30,000/(5*12) = $500 per month

U₂ = $18,000/(6*12) = $250 per month

U₁ = P*(crf-0.583%-360)

= 200,000*(0.00665)

= $1331

For months in years 1-5: payments are $1331 + $250 + $500 = $2081

For months in year 6: payments are $1331 + $250 = $1581

For months in 7-30: payments are $1331

Interest: $1331*360 - $200,000 = $279,018

c) 20-year loan:

U₁ = P*(crf-0.583%-240)

= 200,000*(0.00775)

= $1551

For months in years 1-5: payments are $1551 + $250 + $500 = $2301

For months in year 6: payments are $1551 + $250 = $1801

For months in 7-30: payments are $1551

Interest: $1551*240 - $200,000 = $172,143

d) 15-year loan:

U₁ = P*(crf-0.583%-180)

= 200,000*(0.00899)

= $1798

For months in years 1-5: payments are $1798 + $250 + $500 = $2548

For months in year 6: payments are $1798 + $250 = $2048

For months in 7-30: payments are $1798

Interest: $1798*180 - $200,000 = $123,578

Self-check learning activities:

a) What would be the impact of making payments every four weeks rather than one a month for a fifteen-year $200,000 loan at seven percent compounded on a four- week basis? (Answer: 4-week payment: $1659 Interest: $123,538)

b) For an interest rate of six percent compounded monthly, what would be the monthly payment on a $200,000 loan for fifteen years? (Answer: $1688)

c) For the previous part’s data (b), what would be the total interest paid? (Answer: $103,788)

42.8.7. Saving One-Step-at-a-Time for a Child’s College Education

Problem Statement:

Jack and Jane got married just before Betty Lou arrived. Jane suggested that they start a college fund for Betty Lou. Each year from her first birthday to her eighteenth birthday, Jack and Jane will deposit the same amount. They wanted Betty Lou to have $25,000 on each her eighteenth, nineteenth, twentieth, twenty-first, and twenty-second birthdays to attend college. Assume that the account where they deposited the money earns nine percent compounded annually.

Required activities and questions:

a) Draw the financial-time diagram for Betty Lou’s account.

b) How much do Jack and Jane have to deposit on each of Betty Lou’s birthdays?

Solution:

a) See Figure 44-18.

b) P₁₈ = U₂*(pwf-9%-4)

= 25,000*(3.23972)

= 80,993

U₁ = S₁₈*(SFF-9%-18)

= 80,993*(0.02421)

= $1,961

Self-check learning activities:

a) If Jack and Jane had deposited $5,000 when Betty Lou was born, how much less would each of the birthday deposits have been? (Answer: $571)

b) Suppose Jane wants Betty Lou to have $50,000 for her wedding which is assumed to occur on her twenty-second birthday. How much addition money does Jack need to deposit on each of Betty Lou’s birthdays? (Answer: $787)

42.8.8. Saving Up-Front for a Child’s College Education

Problem statement:

Jill gets married and later starts a family. She and her husband, Dick, decide to make a single deposit on the day baby Ralph is born that will cover Ralph’s college education. Jill and Dick want Ralph to be able to withdraw $25,000 on his nineteenth, twentieth, twenty-first, and twenty-second birthdays. Assume that the account where they deposited the money earns nine percent interest compounded annually.

Required activities and questions:

a) Draw the financial-time diagram for Ralph’s account.

b) How much did Jill and Dick have to deposit when Ralph was born?

Solution:

a) See Figure 44-19.

b) S_RALPH = U*(caf-9%-4)

= 25,000*(4.57313)

= 114,328

P_RALPH = S*(pwf’-9%-22)

= 114,328(0.15018)

= $17,170

Self-check learning activities:

a) If the interest rate changed from nine percent to twelve percent on Ralph’s tenth birthday, how much more money would Ralph have to celebrate his twenty-first birthday? (Answer: $34,713)

b) How much additionally will Jill and Dick have had to deposit when Ralph was born if they wanted him to have $35,000 on his twenty-second birthday to buy a new car? (Answer: $5,256)

42.8.9. Reverse Home Loan

Problem statement:

Jack and Jill are now approaching retirement age. Since Jack did not do as well in planning for his retirement, he and Jane are considering a reverse home loan. This type of loan pays one a certain amount of money each month. This continues until the home is sold or the value of the home has been paid out to the homeowner. If the home is sold before the value of the home has been paid out (borrowed), the amount of the reverse loan must be paid from the selling price. If the full value of the home has been borrowed, the owner must: (1) find somewhere else to live, (2) have other estate items for collateral, or (3) Jack and Jane’s heirs must agree to pay the balance. Assume that: (1) Jack and Jane’s home is worth $250,000, (2) Jack and Jane need an additional $3,000 per month from the reverse home loan, and (3) the interest rate is six percent compounded monthly.

Required activities and questions:

a) Draw the financial-time diagram.

b) How many years can Jack and Jane plan to live in their home until they have borrowed its full value?

Solution:

a) See Figure 44-20.

b) One way to solve this problem is by trial and error.

$250,000 = U*(pwf-0.5%-n)

$250,000 = $3,000*(pwf-0.5%-n)

(pwf-0.5%-n) = 250,000/3,000 = 83.3333

Thus, one needs to find the value of “n” (months) for which the present worth factor equals 83.3333. As shown in Table 44-8, the first estimate is 10 years, which is too large. The second estimate is 8 years, which is too small. The third estimate is 9 years, which is nearly correct. The next estimate is 9 years and 1 month, which is slightly too large. This is the smallest increment that can be directly calculated easily.

Table 44-8, Iterative process
Year	Month	(pwf-0.5%-n)
10 yr	120 months	90.073453
8 yr	96 months	76.095218
9 yr	108 months	83.293424
9 yr – 1 month	109 months	83.874054

Finally, interpolation will be used. The two numbers written to the right of the two brackets shown below are the difference between the top number and the bottom number as shown in Figure 44-21.

The result of taking the ratio of these two numbers indicates what portion of the range should be included:

0.03991/0.58063 = 0.068736

Assume that a month has thirty days. The 0.068736 corresponds to 2.06 days (0.068736 * 30). Therefore, the correct answer is 9 years, 0 months and 2 days.

Self-check learning activities:

a) Assume Jack and Jane plan to live in their home for ten years and wish to borrow the full value of the home over that period of time. How much would they receive per month? (Answer: $2776 per month)

b) Assume that Jack and Jane go with the original option (a) and find it necessary to move to assisted care after five years, how much money will they receive from the sale of their home? (Answer: $127,902)

42.8.10. Equipment Life-Time Costs

Problem statement:

Amalgamated Engineering Products (AEP) has decided that it is necessary to invest in a new piece of major equipment. Two alternatives have been identified and the necessary information is given in Table 44-9.

Required activities and questions:

a) Draw the financial-time diagram for each alternative

b) If the interest rate is twelve percent compounded annually, which alternative is cheaper?

c) At what interest rate compounded annually are the two alternatives equal in cost?

Table 44-9. AEP equipment alternatives
Item	Alternative A	Alternative B
Equipment life	20 years	30 years
Initial cost	$100,000	$125,000
Annual operating &
Maintenance cost	$2,500	$3,000
Major maintenance cost
5-year maintenance	$10,000
10-year maintenance	$10,000	$15,000
15-year maintenance	$10,000
20-year maintenance		$15,000
Salvage value	$15,000	$12,500

Solution:

a) See Figures 42-22 and 42-23.

b)

P_A1 = $100,000 +$10,000*(pwf’-12%-5) + $10,000*(pwf’-12%-10) +

$10,000*(pwf’-12%-15) + $2,500*(pwf-12%-20) - $15,000*(pwf’-12%-20)

P_A1 = $100,000 +$10,000*(0.567427) + $10,000*(0.321973) + $10,000*(0.182696)

+ $2,500*(7.469444) - $15,000*(0.103667)

P_A1 = $100,000 + $5,674 + $3,220 + $1,827 + $18,674 - $1,555

P_A1 = $127,840

P_A = $127,840*[1 + (pwf’-12%-20) + (pwf’-12%-40)]

P_A = $127,840*[1 + 0.103667 + 0.010747]

P_A = $127,840*1.114414

P_A = $142,466

P_B1 = $125,000 + $15,000*(pwf’-12%-10) + $15,000*(pwf’-12%-20) +

$3,000*(pwf-12%-30) - $12,500*(pwf’-12%-30)

P_B1 = $125,000 + $15,000*(0.321973) + $15,000*(0.103667) +

$3,000*(8.055184) - $12,500*(0.033378)

P_B1 = $125,000 + $4,830 + $1,555 + $24,166 - $417

P_B1 = $155,133

P_B = $155,133*[1 + (pwf’-12%-30)]

P_B = $155,133*[1 + 0.033378]

P_B = $155,133*1.033378

P_B = $160,311

Alternative A is favored by $17,845

c) The procedure is to use the calculations for part b) and varying the interest rate until the difference is approximately zero. Table 44.10 shows the answers to the nearest tenth of a percent on the interest rate. The final answer is 3.4%.

Table 44-10. Find interest rate iteratively
Percent Interest Rate	Alternative A	Alternative B	B - A
12%	$142,466	$160,311	$17,845
8%	$171,334	$184,363	$13,029
4%	$245,802	$248,582	$2,780
3%	$281,591	$279,740	-$1,851
3.30%	$269,688	$269,373	-$315
3.40%	$265,959	$266,125	$166

Self-check learning activities:

a) What is the equivalent annual cost of alternatives A and B for an interest rate of 12% compounded annually? (Answer: A: $17,115; B: $19,259)

b) How much annual income would alternatives A and B each have to produce a 5% profit if the interest rate is 12% compounded annually?

(Answer: A: $17,971; b: $20,222)

42.8.11. Equipment Efficiency

Problem statement:

A manufacturer wishes to compare two electric motors for a give activity. This activity requires a 100-horsepower motor that is to be operated at full load 800 hours per year and at half load 1,600 hours per year. Super Motors sells a 100-horsepower motor with a guaranteed full-load efficiency of 91% and a half-load efficiency of 87%; this motor sells for $5,040. Spectacular Motors offers a 100-horsepower motor with a guaranteed full-load efficiency of 89.5% and a half-load efficiency of 85%; this motor sells for $4,420. Efficiency is the ratio energy output to energy input. Motors are rate on output energy. One horsepower is equal to 0.746 kilowatts. Energy costs 6.6 cents per kW-hr. Assume each motor exactly meets its guaranteed efficiency and each has a twenty-five year life with no salvage value. The motors are expected to be equal in all respects except for first cost and efficiency.

Required activities and questions:

a) Determine the equivalent annual cost (U) of each motor assuming a ten-percent compounded annually interest rate.

b) Determine the present worth (P) of each motor assuming a twelve-percent compounded annually interest rate.

Solution:

a) The operating costs are computed on Table 44-11 where the product of the output power and the number of hours is divided by the efficiency to obtain the input power. The product of the input power and the electric rate is the operating cost. Super Motors is referred to as alternative A and Spectacular Motors as alternative B.

U_AT = P*(crf-10%-25) + U_AE

= 5,040*(0.110168) + 8,856

= 555 + 8,856

= $9,411 per year

U_BT = P*(crf-10%-25) + U_BE

= 4,420*(0.110168) + 9,035

= 487 + 9,035

= $9,522 per year

Table 44-11. Motor operating cost
Quantity	Super Motors	Spectacular Motors
Output horsepower	100	100
Output power (kW)	74.6	74.6
Electricity rate ($/kW)	$0.066	$0.066
Full load – hours	800	800
Full load - efficiency	0.91	0.895
Full load - input (kW)	65582	66682
Full load – cost ($)	$4,328	$4,401
Half load - hours	1600	1600
Half load - efficiency	0.87	0.85
Half load - input (kW)	68598	70212
Half load – cost ($)	$4,527	$4,634
Electricity cost ($)	$8,856	$9,035

b) P_AT = 5,040 + 8,856*(pwf-12%-25)

= 5,040 + 8,856*(7.843139)

= 5,040 + 69,458

= $74,498

P_BT = 4,420 + 9,035*(pwf-12%-25)

= 4,420 + 9,035*(7.843139)

= 4,420 + 70,282

= $75,282

Self-check learning activities:

a) What would be the equivalent annual cost if full-load requirement is changed to 1,200 hours and the half-load requirement is changed to 1,200 hours also? Use ten percent interest rate compounded annually.

(Answer: A: $10,443, B: $10,564)

b) How much is a 1% improvement in both full-load efficiency and half-load efficiency worth in terms of first cost for each motor? Use ten percent interest rate compounded annually. (Answer: A: 17.76%, B: 21.15%)

42.8.12. Perpetual Facility

Problem statement:

In some situations, it is desirable to determine the capitalized cost of perpetual (forever) service of a facility, such as a high school football stadium. Assume an initial investment of $10,000,000. Certain elements, such as land and earth fill, are assumed to have perpetual life. Estimated future disbursements are as shown in Table 44-12.

Table 44-12. Stadium disbursements
Item	Life	Cost
Maintenance	Annual	$100,000
Seat replacement	20 years	$2,500,000
Structural framework	40 years	$4,000,000

Required activities and questions:

Using an interest rate of twelve percent compounded annually, compute the capitalized cost of perpetual service of the stadium.

Solution:

Initial investment $10,000,000

Annual maintenance: $100,000/0.12 $833,333

Seat replacement: $2,500,000*(sff-12%-20)/i

$2,500,000*(0.013879)/0.12 $289,141

Structural framework: $4,000,000*(sff-12%-40)/i

$4,000,000*(0.001304)/0.12 $43,454

__________

Total $11,165,929

Self-check learning activities:

What would have been the perpetual cost if the interest rate were reduced to ten percent compounded annually? (Answer: $11,526,867)

44.9. Summary

Hopefully, after reading this chapter and working its problems, you now have a better understanding of engineering economics and how they can be utilized. Remember that time is money both in business and in your personal life.