Chapter 44
An Overview of
Engineering Economics
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 fouryear period, you will have paid a total of $13,200 for the loan of the $10,000. Table 441 illustrates this case as plan number one. The table should be selfexplanatory.
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 fouryear 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 441.
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 onefourth 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 441).
All of the multiyear 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 441 Simple
and compound interest examples 



Plan 
End of Year 
Interest Due (8% of the Money owed at start of year) 
Total Money Owed Before YearEnd Payment 
YearEnd Payment 
Money Owed After YearEnd 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 


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 442. 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.
Financialtime 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 442 is provided solely as a means
of illustrating all of the terms except i in Table
442. 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 onetime 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
443. This might represent plan number
one Table 441 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 financialtime diagram to represent it? The result of an athome 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 444.


Table 442 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 endofperiod 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 endofperiod 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 financialtime diagrams for plan numbers two, three, and four in
Table 441. The correct answers are
shown in Figures 425 through Figure 447.
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*(sffi%n)
Given P, to find U: U = P*[{i*(1+i)^{n}}/{(1+i)^{n} – 1}]
= P*(uniform series capital recovery factor)
= P*(crfi%n)
Given U, to find S: S = U*[{(1+i)^{n} –
1}/i]
= U*(uniform series compound amount factor)
= U*(cafi%n)
Given U, to find P: P = U*[{(1+I)^{n} –
1}/{i*(1+i)^{n}}]
= U*(uniform series present worth factor)
= U*(pwfi%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*(gsufi%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*(gsrfi%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 443 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 444.
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.
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 twentyfour years as analysis period, since that is the smallest
number that both six and eightyears will both divide into without a remainder. The original financialtime 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 443. Compounding
effects 



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



Table 444. 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.
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 coop 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 twentytwo 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 costofliving salary increase each year.
·
Jack
will receive a promotion at the beginning of his eleventh and twentyfirst years
of service.
·
Jill
will work hard during her professional career and receive a promotion at the
beginning of her ninth, seventeenth, and twentyfifth years of service.
·
Each
time a promotion is received the recipient will receive a onetime five percent
salary increase that includes their costofliving 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 pretax 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 financialtime 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
445.
b)
$6,959,719
 $5,499,187 = $1,460,532
c)
See
Figures 428 and 429.
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 446 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 445 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 

Selfcheck learning activities:
a)
Change
the annual costofliving 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 eightyear 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
fouryear period. Both Jack and Jill
chose to make equal monthly payments.
Required activities and questions:
a)
Draw
the financialtime 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 4210 and 4211.
b)
U_{JACK}
= P*(crf1%96)
= 50,000*(0.01625)


Table 446 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*(crf1%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*(pwf1%96) = 812.64*(61.52770) = $50,000
P_{JILL} = U*(pwf1%48)
=1,316.69*(37.97396) = $50,000
This is what should be expected.
Selfcheck 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 fouryear 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
companymatched 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 pretax 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 stateofthe 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 445. The following activities are to be performed
and the following questions are to be answered:
a)
Draw
the financialtime 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
financialtime diagrams are shown in Figures 4212 and 4213. Jill’s data will
be based on the data in Table 447.
b)
Jack’s
supplemental retirement account final amount is relatively simple to determine;
it is
S_{JACK} = U*(caf0.5%456)
= 100*(1744.259173)
= $174,426


Table 447 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 447. Jill’s
final amount is $1,836,566 as shown in Table 447.
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.
Selfcheck 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
audiovisual 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 audiovisual system on sale for $800, which she
paid cash for from her $3,200 furniture savings. She then went to Fashion Interiors, a highquality
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 financialtime 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 outofpocket 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 4214 and 4215.
b)
U_{JACK}
= P*(crf1.5%36)
=
7500*(0.03615)
= $271.14 per month
c)
U_{JILL}
= P*(crf0.75%36)
=
7500*(0.03180)
= $238.50
d)
($271.14
 $238.50)*12 = $32.64*12 = $391.74
Selfcheck 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)
Problem statement:
Jack is considering a threeyear 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 financialtime diagram.
b)
What
is his monthly payment for the threeyear lease?
c)
Suppose
Jack bought the car with $5,000 down and a threeyear loan of ten percent
compounded quarterly, what would his monthly payment be?
Solution:
a)
See Figure
4416.
b)
$35,000
= $5,000 +U*(pwf2.5%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*(pwf2.5%12)
$30,000 = U*(10.257765)
U = $30,000/10.257765
=
$2925
Monthly payment = $2925/3 = $975 per month
Selfcheck 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 financialtime diagram for their home purchase assuming a thirtyyear loan.
b)
For a
thirtyyear 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 twentyyear loan.
d)
Repeat
part a) for a fifteenyear loan.
Solution:
a)
See
Figure 4617.
b)
30tear
loan:
U_{3} = $30,000/(5*12) =
$500 per month
U_{2} = $18,000/(6*12) =
$250 per month
U_{1} = P*(crf0.583%360)
= 200,000*(0.00665)
= $1331
For months in years 15: payments are $1331
+ $250 + $500 = $2081
For months in year 6: payments are $1331 + $250 = $1581
For months in 730: payments are $1331
Interest: $1331*360  $200,000 = $279,018
c)
20year
loan:
U_{1} = P*(crf0.583%240)
= 200,000*(0.00775)
= $1551
For months in years 15: payments are $1551
+ $250 + $500 = $2301
For months in year 6: payments are $1551 + $250 = $1801
For months in 730: payments are $1551
Interest: $1551*240  $200,000 = $172,143
d)
15year
loan:
U_{1} = P*(crf0.583%180)
= 200,000*(0.00899)
= $1798
For months in years 15: payments are $1798
+ $250 + $500 = $2548
For months in year 6: payments are $1798 + $250 = $2048
For months in 730: payments are $1798
Interest: $1798*180  $200,000 = $123,578
Selfcheck learning activities:
a) What would be the impact of making payments
every four weeks rather than one a month for a fifteenyear $200,000 loan at
seven percent compounded on a four week basis? (Answer: 4week 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
OneStepataTime 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,
twentyfirst, and twentysecond 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 financialtime 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
4418.
b)
P_{18}
= U_{2}*(pwf9%4)
= 25,000*(3.23972)
= 80,993
U_{1} = S_{18}*(SFF9%18)
= 80,993*(0.02421)
= $1,961
Selfcheck 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 twentysecond birthday. How much
addition money does Jack need to deposit on each of Betty Lou’s birthdays? (Answer: $787)
42.8.8.
Saving
UpFront 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,
twentyfirst, and twentysecond birthdays.
Assume that the account where they deposited the money earns nine
percent interest compounded annually.
Required activities and questions:
a)
Draw
the financialtime diagram for Ralph’s account.
b)
How
much did Jill and Dick have to deposit when Ralph was born?
Solution:
a)
See Figure
4419.
b)
S_{RALPH}
= U*(caf9%4)
=
25,000*(4.57313)
=
114,328
P_{RALPH}
= S*(pwf’9%22)
= 114,328(0.15018)
= $17,170
Selfcheck 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 twentyfirst
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 twentysecond birthday to buy a new car? (Answer: $5,256)
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 financialtime 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
4420.
b)
One
way to solve this problem is by trial and error.
$250,000 = U*(pwf0.5%n)
$250,000 = $3,000*(pwf0.5%n)
(pwf0.5%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 448, 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 448,
Iterative process 



Year 
Month 
(pwf0.5%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 4421.
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.
Selfcheck 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
LifeTime 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 449.
Required activities and questions:
a)
Draw
the financialtime 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 449. 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 


5year maintenance 
$10,000 

10year maintenance 
$10,000 
$15,000 
15year maintenance 
$10,000 

20year maintenance 

$15,000 
Salvage value 
$15,000 
$12,500 

Solution:
a)
See
Figures 4222 and 4223.
b)
P_{A1} = $100,000 +$10,000*(pwf’12%5)
+ $10,000*(pwf’12%10) +
$10,000*(pwf’12%15) + $2,500*(pwf12%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*(pwf12%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 4410. 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 

Selfcheck 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)
Problem statement:
A manufacturer wishes to compare two
electric motors for a give activity.
This activity requires a 100horsepower 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 100horsepower motor
with a guaranteed fullload efficiency of 91% and a halfload efficiency of
87%; this motor sells for $5,040.
Spectacular Motors offers a 100horsepower motor with a guaranteed
fullload efficiency of 89.5% and a halfload 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 kWhr. Assume each motor exactly meets its
guaranteed efficiency and each has a twentyfive 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 tenpercent compounded
annually interest rate.
b)
Determine
the present worth (P) of each motor assuming a twelvepercent compounded
annually interest rate.
Solution:
a)
The
operating costs are computed on Table 4411 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*(crf10%25) + U_{AE}
= 5,040*(0.110168) + 8,856
= 555 + 8,856
= $9,411 per year
U_{BT} = P*(crf10%25) + U_{BE}
= 4,420*(0.110168) + 9,035
= 487 + 9,035
= $9,522 per year


Table 4411. 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*(pwf12%25)
= 5,040 + 8,856*(7.843139)
=
5,040 + 69,458
=
$74,498
P_{BT} = 4,420 + 9,035*(pwf12%25)
=
4,420 + 9,035*(7.843139)
=
4,420 + 70,282
=
$75,282
Selfcheck learning activities:
a)
What
would be the equivalent annual cost if fullload requirement is changed to
1,200 hours and the halfload 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 fullload efficiency and halfload efficiency
worth in terms of first cost for each motor?
Use ten percent interest rate compounded annually. (Answer: A: 17.76%, B: 21.15%)
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 4412.


Table 4412. 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*(sff12%20)/i
$2,500,000*(0.013879)/0.12 $289,141
Structural
framework: $4,000,000*(sff12%40)/i
$4,000,000*(0.001304)/0.12 $43,454
__________
Total $11,165,929
Selfcheck learning activities:
What would have been the perpetual cost if
the interest rate were reduced to ten percent compounded annually? (Answer: $11,526,867)
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.