Lecture
3
Engineering Economics
ENGR 3300
Department of Mechanical Engineering
Inter American University of Puerto Rico
Bayamon Campus
Dr. Omar E. Meza Castillo
omeza@bayamon.inter.edu
http://www.facultad/bayamon.inter.edu/omeza
Lecture
3
Time Value of Money
More Interest Formulas: Uniform
Series, Arithmetic Gradient,
Geometric Gradient, Nominal
and Effective Interest
Inter - Bayamón
ENGR 3300 Engineering Economics
Uneven-Payment Series
How much do you need to deposit today (P)
to withdraw $25,000 at n=1, $3,000 at n=2, and
$5,000 at n=4, if your account earns 10%
annual interest?
Lecture 3
ENGR 3300 Engineering Economics
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Inter - Bayamón
Equal-Payment (Uniform) Series
Lecture 3
ENGR 3300 Engineering Economics
Inter - Bayamón
Equal-Payment (Uniform) Series
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Compound-Amount Factor
Find F, given A, i, and N
Lecture 3
ENGR 3300 Engineering Economics
Lecture 3
Inter - Bayamón
Inter - Bayamón
ENGR 3300 Engineering Economics
 Given: A=$5,000; N=5 years; and i=6%
 Find: F
 Solution: F=$5,000(F/A,6%,5)=$28,185.46
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Sinking-Fund Factor
Find A, given F, i, and N
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
You want to setup a college saving plan for your
daughter. She is currently 10 years old and will go to
the college at age 18. You assume tat when she stars
college, she will need at least $100,000 in the bank.
How much do you need to save each year in order to
have the necessary funds if the current rate of interest
is 7%? Assume that end-of-year deposits are made.
 Given: F=$100,000; N=8 years; and i=7%. Find: A
 Solution: A=$100,000(A/F,7%,8)=$9.746.78
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Capital-Recovery Factor (Annuity Factor)
Find A, given P, i, and N
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
You borrowed $21,061.82 to finance the educational
expenses for your senior year of college. The loan will
be paid off over five years. The loan carries an interest
rate of 6% per year and is to be repaid in equal
annual installments over the next five years. Assume
that the money was borrowed at the beginning of your
senior year and that the first installment will be due a
year later.
 Given: P=$21,061.82; i=6% per year and N=5 years
P
 Find: A
i=6%
N=5
 Solution: A=$21,061.82(A/P,6%,5)=$5,000
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Dealing with Gradient Series
Engineers frequently encounter situations
involving periodic payments that increase or
decrease by a constant amount G or constant
percentage (growth rate) from period to
period.
 Linear Gradient Series
 Geometric Gradient Series
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Gradient Series as a Composite Series
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Handling Linear Gradient Series
Present-Worth Factor: Linear Gradient: Find
P, given G, N, and i
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
You borrowed $10,000 from a local bank, with the
agreement that you will pay back the loan according
to a graduated payment plan. If your first payment is
set at $1,500, what would the remaining payment look
like at borrowing rate of 10% over five years?
 Given: P=$10,000; A1=$1,500, N=5 year and i=10%
$10,000
 Find: G
5
$1,500
G
2G
3G
4G
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
 Solution:
 Since the loan payment series consists of two




parts (1) a $1,500 equal-payment series (2) a
strict gradient series (unknown, yet to be
determined) we can calculate the present value
of each series and equate then with $10,000:
$10,000 = $1,500(P/A, 10%,5) + G(P/G, 10%,5)
$10,000 = $1,500(3.7908) + G(6.8618)
$10,000 = $5,686.18 + 6.8618 G
G = $628.67
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Handling Linear Gradient Series
Present-Worth Factor: Find P, given A1, g, i,
and N
 Another kind of gradient series is formed when
the series in a cash flow is determined, not by
some fixed amount like $500, but by some fixed
rate expressed as a percentage. Many
engineering economic problems, particularly
those relating
to
construction
costs
or
maintenance costs, involve cash flows that
increase or decrease over time by a constant
percentage (geometric), a process that is called
compound growth. Price changes caused by
inflation are good example of such a geometric
series.
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
 If we use g to designate the percentage change
in a payment from one period to the next, the
magnitude of nth payment An is related to the first
payment A1 as follows:
 The g can take either a positive or a negative
sign, depending on the type of cash flow. If g>0,
the series will increase, if g<0, the series will
decrease.
Lecture 3
ENGR 3300 Engineering Economics
Lecture 3
Inter - Bayamón
Inter - Bayamón
ENGR 3300 Engineering Economics
 1  1  g N 1  i  N 
 A1 


ig

P 
 A  N 
if i  g
1
  1  i 
if i  g ,
 Or we can write
P=A1(P/A1,g,i,N)
Lecture 3
Inter - Bayamón
ENGR 3300 Engineering Economics
Suppose that your retirement benefits during your first
year of retirement are $50,000. Assume that this
amount is just enough to meet your cost of living
during the first year. However, your cost of living is
expected to increase at an annual rate of 5%, due to
inflation.
Suppose you do not expect to receive any cost-ofliving adjustment in your retirement pension. Then,
some of your future cost of living has to come from
your savings other than retirement pension. If your
saving account earns 7% interest a year, how much
should you set aside in order to meet this future
increase in the cost of living over 25 years?
 Given: A1=$50,000, g=5%, i=7%, N=25 years
 Find: P
Lecture 3
ENGR 3300 Engineering Economics
$940,167.22
25
Lecture 3
Inter - Bayamón
Inter - Bayamón
ENGR 3300 Engineering Economics
 Find the equivalent amount of total benefits
paid over 25 years:
 P=$50,000(P/A,7%,25)=$582,679
 The required additional saving to meet the
future increase in cost of living will be:
 $940,167.22 - $582,679= $358,017
26
Lecture 3
ENGR 3300 Engineering Economics
Lecture 3
Inter - Bayamón
Inter - Bayamón
ENGR 3300 Engineering Economics
Homework 2  www.bc.inter.edu/facultad/omeza
Lecture 3