Lecture
2
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
2
Time Value of Money
Interest and Equivalence: Time
Value of Money, Simple Interest,
Equivalence, Single Payment
Compounded Interest
ENGR 3300 Engineering Economics
Lecture 2
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ENGR 3300 Engineering Economics
Time Value of Money
 The economic value of a sum of money
depends of WHEN it is received.
 A dollar in our hands today is worth more
than a dollar promised to us in the future.
o The dollar we have today is ours for
certain.
o The dollar can be “consumed”
immediately if desired.
o The dollar has EARNING POWER – we can
invest or lend the dollar and receive
INTEREST.
Lecture 2
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ENGR 3300 Engineering Economics
Question 1 – College Loan
 Joe Cool’s parents are offering him a
college loan of $5K per year for 4
years to get a UG degree in
Engineering.
 Beginning one year after graduation,
Joe has to repay the loan at $5K per
year. Is this a good deal for Joe?
Lecture 2
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ENGR 3300 Engineering Economics
Question 2 - Power-Ball Lottery
 A suburban Chicago couple won the
Power-ball.
 They had to choose between a single
lump sum $104 million, or $198 million
paid out over 25 years (or $7.92 million
per year).
 The winning couple opted for the
lump sum.
 Did they make the right choice?
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ENGR 3300 Engineering Economics
What Do We Need to Know?
 To make such comparisons (the lottery
decision problem), we must be able
to compare the value of money at
different point in time.
 To do this, we need to develop a
method for reducing a sequence of
benefits and costs to a single point in
time. Then, we will make our
comparisons on that basis.
Lecture 2
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ENGR 3300 Engineering Economics
Time Value of Money
 Money has a time value because it
can earn more money over time
(earning power).
 Time value of money is measured in
term of interest rate.
 Interest is the cost of money – a cost
to the borrower and an earning to the
lender.
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ENGR 3300 Engineering Economics
Elements of Transactions Involving
Interest
 The initial amount of money invested or
borrowed in a transaction is called the
principal (P).
 The interest rate (i) measures the cost or
price of money and is expressed as a
percentage per period of time.
 A period of time called the interest period
(n) determines how frequently interest is
calculated.
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ENGR 3300 Engineering Economics
….
 A specified length of time marks the
duration of the transaction and thereby
establishes a certain number of interest
periods (N).
 A plan for receipts or disbursements (An)
yields a particular cash flow pattern over a
specified length of time.
 A future amount of money (F) results from
the cumulative effects of the interest rate
over a number of interest periods.
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ENGR 3300 Engineering Economics
 An = A discrete payment or receipt occurring at





the end of some interest period.
i = The interest rate per interest period.
N = The total number of interest periods.
P = a sum of money at a time chosen for
purposes of analysis as time zero, sometimes
referred to as the present value or present worth.
F = A future sum of money at the end of the
analysis period. This sum may be specified as Fn.
A = An end-of-period payment or receipt in a
uniform series that
continues for N periods. This
is a special situation where A1 = A2 = …= AN.
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ENGR 3300 Engineering Economics
Example of an Interest Transaction
 An electronics manufacturing company buys a
machine for $25,000 and borrows $20,000 from a
bank at a 9% annual interest rate. In addition, the
company pays a $200 loan origination fee when
the loan commences.
Lecture 2
Lecture
2
Cash Flow Diagrams
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ENGR 3300 Engineering Economics
Cash Flow Diagrams
 Represent time by a horizontal line
marked off with the number of interest
period specified.
 The cash flows over time are represented
by arrows at relevant periods: Upward
arrows denote positive flows (receipts),
downward
arrows
negative
flows
(disbursements).
 Cash flow diagrams function in a manner
similar to free-body diagrams or circuit
diagrams,
which
most
engineers
frequently use.
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ENGR 3300 Engineering Economics
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Cash Flow Diagrams
Lecture 2
Lecture
2
Simple and
Compound Interest
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ENGR 3300 Engineering Economics
Methods of Calculating Interest
 Simple interest is interest earned on only
the principal amount during each interest
period.
 In general, for a deposit of P dollars at a
simple interest rate of i for N periods, the
total earned interest would be
 The total amount available at the end of
N periods thus would be
Lecture 2
ENGR 3300 Engineering Economics
LET’S PRACTICE
Lecture 2
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ENGR 3300 Engineering Economics
Question #1
 Suppose a bank is offering its customers 3%
interest on savings accounts. If a customer
deposits $1500 in the account, how much interest
does the customer earn in 5 years?
 In this problem, we are given the interest rate (i),
the amount put into the account (P), and the
amount of time (N). However, before we can put
these values into our formula, we must change
the 3% to a decimal and make it 0.03. Now we
are ready to go to the formula.
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ENGR 3300 Engineering Economics
I  (0.03 * 1,500) * 5
I  225
So after 5 years, the account has earned $225 in
interest.
If we want to find out the total amount in the
account, we would need to add the interest to the
original amount.
In this case, there would be $1725 in the account.
Lecture 2
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ENGR 3300 Engineering Economics
Question #2
 Jamie wants to earn $500 in interest so she’ll have
enough to buy a used car. She puts $2000 into an
account that earns 2.5% interest. How long will
she need to leave her money in the account to
earn $500 in interest?
I  (i * P ) * N
I
N
(i * P )
500
N
(0.025 * 2000)
N  10 years
Lecture 2
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ENGR 3300 Engineering Economics
Question #3
 A local bank is advertising that you can double
your money in eight years if you invest with them.
Suppose you have $1000 to invest. What interest
rate is the bank offering?
I  (i * P ) * N
I
i
( N * P)
1000
i
(8 * 1000)
i  12.5%
Lecture 2
ENGR 3300 Engineering Economics
TRY THESE
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ENGR 3300 Engineering Economics
Question #1
 Kelly plans to put her graduation money into an
account and leave it there for 4 years while she
goes to college. She receives $750 in graduation
money that she puts it into an account that earns
4.25% interest. How much will be in Kelly’s
account at the end of four years?
A. $127.50
B. $754.0425
C. $877.50
D. $1275
Lecture 2
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ENGR 3300 Engineering Economics
Question #2
 Randy wants to move his savings account to a
new bank that pays a better interest rate of 3.5%
so that he can earn $100 in interest faster than at
his old bank. If he moves $800 to the new bank,
how long will it take for him to earn the $100 in
interest?
A. 3.57 years
B. 0.357 years
C. 0.28 years
D. 2.8 years
Lecture 2
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ENGR 3300 Engineering Economics
Methods of Calculating Interest
 Compound interest, the interest earned in
each period is calculated on the basis of
the total amount at the end of the
previous period. This total amount includes
the
original
principal
plus
the
accumulated interest that has been left in
the account.
 In general, if you deposited (invested) P
dollars at the interest rate i, you would
have P + iP = P(1 + i) dollars at the end of
one period.
Lecture 2
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ENGR 3300 Engineering Economics
 If the entire amount (principal and
interest) is reinvested at the same rate i for
another period, at the end of the second
period you would have
P(1 + i) + i[P(1 + i) ] = P(1 + i) (1 + i) = P(1 + i)2
 Continuing, we see that the balance after
the third period is
P(1 + i)2 + i[P(1 + i)2 ] = P(1 + i)3
 This interest-earning process repeats, and
after N periods the total accumulated
value (balance) F will grow to F = P(1 + i)N
Lecture 2
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ENGR 3300 Engineering Economics
If the interest is compounded annually
F  P(1  i )
N
If the interest is compounded a number of interest
periods different to 1 (annually). Where n is the
number of interest periods per year.
i nN
F  P(1  )
n
If the interest rate is compounded monthly, then n=12
If the interest rate is compounded quarterly, then n=4
If the interest rate is compounded semi-annually, then n=2
If the interest rate is compounded annually, then n=1
Lecture 2
ENGR 3300 Engineering Economics
LET’S PRACTICE
Lecture 2
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ENGR 3300 Engineering Economics
Question #1
 Suppose you deposit $1000 in a bank savings
account that pays interest at a rate of 10%
compounded annually. Assume that you don’t
withdraw the interest earned at the end period
(one year), but let it accumulate. How much
would you have at the end of year 3?
F  P(1  i ) N
F  1000 * (1  0.1)3
F  $1,331
Lecture 2
ENGR 3300 Engineering Economics
Inter - Bayamón
Question #1
Lecture 2
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ENGR 3300 Engineering Economics
Question #2
 If you deposit $4,000 into an account paying 6%
annual interest compounded quarterly, how
much money will be in the account after 5 years?
i nN
F  P (1  )
n
F
F
F
F
0.06 4*5
 4000 * (1 
)
4
 4000 * (1.015) 20
 4000 * (1.346855007)
 5387.42
After 5 years there will be $5,387.42 in the account
Lecture 2
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ENGR 3300 Engineering Economics
Question #3
 How much money would you need to deposit
today at 9% annual interest compounded
monthly to have $12,000 in the account after 6
years?
i
F  P (1  )
n
nN
0.09 12*6
12,000  P * (1 
)
12
12,000  P * (1.0075) 72
12,000  P * (1.712552707)
P  7007.08
You would need to deposit $7007.08 in the account
Lecture 2
ENGR 3300 Engineering Economics
TRY THESE
Lecture 2
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ENGR 3300 Engineering Economics
Question #1
 If you deposit $5,000 into an account paying 6%
annual interest compounded monthly, how long
until there is $8,000 in the account?
A. 6 years
B. 7.9 years
C. 8.1 years
D. 10 years
Lecture 2
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ENGR 3300 Engineering Economics
Question #2
 If you deposit $6,500 into an account paying 8%
annual interest compounded monthly, how much
money will be in the account after 7 years?
A. $8,534.12
B. $10,358.24
C. $11,358.24
D. $12,000
Lecture 2
Lecture
2
Economic
Equivalence
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ENGR 3300 Engineering Economics
Economic Equivalence
 What do we mean by economic
equivalence?
 Why do we need to establish an
“economic equivalence?”
 How do we establish an
“economic equivalence?”
Lecture 2
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ENGR 3300 Engineering Economics
Economic Equivalence
 Various dollar amounts that will be
economically equivalent to $3,000 in 5
years, given an interest rate of 8%.
Lecture 2
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ENGR 3300 Engineering Economics
Principles of Economic Equivalence
 Equivalence
calculations made to
compare
alternatives
require
a
common time basis.
 Equivalence depends on interest rate.
 Equivalence calculations may require
the conversion of multiple payment
cash flows to a single cash flow.
 Equivalence in maintained regardless
of point-of-view as long we use the
same interest rate.
Lecture 2
ENGR 3300 Engineering Economics
Inter - Bayamón
Types of Cash Flows
Lecture 2
Lecture
2
Interest Formulas for
Single Cash Flows
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ENGR 3300 Engineering Economics
Compound-Amount Factor
Single Amounts: Find F, given P, i, N
 Single-Payment
Compound-Amount
Factor (growth factor)
 Given:
i  10%
N  8 years
P  $2,000
Find F
Lecture 2
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Continued …
 Solution:
ENGR 3300 Engineering Economics
F  $2,000 * (1  0.10)8
 $2,000( F / P,10%,8)
 $2,000 * (2.1436)  $4,287.18
Calculator
Appendix B
Microsoft Excel
Lecture 2
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ENGR 3300 Engineering Economics
Present-Worth Factor
Single Amounts: Find P, given F, i, N
 Single-Payment Present-Worth Factor
(discounting factor)
 Given:
i  12%
N  5 years
F  $1,000
Find P
P  F 1  i 
P  F P / F , i, N 
N
F
0
P
Lecture 2
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Continued …
 Solution:
ENGR 3300 Engineering Economics
P  $1,000 * (1  0.12) 5
 $1,000( P / F ,12%,5)
 $1,000 * (0.5674)  $567.40
Calculator
Appendix B
Microsoft Excel
Lecture 2
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ENGR 3300 Engineering Economics
Solving for Time and Interest Rates
Solving for i
 Given:
P  $3,000
F  $4,000
N  5 years
Find i
F
i?
0
P
Lecture 2
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Continued …
 Solution:
ENGR 3300 Engineering Economics
$4,000  $3,000 * (1  i ) 5
4
 ( F / P, i %,5)
3
1.3333  ( F / P, i %,5)  i  6%
Calculator
Appendix B
Microsoft Excel
Lecture 2
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ENGR 3300 Engineering Economics
Solving for Time and Interest Rates
Single Amounts: Find N, given P, F, i
 Given:
P  $3,000
F  $5,000
i 12% APR
Find N
F
i
0
N
P
Lecture 2
Inter - Bayamón
Continued …
 Solution:
Calculator
ENGR 3300 Engineering Economics
$5,000  $3,000 * (1  0.12) N
5
log5 / 3
N
 (1.12) or N 
 4.5 years
3
log1.12 
Microsoft Excel
Lecture 2
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ENGR 3300 Engineering Economics
Homework 1  www.bc.inter.edu/facultad/omeza
Lecture 2