Engineering Economics
1
Money has a time value
because it can earn more
money over time (earning
power).
 Money has a time value
because its purchasing power
changes over time (inflation).
 Time value of money is
measured in terms of market
interest rate which reflects
both earning and purchasing
power in the financial market.

The Interest Rate


Interest is the cost of
money—a cost to the
borrower and an earning to
the lender.
Elements of Transactions
Involving Interest:
Principal
 Interest rate
 Interest period
 Total number of interest
periods
 A plan for receipts or
disbursements
 A future amount of
money
◦ INVESTMENT
 INTEREST = VALUE NOW - ORIGINAL AMOUNT
◦ LOAN
 INTEREST = TOTAL OWED NOW - ORIGINAL
AMOUNT
4
Example
You borrowed $10,000 for one full year.
Must pay back $10,700 at the end of one
year, find the interest amount and the
interest rate:
Solution
Interest Amount (I) = $10,700 - $10,000
Interest Amount = $700 for the year
Interest rate (i) = 700/$10,000 = 7%/Yr
5
•Notations
I = the interest amount in $
i = the interest rate (%/interest period)
N = No. of interest periods (1 for this
problem)
P = The actual borrowed amount or the
principal ($10,000 in this problem)
6
•If you borrow $20,000 for 1 year at 9% interest per
year calculate the total amount of money that you
will pay by the end of that year.
Solution
P= $20,000
i = 0.09 per year and N = 1 Year
Interest Amt (I) = i*P*N =(0.09)($20,000) = $1,800
Total Amt Paid after one year (Future amount of
money)= P + I = $21,800
Create an Excel Sheet
7
•On the other hand if you invest $20,000 for one year in a
venture that will return to you, 9% per year.
At the end of one year, you will have:
Original $20,000 back
Plus……..
The 9% return on $20,000 = $1,800
We say that you earned 9%/year on the investment!
This is your RATE of RETURN on the investment
8
•Where a country’s currency becomes
worth less over time thus requiring more of
the currency to purchase the same amount
of goods or services in a time period
9
•Inflation impacts:
•Purchasing Power (reduces)
•Operating Costs (increases)
•Rate of Returns on Investments
(reduces)
10
•Taxes represent a significant negative cash
flow
•A realistic economic analysis must assess
the impact of taxes
•Called an AFTER-TAX cash flow analysis
•Not considering taxes is called a BEFORETAX Cash Flow analysis
11

Cash Flows
◦ Inflows (Receipts) =====> Revenues or Benefits
(+ positive)
◦ Outflows (Disbursements) =====> Costs (negative)
Positive +
T=0
$20,000 is
received here
$21,800 paid
t = 1 Yr back here
Negative -
Methods of Calculating Interest
• Simple interest: the practice of charging
an interest rate only to an initial sum
(principal amount).
• Compound interest: the practice of
charging an interest rate to an initial
sum and to any previously accumulated
interest that has not been withdrawn.
Simple Interest
• P = Principal
amount
• i = Interest rate
• N = Number of
interest periods
• Example:
P = $1,000
i = 8%
N = 3 years
End of
Year
Beginnin
g
Balance
Interest
earned
0
Ending
Balance
$1,000
1
$1,000
$80
$1,080
2
$1,080
$80
$1,160
3
$1,160
$80
$1,240
F  P  (iP ) N
where
P = Principal amount
i = simple interest rate
N = number of interest periods
F = total amount accumulated at the end of period N
F  $1, 000  (0.08)($1, 000)(3)
 $1, 240
Compound Interest
• Compound interest: the practice of
charging an interest rate to an initial
sum and to any previously accumulated
interest that has not been withdrawn.
Compound Interest
• P = Principal
amount
• i = Interest rate
• N = Number of
interest periods
• Example:
P = $1,000
i = 8%
N = 3 years
End
of
Year
Beginning
Balance
Interest
earned
0
Ending
Balance
$1,000
1
$1,000
$80
$1,080
2
$1,080
$86.40
$1,166.40
3
$1,166.40
$93.31
$1,259.71
$1,000
0
$1,080
1
$1,166.40
$1,080
2
$1,166.40
3
$1,259.71
n = 0: P
n = 1: F1 = P(1+i)
n = 2: F2 = F1(1+i)= P(1+i)2
.
.
.
n = N: FN = P(1+i)N
$1,000
0
1
2
3
F  $1, 000(1  0.08)3
 $1, 259.71
$1,259.71
$1,259.71 in three years is the economic equivalence of
$1,000 now with interest rate 8% with compound calculation
Example
•You travel at 68 miles per hour
•Equivalent to 110 kilometers per hour
•Thus:
•68 mph is equivalent to 110 kph
•Using two measuring scales
•Miles and Kilometers
•Is “68” equal to “110”?
•No, not in terms of absolute numbers
•But they are “equivalent” in terms of the
two measuring scales
•Miles
•Kilometers
•Economic Equivalence
•Two sums of money at two different points
in time can be made economically
equivalent if:
•We consider an interest rate and,
•Number of Time periods between the
two sums
Equality in terms of Economic Value
•Returning to the Example from previous slides
•Diagram the loan (Cash Flow Diagram)
•The company’s perspective is shown
$20,000 is
received here
T=0
t = 1 Yr
$21,800 paid
back here
$20,000 now is economically equivalent to $21,800
one year from now IF the interest rate is set to
equal 9%/year
•$20,000 now is not equal in magnitude to
$21,800 1 year from now
•But, $20,000 now is economically
equivalent to $21,800 one year from now if
the interest rate in 9% per year.
•To have economic equivalence you must
specify:
•Timing of the cash flows
•An interest rate (i% per interest period)
•Number of interest periods (N)
At 8% interest, what is the equivalent worth of $2,042 now 5 years from
now?
If you deposit $2,042 today in a savings
account that pays 8% interest annually.
how much would you have at the end of
5 years?
0
1
2
33
4
5
$2,042
=
0
5
F
F  $2, 042(1  0.08)
 $3, 000
5
At what interest rate would these two amounts be
equivalent in 5 years period of time?
$2,042
0
i=?
$3,289
5
Equivalence Between Two Cash Flows
• Step 1: Determine the $2,042
base period, say, year
5.
• Step 2: Identify the
interest rate to use.
0
• Step 3: Calculate
equivalence value.
$3,289
5
i  6%, F  $2,042(1  0.06)5  $2,733
i  8%, F  $2,042(1  0.08)5  $3,000
i  10%, F  $2,042(1  0.10)5  $3,289
Find the present dollar amount that will be
economically equivalent to $5,877.32 in 5
years, given an interest rate of 8%.
P
F
5877.32

 4000
N
5
(1  i )
(1  .08)
P
F
$4,000 $4,320
0
1
$4,665.6 $5,038.85 $5,441.96
2
3
4
$5877.32
5
EXAMPLE OF EQUIVALENT CASH FLOWS
$2,042 today was
equivalent to receiving
$3,000 in five years, at an
interest rate of 8%. Are
these two cash flows are
also equivalent at the
end of year 3?

Equivalent cash flows
are equivalent at any
common point in time, as
long as we use the same
interest rate (8%, in our
example).

PRACTICE PROBLEM
Compute the
equivalent value of the
cash flow series at n = 3,
using i = 10%.
 Solution:

Compounding Process: $511.90
V3  100(1  0.10)3  $80(1  0.10)2  $120(1  0.10)  $150
 $200(1  0.10)1  $100(1  0.10)2
Discounting Process: $264.46
 $511.90  $264.46
 $776.36
V3
V
$200
$150
$120
$100
$100
$200
=
$120
$80
0
1
2
3
4
5
0
1
2
3
4
5
$150
$100
$80
$100
Contemporary Engineering Economics, 5th edition, © 2010
0
1
2
3
4
5

How many years
would it take an
investment to
double at 10%
annual interest?
2P
0
N=?
P
F  2 P  P (1  0.10) N
2P
0
N=?
P
2  1.1N
log 2  N log1.1
log 2
N
log1.1
 7.27 years
 Approximating
72
how long it will N 
interest rate (%)
take for a sum
of money to
72

double
10
 7.2 years

Single payment
compound
amount factor
(growth factor)
F  P(1  i)
F  P( F / P, i, N )
N
F
0
N
P
Practice Problem
• If you had $2,000 now and invested it at
10%, how much would it be worth in 8
years?
F=?
i = 10%
0
8
$2,000
Given:
P  $2, 000
i  10%
N  8 years
Find: F
F  $2, 000(1  0.10)8
 $2, 000( F / P,10%,8)
 $4, 287.18
EXCEL command:
=FV(10%,8,0,2000,0)
=$4,287.20
Single Cash Flow Formula
• Single payment
present worth factor
(discount factor)
• Given:
i  12%
P  F(1  i)  N
P  F( P / F, i, N )
0
N  5 years
N
F  $1,000
• Find:
P  $1,000(1  0.12)
5
 $1,000( P / F,12%,5)
 $567.40
F
P

You want to set aside a lump sum amount
today in a savings account that earns 7%
annual interest to meet a future expense in
the amount of $10,000 to be incurred in 6
years. How much do you need to deposit
today?
$10,000
0
6
6
P
P  $10, 000(1  0.07)
 $10, 000( P / F , 7%, 6)
 $6, 663

$25,000
$5,000
$3,000
0
1
P
2
3
4
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?
$25,000
Uneven
Payment Series
$5,000
$3,000
0
1
2
3
4
P
$25,000
$5,000
$3,000
0
1
2
3
4
+
0
1
2
3
4
+
P2
0
1
2
3
4
P3
P1
P1  $25, 000( P / F ,10%,1)
 $22, 727
P2  $3, 000( P / F ,10%, 2)
 $2, 479
P  P1  P2  P3  $28,622
P3  $5, 000( P / F ,10%, 4)
 $3, 415
0
1
2
3
4
Beginning
Balance
0
28,622
6,484.20
4,132.62
4,545.88
Interest
Earned
(10%)
0
2,862
648.42
413.26
454.59
Payment
+28,622
-25,000
-3,000
0
-5,000
Ending
Balance
$28,622
6,484.20
4,132.62
4,545.88
0.47
Rounding error
•P = value or amount of money at a time
designated as the present or time 0.
•Also P is referred to as present worth (PW),
present value (PV), net present value (NPV),
discounted cash flow (DCF), and capitalized
cost (CC); dollars
• F = value or amount of money at some
future time.
•Also F is called future worth (FW) and
future value (FV); dollars
•A = series of consecutive, equal,
end-of-period amounts of money.
•Also A is called the annual worth (AW) and
equivalent uniform annual worth (EUAW);
dollars per year, dollars per month
•n = number of interest periods; years,
months, days
• i = interest rate or rate of return per
time period; percent per year, percent per
month
• t = time, stated in periods; years,
months, days, etc