1. Engineering
Management, Economics and
Risk Analysis
Dr Jeb Gholinezhad

3. Key Words and Concepts
• Capital: refers to wealth in the form of money or property that can be used to
produce more wealth.
• Interest: The return on capital or invested money and resources
• Cash flow diagram: The pictorial description of when dollars are received
and spent.
• Equivalence Occurs when different cash flows at different times are equal in
economic value at a given interest rate.
• Present worth A time 0 cash flow that is equivalent to one or more later cash
flows.
• Nominal interest rate The rate per year without adjusting for the number of
compounding periods.
• Effective interest rate The rate per year after adjusting for the number of
compounding periods

4. Time Value of Money
❑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 interest rate.
❑Interest is the cost of money—a cost to the borrower and an earning
to the lender

5. Why Interest exist?
Taking the lender’s point of view:
• Risk: Possibility that the borrower will be unable to pay
• Inflation: Money repaid in the future will “value” less
• Transaction Cost: Expenses incurred in preparing the loan agreement
• Opportunity Cost: Committing limited funds, a lender will be unable to
take advantage of other opportunities.
• Postponement of Use: Lending money, postpones the ability of the
lender to use or purchase goods.
From the borrowers perspective …. Interest represents a cost !

6. Return to capital in the form of interest and
profit is an essential ingredient of engineering
economy studies.
• Interest and profit pay the providers of capital for
forgoing its use during the time the capital is being
used.
• Interest and profit are payments for the risk the
investor takes in letting another use his or her
capital.
• Any project or venture must provide a sufficient
return to be financially attractive to the suppliers
of money or property.

7. Simple Interest
Simple Interest is also known as the Nominal Rate of Interest
Annualized percentage of the amount borrowed (principal) which is paid for the use of the
money for some period of time.
Suppose you invested $1,000 for one year at 6% simple rate; at the end of one year the investment
would yield:
This means that each year interest gives $60
How much will you get after 3 years?
➢ Note that each year the interest are calculated only over $1,000.
➢ Does it mean that you could draw the $60 at the end of each year?

8. Simple Interest
Simple Interest is also known as the Nominal Rate of Interest
Annualized percentage of the amount borrowed (principal) which is paid for the use of the
money for some period of time.
Suppose you invested $1,000 for one year at 6% simple rate; at the end of one year the investment
would yield:
$1,000 + $1,000(0.06) = $1,060
This means that each year interest gives $60
How much will you get after 3 years?
$1,000 + $1,000(0.06) + $1,000(0.06) + $1,000(0.06) = $1,180
➢ Note that each year the interest are calculated only over $1,000.
➢ Does it mean that you could draw the $60 at the end of each year?

9. Simple Interest
When the total interest earned or charged is linearly
proportional to the initial amount of the loan (principal),
the interest rate, and the number of interest periods, the
interest and interest rate are said to be simple.
The total interest, I, earned or paid may be computed using
the formula below.
I = (P)(N)(i)
P = principal amount lent or borrowed
N = number of interest periods (e.g., years)
i = interest rate per interest period
The total amount repaid at the end of N interest periods is:
P + I = P(1+Ni).

10. Example:
If $5,000 were loaned for five years at a
simple interest rate of 7% per year, the
interest earned would be
So, the total amount repaid at the end
of five years would be

11. Example:
If $5,000 were loaned for five years at a
simple interest rate of 7% per year, the
interest earned would be
So, the total amount repaid at the end
of five years would be the original
amount ($5,000) plus the interest
($1,750) = 5000+1750= $6750.
I = (P) (N) (i)
I = 5000 x 5 x 7/100 = 1750

12. Compound Interest
If the percentage is not paid at the end of the period, then, this amount is added to the original amount
(principal) to calculate the interest for the second term.
This “adding up” defines the concept of Compounded Interest
Now assume you invested $1,000 for two years at 6% compounded annually; at the end of one year
the investment would yield:
Since interest is compounded annually, at the end of the second year the investment would be worth:
How much this investment would yield at the end of year 3?

13. Compound Interest
If the percentage is not paid at the end of the period, then, this amount is added to the original amount
(principal) to calculate the interest for the second term.
This “adding up” defines the concept of Compounded Interest
Now assume you invested $1,000 for two years at 6% compounded annually; at the end of one year
the investment would yield:
$1,000 + $1,000 ( 0.06 ) = $1,060 or $1,000 ( 1 + 0.06 )
Since interest is compounded annually, at the end of the second year the investment would be worth:
[ $1,000 ( 1 + 0.06 ) ] + [ $1,000 ( 1 + 0.06 ) ( 0.06 ) ] = $1,124
Principal and Interest for First Year Interest for Second Year
Factorizing:
$1,000 ( 1 + 0.06 ) ( 1 + 0.06 ) = $1,000 ( 1 + 0.06 )2 = $1,124
How much this investment would yield at the end of year 3?
$1,000 ( 1 + 0.06 )3 = $1,191

14. Compound interest reflects both the remaining principal
and any accumulated interest. For $1,000 at 10%…
Period
(1)
Amount owed
at beginning of
period
(2)=(1)x10%
Interest
amount for
period
(3)=(1)+(2)
Amount
owed at end
of period
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331
Compound interest is commonly used in personal and professional financial transactions.

15. Compounding Process
$1,000
$1,100
$1,100
$1,210
$1,210
$1,331
0
1
2
3
i=10%

17. • Notation used in formulas for compound interest
calculations.
i = effective interest rate per interest period
N = number of compounding (interest) periods
P = present sum of money; equivalent value of one or more
cash flows at a reference point in time; the present
F = future sum of money; equivalent value of one or more
cash flows at a reference point in time; the future
A = end-of-period cash flows in a uniform series continuing
for a certain number of periods, starting at the end of the
first period and continuing through the last

18. Future Value
Example: Find the amount which will accrue at the end of Year 6 if
$1,500 is invested now at 6% compounded annually.
Method #1: Direct Calculation
(F/P,i,n)
F = P (1+i)n
Given Find F
n =
P =
i =

19. Future Value
Example: Find the amount which will accrue at the end of Year 6 if
$1,500 is invested now at 6% compounded annually.
Method #1: Direct Calculation
(F/P,i,n)
F = P (1+i)n
Given Find F
n = 6 years F = (1,500)(1+0.06)6
P = $ 1,500 F = $ 2,128
i = 6.0 %

20. Future Value
Example:
Find the amount which will accrue at the end of Year 6 if $1,500 is invested now at
6% compounded annually.
Method #2: Tables
• The value of (1+i)n = (F/P,i,n) has been tabulated for various i and n.
• The first step is to layout the problem as follows:
F = P (F/P,i,n)
F = 1500 (F/P, 6%, 6) F = 1500 ( ) → F = 1500 (1.4189) = $2,128
https://global.oup.com/
us/companion.website
s/9780199778126/pdf/
Appendix_C_CITables.
pdf

21. Economic equivalence allows us to compare
alternatives on a common basis.
• Each alternative can be reduced to an equivalent
basis dependent on
• interest rate,
• amount of money involved, and
• timing of monetary receipts or expenses.
• Using these elements we can “move” cash flows so
that we can compare them at particular points in
time.

22. Present Value of a Single Amount
Instead of asking what is the future value of a current
amount, we might want to know what amount we must
invest today to accumulate a known future amount.
This is a present value question.
Present value of a single amount is today’s equivalent to a
particular amount in the future.

23. Present Value
If you want to find the amount needed at present in order to accrue a certain
amount in the future, we just solve Equation 1 for P and get:
P = F / (1+i)n (2)
Example: If you will need $25,000 to buy a new truck in 3 years, how much
should you invest now at an interest rate of 10% compounded annually?
(P/F,i,n)
Given Find P
F =
n =
i =

24. Present Value
If you want to find the amount needed at present in order to accrue a certain
amount in the future, we just solve Equation 1 for P and get:
P = F / (1+i)n (2)
Example: If you will need $25,000 to buy a new truck in 3 years, how much
should you invest now at an interest rate of 10% compounded annually?
Given Find P
F = $25,000 P = F / (1+i)n
n = 3 years P = (25,000) /(1 + 0.10)3
i = 10.0% = $18,783

25. Present Value
Example: If you will need $25,000 to buy a new truck in 3 years, how much should you invest now at
an interest rate of 9.5% compounded annually?
Method #1: Direct Calculation: Straight forward - Plug and Crank
Method #2: Tables: Interpolation
Which table in the appendix will be used? Tables i = 9% & 10%
What is the factor to be used? (9%): 0.7722
(10%): 0.7513
9.5%: 0.7618
P = F(P/F,i,n) = (25,000)(0.7618) = $ 19,044
Go to the interest tables and try to attain the numbers then calculate for your self the
interest at 9.5%

27. We can apply compound interest formulas to “move”
cash flows along the cash flow diagram.
Using the standard notation, we find that a present amount,
P, can grow into a future amount, F, in N time periods at
interest rate i according to the formula below.
In a similar way we can find P given F by

28. It is common to use standard notation for interest factors.
This is also known as the single payment compound amount
factor. The term on the right is read “F given P at i% interest per
period for N interest periods.”
is called the single payment present worth factor.

29. We can use these to find economically equivalent values at
different points in time.
$2,500 at time zero is equivalent to how much after six years if the interest
rate is 8% per year? Future worth
$3,000 at the end of year seven is equivalent to how much today (time zero) if
the interest rate is 6% per year? Present worth

30. We can use these to find economically equivalent values at
different points in time.
$2,500 at time zero is equivalent to how much after six years if the interest
rate is 8% per year? Future worth
$3,000 at the end of year seven is equivalent to how much today (time zero) if
the interest rate is 6% per year? Present worth

31. Nominal and effective interest rates.
• More often than not, the time between successive compounding, or the
interest period, is less than one year (e.g., daily, monthly, quarterly).
• The annual rate is known as a nominal rate.
• A nominal rate of 12%, compounded monthly, means an interest of 1%
(12%/12) would accrue each month, and the annual rate would be effectively
somewhat greater than 12%.
• The more frequent the compounding the greater the effective interest.

32. i = Nominal rate of interest
r = Effective interest rate per period
• When the compounding frequency is annually: r = i
• When compounding is performed more than once per year,
the effective rate (true annual rate) always exceeds the
nominal annual rate: r > i
Nominal and effective interest rates

33. Making Interest Rates Comparable
❑The annual percentage rate (APR) indicates the amount of interest
paid or earned in one year with a pre-specified compounding
frequency. APR is also known as the nominal or stated interest rate.
This is the rate required by law.
❑We cannot compare two loans based on APR if they do not have the
same compounding period.
❑To make them comparable, we calculate their equivalent rate using
an annual compounding period. We do this by calculating the
effective annual rate (EAR) .
❑The two should generate the same amount of money in one year.
33

34. The effect of more frequent compounding can be
easily determined.
Let i be the nominal, annual interest rate and m the number of compounding
periods per year. We can find, r, the effective interest by using the formula below.
For an 18% nominal rate, compounded quarterly, the effective interest is.
For a 7% nominal rate, compounded monthly, the effective interest is.
𝑟 = 1 +
𝑖
𝑚
𝑚
− 1
𝑟 = 1 +
0.18
4
4
− 1 = 19.25%
𝑟 = 1 +
0.07
12
12
− 1 = 7.23%

35. Example: If a student borrows $1,000 from a finance company which charges interest at a
compound rate of 2% per month:
• What is the nominal interest rate:
• What is the effective annual interest rate:
Example: Calculate the EAR for a loan that has a 5.45% quoted annual interest rate
compounded monthly.

36. Example: If a student borrows $1,000 from a finance company which charges interest at a
compound rate of 2% per month:
• What is the nominal interest rate:
i = (2%/month) x (12 months) = 24% annually
• What is the effective annual interest rate:
r = (1 + i/m)m – 1
r = (1 + .24/12)12 – 1 = 0.268 (26.8%)
Example: Calculate the EAR for a loan that has a 5.45% quoted annual interest rate
compounded monthly.
➢Monthly compounding implies 12 compounding per year. m=12.
➢EAR = (1+APR/m)m - 1 = (1+.0545/12)12 - 1
= 1.0558 – 1= .05588 or 5.59%

37. Compound Interest Formula
Where n = mt, and
F = Accumulated amount at the end of t years=Future value
P = Principal= present worth
i = Nominal interest rate per year
m = Number of conversion periods per year
t = Term (number of years)
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
◆ There are n = mt periods in t years, so the accumulated amount at the end
of t years is given by

38. Example
• Find the accumulated amount after 3 years if $1000
is invested at 8% per year compounded
a. Annually
b. Semiannually
c. Quarterly
d. Monthly
e. Daily

39. Solution
a. Annually.
Here, P = 1000, i = 0.08,
and m = 1.
Thus,
n = 3, so
$1259.71.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 1 +
0.08
1
3
= 1000(1.08)3
≈ 1259.71
Solution
b. Semiannually.
Solution
c. Quarterly.

40. Solution
a. Annually.
Here, P = 1000, i = 0.08,
and m = 1.
Thus,
n = 3, so
$1259.71.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 1 +
0.08
1
3
= 1000(1.08)3
≈ 1259.71
Solution
b. Semiannually.
Here, P = 1000, i = 0.08,
and m = 2.
Thus,
n = (3)(2) = 6, so
$1265.32.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 1 +
0.08
2
6
= 1000(1.04)6
≈ 1265.32
Solution
c. Quarterly.
Here, P = 1000, r = 0.08,
and m = 4.
Thus,
n = (3)(4) = 12, so
$1268.24.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 ቆ1
= 1000(1.02)12
≈ 1268.24

41. Solution
d. Monthly.
Solution
e. Daily.

42. Solution
d. Monthly.
Here, P = 1000, i = 0.08, and m = 12.
Thus, n = (3)(12) = 36, so
$1270.24.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 1 +
0.08
12
36
= 1000 1 +
0.08
12
36
≈ 1270.24
Solution
e. Daily.
Here, P = 1000, i = 0.08, and m = 365.
Thus, n = (3)(365) = 1095, so
$1271.22.
𝐹 = 𝑃 1 +
𝑖
𝑚
𝑛
= 1000 1 +
0.08
365
1095
= 1000 1 +
0.08
365
1095
≈ 1271.22

43. Interest
Compounded
F=P(1+i)n
Continuous
F=Pe(i
I
)(n)
Discrete
Time unit is one year
effective interest ie
Fn=P(1+ie)n
Time unit less than
one year nominal
interest iI
Fn=P(1+iI/m)mn
Simple
F=P(1+in)