Engineering economics Slides for Postgraduates

Engineering Management,
Economics and Risk Analysis
(EMERA)
Dr Jeb Gholinezhad

2
Engineering
is the profession in which
• Engineering
knowledge (math and natural sciences
gained by study, experience and practice) is
applied with judgment to develop ways to
utilize, economically, the materials and
forces of nature for the benefit of mankind.
Two environments are needed for the work of an engineer
– physical
– economical

3
Engineering Process
• Determination of objectives
• Identification of strategic factors
• Determination of means (engineering proposals)
• Evaluation of engineering proposals
• Assistance in decision making

4
Engineering Economy
• It deals with the concepts and techniques of analysis
useful in evaluating the worth of systems, products,
and services in relation to their costs
• It is used to answer many different questions
– Which engineering projects are worthwhile?
• Has the mining or petroleum engineer shown that the mineral or oil deposits is worth
developing?
– Which engineering projects should have a higher priority?
• Has the industrial engineer shown which factory improvement projects
should be funded with the available dollars?
– How should the engineering project be designed?
• Has civil or mechanical engineer chosen the best thickness for insulation?

5
Basic Concepts
• Cash flow
• Interest Rate and Time value of money
• Equivalence technique

6
Cash Flow
• Engineering projects generally have economic
consequences that occur over an extended period
of time
– For example, if an expensive piece of machinery is installed in a
plant were bought on credit, the simple process of paying for it
may take several years
– The resulting favourable consequences may last as long as the
equipment performs its useful function
• Each project is described as cash receipts or
disbursements (expenses) at different points in
time

7
Some definitions
• First cost: expense to build or to buy and install
• Operations and maintenance (O&M): annual expense, such
as electricity, labour, and minor repairs
• Salvage value: receipt at project termination for sale or
transfer of the equipment (can be a salvage cost)
• Revenues: annual receipts due to sale of products or
services
• Overhaul: major capital expenditure that occurs during the
asset’s life

8
Cash Flow Diagrams
• The costs and benefits of engineering projects over time
are summarized on a cash flow diagram (CFD).
• Specifically, CFD illustrates the size, sign, and timing of
individual cash flows, and forms the basis for engineering
economic analysis
• A CFD is created by first drawing a segmented time-based
horizontal line, divided into appropriate time unit.
• Each time when there is a cash flow, a vertical arrow is
added − pointing down for costs and up for revenues or
benefits. The cost flows are drawn to relative scale

Cash Flow Diagrams
 In a cash flow diagram (CFD) the end of period t is the same
as the beginning of period (t+1)
 Beginning of period cash flows are: rent, lease, and insurance
payments
 End-of-period cash flows are: O&M, salvages, revenues,
overhauls
 The choice of time 0 is arbitrary. It can be when a project is
analysed, when funding is approved, or when construction
begins
 One person’s cash outflow (represented as a negative value) is
another person’s inflow (represented as a positive value)
 It is better to show two or more cash flows occurring in the
same year individually so that there is a clear connection from
the problem statement to each cash flow in the diagram 14

10
Example
A man borrowed £1,000 from a bank at 8%/year interest.
Two end-of-year payments: at the end of the first year, he
will repay half of the £1000 principal plus the interest that
is due. At the end of the second year, he will repay the
remaining half plus the interest for the second year.
Construct a cash flow diagram

Example
A man borrowed £1,000 from a bank at 8%/year interest.
Two end-of-year payments: at the end of the first year, he
will repay half of the £1000 principal plus the interest that
is due. At the end of the second year, he will repay the
remaining half plus the interest for the second year.
Construct a cash flow diagram
End of year Cash flow
0 +$1000
1 -$580 (-$500 - $80)
2 -$540 (-$500 - $40) 11

12
Time Value of Money
• Money has value
– Money can be leased or rented
– The payment is called interest
–If you put $100 in a bank at 9% interest for one
time period you will receive back your original
$100 plus $9
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9

13
Compound Interest
• Interest that is computed on the original unpaid debt
and the unpaid interest
• Compound interest is most commonly used in
practice
Interest to be returned = $100 x .09 = $9
In year 2
Interest to be returned = $109 x .09 = $9.81

Annual interest rate
SIMPLE INTEREST
FORMULA
Interest paid
Principal
(Amount of
money invested
or borrowed)
Time (in years)
100
I = PRT
19

15
Example:
If you invested $200.00 in an account that paid
simple interest, find how long you’d need to
leave it in at 4% interest to make $10.00.

If you invested $200.00 in an account that paid
simple interest, find how long you’d need to
leave it in at 4% interest to make $10.00.
enter in formula
as a decimal I = PRT
100
10 = (200)(0.04)T
1.25 yrs = T
Typically interest is NOT simple interest but is paid semi-
annually (twice a year), quarterly (4 times per year), monthly
(12 times per year), or even daily (365 times per year).
16

COMPOUND INTEREST
FORMULA
amount at
the end
Principal
(amount
at start)
annual interest
rate
(as a decimal)
r 
nt
17

n 


A  P1
time
(in years)
number of times per
year that interest in
compounded

Example:
Find the amount that results from $500 invested at
8% compounded quarterly after a period of 2 years.



 

A 5
P
001
.0
r
8
n
4
4
n(
t2)
19
A  $585.83
Find the amount that results from $500 invested at
8% compounded quarterly after a period of 2 years.

Effective rate of interest is the equivalent annual simple
rate of interest that would yield the same amount as
that made compounding.
This is found by finding the interest made when
compounded and subbing that in the simple interest
formula and solving for rate.
20
Effective rate of interest

21
Example:
Find the effective rate of interest for the previous example.
The interest made was $85.83. Use the
simple interest formula and solve for r to
get the effective rate of interest.
I = Prt 85.83=(500)r(2)
r = .08583 = 8.583%

22
Equivalence
• Relative attractiveness of different alternatives can
be judged by using the technique of equivalence
• We use comparable equivalent values of
alternatives to judge the relative attractiveness of
the given alternatives
• Equivalence is dependent on interest rate
• Compound Interest formulas can be used to
facilitate equivalence computations

23
Technique of Equivalence
• Determine a single equivalent value at a point in
time for plan 1.
• Determine a single equivalent value at a point in
time for plan 2.
Both at the same interest rate and at the same time
point.
• Judge the
alternatives
values.
relative attractiveness of the two
from the comparable equivalent

24
Economic Analysis Methods
• Three commonly used economic analysis
methods are
1. Present Worth Analysis
2. Annual Worth Analysis
3. Rate of Return Analysis

25
Present Worth Analysis
• Steps to do present worth analysis for a single
alternative (investment)
– Select a desired value of the return on investment (i)
–Using the compound interest formulas bring all benefits
and costs to present worth
–Select the alternative if its net present worth (Present
worth of benefits – Present worth of costs) > 0

26
Present Worth Analysis
• Steps to do present worth analysis for selecting a
single alternative (investment) from among
multiple alternatives
Step 1: Select a desired value of the return on investment
(i)
Step 2: Using the compound interest formulas bring all
benefits and costs to present worth for each alternative
Step 3: Select the alternative with the largest net present
worth (Present worth of benefits –Present worth of costs)

27
Cost and Benefit Estimates
• Present and future benefits (income) and
costs need to be estimated to determine the
attractiveness (worthiness) of a new product
investment alternative

28
Annual costs and Income for a
Product
• Annual product total cost is the sum of
– annual material,
– labour, and overhead (salaries, taxes, marketing expenses, office
costs, and related costs),
– annual operating costs (power, maintenance, repairs, space costs,
and related expenses),
– and annual first cost minus the annual salvage value.
• Annual income generated through the sales of a product =
number of units sold annually x unit price

29
Rate of Return Analysis
• Single alternative case
• In this method all revenues and costs of the
alternative are reduced to a single percentage
number
• This percentage number can be compared to
other investment returns and interest rates
inside and outside the organization

Supply and Demand
• What is Supply and what is demand
• Laws that govern both
• Curves and equilibrium
• Price and quantity
• Elasticity
30

Supply and Demand
• it is clear that when there is a
decrease in the price of a product,
the demand for the product
increases and its supply decreases
• Also, the product is more in
demand and hence the demand of
the product increases.
• The point of intersection of the supply
curve and the demand curve is known
as the equilibrium point.
• At the price corresponding to this
point, the quantity of supply is equal to
the quantity of demand
Demand refers to how much (quantity) of
a product or service is desired by buyers.
The quantity demanded is the amount of a
product people are willing to buy at a
certain price; the relationship between
price and quantity demanded is known as
the demand relationship. Supply represents
how much the market can offer
Consumers choice: to spend/save money
Producers Choice: Production process = MOST Profitable
36

32
ECONOMICS
• Economics is the science that deals with the production
and consumption of goods and services and the
distribution and rendering of these for human welfare.
The following are the economic goals.
 Economic freedom-Choice
 Economic efficiency-cheap
 Economic equity-Equal
Access Amounts, income
distribution
 Economic security-
Protection against risks
 Full employment- land/labor/capital
 Price stability- NOT going into a
recession and NOT having explosive growth
 Economic growth-standard of living

Flow in an Economy
The circular flow of income is a neoclassical economic model depicting how
money flows through the economy. In its simplest version, the economy is
modelled as consisting only of households and firms. Money flows to workers
in the form of wages, and money flows back to firms in exchange for products.
33

circular economy is an alternative to
a traditional linear economy (make,
use, dispose) in which we keep
resources in use for as long as
possible, extract the maximum value
from them whilst in use, then recover
and regenerate products and materials
at the end of each service life.
Sustainability and Circular Economy
36

ENGINEERING ECONOMICS
Engineering economics, is a subset
of economics for application
to engineering projects. Engineers
• seek solutions to problems,
• and the economic viability of each
potential solution is normally
considered along with the
technical aspects
Energy economics is a broad
scientific subject area which includes
topics related to supply and use of
energy in societies
https://www.youtube.com/watch?v=6T4qJjst_KQ
42
https://www.bp.com/en/global/corporate/energy-economics.html

38
Project Management
• Project Evaluation and Review Technique
• Critical Path Method
• Planned Value (PV), Actual Cost (AC) and Earned Value
(EV) Cost Variance
• Schedule Variance
• Cost Performance Index
• Schedule Performance Index
• Cost-Schedule Index
• Estimated cost to complete (ETC)
• Estimated cost at completion (EAC)

Life Cycle Of An Engineering Project
Acquisition phase: all activities prior to the delivery of products
and services.
•Requirements definition stage—Includes determination of user/customer
needs, assessing them relative to the anticipated system, and preparation of
the system requirements documentation.
•Preliminary design stage—Includes feasibility study, conceptual, and early-
stage plans;
final go–no go decision is probably made here.
Detailed design stage—Includes detailed plans for resources—capital,
human, facilities, information systems, marketing, etc.; there is some
acquisition of assets, if economically justifiable.
-Operation phase: all activities are functioning, products and services are
available.
-Construction and implementation stage—Includes purchases, construction,
and implementation of system components; testing; preparation, etc.
-Usage stage—Uses the system to generate products and services; the
largest portion of the life cycle.
Phase out and disposal phase: covers all activities to transition to a new
system; removal/ recycling/disposal of old system.
39

Types of Efficiency
• Efficiency of a system is
generally defined as the ratio of
its output to input.
• The efficiency can be classified
into technical efficiency and
economic efficiency
Technical Efficiency:
• PROCESS – getting the
MOST output with the
LEAST inputs
• Make it as fast as you can and
as cheap as you can
40
Economic Efficiency:
•Making products wanted or
needed by public
• Looks at cost for rest of society
•What will benefit society as a
whole
Is it Possible to be Technically
efficient, but NOT Economically
Efficient?

Next Week
Read Chapter 1 of the book Engineering Economics R.
Panneerselvam
Find and read the paper attached
Define energy audit and its types
Analyse the paper in terms of good and bad practice of
the economics of Auditing an engineering site
41

Engineering
Economics, Management and
Risk Analysis
Dr Jeb Gholinezhad

What is a Cost?
Cost is a sacrifice of resources.
Let’s start with the basic concept of cost. A cost is a sacrifice
of resources. When we buy one thing, we give up (sacrifice)
the ability to use these resources, for example cash, to buy
something else.
Cost means “the price paid for something”
• Cost ascertainment is computation of actual costs incurred
• Cost estimation is a process of predetermining costs of
goods and service.

Manufacturing cost
• is the sum
consumed in
product.
of costs of
the process
all resources
of making a
• The manufacturing cost is classified into
three categories:
1. direct materials cost,
2. direct labor cost
3. manufacturing overhead.

Direct Manufacturing Costs
A direct Manufacturing cost is a price that can be completely
attributed to the production of specific goods or services.
Direct costs:
Costs that, for a reasonable cost, can
be directly traced to the product.
Direct materials:
Materials directly
traceable to the product
Direct labor:
Work directly traceable to
transforming materials
into the finished product

Indirect Manufacturing Costs
Indirect costs:
Costs that cannot reasonably
be directly traced to the product.
Manufacturing overhead:
All production costs except
direct materials and direct labor.
Indirect materials Other indirect costs
Indirect labor

Prime Cost
• Prime costs are a firm's expenses for the direct
materials and labor used in production.
• It refers to a manufactured product's costs, which
are calculated to ensure the best profit margin for
a company.
• The prime cost calculates the use of raw materials
and direct labor, but does not factor in indirect
expenses, such as advertising .
=Material cost + Labor cost + direct expenses

Factory Cost = Prime Cost + Factory overheads
Cost of production= factory cost + Administration
Overheads
Cost of goods sold (COGS) is the direct costs attributable to
the production of the goods sold in a company. This amount includes the
cost of the materials used in creating the good along with the direct labor
costs used to produce the good. It excludes indirect expenses such as
distribution costs and sales force costs.
COGS =
Beginning Inventory +Purchases during the period –Ending Inventor

Cost of sales =
Cost of goods sold +selling and distribution
overheads
Sales= Cost of sales + Profit
Selling price per unit = Sales/ Quantity sold

Fixed and Variable Costs
total cost (TC) describes the total economic cost of production and is made
up of
• variable costs, which vary according to the quantity of a good produced
and include inputs such as labour and raw materials,
• plus fixed costs, which are independent of the quantity of a good produced
and include inputs (capital) that cannot be varied in the short term, such as
buildings and machinery.
• Total Costs
• TC = TFC + TVC
– Total Fixed Costs (TFC)
– Total Variable Costs (TVC)
– Total Costs (TC)

Average Costs
Fixed cost FC
Quantity Q
AFC  
AVC 
Variable cost

VC
Quantity Q
ATC 
Total cost

TC
Quantity Q
• In economics, average cost and/or unit cost is equal to total
cost (TC) divided by the number of goods produced (the output
quantity, Q).
• Average costs may be dependent on the time period considered
(increasing production may be expensive or impossible in the short
term, for example). Average costs affect the supply curve
• Average Fixed Costs (AFC)
• Average Variable Costs (AVC)
• Average Total Costs (ATC)
ATC = AFC + AVC

Marginal Costs/Revenues
• Marginal Cost the cost added by producing
one additional unit of a product or service.
– helps answer the following, How much does it
cost to produce an additional unit of output?
MC 
(change in total cost)

TC
(change in quantity) Q
• Marginal Revenue the revenue gained by
producing one additional unit of a product
or service.

https://www.youtube.com/watch?v=17Aeg9a2sZM

Break-Even Analysis
• Technique used to examine the relationship
between cost and price and to determine
what sales volume must be reached at a
given price before the company will
completely cover its total costs and past
which it will begin making a profit
• All costs are covered but there isn’t a penny
left over

• Plot the total revenue line and
the total cost line
• Intersection is break-even
• Sensitivity analysis can be
done to examine changes in all
of the assumptions made
© 2007 Wiley
Break-Even Analysis: Graphical Approach
• Compute quantity of goods
that must be sold to break-even
• Compute total revenue at an
assumed selling price
• Compute
variable
quantities
fixed cost and
cost for several

Break-Even Analysis
• Total cost = fixed costs + variable costs (quantity):
TC  F  VCQ
• Revenue = selling price (quantity)
• Break-even point is where total costs = revenue:
R  SPQ
F VCQSPQ
SPVC
F
or Q
TC  R or

Profit
• Profit = total revenue – total costs
TC  F  VCQ
R  SPQ

Example
A firm estimates that the fixed cost of producing
a line of footwear is $52,000 with a $9 variable
cost for each pair produced. They want to Know
– If each pair sells for $25, how many pairs must
they sell to break-even?
– If they sell 4000 pairs at $25 each, how much
money will they make?

Example Solved
• Break-even point:

$52,000
 3,250 pairs
SPVC $25$9
• Profit = total revenue – total costs
P  SPQ F  VCQ
 $254,000  $52,000  $94,000
 $12,000
F
Q 

Break-even calculation: A company is planning to establish a chain of
movie theaters. It estimates that each new theater will cost approximately $1
Million. The theaters will hold 500 people and will have 4 showings each day
with average ticket prices at $8. They estimate that concession sales will
average $2 per patron. The variable costs in labor and material are estimated
to be $6 per patron. They will be open 300 days each year.
 Calculate the Break Even Point
 What is the gross profit if they sell 300,000 tickets
 If concessions average $0.50/patron, what is break-even Q now?
(sensitivity analysis)

Break-even calculation: A company is planning to establish a chain of
movie theaters. It estimates that each new theater will cost approximately $1
Million. The theaters will hold 500 people and will have 4 showings each day
with average ticket prices at $8. They estimate that concession sales will
average $2 per patron. The variable costs in labor and material are estimated
to be $6 per patron. They will be open 300 days each year.
 Calculate the Break Even Point
Total revenues = Total costs @ break-even point Q
Selling price*Q = Fixed cost + variable cost*Q
($8+$2)Q= $1,000,000 + $6*Q
Q = 250,000 patrons (42% occupancy)
 What is the gross profit if they sell 300,000 tickets
Profit = Total Revenue – Total Costs
P = $10*300,000 – (1,000,000 + $6*300,000)
P = $200,000
 If concessions average $0.50/patron, what is break-even Q now?
(sensitivity analysis)
($8.50)Q = 1,000,000 + $6*Q
Q = 400,000 patrons (67% occupancy)
500 people
4 showings
300days

Margin of safety
• Margin of safety (safety margin) is the
difference between the intrinsic value of a stock
and its market price.
• Another definition: In break-even analysis, from
the discipline of accounting, margin of safety is
how much output or sales level can fall before a
business reaches its break-even point.
• The margin of safety is a financial ratio that
measures the amount of sales that exceed
the break-even point.

• It’s called the safety margin because it’s kind of
like a buffer.
• This is the amount of sales that the company or
department can lose before it starts losing money.
• As long as there’s a buffer, by definition the
operations are profitable.
• If the safety margin falls to zero, the operations
break even for the period and no profit is realized.
If the margin becomes negative, the operations
lose money.
Margin of safety

This formula shows the total number of sales above the
breakeven point. In other words, the total number of
sales dollars that can be lost before the company loses
money
This version of the margin of safety equation expresses the
buffer zone in terms of a percentage of sales. Management
typically uses this form to analyze sales forecasts and
ensure sales will not fall below the safety percentage.

This equation measures the profitability buffer zone in
units produced and allows management to evaluate
the production levels needed to achieve a profit

Paper Review
(Coursework Preparation)

Engineering
Management,
Economics and
Risk Analysis
Dr Jeb Gholinezhad

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

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

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 !

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.

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?

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?

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).

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

Example:
If $5,000 were loaned for five years at a
simple interest rate of 7% per year, the
interest earned would be
I = (P) (N) (i)
I = 5000 x 5 x 7/100 = 1750
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.

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?

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
Principaland 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

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.

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

• 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

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
Find F
Given
n =
P =
i =

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
F = (1,500)(1+0.06)6
F = $ 2,128
n = 6 years
P = $ 1,500
i = 6.0 %

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

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.

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.

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 =

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 Fi
nd
P
F = $25,000 P = F / (1+i)n
n =
i =
3 years
10.0%
P = (25,000) /(1 +
0.10)3
= $18,783

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%):
(10%):
0.7722
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%

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

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.

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

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.

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

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

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%

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.

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%

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)
𝑖
𝐹 = 𝑃 1 +
𝑚
𝑛
 There are n = mt periods in t years, so the accumulated amount at the end
of t years is given by

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

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

Solution
a. Annually.
Here, P = 1000, i = 0.08,
and m = 1.
Thus,
n = 3, so
$1259.71.
𝑖
𝐹 = 𝑃 1 +
𝑚
𝑛
0.08
= 1000 1 +
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.
𝑖
𝐹 = 𝑃 1 +
𝑚
𝑛
0.08
= 1000 1 +
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.
𝑖
𝐹 = 𝑃 1 +
𝑚
𝑛
= 1000ቆ
1
= 1000(1.02)12
≈ 1268.24

Solution
d. Monthly.
Solution
e. Daily.

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

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)

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)
Recap

Recap
 Single-payment compound amount
 F = P (F/P,i,n)
 (1+i)n = (F/P,i,n)
Single-payment present worth
 P = F(P/F,i,n)
 (1+i)-n = (P/F,i,n)

EXAMPLE
An engineer received a bonus of $10,000 that he will invest now. He
calculate the equivalent value after 20 years, when he plans to use a
money as the down payment on an island vacation home. Assume a
of 8% per year for each of the 20 years. Find the amount he can pay
direct calculation and tabulated factor.

EXAMPLE
An engineer received a bonus of $10,000 that he will invest now. He
calculate the equivalent value after 20 years, when he plans to use a
money as the down payment on an island vacation home. Assume a
of 8% per year for each of the 20 years. Find the amount he can pay
direct calculation and tabulated factor.
P = $10,000
i = 8% per year
n = 20 years
F = ?
F = P(1 + i)n = 10,000(1.08)20 = 10,000(4.661
F = P(F/P,8%,20) = 10,000(4.6610) = $46

Annuity (uniform series)
• Annuity: A series of equal payments or receipts occurrin
specified number of periods.
• Ordinary annuity: A series of equal payments or receipt
over a specified number of periods with the payments o
occurring at the end of each period.
• Annuity due: A series of equal payments or receipts occ
specified number of periods with the payments or recei
at the beginning of each period.

An annuity with payments at the end of
period is known as an ordinary annu
Ordinary Annuity
1 2 3
$10,000 $10,000 $10,000 $
End of year 1
Today
End of year 2
End of year 3
E

Theuniform seriesfactorsthatinvolve FandAarederivedasfol
(1) Cashflowoccursin consecutiveinterest periods
(2) Lastcashflow occursin sameperiodasF
0 1 2 3 4 5
F=?
A=Given
0 1 2 3
A=?
Note:FtakesplaceinthesameperiodaslastA
Cashflowdiagramsare:
The Future Value of an Ordinary Annuity
(F/A, i, n) is equal-payment
series compound amount factor

2 3 4
A=?
0 1
P=Given
Thecashflowdiagramsare:
Note:Pis oneperiodAheadoffirstAvalue
Theuniform series factorsthatinvolve PandAarederivedasfo
(1) Cashflowoccursin consecutiveinterest periods
(2) Cashflowamountis samein eachinterest period
0 1 2 3 4 5
A=Given
P=?
Present Worth of an Ordinary Annuity
(P/A, i, n) is equal-payment
series present worth factor

Example: How much will you have in 40 years if you
each year and your account earns 8% interest each yea
Finding F when given A

Finding A when given F
Example: How much would you need to set aside each
years, at 10% interest, to have accumulated $1,000,000
the 25 years?
1 2 3 4 5 6 7

Example: How much is needed today to provide an an
of $50,000 each year for 20 years, at 9% interest each
P =
1 2 3 4 5 6 7 8 A =?
Finding P when given A

Example: If you had $500,000 today in an account ear
year, how much could you withdraw each year for 25 y
Finding A when given P

Amortized Loans
An amortized loan is a loan paid off in equal payments –
consequently, the loan payments are an annuity.
Examples: Home mortgage loans, Auto loans
In an amortized loan, the present value can be thought
amount borrowed, n is the number of periods the loan la
the interest rate per period, and payment is the loan pay
made.
Size of each payment remains the same.
However, Interest payment declines each year as the amount
and more of the principal is repaid.

Perpetuities
A perpetuity is an annuity that continues forever or has no m
For example, a dividend stream on a share of preferred stock
are two basic types of perpetuities:
Growing perpetuity in which cash flows grow at a constant rate, g,
period to period.
Level perpetuity in which the payments are constant rate from per
period.
Even if the cash flows are infinite, present values can be fini
discount rate is higher than the growth rate.
To establish a perpetuity based on capitalized cost we should h
accumulated amount of money K in order to provide funds

A company has to replace a present facility after 15 years at an outlay o
to deposit an equal amount at the end of every year for the next 15 yea
of 18% compounded annually. Find the equivalent amount that must
end of every year for the next 15 years.
Annuity - Another Example

A company has to replace a present facility after 15 years at an outlay o
to deposit an equal amount at the end of every year for the next 15 yea
of 18% compounded annually. Find the equivalent amount that must
end of every year for the next 15 years.
Annuity - Another Example
F = $500,000
n = 15 years
i = 18%
A = ?
𝑖
A = F = F (A/F, i, n)
(1+𝑖)𝑛−1
= 500,000 (A/F, 18%, 15)
= 500,000 (0.0164)
= $8,200
The annual equal amount which must be deposited for 15 years is $8,200.

Gradient Series
• Thus far, the discussion has focused on uniform-series
• A great many investment problems in the real world invo
unequal cash flow series and can not be solved with the
previously introduced
• As such, independent and variable cash flows can only be
the repetitive application of single payment equations
• Mathematical solutions have been developed, however, fo
of unequal cash flows:
• Uniform Gradient Series
• Geometric Gradient Series

Arithmetic Gradient Factors (P/G, A
• Cash flows that increase or decrease by a constant a
are considered arithmetic gradient cash flows. The am
of increase (or decrease) is called the gradient.
$125
$100
$150
$175
0 1 2 3 4
G = $25
Base = $100
$2000
$1500
$1000
$500
0 1 2 3 4
G = -$500
Base = $2000

Uniform (Arithmetic) Gradient of Cash Flo
• Find P when given G: P = G ( P/G, i%, n )
• Find the present equivalent value when given the uniform g
• Find F when given G: F = G ( F/G, i%, n )
𝑃 = 𝐺 𝑛
−
1 1 + 𝑖 𝑛 − 1 𝑛
𝑖 𝑖 1 + 𝑖 1 + 𝑖 𝑛
∴ 𝐹 = 𝐺
1 1 + 𝑖 𝑛 − 1
𝑖 𝑖
− 𝑛

Uniform (Arithmetic) Gradient of Cash Flo
Find A when given G:
• Find the annual equivalent value when given the unifo
amount
• Functionally represented as A = G ( A / G, i%, n )
𝑛
1
𝐴 = 𝐺 −
𝑖 1 + 𝑖 𝑛 − 1

2
Example
A sport apparel company has initiated a logo-licensing program. It
revenue of $80,000 in fees next year from the sale of its logo. Fees are
uniformly to a level of $200,000 in 9 years. Determine the arithmetic g
the cash flow diagram.

Example
A textile mill has just purchased a lift truck that has a useful life of
estimates that the maintenance costs for the truck during the first
Maintenance costs are expected to increase as the truck ages at a rate of
remaining life. Assume that the maintenance costs occur at the end of eac
to set up a maintenance account that earns 12% annual interest. A
expenses will be paid out of this account. How much should the firm
now?

P = P1 + P2
P = A1(P/A, 12%, 5) + G(P/G, 12%, 5)
= 1000(3.6048) + 250(6.397)
= 5,204 $
Solution
Given: A1 = 1000, G = 250, i = 12% per year, and n = 5 years
Find: P
The cash flow may be broken into its two components
The first component is an equal-payment series (A1), and the
second is the linear gradient series (G). 1
0 2
G = 250
P = ??
A1 =1000

Example: An engineer is planning for a 15 year retirement. In order
pension and offset the anticipated effects of inflation and increased ta
withdraw $5,000 at the end of the first year, and to increase the withdr
the end of each successive year. How much money must the eng
account at the start of his retirement, if the money earns 6% per y
annually?

Example: An engineer is planning for a 15 year retirement. In order
pension and offset the anticipated effects of inflation and increased ta
withdraw $5,000 at the end of the first year, and to increase the withdr
the end of each successive year. How much money must the eng
account at the start of his retirement, if the money earns 6% per y
annually?
Want to Find: P Given: A1 =5000 $
G =1000 $
i= 6%. and
T = 0
P
1
$5000
$6000
2 14 15
$18000
$19000
$7000
$8000
3 4
AT = (A1) + (A2)
A2 = G(A/G,i,n) = $1000 (A/G,6%,15) =
AT = $5000 + $5926 = $10,926
P = AT (P/A,i,n) = $10,926 (P/A,6%,15) =
$106,120

A company expects to realize revenue of £100000 during the first year
the sale of a new product. Sales are expected to decrease gradually
following a uniform gradient, to a level of £ 47500 during the eight
Draw the cash-flow diagram following the end-of -period convention.
invested at a rate of 10 % what will be the compound amount eight years
Example

1 2 3 4
0
5
i= 10 %
6 7
years
8
F = ??
100 000 m.u.
47 500 m.u.
G = - ((105-47500)/7) = - 7500 £
A1 = 105 £
n = 8
i=10%
Solution
Using tabulated f
Direct calculation: A = A1 + A2 = A1
F = F1 + F2 = A1 (F/A, 10%, 8) + G (F/G, 10%, 8) = 105
= 105 ((1.18-1)/0.1) – 7500 / 0.1 (((1.18 – 1) / 0.1) – 8) = 774
= 885897.2 £ F = A (F/A, 10%,
= 77466.25 (11.
= 885896.29 £

What is the present value of cash flows of $300 at the end of y
cash flow of negative $600 at the end of year 6, and cash flows
of years 7-10 if the appropriate discount rate is 10%?
Complex Cash Flow Stream – Exercise for Stud

Finding N - Exercise for Students
Bla borrowed $100,000 from a local bank, which charges them
7% per year. If Bla pays the bank $8,000 per year, now man
to pay off the loan?

Finding i - Exercise for Students
Jill invested $1,000 each year for five years in a local company
interest after five years for $8,000. What annual rate of return d

Starter
• You own a business. You have a meeting with your Marketing
Director, your Production Manager and your Research Team Manager,
They know you have £1million retained profit which can only be spent
on one project. They each pitch their idea to you, which do you
choose:
• Project 1: A New warehouse which will mean you can double your
capacity
• Project 2: A new marketing campaign which will double sales
• Project 3: Research into new products which will give you an edge in a
highly competitive technical market place

Investment appraisal defined
Investment appraisal attempts to determine the value of capital
expenditure projects. It enables the business and its investors to
compare projects so that the business can expand and meet their
objectives – usually profit maximisation and efficiency.

Investment appraisal – planning process
• Investment appraisal is the
planning process used to
determine whether the long
term investments will give the
best return.
• Projects such as;
• new machinery
• new premises
• research and development
projects

Investment appraisal – decision making
A business will have lots of ideas for projects, and proposals of ways to grow
their business. The business will only have a finite amount of money and so
perhaps only one project will get the funding it needs to go ahead.
Investment appraisal is used to work out which project should get the funds.
Proposal 1: new premises
Proposal 2: new machinery
Proposal 3: Research

Investment appraisal: objective, inputs and process
Investment appraisal

Good Decision Criteria
▶All cash flows considered?
▶TVM considered?
▶Risk-adjusted?
▶Ability to rank projects?
▶Indicates added value to the firm?

Independent versus Mutually Exclusive Projects
• Independent
• The cash flows of one project are unaffected
by the acceptance of the other.
• Mutually Exclusive
• The acceptance of one project precludes
accepting the other. You can choose to
attend graduate school next year at either
Portsmouth or Texas, but not both

Comparison of Alternatives
In most of the practical decision environments, executives
will be forced to select the best alternative from a set of
competing alternatives.
There are several bases for comparing the worthiness of
the projects. These bases are:
1. Present worth method
2. Future worth method
3. Annual equivalent method
4. Rate of return method

Net Present Value
How much value is created from undertaking an investment?
Step 1: Estimate the expected future cash flows.
Step 2: Estimate the required return for projects of this risk level.
Step 3: Find the present value of the cash flows and subtract the initial investment to arrive
at the Net Present Value.

Net Present Value
Sum of the PVs of all cash flows
Initial cost often is CF0 and is an outflow.
n
n
t = 0
CFt
NPV =∑ (1 + R)t
t = 1
CFt
NPV =∑ (1 + R)t
- CF0
NOTE: t=0

NPV – Decision Rule
• If NPV is positive, accept the project
• NPV > 0 means:
• Project is expected to add value to the firm
• Will increase the wealth of the owners
• NPV is a direct measure of how well this project will meet the goal of increasing
shareholder wealth.

Sample Project Data
• You are looking at a new project and have estimated the following cash flows:
• Year 0: CF = -165,000
• Year 1: CF = 63,120
• Year 2: CF = 70,800
• Year 3: CF = 91,080
• Your required return for assets of this risk is 12%.

Computing NPV for the Project
• Using the formula:
t0 (1 R)t
n
NPV  
CFt

Computing NPV for the Project
• Using the formula:
NPV = -165,000/(1.12)0 + 63,120/(1.12)1 + 70,800/(1.12)2 + 91,080/(1.12)3 = 12,627.41
Capital Budgeting Project NPV
Required Return = 12%
Year CF Formula Disc CFs
0 -165,000.00 =(-165000)/(1.12)^0 = -165,000.00
1 63,120.00 =(63120)/(1.12)^1 = 56,357.14
2 70,800.00 =(70800)/(1.12)^2 = 56,441.33
3 91,080.00 =(91080)/(1.12)^3 = 64,828.94
12,627.41
t0 (1 R)t
n
NPV  
CFt

Rationale for the NPV Method
• NPV = PV inflows – Cost
NPV=0 → Project’s inflows are “exactly sufficient to repay the invested
capital and provide the required rate of return”
• NPV = net gain in shareholder wealth
• Rule: Accept project if NPV > 0

There are two basic types
of alternatives
Investment Alternatives
Those with initial (or front-end) capital investment that produces
positive cash flows from increased revenue, savings through reduced
costs, or both.
Cost Alternatives
Those with all negative cash flows, except for a possible positive cash
flow from disposal of assets at the end of the project’s useful life.

MEA - Select the
alternative that gives you
the most money!
When comparing mutually-exclusive alternatives
(MEA):
• For investment alternatives the PW of all cash
flows must be positive, at the MARR, to be
attractive. Select the alternative with the largest
PW.
• For cost alternatives the PW of all cash flows will
be negative. Select the alternative with the largest
(smallest in absolute value) PW.

Investment alternative example
Use a MARR of 10% and useful life of 5 years to select
between the investment alternatives below.
Alternative
A B
Capital investment -$100,000 -$125,000
Annual revenues less expenses $34,000 $41,000

Investment alternative example
between the investment alternatives below.
Alternative
A B
Annual revenues less expenses $34,000 $41,000
Both alternatives are attractive, butAlternative B provides
a greater present worth, so is better economically.

Cost alternative example
between the cost alternatives below.
Alternative
C D
Annual expenses -$25,000 -$30,000

Cost alternative example
between the cost alternatives below.
Alternative
C D
Annual expenses -$25,000 -$30,000
Alternative D costs less thanAlternative C, it has a greater
PW, so is better economically.

Pause and solve
Your local foundry is adding a new furnace. There are
several different styles and types of furnaces, so the foundry
must select from among a set of mutually exclusive
alternatives. Initial capital investment and annual expenses
for each alternative are given in the table below. None have
any market value at the end of its useful life. Using a
MARR of 15%, which furnace should be chosen?
Furnace
F1 F2 F3
Investment $110,000 $125,000 $138,000
Useful life 10 years 10 years 10 years
Total annual expenses $53,800 $51,625 $45,033

Solution
Using a MARR of 15%, the PW is shown for each of the
three alternatives in the table below.
Furnace
F1 F2 F3
Investment $110,000 $125,000 $138,000
Useful life 10 years 10 years 10 years
Total annual expenses $53,800 $51,625 $45,033
Present Worth @ 15% -$380,010 -$384,094 -$364,010
The largest value is -$364,010, indicating that Furnace
F3 is the best alternative.

Example – Future Worth Method
A man owns a corner plot. He must decide which of the several alternatives to select in
trying to obtain a desirable return on his investment. After much study and calculation,
he decides that the two best alternatives are as given in the following table:
Evaluate the alternatives based on the future worth method at i = 12%.

Alternative 1—Build gas station
First cost = Rs. 20,00,000
Net annual income =Annual income –Annual property tax
= Rs. 8,00,000 – Rs. 80,000
= Rs. 7,20,000
Life = 20 years
Interest rate = 12%, compounded annually
Solution
The future worth of alternative 1 is computed as
FW1(12%) = –20,00,000 (F/P, 12%, 20) + 7,20,000 (F/A, 12%, 20)
= –20,00,000(9.646) + 7,20,000 (72.052)
= Rs. 3,25,85,440

Alternative 2—Build soft ice-cream stand
First cost = Rs. 36,00,000
Net annual income =Annual income –Annual property tax
= Rs. 9,80,000 – Rs. 1,50,000
= Rs. 8,30,000
Life = 20 years
Interest rate = 12%, compounded annually
Solution
The future worth of alternative 2 is calculated as
FW2(12%) = – 36,00,000(F/P, 12%, 20) + 8,30,000 (F/A, 12%, 20)
= –36,00,000 (9.646) + 8,30,000 (72.052)
= Rs. 2,50,77,560
The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.

 Also called equivalent annual worth (EAW), equivalent annual cost (EAC), annual
equivalent (AE), and equivalent uniform annual cost (EUAC).
 An application ofAW analysis is life-cycle cost (LCC) analysis.
 Since theAW value is the equivalent uniform annual worth of all estimated receipts
and disbursements during the life cycle of the project or alternative, AWis easy to
understand by any individual acquainted with annual amounts.
 In the annual equivalent method of comparison, first the annual equivalent cost or
the revenue of each alternative will be computed. Then the alternative with the
maximum annual equivalent revenue in the case of revenue-based comparison or
with the minimum annual equivalent cost in the case of cost-based comparison will
be selected as the best alternative.
Annual Equivalent Method

A company is planning to purchase an advanced machine centre. Three original
manufacturers have responded to its tender whose particulars are tabulated as follows:
Example – Annual Equivalent Method
Determine the best alternative based on the annual equivalent method by assuming
i = 20%, compounded annually.

Alternative 1
Down payment, P = Rs. 5,00,000
Yearly equal instalment,A= Rs. 2,00,000
n = 15 years
i = 20%, compounded annually
Solution
The annual equivalent cost expression of the above cash flow diagram is
AE1(20%) = 5,00,000(A/P, 20%, 15) + 2,00,000
= 5,00,000(0.2139) + 2,00,000
= 3,06,950

Alternative 2
n = 15 years
Solution
AE2(20%) = 4,00,000(A/P, 20%, 15) + 3,00,000
= 4,00,000(0.2139) + 3,00,000
= Rs. 3,85,560

Alternative 3
n = 15 years
Solution
AE3(20%) = 6,00,000(A/P, 20%, 15) + 1,50,000
= 6,00,000(0.2139) + 1,50,000
= Rs. 2,78,340.
The annual equivalent cost of manufacturer 3 is less than that of manufacturer 1 and manufacturer 2. Therefore,
the company should buy the advanced machine centre from manufacturer 3.

Rate of Return Method
 Rate of Return (ROR) is also called internal rate
of return (IROR) and return on investment (ROI).
 The rate of return of a cash flow pattern is the
interest rate at which the present worth of that
cash flow pattern reduces to zero.
 The rate paid on the unpaid balance of borrowed money, or the rate earned on the
unrecovered balance of an investment, so that the final payment or receipt brings
the balance to exactly zero with interest considered.
 In this method of comparison, the rate of return for each alternative is computed.
Then the alternative which has the highest rate of return is selected as the best
alternative.

A person is planning a new business. The initial outlay and cash flow pattern for the
new business are as listed below. The expected life of the business is five years. Find
the rate of return for the new business.
Example – Rate of Return Method

Initial investment = Rs. 1,00,000
Annual equal revenue = Rs. 30,000
Life = 5 years
Solution
The present worth function for the business is:
PW(i) = –1,00,000 + 30,000(P/A, i, 5)
When i = 10%,
PW(10%) = –1,00,000 + 30,000(P/A, 10%, 5)
= –1,00,000 + 30,000(3.7908)
= Rs. 13,724.
When i = 15%,
PW(15%) = –1,00,000 + 30,000(P/A, 15%, 5)
= –1,00,000 + 30,000(3.3522)
= Rs. 566.
When i = 18%,
PW(18%) = –1,00,000 + 30,000(P/A, 18%, 5)
= –1,00,000 + 30,000(3.1272)
= Rs. – 6,184

IRR - Advantages
▶Preferred by executives
▶Intuitively appealing
▶Easy to communicate the value of a project
▶Considers all cash flows
▶Considers time value of money
IRR - Disadvantages
• Can produce multiple answers
• Cannot rank mutually exclusive projects

The accuracy of the IRR calculated may depend on the size of the gap between
the discount rates used in the interpolation calculation
The accuracy of the IRR calculated

 Payback Period
Payback analysis determines the required minimum life of an asset, process,
or system to recover the initial investment. Payback analysis should not be
considered the final decision maker; it is used as a screening tool or to
provide supplemental information for a PW,AW, or other analysis.
 Profitability Index
The profitability index allows investors to quantify the amount of value
created per unit of investment.
Other Appraisal Methods

Payback Period
• How long does it take to recover the initial cost of a project?
• Computation
• Estimate the cash flows
• Subtract the future cash flows from the initial cost until initial investment is
recovered
• A “break-even” type measure
• Decision Rule – Accept if the payback period is less than some preset
limit

Calculate Payback Period
• Same project as before
• Year 0: CF = -165,000
• Year 1: CF = 63,120
• Year 2: CF = 70,800
• Year 3: CF = 91,080
C u m . C F s
C a p i t a l B u d g e t i n g P r o j e c t
Y e a r C F
0 $ ( 1 6 5 , 0 0 0 ) $ ( 1 6 5 , 0 0 0 )
( 1 0 1 , 8 8 0 )
( 3 1 , 0 8 0 )
6 0 , 0 0 0
1 $
6 3 , 1 2
0
$
2 $
7 0 , 8 0
0
$
3 $
9 1 , 0 8
0
$
P a y b a c k = y e a r 2 +
+ ( 3 1 0 8 0 / 9 1 0 8 0 )
P a y b a c k = 2 . 3 4 y e a r s
-165000+63120= -101880
-101880+70800= -31080
-31080+91080= + 60000

Payback – Another Example
• A business needs to decide
which project or proposal to
invest in.
• It cannot afford all four so it will
have a number of methods to
work out which one it should
spend the money on.
• The table starts at year 0
because this is the amount
invested in the project at the
start
• For example, proposal 1 will cost
£120,000
Cash
flows
(£000
s)
Proposal
1
Proposal
2
Proposal
3
Proposal
4
Year 0
-£120 -£95 -£80 -£160
Year 1
£80 £10 £30 £30
Year 2
£60 £40 £40 £50
Year 3
£40 £40 £30 £90
Year 4
£20 £60 £30 £80
Year 5
£40 £50 £20 £60

Very simple Payback calculation
Cash
flows
(£000
s)
Proposal
1
Proposal
2
Proposal
3
Proposal
4
Year 0
-£120 -£95 -£80 -£160
Year 1
£80 £10 £30 £30
Year 2
£40 £40 £40 £50
Year 3
£40 £40 £30 £90
Year 4
£20 £60 £30 £80
Year 5
£40 £50 £20 £60
Lets look at proposal 1
• Year 0 is the investment so it will
cost £120,000 to get the project
going
• When will this be paid back?
• In year 2 the project made
£80,000 from year one plus
£40,000 from year 2.
• The project pays back in year 2
Obviously this is too simple – so next slide find out what happens
when its part way through a year

Payback calculation
Cash
flows
(£000s)
Proposal
1
Proposal
2
Proposa
l 3
Proposal
4
Year 0
-£120 -£95 -£80 -£160
Year 1
£80 £10 £30 £30
Year 2
£40 £40 £40 £50
Year 3
£40 £40 £30 £90
Year 4
£20 £60 £30 £80
Year 5
£40 £50 £20 £60
Lets look at proposal 2
Its going to cost the business
£95,000 to get this project up and
running
• Year 1 the project makes £10,000
• So far it has paid back £90,000
• Now we need £5,000 more from
year 4 so:
5,000
60,000
=
X 12 0.9 Round this up
to one month
Final answer is proposal 2
will payback in 3 years
and 1 month

Calculate the payback periods
Cash inflows
(£000s) Proposal 1 Proposal 2 Proposal 3 Proposal 4
Year 0 -£120 -£95 -£80 -£160
Year 1 £80 £10 £30 £30
Year 2 £40 £40 £40 £50
Year 3 £40 £40 £30 £90
Year 4 £20 £60 £30 £80
Year 5 £40 £50 £20 £60
When does the
project payback?

Calculate the payback periods
Cash inflows
(£000s) Proposal 1 Proposal 2 Proposal 3 Proposal 4
Year 0 -£120 -£95 -£80 -£160
Year 1 £80 £10 £30 £30
Year 2 £40 £40 £40 £50
Year 3 £40 £40 £30 £90
Year 4 £20 £60 £30 £80
Year 5 £40 £50 £20 £60
When does the
project payback?
2 years
3 years and 1
month
2 years and 4
months
2 years and 11
months

Advantages and Disadvantages of Payback
• Advantages
• Easy to understand
• Biased towards liquidity
• Disadvantages
• Ignores the time value of money
• Requires an arbitrary cutoff point
• Ignores cash flows beyond the cutoff date

Profitability Index (PI)
Write down the CF stream
cash inflows
cash outflows (investment)
Use the risk adjusted cost of capital to calculate:
NPV = PV (cash inflows) - PV (cash outflows)
Note that PI is a ratio while NPV is a difference
47
PV (cash inflows)
PV (cash outflows)
PI =

The Profitability Index (PI) rule
PI (ratio) rule intuition: look for projects with
PV(cash inflows) > PV(cash outflows)
If PI > 1  Accept project
If PI = 1  indifference
If PI < 1  Reject project
48
PV (cash inflows)
PV (cash outflows)
PI =

Profitability Index
• Measures the benefit per unit cost, based on the time value of money
• A profitability index of 1.1 implies that for every $1 of investment, we create an additional $0.10 in
value
• Can be very useful in situations of capital rationing
• Decision Rule: If PI > 1.0  Accept
A B
CF(0) $
(10,00
0)
$
(100,00
0)
PV(CF) $
15,00
0
$
125,00
0
PI $
1.
50
$
1.
25

Calculate the profitability index assuming 10% discount rate and $200 million
investment using the cash flows in the following table:
Example

Summary
Internal rate of return =
• Discount rate that makes NPV = 0
• Accept if IRR > required return
• Same decision as NPV with
conventional cash flows
• Unreliable with:
• Non-conventional cash flows
• Mutually exclusive projects
Net present value =
• Difference between market value
(PV of inflows) and cost
• Accept if NPV > 0
• No serious flaws
• Preferred decision criterion
Payback period =
• Length of time until initial investment is
recovered
• Accept if payback < some specified target
• Doesn’t account for time value of money
• Ignores cash flows after payback
• Arbitrary cutoff period
Profitability Index =
• Benefit-cost ratio
• Accept investment if PI > 1
• Cannot be used to rank mutually
exclusive projects

Economics and Risk Analysis
Dr Jeb Gholinezhad

Markets
• A market
mechanism, that brings together
is any institutional structure, or
buyers and
sellers of particular goods and services
• Markets exists in many forms
• They determine the price and quantity of a good
or service transacted

Types of Markets
• Perfectly competitive markets have the following two characteristics:
– Goods being sold are all the same
– Both Buyers and sellers are price takers
– Example: farm products (wheat)
• Monopoly is characterized by:
– One seller and many buyers
– Seller sets the price
– Example: standard oil (1870–1911)
• Oligopoly is characterized by
– Few sellers without rigorous competition
– The sellers get together to set a price
– Example: network providers
• Monopolistic competition is characterized by
– Many sellers, each selling a differentiated product
– Sellers have some ability to set the price for their own product
– Example: toothpaste, soft drinks, clothing

Competitive Market Assumptions
1. Fragmented market: many buyers and sellers
 Implies buyers and sellers are price takers
2.Undifferentiated Products: consumers perceive the product
to be identical so don’t care who they buy it from
3.Perfect Information about price: consumers know the price
of all sellers
4.Equal Access to Resources: everyone has access to the same
technology and inputs.
 Free entry into the market, so if profitable for new firms to
enter into the market they will

Demand
• The various amounts of a product that
consumers are willing and able to purchase at
various prices during some specific period
• Demonstrated by demand schedule and
demand curve

Law of Demand
• The inverse relationship between the price and
the quantity demanded of a good or service
during some period of time
• Based on:
1. Income effect
2. Substitution effect
3. Diminishing marginal utility meaning that the first unit
of consumption of a good or service yields more utility than the second and
subsequent units, with a continuing reduction for greater amounts.

Income Effect
• At a lower price, consumers can buy more of a product without
giving up other goods
• A decline in price increases the purchasing power of money/real
income
Substitution Effect
• At a lower price, consumers have the incentive to substitute the
cheaper good for similar goods that are now relatively more
expensive
Diminishing Marginal Utility
States that successive units of a given product yield less and less extra
satisfaction
Therefore, consumers will only buy more of a good if its price is
reduced

Demand Curve
• Shows the inverse relationship between price
and quantity demanded for a good or service
• Derived from a demand schedule showing the
quantity demanded at various prices

The Law of Demand
• The law of demand holds that other things
equal, as the price of a good or service rises, its
quantity demanded falls.
– The reverse is also true: as the price of a good or
service falls, its quantity demanded increases.

Demand Curve
The demand curve has a negative slope, consistent with
the law of demand.

Changes in Demand
• Caused by changes in one or other of the non-
price determinants of demand
• Represented as a shift of the demand curve
either to the right or left
• Represents a change in the quantity demand at
every price, so cannot be related to a change in
price

Changes in Demand
• Tastes or preferences: Preferences change when
– People become better informed
– New goods become available
• Number of buyers
• Income
– Normal or superior goods—demand varies directly with income
– Inferior goods—demand varies inversely with income
• Prices of related goods
– Substitute goods (e.g. Coca-Cola and Pepsi)
– Complementary goods (Computer hardware and software)
• Expectations
• Seasons/weather

Increase in Demand
P 5
Q
0
4
3
2
1
D1
10 20 30 40 50 60 70 80
Quantity demanded
D1
Increase in
Demand
D2
D2
Price
($
per
unit)

D1
Decrease in
Demand
Decrease in Demand
P 5
Q
0
4
3
2
1
10 20 30 40 50 60 70 80
Quantity demanded
D1
D3
D3
Price
($
per
unit)

Changes in Quantity Demanded
• caused by changes in price only
• represented as movement along a demand
curve
• other factors determining demand are held
constant

Movement along a Curve
P 5
Q
0
4
3
2
1
D1
10 20 30 40 50 60 70 80
Quantity demanded
Movement along
a demand curve
D1
Price
($
per
unit)
Change in
quantity demanded

Supply
• The various amounts of a product that
producers are willing and able to supply at
various prices during some specific period
• Demonstrated by the supply schedule and
supply curve

Law of Supply
• Direct relationship between the price and
quantity supplied
• Increased price causes increased quantity
supplied
• Decreased price causes decreased quantity
supplied
• Related to cost-plus pricing model, i.e. as
quantity increases costs often increase so
firm need a higher P to increase Q.

The Law of Supply
• The law of supply holds that other things equal,
as the price of a good rises, its quantity
supplied will rise, and vice versa.
• Why do producers produce more output when
prices rise?
– They seek higher profits
– They can cover higher marginal costs of production

Supply Curve
The supply curve has a positive slope, consistent with
the law of supply.

Change in Supply
• represented as a shift of the supply curve
• caused by changes in determinants of supply
other than price

Non-price determinants of Supply
• Resource price
• Technology
• Prices of other goods
• Expectations
• Number of sellers

Increase in Supply
S1
P
Q
0
5
4
3
2
1
2 4 6 8 10 12 14 16
Quantity supplied (000/week)
Price
($
per
unit)
S1
S2
S2

Decrease in Supply
S1
P
Q
0
5
4
3
2
1
2 4 6 8 10 12 14 16
Price
($
per
unit)
S1
S3
S3

Changes in Quantity Supplied
• Caused by changes in price only
• Represented as a movement along a supply
curve

Movement along a Supply Curve
S1
P
Q
0
5
4
3
2
1
2 4 6 8 10 12 14 16
Price
($
per
unit)
S1
Movement along
a supply curve

Movement along a Supply Curve
S1
P
$5
Q
0
4
3
2
1
2 4 6 8 10 12 14 16
Price
($
per
unit)
S1
Movement along
a supply curve

Equilibrium
• In economics, an equilibrium is a situation in
which:
– there is no inherent tendency to change,
– quantity demanded equals quantity supplied, and
– the market just clears.

Equilibrium
Equilibrium occurs at a price of $3 and a quantity of 30
units.

Shortage (Excess Demand)
• Occurs when the quantity demanded exceeds the
quantity supplied at the current price
• Competition amongst buyers eventually bids up the
price until equilibrium is reached
Surplus (Excess Supply)
• Occurs when the quantity supplied exceeds the
quantity demanded at the current price
• Competition amongst producers eventually causes the
price to decline until equilibrium is reached

Shortages and Surpluses
• A shortage occurs
when quantity
demanded exceeds
quantity supplied.
– A shortage implies
the market price is
too low.
• A surplus occurs
when quantity
supplied exceeds
quantity demanded.
– A surplus implies
the market price is
too high.
D
P
Q
0
5
4
3
2
1
2 4 6 7 8 10 12 14 16 18
Units of X (000/week)
Price
($
per
unit)
Equilibrium pr
surplus
shortage
S

Changes in Demand and Supply
• Changes or shifts will disrupt the equilibrium
• The market will adjust until once again an
equilibrium is reached
• The equilibrium price and quantity traded will
change

Increase in Demand
P
Q
0
D1
D2
D1
D2
S
Equilibrium
price & quantity
rise

Decrease in Demand
D1
S
P
Q
0
D1
D2
D2
Equilibrium
price & quantity
fall

Increase in Supply
S1
P
Q
0
D1
D1
Equilibrium
price falls & quantity
rises
S2
S1
S2

S2
Decrease in Supply
S1
P
Q
0
D1
D1
Equilibrium
price rises & quantity
falls
S2
S1

Both Demand & Supply Increase
Q
0
D1
D1 S1
P D2
D2
S2
S2
S1
Quantity will increase but price change will be in determinant

Price Ceilings & Floors
• A price ceiling is a legal maximum that can be
charged for a good.
– Results in a shortage of a product
– Common examples include apartment rentals and
credit cards interest rates.
• A price floor is a legal minimum that can be
charged for a good.
– Results in a surplus of a product
– Common examples include, minimum wage

Price Ceiling
A price ceiling is set at $2 resulting in a shortage of
20 units.
Creating a limit for how high a
gallon of gas may sell for (a
price ceiling), however, will
cause more harm than good.
This upper limit of $2 will bring
more people to demand and
buy gas, but companies will
supply less gas because they
are not making as much money
from what they sell.

Price Floor
A price floor is set at $4 resulting in a surplus of 20
units.
An example of a price floor is
minimum wage laws; in this case,
employees are the suppliers of
labour and the company is the
consumer. When the minimum
wage is set above the equilibrium
market price for unskilled labour,
employers hire fewer workers.

Example
The demand and supply for a monthly cell phone plan
with unlimited texts can be represented by:
QD= Quantity demanded = 50 − 0.5P
QS= Quantity Supplied = −25 + P
where P is the monthly price, in dollars. Answer the
following questions:
a. If the current price for a contract is $40 per month, is the
market in equilibrium?
b. Would you expect the price to rise, fall, or be unchanged?
c. If so, by how much? Explain.

Example - Answer
a. To solve the problem:
Compute quantity supplied and demanded at a price of $40,
QD = 50 − 0.5P = 50 − 0.5(40) = 30
QS = −25 + P = −25 + 40 = 15
QD > QS, and the market is not in equilibrium as there is excess
demand (shortage).

b. What must happen to price?
Price needs to rise
c. Solve for equilibrium price and quantity
QS = QD = Q*
−25 + P* = 50 − 0.5P*  P* = $50, Q* = 25
Price must rise by $10, and 10 more contracts must be
sold.
Example - Answer

Economics and Risk Analysi
Dr Jeb Gholinezhad

Food for Thought
Suppose that you are the project manager for planning and scheduling the Olympic games to
take place
within one month in a major city. What constraints would you consider ?
• Qualifying rounds obviously have to take place before quarterfinals, semi-finals, and finals. (Precedence
relationships)
• Then, there is the need to spread out the events, in both time and space.
• One concern is to avoid traffic jams that might result if two or more popular events were scheduled in
nearby facilities
at the same time.
• Even when different venues are involved, it is desirable to schedule the most attractive events at
different times, to allow the largest possible audience for the greatest number of events.
• The requirements of live TV coverage of different events for different time zones also have to be
considered. For
instance, interest in soccer matches would be high in Europe, Africa, and South America, but not in North
America.
• There are constraints on the available equipment (such as TV cameras) and personnel (for example,
security).

Project Management
• What is a project?
• A series of related jobs, usually directed toward some major
output and requiring a significant period of time to perform. It
is an endeavour with:
 objectives
 multiple activities
 defined precedent relationships
 a specific time period for completion
• What is project management?
• Planning, directing, and controlling resources (people, equipment,
material, etc.) to meet the technical, cost, and time constraints of
the project.

Defining the Project
• Statement of Work
• A written description of the objectives to be achieved
• Task
• A further subdivision of a project – usually shorter than
several months and performed by a single group or
organization
• Work Package
• A group of activities combined to be assignable to a single
organizational unit

Defining the Project
Continued
• Project Milestone
• Specific events in the life of the project
• Work Breakdown Structure
• Defines the hierarchy of project tasks, subtasks, and work
packages
• Gantt Chart
• A tool used for scheduling and time-planning of the project

Work Breakdown Structure
Overview
Details
Gives a hierarchical structure to a project.

Work Breakdown Structure – Example
Large Optical Scanner Design
Overview
Details

Gantt Chart
• Developed by Henry L. Gantt in 1918
• A popular tool in production and project scheduling
• Has simplicity and clear graphical display
• Useful for simple project scheduling
• Shows the required time for each activity
• Shows if individual activities are ‘on schedule’, ‘behind schedule’
or ‘ahead of schedule’
• Fails to show if the overall project is being delayed!
• Fails to show the precedence relationships!

Gantt Chart - Example
• Indicates the earliest possible starting time for
each activity:
• Activity C cannot start before time 5 since
activity B must be completed before activity C
can begin.
• At any point in time, it is clear which activities
are on schedule and which are not:
• As of week 13, activities D, E, and H are behind
schedule, while G has actually been completed
(all shaded) and hence is ahead of schedule.

Project Management Techniques
 Project management is generally applied for constructing public utilities, large
industrial projects, and organizing mega events.
 Modern projects ranging from building a suburban shopping centre to
putting a man on the moon are very large, complex, and costly. Completing
such projects on time and within the budget is not an easy task.
 In particular, complicated problems of scheduling arise from the
interdependence of activities (precedence relationships). Typically, certain of
the activities may not be initiated before others have been completed.
 In dealing with projects possibly involving thousands of such
dependency relations, it is no wonder that managers seek effective
methods of analysis.

Project Management Techniques
Some of the key questions to be answered are:
1. What is the expected project completion date?
2. What is the potential “variability” in this date?
3. What are the scheduled start and completion dates for each specific
activity?
4. What activities are critical in the sense that they must be completed
exactly as scheduled in order to meet the target for overall project
completion?
5. How long can noncritical activities be delayed before a delay in
the overall completion date is incurred?
CPM (Critical Path Method) and PERT (Project Evaluation and
Review Technique) project network planning tools can answer the
above questions.

Network-Planning Models
A project is made up of a sequence of activities that
form a network representing a project.
The path taking longest time through this network of
activities is called the “critical path.”
The critical path provides a wide range of scheduling
information useful in managing a project.
Critical path method (CPM) helps to identify the
critical path(s) in the project networks.

Network Planning Techniques
• Program Evaluation & Review Technique (PERT):
• Developed to manage the Polaris missile project
• Many tasks pushed the boundaries of science &
engineering (tasks’ duration = probabilistic)
• Critical Path Method (CPM):
• Developed to coordinate maintenance projects in the
chemical industry
• A complex undertaking, but individual tasks are routine
(tasks’ duration = deterministic)

Both PERT and CPM
• Graphically display the precedence relationships &
sequence of activities
• Estimate the project’s duration
• Identify critical activities that cannot be delayed without
delaying the project
• Estimate the amount of slack associated with non-critical
activities

Critical Path Method (CPM)
Identify each activity to be done and
estimate how long it will take.
Determine the required sequence and
construct a network diagram.
Determine the critical path.
Determine the early start/finish and
late start/finish schedule.

Network Diagrams
• Activity-on-Node (AON):
• Uses nodes to represent the activity
• Uses arrows to represent precedence relationships

Example: Step 1-Define the Project: Cables By Us is bringing a new
product on line to be manufactured in their current facility in some
existing space. The owners have identified 11 activities and their
precedence relationships. Develop an AON for the project.
Activity Description
Immediate
Predecessor
Duration
(weeks)
A Develop product specifications None 4
B Design manufacturing process A 6
C Source & purchase materials A 3
D Source & purchase tooling &
equipment
B 6
E Receive&installtooling&equipment D 14
F Receive materials C 5
G Pilot production run E & F 2
H Evaluate product design G 2
I Evaluate process performance G 3
J Write documentation report H & I 4
K Transition to manufacturing J 2

Step 2- Diagram the Network for
Cables By Us
A
ctivity Desc
Predecessor (weeks)
A Developproductspecifications None 4
B Design manufacturing process A 6
C Source & purchase materials A 3
D Source&purchasetooling&equipment B 6
E Receive & install tooling &
equipment
D 14
F Receive materials C 5
G Pilot production run E & F 2
H Evaluate product design G 2
I Evaluate process performance G 3
J Write documentation report H & I 4
K Transition to manufacturing J 2

Step 3 (a)- Add Deterministic Time Estimates and Connected Paths
Step 3 (a) (Continued): Calculate the Path Completion Times
Paths Path duration
ABDEGHJK 40 The longest path (ABDEGIJ
ABDEGIJK 41 duration (project cannot finis
ACFGHJK 22 longest path)
ACFGIJK 23
ABDEGIJK is the project’s critic

Some Network Definitions
• All activities on the critical path have zero slack
• Slack defines how long non-critical activities can be delayed without
delaying the project
• Slack = the activity’s late finish minus its early finish (or its late start
minus its early start)
• Earliest Start (ES) = the earliest finish of the immediately preceding
activity
• Earliest Finish (EF) = is the ES plus the activity time
• Latest Start (LS) and Latest Finish (LF) depend on whether or not
the activity is on the critical path

To Find the Critical Path and Slacks
1. Work from top to bottom in the network, calculating
• ES = earliest start time for an activity
EF = earliest finish time for an activity
ES for activity i = largest EF of the immediate predecessors
ES = 0 if no immediate predecessors
EF = ES + activity duration time
2. Work from bottom to top in the network, calculating
• LS = latest start time for an activity
LF = latest finish time for an activity
LS = LF – activity duration time
LF for activity i = smallest LS of the immediate successors
LF at Finish = EF at Finish if no immediate successors
• Slack = LF - EF = LS - ES
If slack is zero, the activity is on the critical path.

Calculate earliest start (ES) and earliest finish (EF) f
network in the previous example:

Calculate latest start (LS) and latest finish (LF) for the networ
previous example:
A c t i v i t y
L a t e
F i n i s h
E a r l
y
F i n i s
h
S l a c k
( w e e k s
A 4 4 0
B 1 0 1 0 0
C 2 5 7 1 8
D 1 6 1 6 0
E 3 0 3 0 0
F 3 0 1 2 1 8
G 3 2 3 2 0
H 3 5 3 4 1
I 3 5 3 5 0
J 3 9 3 9 0
K 4 1 4 1 0
Nodes on the Critical path have 0 slack

• Use a project network, Activity-on-Node (AON):
• Nodes: activities, or tasks, to be performed
• Arcs: show immediate predecessors to an activity
• Times: duration times of activities are written next to the node
Critical Path
• A path through a project network is a route from START to FINISH.
• The length of path is the sum of the task times (durations) of
the nodes (activities) on the path.
• The critical path is the longest path.
• The project duration is the length of the longest path.
PERT/CPM-2
Project Network

■ Activity time estimates usually cannot be made
with certainty.
■ PERT used for probabilistic activity times.
■ In PERT, three time estimates are used: most
likely time (m), the optimistic time (a), and the
pessimistic time (b); using Beta Distribution.
■ These provide an estimate of the mean and
variance of a beta distribution:
variance: 𝜎2 =
𝑏−
𝑎 6
2
mean (expected time): 𝜇 = 𝑎+4𝑚+𝑏
6
Probabilistic Activity Times

Example: Using Probabilistic Time Estimates
Activity Description
Optimistic
time
Most
likely
time
Pessimistic
time
A Developproductspecifications 2 4 6
B Design manufacturing process 3 7 10
C Source & purchase materials 2 3 5
D Source & purchase tooling &
equipment
4 7 9
E Receive & install tooling &
equipment
12 16 20
F Receive materials 2 5 8
G Pilot production run 2 2 2
H Evaluate product design 2 3 4
I Evaluate process performance 2 3 5
J Write documentation report 2 4 6
K Transition to manufacturing 2 2 2

Calculating Expected Task Times
Activity
Optimistic
time
Most likely
time
Pessimistic
time
Expected
time
A 2 4 6 4
B 3 7 10 6.83
C 2 3 5 3.17
D 4 7 9 6.83
E 12 16 20 16
F 2 5 8 5
G 2 2 2 2
H 2 3 4 3
I 2 3 5 3.17
J 2 4 6 4
K 2 2 2 2
Expected time =
optimistic + 4 most likely + pessimistic
6
=
𝑎 + 4𝑚 + 𝑏
6

Network Diagram with Expected Activity Times
• ABDEGIJK is the expected critical path & the project has an e
of 44.83 weeks
Activities on paths E
ABDEGHJK
ABDEGIJK
ACFGHJK
ACFGIJK
Activity
Expected
time
A 4
B 6.83
C 3.17
D 6.83
E 16
F 5
G 2
H 3
I 3.17
J 4
K 2

Estimating the Probability of Completion Dates
• Using probabilistic time estimates offers the advantage of predicting the
probability of project completion dates
• We have already calculated the expected time for each activity by making three
time estimates
• Now we need to calculate the variance for each activity
• The variance of the beta probability distribution is:
• where b=pessimistic activity time estimate
a=optimistic activity time estimate
σ2 =
𝑏 − 𝑎
6
2

Project Activity Variance
Activity Optimistic Most
Likely
Pessimistic Variance
A 2 4 6 0.44
B 3 7 10 1.36
C 2 3 5 0.25
D 4 7 9 0.69
E 12 16 20 1.78
F 2 5 8 1.00
G 2 2 2 0.00
H 2 3 4 0.11
I 2 3 5 0.25
J 2 4 6 0.44
K 2 2 2 0.00
σ2 =
b − 𝑎
6
2
Path
Number
Activities on Path Path Variance
(weeks)
1 A,B,D,E,G,H,J,k 4.82
2 A,B,D,E,G,I,J,K 4.96
3 A,C,F,G,H,J,K 2.24
4 A,C,F,G,I,J,K 2.38

Calculating the Probability of Completing the Project in Less Than a Specified Time
• When you know:
• The expected completion time
• Its variance
• You can calculate the probability of completing the project in “X”
weeks with the
following formula:
Where x = the specified completion date
𝜇 = the expected completion time of the path
σ = standard deviation of the expected completion time
z = =
specified time − path expected time x − 𝜇
path standard deviation time 𝜎

Example: Calculating the probability of finishing the project in 48 weeks
• Use the z values in Appendix B to determine probabilities
• E.G. for path 1
Path
Number
Activities on Path Path Variance
(weeks)
z-value Probability of
Completion
1 A,B,D,E,G,H,J,k 4.82 1.5216 0.9357
2 A,B,D,E,G,I,J,K 4.96 1.4215 0.9222
3 A,C,F,G,H,J,K 2.24 16.5898 1.000
4 A,C,F,G,I,J,K 2.38 15.9847 1.000
  1.52

4.82
 48 weeks 44.66 weeks
z  


Example: Identify Activities and Construct Network
A(21)
C(7)
B(5) D(2)
F(8)
E(5)
G(2)

CPM: Determine Early Start/Early Finish and Late Start/ Late Finish Schedule
A(21)
C(7)
B(5) D(2)
F(8)
E(5)
0 21
0 21
21
21
28
28
21
21
26
26
26
26
28
28
28
28
36
36
28
31
33
36
36
36
Critical
Path 1:
ACFG
Critical
Path 2:
ABDFG

Network with Time Estimates (PERT)
A(21)
C(7)
B(5) D(2)
F(8)
E(5)
G(2)
21
21
21
21
28
28
21
21
26
26
26
26
28
28
28
28
36
36
28
31
33
36
36
36
38
38
0
0

■A branch reflects an activity of a project.
■A node represents the beginning and end of activities, referred to as
events.
■Branches in the network indicate precedence relationships.
■When an activity is completed at a node, it has been realized.
The Project Network - CPM/PERT Activity-on-Arc (AOA)
Network

The Project Network
House Building Project Data
Number Activity Predecessor Duration
1 Design house and obtain
financing
-- 3 months
2 Lay foundation 1 2 months
3 Order and receive materials 1 1 month
4 Build house 2,3 3 months
5 Select paint 2, 3 1 month
6 Select carper 5 1 month
7 Finish work 4, 6 1 month

■ Activities can occur at the same time (concurrently).
■ Network aids in planning and scheduling.
■ Time duration of activities shown on branches.
The Project Network
Concurrent Activities
Figure: Concurrent activities for house-building project

■ A dummy activity shows a precedence relationshi
reflects no passage of time.
■ Two or more activities cannot share the same start an
nodes.
The Project Network
Dummy Activities
Figure: A dummy activity

Project Cost Management
•“The processes involved in planning, estimating,
budgeting, and controlling costs so that the project can be
completed within the approved budget”
Why Do We Manage Cost?
•Part of triple constraint, can’t
manage one without the
others (scope/budget , time,
and quality)
•Plots of cost and scope against
plan can help spot problems
early
Cumulative
Value
Time
Planned
Value
(PV)
Actual
Costs
(AC)
Earned
Value
(EV)
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