Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Introduction
Dr. Shree Raj Shakya
2022
Course Objectives
This course aims to provide sound and comprehensive
coverage of engineering economics especially the
following:
• To explain how business operates, how engineering project
decisions are made within the business, and how engineering
decisions can affect profit of the firm.
• To build a thorough understanding of theoretical and
conceptual basis of financial analysis of project.
• To help engineer for making correct/informed financial
decisions when acting as a team member or project manager
for an engineering project.
• To introduce use of computer based critical decision-making
tools (software) so that engineers can make correct/informed
decision under different constraints and uncertainty.
Course outline
•
•
•
•
•
•
•
•
•
•
Introduction to engineering economics
Cost concepts and behavior
Understanding financial statements
Time value of money
Project Evaluation Techniques
Depreciation
Income Tax & Discounted Cash-flow models
Project risk analysis
Economic analysis in public sector
Course Presentation
Text books and software
• Chan S. Park. “Contemporary Engineering
Economics”, Fifth Edition. Prentice Hall of India Pvt.
Ltd., New Delhi. 2011
• Other Related books
• Excel inbuilt financial analysis packages, Solver, Crystal
ball Pro etc.
Evaluation Criteria
Internal Evaluation
Assessment :
Case Study Project + Presentation :
(3-4 students in one group)
Attendance :
10
8
Total :
20
2
Engineering Economics
• Engineers have to participate in various decisionmaking process in a business firm or industry.
• They get involved in various economic decisions
related to engineering projects such as development of a
product, purchase of an equipment, construction of plant,
dams, building etc.
• Economics is the study of how people use their
limited resources to try to satisfy unlimited wants.
• Engineering Economics is the study of how to make
economic decisions in engineering projects.
Rational Decision-Making Process:
Example Car to Lease
▪
▪
▪
▪
▪
Recognize the decision
problem
Collect all needed
(relevant) information
Identify the set of feasible
decision alternatives
Define the key objectives
and constraints
Select the best possible
and implementable
decision alternative
•
Need to lease a car
•
Gather technical and
financial data
Select cars to consider
Wanted: small cash outlay,
safety, good performance,
aesthetics,…
Choice/ Select a car (i.e.,
Honda, Saturn or another
brand)
•
•
•
7
Engineering Economic Decisions
Needed e.g. in the following (connected) areas:
Profit! Then continue
at the next stage…
Manufacturing
Design
Financial
planning
Investment
and loan
Marketing
8
What Makes Engineering Economic Decisions
Difficult? Predicting the Future
• Estimating the required
investments
• Estimating product
manufacturing costs
• Forecasting the demand
for a brand new product
• Estimating a “good”
selling price
• Estimating product life
and the profitability of
continuing production
9
The Role of Engineers in Business
Create & Design
• Engineering Projects
Analyze
Evaluate
Evaluate
• Production Methods
• Engineering Safety
• Environmental Impacts
• Market Assessment
• Expected
Profitability
• Timing of
Cash Flows
• Degree of
Financial Risk
• Impact on Financial
Statements
• Firm’s Market Value
• Stock Price
10
Accounting vs. Engineering Economy
Evaluating past performance
Accounting
Evaluating and predicting future events
Engineering Economy
Past
Future
Present
11
Key Factors in Selecting Good
Engineering Economic Decisions
❑
Objectives
❑
❑
Profit Maximization, Cost
Minimization
Capital Investment,
Human Resource, Raw
materials, etc.
❑
Available Resources
❑
Time
❑
Short Term, Medium
Term, Long Term
❑
Uncertainty
❑
Cost (Investment, O&M
cost), Revenue, other
12
externalities
Large-Scale Engineering Projects
These typically
• require a large sum of investment
• can be very risky
• take a long time to see the financial outcomes
• lead to revenue and cost streams that are difficult to
predict
All the above aspects (and some others not listed here)
point towards the importance of Engineering Economic
Analysis
13
Types of Strategic Engineering Economic
Decisions in the Manufacturing Sector
❑ Service Improvement
❑ Equipment and Process Selection
❑ Equipment Replacement
❑ New Product and Product Expansion
❑ Cost reduction or profit maximization can be seen as
generic (common, eventual) objectives
❑ In the most general sense, we have to make decisions
under resource constraints, and in presence of
uncertainty
14
Example 1:
Healthcare Service Improvement
1 Traditional Plan: Patients
visit the service providers
◼ 2 New Strategy: Service
providers visit the patients
Which one of the two plans
is more economical? The
answer typically depends on
the type of patients and the
services offered. Examples?
◼
service providers
patients
1
2
15
Example 2:
Equipment and Process Selection
• How do you choose between using alternative materials
for an auto body panel?
• The choice of material will dictate the manufacturing
process and the associated manufacturing costs
16
Example 3:
Equipment Replacement Problem
• Key question:
When is the right time to
replace an old machine
or equipment?
17
Example 4:
New Product and Product Expansion
• Shall we build or acquire
a new facility to meet the
increased (increasing
forecasted) demand?
• Is it worth spending
money to market a new
product?
18
Example 5: MACH 3 Project
• R&D investment: $750 million(!)
• Product promotion through
advertising: $300 million(!)
• Priced to sell at 35% higher than
the preceding Sensor Excel
model (i.e., about $1.50 extra
per razor)
• Question 1: Would consumers
pay $1.50 extra for a shave with
greater smoothness and less
irritation?
• Question 2: What happens if the
blade consumption drops more
than 10% – due to the longer
blade life of the new razor?...
19
Example 6: Cost Reduction
• Should a company buy
new equipment to
perform an operation that
is now done manually?
• Should we spend money
now, in order to save
more money later?
• The answer obviously
depends on a number of
factors.
20
Further Areas of Strategic Engineering
Economic Decisions in the Service Sector
❑
❑
❑
❑
❑
❑
❑
❑
Commercial Transportation
Logistics and Distribution
Healthcare Industry
Electronic Markets and Auctions
Financial Engineering and Banking
Retail
Hospitality and Entertainment
Customer Service and Maintenance
21
The Four Fundamental Principles of
Engineering Economics
1: An instant dollar is worth more than a distant dollar…
2: Only the relative (pair-wise) difference among the
considered alternatives counts…
3: Marginal revenue must exceed marginal cost, in order
to carry out a profitable increase of operations
4: Additional risk is not taken without an expected
additional return of suitable magnitude
22
Principle 1
An instant dollar is worth more than
a distant dollar…
Today
6 months later
23
Principle 2
Only the cost (resource) difference
among alternatives counts
Option
Monthly
Fuel Cost
Monthly
Maintenance
Cash paid
at signing
(cash
outlay )
Monthly
payment
Salvage
Value at end
of year 3
Buy
$960
$550
$6,500
$350
$9,000
Lease
$960
$550
$2,400
$550
0
The data shown in the green fields are irrelevant items for decision
making, since their financial impact is identical in both cases
24
Principle 3
Marginal (unit) revenue has to
exceed marginal cost, in order to
increase production
Marginal
cost
Manufacturing cost
1 unit
Marginal
revenue
Sales revenue
1 unit
25
Principle 4
Additional risk is not taken without a
suitable expected additional return
Investment Class
Potential
Risk
Expected
Return
Savings account
(cash)
Lowest
1.5%
Bond (debt)
Moderate
4.8%
Stock (equity)
Highest
11.5%
A simple illustrative example. Note that all investments imply
some risk: portfolio management is a key issue in finance
26
Summary
• The term engineering economic decision refers to any
investment or other decision related to an engineering
project
• The five main types of engineering economic decisions are
(1) service improvement, (2) equipment and process
selection, (3) equipment replacement, (4) new product
and product expansion, and (5) cost reduction
• The factors of time, resource limitations and uncertainty
are key defining aspects of any investment project
• All listed decision types can be seen and modeled as a
constrained decision (optimization) problem
27
THANK YOU
Types of Business Organizations
• Proprietorship
• Partnership
• Corporation
• Their negative and positive sides ?
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Cost Concepts and Behavior
Dr. Shree Raj Shakya
2022
Various types of manufacturing costs incurred by a
manufacturer
Costing Systems
Costing systems aim to report costs of products,
services in using the resources of the firms.
In engineering economics, the term cost is used in many
different ways.
• there are many types of costs, each is classified differently
according to the immediate needs of management.
•eg, engineers may want cost data to prepare external
reports, to prepare planning budgets, or to make
decisions.
• also, each different use of cost data demands a different
classification and definition of cost.
•eg, the preparation of external financial reports requires
the use of historical cost data, whereas decision making
may require current cost data or estimated future cost
data.
Manufacturing Costs
• In converting raw materials into finished goods,
a manufacturer incurs various costs associated
with operating a factory.
• Most manufacturing companies divide
manufacturing costs into three broad categories:
– direct raw material costs,
– direct labor costs,
– and manufacturing overhead.
Manufacturing Costs
• Direct Raw Materials:
– Direct raw materials are any materials that are used in
the final product and that can be easily traced to it.
– Eg, wood in furniture, steel in bridge construction, paper
in printing firms, and fabric for clothing manufacturers.
– finished product of one company can become the raw
materials of another company.
• Direct Labor:
– Direct labor incurs costs that go into the production of
a product.
– Eg, labor costs of assembly-line workers, labor costs of
welders in metal-fabricating industries, carpenters or
bricklayers in home building, and machine operators in
various manufacturing operations.
Manufacturing Costs
• Manufacturing Overhead :
– the third element of manufacturing cost, includes all
costs of manufacturing except the costs of direct
materials and direct labor
– not easily traceable to specific units of output
• Eg, costs of indirect materials; indirect labor; maintenance
and repairs on production equipment; heat and light,
property taxes, depreciation, and insurance on
manufacturing facilities; and overtime premiums
– finished product of one company can become the
raw materials of another company.
• Sometimes it may not be worth the effort to trace the costs
of materials that are relatively insignificant in the finished
products. Materials such as solder and glue (indirect materials)
Non-Manufacturing Costs
• Two additional costs incurred in supporting any
manufacturing operation are
1.
2.
Marketing or selling costs and
Administrative costs.
• Overhead: Heat and light, property taxes, and depreciation
or similar items associated with the company’s selling and
administrative functions.
• Marketing: Advertising, shipping, sales travel, sales
commissions, and sales salaries. Marketing costs include all
executive, organizational, and clerical costs associated with
sales activities.
• Administrative functions: Executive compensation,
general accounting, public relations, and secretarial support,
associated with the general management of an organization.
Classifying Costs for Financial Statements
• For purposes of preparing financial statements, we
often classify costs as either period costs or
product costs
• Period costs: costs charged to expenses in the period in
which they are incurred.
– assumption is that associated benefits are received in the same
period the cost is incurred.
– Eg, all general and administrative expenses, selling expenses, and
insurance and income tax expenses. Therefore, advertising costs,
executives’ salaries, sales commissions, public-relations costs, and
other nonmanufacturing costs discussed earlier would all be period
costs.
– not related to the production and flow of manufactured goods, but
are deducted from revenue in the income statement. In other words,
period costs will appear on the income statement as expenses
during the time in which they occur
Classifying Costs for Financial Statements
• Product costs: Some costs are better matched against
products than they are against periods.
– consist of the costs involved in the purchase or manufacture of
goods.
– In the case of manufactured goods, product costs are the costs of
direct materials, direct labor costs, and manufacturing
overhead. Product costs are not viewed as expenses; rather, they
are the cost of creating inventory. Thus, product costs are
considered an asset until the associated goods are sold. At the time
they are sold, the costs are released from inventory as expenses
(typically called cost of goods sold) and matched against sales
revenue.
– Since product costs are assigned to inventories, they are also
known as inventory costs.
Classifying Costs for Financial Statements
How the period costs and product costs flow through financial
statements from the manufacturing floor to sales.
Classifying Costs for Financial Statements
Cost flows and classifications in a manufacturing company
*Based on an annual volume of 185,000 cones.
Product costs: Costs incurred in preparing 185,000 ice cream
cones per year
Period costs: Costs incurred in running the shop regardless
of sales volume
Job and Process Costing System
Job Costing System
In this system, the cost of a product or service is obtained
by assigning costs to a distinct, identifiable product or
service.
Process Costing System
In this system, the cost of a product or service is obtained
by assigning costs to masses of similar units and then
calculating units costs on an average basis.
Job Costing System in Manufacturing
Direct cost
Direct Material
cost
Direct
Manufacturing
labor cost
Indirect cost
Manufacturing
overhead
Process Costing Systems in
Manufacturing
Direct Material cost
Conversion cost
Total costs
Manufacturing Unit cost
Conversion costs: All manufacturing costs other than
direct materials costs.
Cost Behaviors
Fixed cost: A cost that remains Constant, regardless of any
change in a company’s activity (production volume).
Eg: the annual insurance premium, property tax, and license
fee are fixed costs, since they are independent of the
production volume per year for manufacturing company or
number of miles driven per year for transport company, building
rents; depreciation of buildings, machinery, and equipment; and
salaries of administrative and production personnel.
Variable cost: A cost that changes in proportion to a change in
a company’s activity or business.
Eg: Gasoline is a good example of a variable automobile cost,
because fuel consumption is directly related to miles driven.
Similarly, the cost of replacing tires will increase as a vehicle is
driven more.
Cost Behaviors
Mixed cost: Costs are fixed for a set level of production
or consumption, becoming variable after the level is exceeded.
Eg: In automobile example, depreciation (loss of value) is a
mixed cost. On the one hand, some depreciation occurs simply
from the passage of time, regardless of how many miles a car is
driven, and this represents the fixed portion of depreciation. On
the other hand, the more miles an automobile is driven a year,
the faster it loses its market value, and this represents the
variable portion of depreciation.
A typical example of a mixed cost in manufacturing is the cost
of electric power. Some components of power consumption,
such as lighting, are independent of the operating volume, while
other components (e.g., the number of machine-hours
equipment is operated) may vary directly with volume.
Cost Behaviors
Average Unit Cost: We often use the term average cost to
express activity cost on a per unit basis. In terms of unit costs,
the description of cost is quite different:
•Variable cost per unit of volume is a constant.
• Fixed cost per unit varies with changes in volume: As the
volume increases, the fixed cost per unit decreases.
• Mixed cost per unit also changes as volume changes, but
the amount of change is smaller than that for fixed costs.
.
Cost–volume relationships pertaining to
annual automobile costs
Depreciation
Tax,
insurance
Fuel
Average cost per mile of owning and
operating a car
OM 
Cost Driver
Any factor that affects costs is called cost driver.
Variable cost
It is a cost that changes in total in proportion to changes in a cost
driver.
Fixed cost
It is a cost that does not change in total despite changes of a cost
driver.
Capitalized cost
A cost that is first recorded as an asset and then becomes an expense
such as depreciation of machines, computers, equipment etc.
Inventoriable cost
Cost associated with purchase of materials and other manufacturing
inputs.
Period cost
A cost that is reported as an expense in a particular period.
Product cost
A cost of creating inventory and is considered an asset until the associated
good is sold where it is released from inventory as expenses and matched
against sales revenue.
Differential (incremental) costs: Costs that represent differences in total
costs, which result from selecting one alternative instead of another.
Opportunity costs: Benefits that could have been obtained by taking an
alternative action.
Sunk costs: Past costs not relevant to decisions because they cannot be
changed no matter what actions are taken.
Marginal costs: Added costs that result from increasing rates of outputs,
usually by single units.
Cost Concepts Relevant to Decision Making
1. Method Change
Variable costs:
Materials
Current Die
Better Die
Differential Cost
$150,000
$170,000
+ $20,000
Machining labor
85,000
64,000
- 21,000
Electricity
73,000
66,000
- 7,000
Supervision
25,000
25,000
0
Taxes
16,000
16,000
0
Depredation
40,000
43,000
+ 3,000
$389,000
$384,000
- $5,000
Fixed costs:
Total
Cost Concepts Relevant to Decision Making
2. Operation Planning
(Break-Even Volume
Analysis)
• 24,000 initial order
• 4,000 extra
• Option 1 – extra
time or Saturday
running with
12,000
($36 extra per unit)
• Option 2 –
Second shift
operation with
21,000
(13,500 + 31.5 Q)
Cost Concepts Relevant to Decision Making
3. Make or Buy Decision
Make Option
Buy Option
Differential
Cost
Variable cost
Direct materials
100,000
- 100,000
Direct labor
190,000
-190,000
35,000
-35,000
Power and water
Gas filters
340,000
340,000
Fixed costs
Heating light
20,000
20,000
0
Depreciation
100,000
100,000
0
-35,000
-35,000
$445,000
$425,000
-$20,000
$22.25
$21.25
$1.00
Rental income
Total cost,
Unit cost
Marginal versus average cost per kWh
Marginal versus average cost per kWh
Marginal versus average cost per kWh
Profit Maximization Problem : Marginal Analysis
Example 8.7
THANK YOU
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Financial Statement and
Ratio Analysis
Dr. Shree Raj Shakya
2022
The accounting system
and flow of information
Summary of major factors affecting stock prices
Financial Statements
• These are statements of financial information to
the managers and the shareholders.
– Income Statement (Profit and Loss Statement)
– Balance Sheet
– Cash Flow Statement
6/14/2022
4
Financial Status for Business
Beginning of fiscal period
How much profit did the
company make during the
fiscal period?
Income Statement
How much cash did the
company generate and
spend during the period?
Cash Flow Statement
What is the company’s
financial position at
the end of fiscal period?
Balance Sheet
End of fiscal period
Balance Sheet
• It gives snapshot summary of the firm's
financial position at a single point in time.
• The balance sheet shows the net worth of
shareholders at a point in time, whereas
income statement measures changes in net
worth.
• Liabilities indicate what money has been made
available to the firm.
• Assets show how the firm has used the money
made available to it.
6/14/2022
6
Balance Sheet
6/14/2022
7
Balance Sheet
• Current Liabilities are the short -term debt
obligations of a firm, with maturities of less than
one year.
• Fixed liabilities are firm's long-term finance
such as long-term debts from banks and the
public.
• Shareholders' equity is the money invested by
the shareholders and the retained earnings.
6/14/2022
8
Balance Sheet
• Fixed Assets are acquired for long-term
uses in the firm such as plant, building,
land, and equipment.
• Current Assets are cash, accounts
receivables, and inventories of finished
goods and raw materials.
6/14/2022
9
Balance Sheet
• Depreciation is the allocation of cost of an
asset to different time periods.
• Working Capital is composed of firm's
current assets.
• Net Working Capital is the difference
between current assets and current
liabilities.
6/14/2022
10
Balance Sheet
Fixed Assets
6/14/2022
Land
Buildings
Plant and Machinery
Fixtures and fittings
Motor vehicles
11
Balance
Sheet
Cash in hand
Cash in bank
Marketable securities
Accounts receivables
Prepaid expenses
Deposits
CURRENT
ASSETS
Finished Products
Work in progress
Raw materials
Other supplies
6/14/2022
12
Balance Sheet
ASSETS = LIABILITIES + OWNERS’ EQUITY
Balance Sheet
Balance Sheet Statement – Dell Computer Corporation
(in millions)
28-Jan-00
29-Jan-99
Change
Percent
ASSETS
Current assets:
Cash
$ 3,809
$ 1,726
$ 2,803
121%
323
923
(600)
-65%
2,608
2,094
514
25%
Inventories
391
273
118
43%
Other
550
791
(241)
-30%
7,681
5,807
1,874
32%
765
523
242
46%
Long-term investments
1,048
532
516
97%
Equity securities and other investments
1,673
-
1,673
304
15
289
1927%
$ 11,471
$ 6,877
$ 4,594
67%
Short-term investments
Account receivables, net
Total current assets
Property, plant and equipment, net
Goodwill and others
Total assets
Balance Sheet
Balance Sheet Statement – Dell Computer Corporation
(in millions)
28-Jan-00
29-Jan-99
Change
Percent
LIABILITIES AND STOCKHOLDERS'EQUITY
Current liabilities:
Accounts payable
$ 3,538
$ 2,397
$ 1,141
48%
Accrued and other
1,654
1,298
356
27%
5,192
3,695
1,497
41%
Long-term debt
508
512
(4)
-1%
Other
463
349
114
33%
6,163
4,556
1,607
35%
3,583
1,781
1,802
101%
1,260
606
654
108%
465
(66)
531
5,308
2,321
2,987
129%
$ 11,471
$ 6,877
$ 4,594
67%
Total current liabilities
Total liabilities
Stockholders' equity:
Preferred stock
Common stock and capital in excess of $0.01 par
value
Retained earnings
Other
Total stockholders' equity
Total liabilities and stockholders' equity
-
Income Statement
• It shows the record of financial events
between two points in time. It has revenue
from sales and expenses incurred during the
period.
• Net Worth
The value of total assets minus total liabilities or the
value of the owners' claim on the assets.
6/14/2022
16
Income Statements (P/L statement)
• Expenditures are all cash outflows
– Expenses are only those expenditures that
affect net worth of the shareholders and
appear in the Income Statement.
• Receipts are all cash inflows.
– Revenues are only those receipts that affect
net worth and thus appear in the income
statement.
6/14/2022
17
Income Statement
Revenue
Costs of Goods Sold (COGS)
Gross Profit
Expenses
Net Income
6/14/2022
18
Income Statement
Income Statement – Dell Computer Corporation
(in millions, except per share amount)
Net Revenue
28-Jan-00
29-Jan-99
$25,265
$18,243
20,047
14,137
5,218
4,106
2,387
1,788
568
272
2,955
2,060
2,263
2,046
188
38
2,451
2,084
785
624
$1,666
$1,460
Basic
$0.66
$0.58
Diluted
$0.61
$0.53
Basic
2,536
2,531
Diluted
2,728
2,772
606
607
1,666
1,460
Repurchase of common stocks
(1,012)
(1,461
Balances at end of period
$1,260
$606
Cost of revenue
Gross margin
Operating expenses:
Selling, general and administrative
Research, development and engineering
Total operating expenses
Operating income
Other income
Income before income taxes
Provision for income taxes
Net income
Earnings per common share:
Weighted average shares outstanding:
Retained Earnings:
Balances at beginning of period
Net income
Cash Flow Statement
• Source- and -use-of-funds Statement (cash flow
statement) is a summary of the flow of the financial
activity of the firm. It shows where the firm obtains
cash and how it uses it.
• Sources of funds
– Increase in liabilities
– Increase net worth through retained earnings or
capital contribution by the shareholders
– Reduction in assets through sales of assets
• Uses of funds
– reduction in liabilities
– reduction in net worth through payment of dividends
or losses
– increase in assets
6/14/2022
21
Cash Flow Statement
• Cash flow from operating activities
– Net profit
– Depreciation
– Decrease in account receivables
– Increase in accounts payables
• Cash flow from investing activities
– Sales of fixed assets
– Investment in new fixed assets
6/14/2022
22
Cash Flow Statement
• Cash flow from financial activities
– Increase in debt (cash inflow)
– Issuance of new shares (cash inflow)
– Dividend payment (cash outflow)
6/14/2022
23
Cash Flow Statement – Dell Computer Corporation
(in millions)
Net income
28-Jan-00
29-jan-99
$1,666
$1,460
156
103
2,104
873
3,926
2,436
(3,101)
(1,938)
2,319
1,304
397
(296)
(1,183)
(930)
(1,061)
(1,518)
289
212
Proceeds from issuance of long-term debt
20
494
Cash received from sale of equity options and other
63
Repayment of borrowings
(6)
Depreciation and amortization
Changes in working capital
Net cash provided by operating activities
Cash flows from investing activities:
Marketable securities:
Purchase
Sales
Capital expenditures
Net cash used in investing activities
Cash flows from financing activities:
Purchase of common stock
Issuance of common stock under employee plans
Net cash used in financing activities
(695)
(812)
Effect of exchange rate changes on cash
35
(10)
$2,083
$684
1,726
1,042
$3,809
$1,726
Net increase in cash
Cash at beginning of period
Cash at end of period
Ratio Analysis
• In ratio analysis, we relate various items from the
firm's financial statements to each other with the
aim of assessing and analyzing the firm's
financial position.
• By comparing the financial ratios of the same
company over different periods, or by comparing
with that of other companies in the industry, or the
industry average, we can compare the relative
performance of the company.
• Common comparisons are:
1. Same company over several years.
2. An industry leader or "best practice."
3. Industry norms found in publications.
4. A self-developed set of comparable companies.
5. International variations on 2, 3, and 4.
6. Rules of "thumb."
Types of ratios used in evaluating a firm’s financial health
Profitability Ratios
Profitability ratios assess the profitability of the firm.
Profit Margin = Net income / Revenue
Return on assets = Net income / Total Assets
Return on equity (Return on Investment) = Net income /
Owners' Equity
Normally profitability ratios higher than comparables are
good news.
Asset management/ Activity Ratios
These ratios help in evaluating the managerial efficiency of the firm.
Asset Turnover = Revenue / Total Assets
Days' receivables = Accounts receivables / Average Days Sales
Average days sales = Revenue / 365
Any turnover ratio: Normally higher than the comparable ratios are good
news, although for plant and equipment one may worry if necessary
repairs and maintenance are being carried out if substantially higher
turnover than industry.
Number of days sales in a current asset: Normally higher than
comparables is bad news as your conversion of the asset into cash is
slower than the reference industry.
For inventory: higher ratios may indicate old, slow moving inventory;
for accounts receivable: a higher ratio may indicate problems
collecting A/R.
Debt management / Leverage Ratios
Leverage ratios measure the degree to which a firm relies on bank debts
and other debt securities
Debt ratio = Total Liabilities / Total Assets
Debt to equity ratio = Total Liabilities / Owners' Equity
Times interest earned = (Income before taxes + Interest Expense - Equity
in earnings of affiliates) / Interest Expense
(Equity in earnings of affiliates are earnings NOT available to pay interest
with so many analysts remove it before calculating times interest earned)
Days payables = Accounts payable / Average operating expenses
Average operating expenses = Operating expenses / 365
The amount of debt relative to equity is a major management decision that is
discussed in great detail in finance courses. The higher any debt ratio is the
greater the company's risk as fixed interest expenses are increasing. Thus,
higher debt ratios than comparables may indicate more risk than
comparables. However, if the company is well managed, this risk may be
offset with higher returns to shareholders.
Times interest earned ratios that are higher than comparables indicate
greater ease in paying interest, therefore, less risk.
Liquidity ratios
These ratios reflect a firm's short-term ability to pay its debts.
Current ratio = Current assets / Current liabilities
Quick ratio = (Cash + Accounts receivable) / Current liabilities
Higher liquidity ratios than comparables are subject to two
interpretations dependent on context.
Higher liquidity may indicate better ability to pay short term debts
as they come due
or alternatively may indicate inadequate controls over credit
granting (for A/R) or inventory buildup.
Market trend/ ratios
These ratios reflect the data from the financial statements to financial
market data.
These ratios provide some insights into investors' perception of the
firm and its securities.
Price/earnings ratio = Year end market price per share / Fully diluted earnings per share
Dividend yield = Dividend per share / Year end price per share
It is a very controversial ratio not easily subject to interpretation.
Research shown that of all ratios, P/E is most context specific in its
interpretation. For example, a high P/E has been cited as being an
indication of a highly risky stock, a stock with high growth potential, and
a stock with an unusually bad current year that is not "normal."
Exercise
2.3, 2.5, 2.6, 2.7
THANK YOU
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Time Value of Money, Interest Rate and
Economic Equivalence
Dr. Shree Raj Shakya
2022
Time Value of MONEY
• Money has a time value.
• The economic value of an amount depends on when it is
received.
• Money has earning power over time.
2
Lecture 4 - Dr. Shree Raj Shakya
Time Value of MONEY
A rupee received today has more value than a rupee
received at some future time
3
Lecture 4 - Dr. Shree Raj Shakya
Time Value of MONEY
• Gains achieved or losses incurred by delaying
consumption
4
Lecture 4 - Dr. Shree Raj Shakya
Cost of money => Interest
The cost of money is measured by an interest rate,
a percentage that is periodically applied and added
to an amount of money over a specified length of
time
5
Financial terms
• Principal: initial amount of money in transactions
of debt or investment
• Interest rate: measures the cost of money,
expressed as percentage per period of time
• Interest period: a period of time when interest is
frequently calculated (usually one year)
• Number of periods: specified length of time
when transaction is done
• Future amount: amount of money resulted from
cumulative effects of interest rate over a number
of several interest periods
6
Lecture 4 - Dr. Shree Raj Shakya
Some Notations
An = Discrete payment or receipt at the end of
interest period
i = interest rate per interest period
N = number of interest periods
P = A sum of money at a time chosen for analysis,
or called present value (PV) or present worth
F = A future sum of money at the end of analysis
period, or Fn ,at the end of some interest period
A = An end- of -period payment or receipt in a
uniform series
Vn = An equivalent sum of money at the end of
specified period, considering time value of money;
V0 =P ; Vn = F
7
Lecture 4 - Dr. Shree Raj Shakya
Cash flow Diagrams
It is a convenient way of representing problems
involving time value of money in graphic forms.
The cash flow diagram
shows net flows at the
end of interest periods.
8
Lecture 4 - Dr. Shree Raj Shakya
Cash flow Diagrams
Repayment Plans for Example Given in Text (for N = 5 years, and i = 9%)
9
Lecture 4 - Dr. Shree Raj Shakya
Simple Interest
Under simple interest, the interest earned during
each interest period will not earn an additional
interest on interest amount in the remaining
periods.
I=(iP)N
F =P+I=P(1+iN)
10
Lecture 4 - Dr. Shree Raj Shakya
Compound Interest
Under this interest, the interest earned in each period
is based on the total amount owed at the end of the
previous period.
This means interest will be earned on the interest
charged in the previous period as well.
At the end of first year, F = P + iP = P(1+i)
At the end of 2nd year, F = P (1+i) + i [ P (1+i) ]
= P (1+i)2
A the end of n periods, F = P (1+i)n
11
Relationship between simple interest and compound
interest
12
Lecture 4 - Dr. Shree Raj Shakya
Economic Equivalence
Economic equivalence exists when cash flows
have the same economic effect and thus could
be traded for one another in the financial market.
13
Lecture 4 - Dr. Shree Raj Shakya
Suppose you are offered the alternative of receiving
either Rs 3,000 at the end of 5 years or Rs P today.
There is no question that the Rs 3,000 will be paid in full
(no risk). Having no current need for the money, you would
deposit the P rupees in an account that pays 8% interest.
What value of P would make you indifferent in your choice
between P rupees today and the promise of Rs 3,000 at
the end of 5 years from Now?
14
You borrowed $ 1,000 from a bank for 3 years at 10%
annual interest.
The bank offered you two options
(a) repaying the loan all at once at the end of 3 years,
or
(b) repaying the interest charges for each year at the
end of that year.
15
Lecture 4 - Dr. Shree Raj Shakya
Since two interest payments are equivalent, the bank
would be economically indifferent to a choice between
the two plans
16
Five Types of Cash flows
•
•
•
•
•
Single cash flow
Uniform Cash flows
Linear Gradient Series
Geometric Gradient Series
Irregular series
17
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow
18
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow
F =P(1+i)N =P (F/P, i, N)
where, (1+i)N is called single payment
compound amount factor.
P =F(1+i)-N=F (P/F, i, N)
where, (1+i)-N is called single payment
present value (worth) factor.
19
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow (Example)
If you had $2,000 now and invested it at 10%, how
much would it be worth in eight years?
20
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow (Example)
21
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow (Example)
22
Lecture 4 - Dr. Shree Raj Shakya
Single cash flow (Example)
23
Lecture 4 - Dr. Shree Raj Shakya
Uniform Cash flows
24
Uniform Cash flows
= A(F/A, i, N)
= F(A/F, i, N)
where [(1+i)N –1] / i is called equal payment series
compound amount factor or uniform series
compound amount factor.
where i/ [(1+i)N –1] is called equal payment series
25
sinking - fund factor.
Uniform Cash flows
= A (P/A, i, N)
= P (A/P, i, N)
where [(1+i)N –1] / i(1+i)N is called equal payment
series present value (worth) factor.
where i(1+i)N / [(1+i)N –1] is called equal payment
series capital recovery factor or annuity factor. 26
A person now 35 years old. he plan to invest an equal
sum of Rs. 10,000 at the end of every year for next 25
years starting from the end of the next year. The bank
gives 20% interest rate compounded annually.
Find maturity value when he is 60 years old.
uniform series
compound
amount factor
27
Linear Gradient Series
28
Linear Gradient Series
29
Linear Gradient Series
Gradient series is to be taken as composite
series, or as a set of two cash flows – one as
uniform series and another as gradient series.
P =P1+ P2
P2 =G [(1+i)N –iN-1]/ i2(1+i)N =G(P/G,i,N)
where (P/G, i, N) is called gradient series
present value (worth) factor.
30
Linear Gradient series
A
= G [(1+i)N - iN -1] / i[(1+i)N -1]
=G(A/G, i, N)
where (A/G, i, N) is called Gradient to equal
payment series conversion factor.
31
Lecture 4 - Dr. Shree Raj Shakya
A person has 10 years of service. He would like to deposit
20% of his salary, which is Rs. 4,000, at the end of the
first year, and thereafter he wishes to deposit the amount
with an annual increase of Rs. 500 for the next 9 years
with an interest rate of 15%.
Find the total amount at the end of the 10th year of the
above series?
32
Lecture 4 - Dr. Shree Raj Shakya
Geometric Gradient Series
33
Geometric Gradient Series
34
Geometric Gradient Series
Geometric gradient
Present worth factor
(P/A1, g, i, N)
35
Geometric Gradient Series
F
= A1[(1+i)N - (1+g) N ] / (i - g)
if i≠g
=N A1 / (1+i)N-1
if i=g
= A1(F/ A1, g, i, N)
where (F/ A1, g, i, N) is called future value
(worth) equivalent of geometric gradient series.
36
Lecture 4 - Dr. Shree Raj Shakya
37
38
Irregular series
39
Lecture 4 - Dr. Shree Raj Shakya
Irregular series
40
Irregular series
41
Assignment L4
3.3, 3.5, 3.6, 3.17, 3.21, 3.30, 3.38,
3.45, 3.51, 3.68
Lecture 4 - Dr. Shree Raj Shakya
42
Practice
Lecture 4 - Dr. Shree Raj Shakya
43
Lecture 4 - Dr. Shree Raj Shakya
44
Practice
Lecture 4 - Dr. Shree Raj Shakya
45
Lecture 4 - Dr. Shree Raj Shakya
46
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Effective Interest and Cash Flow
Dr. Shree Raj Shakya
2022
Lecture 5
Nominal Interest rate
• If a financial institutions uses a unit of
time other than a year, i.e. a quarter, a
month, half-year, then it quotes interest
rate on annual basis such as r%
compounded monthly, quarterly, or
half-yearly
• The interest rate or Annual Percentage
Rate (APR) is called the nominal interest
rate.
Lecture 5 - Dr.Shree Raj Shakya
2
Effective Interest rate
• The effective interest rate is the one rate
that truly represents the interest earned
in a year.
ia = (1+r/M)M -1
where,
ia = effective annual interest rate
M =the number of compounding period per year
r/M = the interest rate per compounding period
Lecture 5 - Dr.Shree Raj Shakya
3
Nominal and Effective Interest Rates with Different
Compounding Periods
Lecture 5 - Dr.Shree Raj Shakya
4
Effective Interest rate per payment
period
• If the transaction occurs more than one
time a year then, effective interest rate
per payment period becomes.
i = (1+r/(CK))c -1
where,
i = effective interest rate per payment period
C = No. of compound period per payment period
K = No. of payment period per year
r/CK = the interest rate per compounding period
Lecture 5 - Dr.Shree Raj Shakya
5
Continuous compounding interest
rate
• As the number of compounding periods
'M' becomes large, then r/M becomes too
small, hence as M approaches infinity, r/M
tends to '0', we comes to the situation of
continuous compounding.
ic = er/k -1
where,
ic = effective interest rate per payment period
r = nominal interest rate
K = No. of payment period per year
Lecture 5 - Dr.Shree Raj Shakya
6
Calculating an Effective Interest Rate with Quarterly
Payment
i = (1+r/(CK))c -1
Lecture 5 - Dr.Shree Raj Shakya
7
Calculating an Effective Interest Rate with Quarterly
Payment
Lecture 5 - Dr.Shree Raj Shakya
8
Calculating an Effective Interest Rate with Quarterly
Payment
Lecture 5 - Dr.Shree Raj Shakya
9
Examples 1
Suppose you make equal quarterly deposits
of Rs 1,000 into a fund that pays interest at a
rate of 12% compounded monthly.
Find the balance at the end of year '1'
Quarterly Amount deposited, A = Rs 1,000
Nominal interest rate, r = 12% per year
No. of compounding period per year, M = 12
No. of payment period per year, K = 4
Compounding period per payment period, C = M/K = 3
Effective interest rate per payment period, i = (1+r/CK)C -1
No of payment period, N = K x 1 = 4
F = A x (F/A, i, N)
Lecture 5 - Dr.Shree Raj Shakya
10
Examples 1
F = A x (F/A, i, N)
Lecture 5 - Dr.Shree Raj Shakya
11
Lecture 5 - Dr.Shree Raj Shakya
12
Interpolate for 3.03%
Lecture 5 - Dr.Shree Raj Shakya
13
Examples 2
A series of equal quarterly receipts of Rs 500
extends over a period of 5 years. What is the
present value of this quarterly payment series at
8% interest compounded continuously?
Equal quarterly receipt, A = Rs 500
Nominal interest rate, r = 8%
No. of compounding period per year = continuous
No. of receipts per year, K = 4
Effective interest rate per quarter, Ic = er/k -1
No of quarterly receipt, N = 4 qtr/yr x 5yrs = 20
P = A(P/A, ic, N)
Lecture 5 - Dr.Shree Raj Shakya
14
Examples 2
P = A(P/A, ic, N)
Lecture 5 - Dr.Shree Raj Shakya
15
Lecture 5 - Dr.Shree Raj Shakya
16
Lecture 5 - Dr.Shree Raj Shakya
17
Examples 3
Suppose you make Rs 1,000 monthly deposit to a
registered retirement savings plan that pays
interest at a rate of 10% compounded quarterly.
Compute the balance at the end of 8 years
Monthly Amount deposited, A = Rs 1,000
Nominal interest rate, r = 10%
No. of compounding period per year, M = 4
No. of payment period per year, K = 12
Compounding period per payment period, C = M/K = 1/3
Effective interest rate per payment period, i = (1+r/CK)C -1
No of payment period, N = K x 8 = 96
F = A x (F/A, i, N)
Lecture 5 - Dr.Shree Raj Shakya
18
Examples 3
Equivalent monthly interest
i = (1+r/CK)C -1
= (1+0.10/(1/3)12)1/3 -1
= 0.826% per month
F = A x (F/A, i, N)
96
Lecture 5 - Dr.Shree Raj Shakya
19
Lecture 5 - Dr.Shree Raj Shakya
20
Lecture 5 - Dr.Shree Raj Shakya
21
Cash Flow Example 1
If you win a lottery you have the choice of either
accepting
• Rs I million now or
• taking the 20 annual installments (Rs 100,000
each year, a total of Rs 2 million).
A local bank offers an annual interest rate of 10%
Which option would be more desirable?
Lecture 5 - Dr.Shree Raj Shakya
22
Cash Flow Example 2
To help you reach Rs 5,000 goal in 5 years from
now,
your father 'offers to give you Rs 500 now.
You plan to get a part-time job and make additional
deposits at the end of each year. ( The first deposit
is made at the end of first year.).
If all your money is deposited in a bank that pays
7% interest,
how large must your annual deposits be?
Lecture 5 - Dr.Shree Raj Shakya
23
Cash Flow Example 3
A small firm has borrowed Rs 100,000 to
purchase certain equipment.
The loan carries an interest rate of 15% and
is to be repaid in equal installments over the
next 5 years.
Compute the amount of this annual
installment.
Lecture 5 - Dr.Shree Raj Shakya
24
Cash Flow Example 4
Ram and Sita just opened two savings accounts at
a bank.
The accounts earn 10% annual interest.
Ram wants to deposit Rs 1000 in his account at the
end of first year and increase this amount by Rs
300 for each of the following 5 years.
Sita wants to deposit an equal amount each year
for next 6 years.
What should be the size of Sita's annual
deposit so that two accounts would have equal
balances at the end of 6 years?
Lecture 5 - Dr.Shree Raj Shakya
25
Cash Flow Example 5
A self-employed individual, Krishna, is opening a retirement
account at a bank.
His goal is to accumulate Rs 1,000,000 in the account by
the time he retires from work in 20 years.
A local bank is willing to open such a retirement account
that pays 8% interest compounded annually, throughout the
20 years.
Krishna expects his annual income will increase at a 6%
annual rate during his working career.
He wishes to start with a deposit at the end of year '1' of A,
and increase the deposit at a rate of 6% each year
thereafter.
What should be the size of his first deposit A?
The first deposit will occur at the end of year ‘1', and the
subsequent deposits will he made at the end of each year.
The last deposit will be made at the end of year'20'.
Lecture 5 - Dr.Shree Raj Shakya
26
Lecture 5 - Dr.Shree Raj Shakya
27
Lecture 5 - Dr.Shree Raj Shakya
28
Assignment L5
• Problems from Chan S Park,
Contemporary Engineering Economics,
Fourth Edition Book
➢4.1, 4.2,4.3, 4.9, 4.14, 4.16, 4.26, 4.28, 4.31,
4.51,
Lecture 5 - Dr.Shree Raj Shakya
29
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Project Evaluation Techniques
Dr. Shree Raj Shakya
2022
Lecture 6
Project Cash flows
In most of the engineering economic decision
problems, we usually make an initial investment
at the beginning of the project. Then this
investment will make a series of cash benefits
over a period of future years.
The representation of future earnings along with
the initial capital expenditure and annual expenses
such as wages, raw material costs, operating
costs, maintenance costs, and income taxes is
called the project cash flows.
Project Evaluation techniques
With the background of knowledge of
interest rate and time value of money,
we can evaluate whether a project is
feasible or not on the basis of cash
flow equivalence methods.
These techniques are also called Capital
Budgeting techniques in finance.
Evaluation of Project
We can evaluate a project or select a project
from different alternatives by the following
methods:
a) Equivalent Present Value (worth)
method
b) Equivalent Future Value (worth)
method
c) Equivalent Annual Value (worth)
method
Equivalent Present Value (worth)
method
Equivalent Future Value (worth)
method
Equivalent Annual Value (worth)
method
Apart from the above, we can evaluate a
project on the basis of
Payback Period
and
Internal Rate of Return (IRR)
Payback Period Method
The payback period is the number of years
required to recover the investment made in a
project.
Example:
Initial Investment = Rs 300,000
Annual Net benefits = Rs 75,000
Payback period
= Initial investment/ Uniform Annual benefit
=4 years
Benefits & flaws of Payback period
method
Benefit: simplicity
Flaws: failure to measure profitability
of project & no time value of money
consideration
Discounted Payback Period
To modify the payback period method, we may
consider the cost of fund (interest) used to
support the project. This modified method is
called discounted payback period.
Discounted Payback period N occurs, when
N
 Discounted Annual Benefit  Initial Investment
0
By hit and trial method we determine the value of N
Present Value Analysis
Until 1950s, the payback period method was
widely used as a means of making investment
decisions. As there are flaws in this method,
businessmen began to search methods for
improving the project evaluations. This led to the
development of Discounted Cash Flow (DCF)
techniques, which take account of time value of
money. One of the DCF methods is Net Present
Value (NPV) method.
Net Present Value (NPV) method
Under NPV method, PV of all cash inflows
are compared with PV of all cash outflows.
The difference between these PV's is
called Net Present Value (NPV).
NPV Criterion
The basic procedure for applying this criterion:
1. Determine the required rate of interest the firm
wants. This interest rate is called required rate
of return (ROR) or minimum attractive rate of
return (MARR).
2. Estimate the service life of the project
3. Estimate the cash inflow for each period of the
service life
4. Estimate the cash outflow for each period of
service life
5. Determine the net cash flows
6. Find the present value of each net cash flow
discounted at the MARR.
7. Add up the PV's including the initial
investment. Their sum is defined as the NPV of
the project
NPV = A0/(1+i)0 + A1/(1+i)1 +
2
n
A2/(1+i) + ….. +An/(1+i)
N
NPV =  An (P / F, i, n )
n =0
If NPV > 0, accept the investment
If NPV = 0, remain indifferent
If NPV< 0, reject the investment.
A micro-hydro project example
Calculate the NPV for the scheme assuming a 12% discount factor
Revenue: NRs. 300,000 each year for 15 years
Initial Investment: NRs. 1,200,000.
Yearly expenditure (O & M, salary and others): NRs. 80,000.
Solution,
The annual net income is NRs. 220,000.
The total present value (PV) of receiving this annuity amount for 15
years is
n
 (1 + i) - 1
PV = A  
= A ( P / A, i, n)
n 
 i (1 + i) 
= NRs.1,496,000
NPV (r=12%) = PV – investment
= NRs.1,496,000 – 1,200,000 = NRs. 296,000 > 0
Example 1
If a project's initial investment is Rs 300,000 and
gives an equal annual savings of Rs 75,000 for
next 10 years. If the company wants MARR of
15%, determine the NPV of the project.
Example 2
Tiger Machine Tool Company is considering the
proposed acquisition of a new metal-cutting
machine. The required initial investment of Rs
750,000 and the projected cash benefits for three
years are as follows:
End of year
Cash flow
0
- Rs 750,000
1
Rs 244,000
2
Rs 273,400
3
Rs 557,600
If the company wants MARR is 15%, determine the
NPV of the project?
Example 3
A textile company is considering two independent
investment proposals. Their expected cash flow
streams are given as follows:
Year
Proj. A
Proj. B
Year
Proj. A
Proj. B
0
-250,000
-350,000
6
72,500
70,000
1
72,500
60,000
7
70,000
2
72,500
60,000
8
70,000
3
72,500
60,000
9
70,000
4
72,500
70,000
10
70,000
5
72,500
70,000
If the company wants MARR is 12%, which proposals
should be accepted to the company? Perform NPV
analysis
Practice
• 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.9
Lecture 7 - Dr.Shree Raj Shakya
24
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Project Evaluation Techniques
Dr. Shree Raj Shakya
2022
Lecture 7
Future Value Analysis
The NPV or NPW measures the surplus in an
investment project at time '0'.
Sometimes we might need to find the equivalent
worth or value of a project at the end of the
investment period.
Hence, the Net Future Value (NFV) or Net Future
Worth (NFW) measures the surplus at the end
of the investment period.
Lecture 7 - Dr.Shree Raj Shakya
2
NFV Criterion
NFV = A0(1+i)n + A1(1+i)n-1 + A2/(1+i) n-2 +
….. +AN
=
N
A
n =0
n
(1 + i )
N −n
NFV =  An (F / P, i, N − n )
If NFV >0, accept the project
If NFV = 0, remain indifferent
If NFV<0, reject the project
Lecture 7 - Dr.Shree Raj Shakya
3
NET FUTURE VALUE
Lecture 7 - Dr.Shree Raj Shakya
4
NET FUTURE VALUE
Lecture 7 - Dr.Shree Raj Shakya
5
Robot manufacturing facility
Compute the equivalent worth of this investment at the start
of operation.
Assume that the company's expected MARR is 15%.
Lecture 7 - Dr.Shree Raj Shakya
6
Method 1
Method 2
Horton Corporation would set the price of the plant at $43.98 million
($18.40 + $25.58) at a minimum.
7
Example 1
An investment company is considering two independent
investment proposals. Their expected cash flow streams are
given as follows:
Year
Proj. A
Proj. B
Year
Proj. A
Proj. B
0
-250,000
-350,000
6
72,500
70,000
1
72,500
60,000
7
70,000
2
72,500
60,000
8
70,000
3
72,500
60,000
9
70,000
4
72,500
70,000
10
70,000
5
72,500
70,000
If the company wants MARR is 13%, which proposals should
be accepted to the company? Perform NFV analysis
Lecture 7 - Dr.Shree Raj Shakya
8
Lecture 7 - Dr.Shree Raj Shakya
9
Capitalized Equivalent Method
Another method of PV criterion is useful when the
life of project is perpetual or planning horizon is
very long (say, 40 years or more).
Capitalized Equivalent (CE) for Perpetual Service
life Project
CE(i) = A / i
The process of calculating PV cost for infinite
period is called capitalization of project cost.
The cost is known as the Capitalized cost i.e. the
amount of money to be invested now to get a
certain return 'A' at the end of each and every
Lecture 7 - Dr.Shree Raj Shakya
10
year forever.
Lecture 7 - Dr.Shree Raj Shakya
11
Lecture 7 - Dr.Shree Raj Shakya
12
Hydro Power Plant
Service life = 50, MARR = 8%
Lecture 7 - Dr.Shree Raj Shakya
13
Lecture 7 - Dr.Shree Raj Shakya
14
Mutually Exclusive Projects
Mutually exclusive means
that any one of several alternatives will
fulfill the same need and
that selecting one alternative means that
others will be excluded.
Lecture 7 - Dr.Shree Raj Shakya
15
Revenue Projects and Service Projects
Revenue projects are those projects whose
revenues depend on the choice of the alternative.
Service projects are those projects whose
revenues do not depend on the choice of the
project.
For revenue projects, we use NPV of revenues and choose
the project which has the highest NPV.
For service projects, we use the NPV of costs and choose
the project which has the least negative NPV.
Lecture 7 - Dr.Shree Raj Shakya
16
Analysis Period
It is the time-span over which the economic effects
of an investment will be evaluated. It is also called
as study period or planning horizon. It may be
taken as the required service period.
Situations when analysis period and project life differ
1. Equal lives
2. Project life longer than analysis period
3. Project life shorter than analysis period
4. Analysis period not specified
Lecture 7 - Dr.Shree Raj Shakya
17
1. Equal Project Lives
Automation option
Given: Cash flows for three projects, i = 12% per year
Find: The NPW of each project, select best option
Lecture 7 - Dr.Shree Raj Shakya
18
Option 3 has the greater PW and thus would be preferred
Lecture 7 - Dr.Shree Raj Shakya
19
2.
Project life longer than analysis period
Ripper-bulldozer to dig and load radio active material
within two year
Lecture 7 - Dr.Shree Raj Shakya
20
Model A has the greater PW
(least negative) and thus would
be preferred
Lecture 7 - Dr.Shree Raj Shakya
21
3.
Project life shorter than analysis period
Installation of an automatic mailing system to handle
product announcements and invoices
Semi-Automatic
Full-Automatic
Given: Cash flows for two alternatives as shown, analysis period of 5
years, i = 15%
Find: The NPW of each alternative, select which option
Lecture 7 - Dr.Shree Raj Shakya
22
Operating Cost
Leasing Cost
Since these are service projects, model B is the better choice.
Lecture 7 - Dr.Shree Raj Shakya
23
4.
Analysis period not specified
Lowest common
multiple criteria
(3 x 4)
Lecture 7 - Dr.Shree Raj Shakya
24
Model B is a better
choice
25
Exercise 1
Consider the following two mutually exclusive investment
projects, each with MARR = 15%:
(a) On the basis of the NPW criterion, which project would be
selected?
(b) Study for MARR = 7%, 30% and 50%.
Lecture 7 - Dr.Shree Raj Shakya
26
Lecture 7 - Dr.Shree Raj Shakya
28
Lecture 7 - Dr.Shree Raj Shakya
29
Lecture 7 - Dr.Shree Raj Shakya
30
Exercise 2
Consider the following two mutually exclusive investment
projects
On the basis of the NPW criterion, which project would be
selected if you use an infinite planning horizon with project
repeatability (the same costs and benefits) likely?
Assume that i = 12%.
Lecture 7 - Dr.Shree Raj Shakya
31
Practice
• 5.10, 5.11, 5.14, 5.17, 5.27, 5.29, 5.30,
5.31, 5.36, 5.38, 5.40, 5.46, ST5.1.
Lecture 7 - Dr.Shree Raj Shakya
33
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Project Evaluation Techniques
Dr. Shree Raj Shakya
2022
Lecture 8
Annual Equivalent Value Analysis
The annual equivalent value (AE) criterion is a
basis for measuring investment value by
determining equal payments on an annual basis.
First, we have to find the NPV of the project and
then convert it to equal annual payments.
AE(i) = PV (i)(A / P, i, n )
If AE>0, accept the project
If AE =0, remain indifferent
If AE <0, reject the project
Lecture 8 - Dr.Shree Raj Shakya
2
Annual Equivalent Value
Lecture 8 - Dr.Shree Raj Shakya
3
Benefits of AE analysis
1. Consistency of report format: Financial and
engineering managers may prefer to work on
yearly costs rather than overall costs.
2. Need for unit costs: In many situations,
project must be broken down into unit costs
for comparison and ease.
3. Unequal project lives: Comparing projects
with unequal service lives is complicated in
calculations, but using AE analysis, this
problem can be easily solved.
Lecture 8 - Dr.Shree Raj Shakya
4
Example: Make or Buy Option
Ampex Corporation currently produces both videocassette cases and
metal particle magnetic tape for commercial use.
An increased demand for metal particle tapes is projected, and Ampex is
deciding between
• increasing the internal production of empty cassette cases and
magnetic tape or
• purchasing empty cassette cases from an outside vendor.
If it purchases the cases from a vendor, the company must also buy
specialized equipment to load the magnetic tapes, since its current
loading machine is not compatible with the cassette cases produced by
the vendor under consideration.
The projected production rate of cassettes is 79,815 units per week
for 48 weeks of operation per year. The planning horizon is seven
years. MARR is 14%, calculate the unit cost under each option.
Lecture 8 - Dr.Shree Raj Shakya
5
Example: Make or
Buy Option
Lecture 8 - Dr.Shree Raj Shakya
6
Example: Make or Buy Option
Lecture 8 - Dr.Shree Raj Shakya
7
Example: Make or Buy Option
CR(i%) = (I - S)*(A/P, i%, n) + i * S
Lecture 8 - Dr.Shree Raj Shakya
8
Example: Make or Buy Option
Lecture 8 - Dr.Shree Raj Shakya
9
Example: Unit cost/saving
Tiger Machine Tool Company is considering the proposed
acquisition of a new metal-cutting machine. The required
initial investment of $75,000 and the projected cash benefits'
over the project's 3-year life are as follows. Suppose that the
machine will be operated for 2000 hours per year. Compute
the equivalent savings per machine hour at i = 15%.
Given: NPW = $3553, N = 3 years, i = 15% per year, 2000
machine hours per year
Find: Equivalent savings per machine hour
Solution,
NPW(15%) = -75,000
+ 24,400 (P/F, 15%, 1)
+ 27,340 (P/F, 15%, 2)
+ 55,760 (P/F, 15%, 3)
= 3,553
Lecture 8 - Dr.Shree Raj Shakya
10
If Annual Operating Hours Fluctuate,
then we have to find out the effective Annual Operating
Hours
Suppose: 1st Year= 1,500, 2nd Year= 2,500 and 3rd Year =
2,000
Effective Annual Operating Hour = [ (1,500)(P/F, 15%, 1) +
(2,500)(P/F, 15%, 2) + (2,000)(P/F, 15%, 3) ] (A/P, 15%, 3)
=1,976.16
Lecture 8 - Dr.Shree Raj Shakya
11
Operating costs and capital costs
Operating costs are incurred by the
operations of the plant or factory.
Capital costs are incurred during the start of
the project like initial investment or initial
borrowing.
s
Capital costs are incurred only one time in
the project life, where as operating costs
incur annually.
The annual equivalent of the capital cost is
capital recovery cost 'CR'.
Initial Investment
CR(i) = P(A/P,i,n) – S(A/F,i,n)
CR(i) = (P-S)(A/P,i,n) + i x S
Lecture 8 - Dr.Shree Raj Shakya
12
Example 1
Consider a machine that costs $5000 and has a
5-year useful life. At the end of the 5 years, it can
be sold for $1000 after tax adjustment. If the firm
could earn an after-tax revenue of $1100 per year
with this machine, should it be purchased at an
interest rate of 10%? Find: AE, and determine
whether to purchase
Given: I = $5000,
S = $1000, A = $1100,
N = 5 years, i = 10% per
year
Lecture 8 - Dr.Shree Raj Shakya
13
Method 1:
Lecture 8 - Dr.Shree Raj Shakya
14
Lecture 8 - Dr.Shree Raj Shakya
15
Method 1:
Lecture 8 - Dr.Shree Raj Shakya
16
Method 2:
Lecture 8 - Dr.Shree Raj Shakya
17
Lecture 8 - Dr.Shree Raj Shakya
18
Method 2:
Lecture 8 - Dr.Shree Raj Shakya
19
Lecture 8 - Dr.Shree Raj Shakya
20
Example: Break-even point-per unit of Equipment use
Sam purchased van worth $11,000 for office work use. If I = 6%, what
should be the reimbursement rate per kilometer that should be claimed to his
office so that Sam can break even?
21
Suppose Company pays Sam $X per kilometer for his personal car
the annual equivalent reimbursement would be AE(6%)
22
Example: Comparison of projects
i = 15% per year Find: AE, and which alternative is preferred
Lecture 8 - Dr.Shree Raj Shakya
23
Example: Comparison of projects
i = 15% per year Find: AE, and which alternative is preferred
Model B
Model A
Lecture 8 - Dr.Shree Raj Shakya
24
Lecture 8 - Dr.Shree Raj Shakya
26
Lecture 8 - Dr.Shree Raj Shakya
27
Solve
Lecture 8 - Dr.Shree Raj Shakya
28
Lecture 8 - Dr.Shree Raj Shakya
29
Practice
• 6.3. 6.5, 6.6, 6.11, 6.16, 6.20, 6.24, 6.27,
6.28, 6.39
Lecture 7 - Dr.Shree Raj Shakya
30
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Project Evaluation Techniques
Dr. Shree Raj Shakya
2022
Lecture 9
Limitation of NPV only
A1
A2
Initial
investment
- 1,00,000
- 5,00,000
n
5
5
NPV(10%)
50,000
1,00,000
NPV high so we
select
What about high Rate
of return?
2
Internal Rate of Return (IRR)
IRR is the interest rate earned on the
unrecovered project balance of investment
such that, when the project terminates, the
unrecovered project balance will be zero.
NPV = A0/(1+i*)0 + A1/(1+i*)1 + A2/(1+i*)2
…….+ An/(1+i*)n =0
If IRR>MARR, accept the project
If IRR = MARR, remain indifferent
If IRR < MARR, reject the project
Lecture 9 - Dr. Shree Raj Shakya
3
Generalized cash flow diagram
Present worth (NPV) function graph
4
Simple Investments and
Non-simple investment
Simple Investments
A simple investment is defined as that
investment, when the sign change in the project
cash flow occurs only once.
Non-simple Investments
A non-simple investment is that investment
where the sign change in the project cash flow
occurs more than once
Lecture 9 - Dr. Shree Raj Shakya
5
NPV profile for a simple investment
Single IRR
Lecture 9 - Dr. Shree Raj Shakya
6
NPW profile for a typical non-simple investment
Multiple IRR
Lecture 9 - Dr. Shree Raj Shakya
7
Multiple IRR
When there are multiple values of IRR, we
can predict unique value of IRR by
examining its cash flows.
1.Net cash flow rule of sign
2.Accumulated net cash flow rule of sign
Lecture 9 - Dr. Shree Raj Shakya
8
Net Cash flow rule of sign
The number of real i* that are greater than -100%
for a project with 'n' periods is never greater than
the number of sign changes in the sequence of
the An values.
There are 3 sign
changes in the
cash flow
sequence, so
there are 3 or
fewer real
positive i*s. 9
Accumulated Cash flow rule of sign
If the net cash flow sign test shows multiple
values of i*, then we should proceed to this
sign test.
If the series of cumulative cash flows start
negatively and changes the sign only once,
then there exists a unique positive i*.
10
Predict the number of real positive rate(s) of
return (IRR) for each cash flow series:
Lecture 9 - Dr. Shree Raj Shakya
11
The cash flow rule of signs indicates the following
possibilities for the positive values of IRR (i*)
For project A there is only once sign change therefore
there is only one real IRR value
Lecture 9 - Dr. Shree Raj Shakya
12
For cash flows B, C, and D, we would like to apply the
more discriminating cumulative cash flow test to see if
we can specify a smaller number of possible values of
real IRR (i*):
Only project D begins negatively and passes the test; we
may predict a unique i* value, rather than 2, 1, or 0 as
predicted by the cash flow rule of signs. Projects B and C fail
the test and we cannot eliminate the possibility of multiple
Lecture 9 - Dr. Shree Raj Shakya
13
i*s.
Methods for determining IRR
1. The direct-solution method
2. The trial-and-error method, and
3. The graphic method
Lecture 9 - Dr. Shree Raj Shakya
14
1. The direct-solution method
For project with two-flow transaction (Investment with
single future payment) or project with a service life of 2
years of return.
Use direct Mathematical Solution
Lecture 9 - Dr. Shree Raj Shakya
15
For Project 1
If PW(i*) = 0 then FW(i*) = 0
PW(i*) = - $1000 + $1500 ( P/F, i*, 4 )
=0
$1000 = $1500 / ( 1 + i* ) 4
$1000 (1 + i* ) 4
(1 + i* )4
= $1500
= 1.5
1.5 = (1 + i* ) 4
Taking a natural log on both sides, we obtain
Lecture 9 - Dr. Shree Raj Shakya
16
For Project 2:
17
Lecture 9 - Dr. Shree Raj Shakya
18
2. The trial-and-error method
Agrotech Corporation is considering a proposal to
establish a facility to manufacture an electronically
controlled "intelligent" crop sprayer. This project
would require an investment of $10 million in
assets and would produce an annual after-tax net
benefit of $1.8 million over a service life of 8
years. When the project terminates, the net
salvage value of the assets would be $1 million.
Compute the rate of return of this project.
Lecture 9 - Dr. Shree Raj Shakya
19
Given: Initial investment (I) = $10 million,
A = $1.8 million, S = 1 million, N = 8 years
Find: Internal Rate of Return (i*)
Lecture 9 - Dr. Shree Raj Shakya
20
We start with a guessed interest rate of 8%.
The present worth of the cash flows in millions of dollars is
Since this NPV is positive, we must raise the interest
rate to bring this value toward zero.
When we use an interest rate of 12%, we find that
PW(i) will be zero at i somewhere between 8% and 12%.
Using straight-line interpolation, we approximate
Lecture 9 - Dr. Shree Raj Shakya
21
If we compute the present worth at this interpolated
value, we obtain
As this is not zero, we may recompute the i* at a lower
interest rate, say 10%.
With another round of linear interpolation, we approximate
At this interest rate,
22
3. The graphic method
23
Lecture 9 - Dr. Shree Raj Shakya
24
Solve
Consider two independent investments, A and B, with
the following sequences of cash f lows:
Compute the IRR (i*) for each investment.
Lecture 9 - Dr. Shree Raj Shakya
25
Incremental IRR Analysis
• General IRR Ranking ignores the scale of the
investment.
• Incremental IRR Method address this
n
A1
A2
0
- 1,000
- 5,000
1
2,000
7,000
IRR
100%
40%
PW(10%)
$818
$1364
Initial Investment
high for A2
A1 is better with
respect to IRR
A2 is better with
respect to NPV or
PW
MARR = 10%
Lecture 9 - Dr. Shree Raj Shakya
26
Incremental IRR Analysis
• General IRR Ranking ignores the scale of the
investment.
• Incremental IRR Method address this
n
A1
A2
A2 - A1
0
- 1,000
- 5,000
-4,000
1
2,000
7,000
5,000
IRR
100%
40%
25%
PW(10%)
$818
$1364
$545
MARR = 10%
Incremental IRR Method measures the rate of return
value for difference in the investment amount
Lecture 9 - Dr. Shree Raj Shakya
27
Incremental IRR Testing Procedure
• Generate the Incremental cash flow by
subtracting higher initial investment project cash
flow (B) with smaller one's (A) so that the initial
investment cash flow becomes negative (B-A).
• Calculate the IRRB-A
• Selection Criteria
If IRRB-A  MARR, select B
If IRRB-A = MARR, select either one
If IRRB-A  MARR, select A
Lecture 9 - Dr. Shree Raj Shakya
28
Example
n
B1
B2
B2 – B1
0
1
2
- 3,000
1,350
1,800
- 12,000
4,200
6,225
- 9,000
2,850
4,425
3
1,500
6,330
4,830
MARR = 10%
NPV ( i) = - 9000 + 2850(P/F, I, 1) + 4425(P/F, I, 2) + 4830(P/F, I, 3)
0
= - 9000 + 2850 / (1 + i*) + 4425 / (1 + i*)2
+ 4830 / (1 + i*)3
Solving, i*B-A = 15%
Select project B
Lecture 9 - Dr. Shree Raj Shakya
29
Perform Incremental IRR Analysis
and Select the best option
n
D1
D2
D3
0
-2000
-1000
-3000
1
1500
800
1500
2
1000
500
2000
3
800
500
1000
IRR
34.37%
47.76%
24.81%
Lecture 9 - Dr. Shree Raj Shakya
30
Practice
• 7.6, 7.8, 7.11, 7.20, 7.27, 7.29, 7.37, 7.39,
7.41, 7.45, 7.49
Lecture 9 - Dr. Shree Raj Shakya
31
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Depreciation
Dr. Shree Raj Shakya
2022
Lecture 10
1
Asset Depreciation
Depreciation is the gradual decrease in
utility of fixed assets with use and time.
Physical Depreciation
It is the reduction in an asset's capacity to
perform its service due to physical
impairment.
Functional Depreciation
It occurs as a result of changes in the
organization or in technology that decrease
or eliminate the need for an asset.
2
Economic Depreciation
Economic Depreciation
=
purchase price - market value
Both physical and functional depreciation are
categories of economic depreciation.
3
Accounting Depreciation
It is the systematic allocation of the initial cost
of an asset (machine or equipment) in parts
over a time known as its depreciable life.
(Financial statements, Balance Sheet and Income Statement)
In engineering economic analysis, we use the
concept of accounting depreciation exclusively.
This is because accounting depreciation provides
a basis for determining the income taxes
associated with any project undertaken.
4
Depreciable Property
For the purposes of tax, any depreciable property has the
following Characteristics.
1. It must be used in business or must be held for the
production of income.
2. It must have a definite service life, and that life must be
longer than 1 year.
3. It must be something that wears out, decays, gets used
up, becomes obsolete, or loses value from natural causes.
Depreciable property includes buildings, machinery, equipment, and
vehicles.
Inventories are not depreciable property, because they are held
primarily for sale to customers in the ordinary course of business.
If an asset has no definite service life, the asset cannot
5
be depreciated (For example, Land)
Physical Depreciation
Economic
Depreciation
A gradual decrease in
utility of asset
Functional
Depreciation
Depreciation
Book Depreciation
Accounting
Depreciation
The systematic
allocation of an asset's
value in parts over its
depreciable life
Tax Depreciation
6
Net Income
When a project's revenue exceeds its expenses, we
say that the project generated a profit or income. If
the project's revenue is less than its expenses, then
we say that the project resulted in a loss.
Revenue
- Expenses (cost of goods sold)
Gross Profit
- Operating expenses
- Depreciation
Taxable Income (Income before tax)
- Income Tax
Retained Income
Net Income
Cash flow = net income + depreciation
7
Depreciation Methods
The most widely used methods are:
1. Straight-line Method
2. Declining Balance Method, and
3. Sum-of-years'- digit method
8
1. Straight-Line Method
In this method, it is assumed that the fixed asset is
depreciated in a uniform way.
Where
D
n
P −S
=
N
Dn = the depreciation charge during n year
P = the cost of the asset, including installation expenses
S = salvage value at the end of the useful life of asset
N = the useful life
The book value = cost base - total depreciation charges
Bn = P - (D1 + D2 +……….+ Dn)
9
2. Declining Balance Method
In this method, a fixed fraction of the initial book
balance is deducted each year. The fraction or
declining balance rate is obtained by
d = 1/N
The most common multiplier is '1'. If this is '2', then
it is called double-declining balance method.
D1 = dP
D2 = d(P - D1)
D3 = d(P - D1- D2)
For 'n' year,
= dP(1-d)
=dP(1-d)2
Dn = dP(1-d)n-1
10
We can also compute the total DB depreciation
at the end of 'n' years
TDB = D1 + D2 + D3 + D4 + …….. + Dn
= dP +dP(1 -d) +dP(1-d)2+ ……. +(1-d)n-1
TDB = P[1-(1-d)n]
11
3. Sum-of-years'-Digit Method
(SOYD)
In this method,
SOYD = 1 +2 +……. +N =N(N +1)/2
Where,
N = the useful life
Dn =(N - n +1)(P - S)/SOYD
12
EXAMPLE: Straight-Line Depreciation
13
EXAMPLE: Double Declining-Balance Depreciation
declining balance rate d = 2/N , For 'n' year, Dn = dP(1-d)n-1
14
EXAMPLE: Double Declining-Balance Depreciation
for Final Book Value (Bn)  Salvage Value (S)
15
Useful Life
16
Tax Depreciation Rates
Houses & Building
5%
Transportation equipment
Car, Jeep, Van & Motorcycle
Cycle
15 %
20 %
Furniture
Metal
Wooden
10 %
15 %
Equipment & Machinery
Machinery
Computers
Laboratory equipment
X-Ray machine
Typewriter, photocopy machine
15 %
20 %
15 %
20 %
15 %
17
Equivalence between declining balance
and straight-line depreciation method
Declining
5%
7%
10 %
15 %
20 %
25 %
50 %
Straight line
1.65%
2.60 %
3.40 %
5.30 %
7.30 %
9.60 %
19.20 %
Corporate Income Tax (manufacturing) 20 %
Other Industries
25 %
Value Added Tax
13 %
18
https://thepoabookworm.wordpress.com/2019/01/08/depreciation/
19
Example
https://www.dummies.com/article/business-careers-money/business/accounting/general-accounting/depreciation-methods-175280/
20
Example
https://learn.financestrategists.com/explanation/depreciation-and-disposal-of-fixed-assets/comparison-of-various-depreciationmethods/
21
Example
https://learn.financestrategists.com/explanation/depreciation-and-disposal-of-fixed-assets/comparison-of-various-depreciationmethods/
22
Calculate depreciation in Excel
https://www.journalofaccountancy.com/issues/2021/may/how-to-calculate-depreciation-in-excel.html
23
Calculate depreciation in Excel
24
Depreciation formula
25
Practice
9.1, 9.2, 9.6, 9.7, 9.8, 9.9, 9.10,
9.11, 9.12, 9.13
26
Department of Mechanical Engineering,
Pulchowk campus, Institute of Engineering,
Tribhuvan University
ENGINEERING ECONOMICS
Benefit-Cost Analysis
Dr. Shree Raj Shakya
2022
Lecture 11
Benefit-Cost Analysis
The objective of private investment is to increase
the net worth of the company.
In the public sector, government spend a lot of
money in projects such as education, road
construction, hydro-power plants etc. to increase
net public benefits.
Benefit-Cost Analysis is a decision-making tool
for systematically developing useful information
about the desirable or undesirable effects of a
public project.
Benefit-Cost Analysis tries to determine whether
the social benefits are greater than social costs.
Lecture 11 - Dr. Shree Raj Shakya
2
There are three types of benefit-Cost
Analysis problems
1. to maximize benefits over any costs
2. to maximize net benefits when
benefits and costs vary
3. to minimize costs to get a certain
level of benefits
Lecture 11 - Dr. Shree Raj Shakya
3
Benefit-Cost Ratio Analysis
1.
2.
3.
4.
5.
Identify users' benefits from the project
Quantify, as much as possible, in Rupee term
Identify sponsors' costs
Quantify, as much as possible, in Rupee term
Determine the equivalent benefits and costs at the
base period using the social interest rate
Lecture 11 - Dr. Shree Raj Shakya
4
Social benefits and Social costs
Social Benefits
1.
2.
3.
Benefits due to reduction of deaths
Benefits due to reduction of damage to property
Primary users' disbenefits
B = B + B − DB
2
1
1
Social Costs
1.
2.
Primary Sponsors' costs
Primary Sponsors' savings
C = C −S
1
1
B - C should be positive
Lecture 11 - Dr. Shree Raj Shakya
5
Benefit-Cost Ratio Analysis
sponsor's costs (C) consist of the equivalent capital expenditure ( I )
and the equivalent annual operating costs (C') accrued in each
successive period
6. If BC (i) >1 we should accept a public project
Lecture 11 - Dr. Shree Raj Shakya
6
Net B/C Ratio
• considers only the initial capital expenditure as a
cash outlay, and uses annual net benefits.
If we are to accept a project, the B'C(i) must be greater
than 1
some analysts prefer this measure because, it indicates
the net benefit (B') expected per dollar invested
Lecture 11 - Dr. Shree Raj Shakya
7
Example
A public project being considered by a local government
has the following estimated benefit-cost profile
Assume that i = 10%, N = 5.
Compute B, C, I, C', BC(10%) and B'C(10%).
Lecture 11 - Dr. Shree Raj Shakya
8
Lecture 11 - Dr. Shree Raj Shakya
9
Benefit-Cost Ratio
Net Benefit-Cost Ratio
Project is economically viable or profitable
Lecture 11 - Dr. Shree Raj Shakya
10
Relationship Between
B/C Ratio and NPW
Both gives same result
Lecture 11 - Dr. Shree Raj Shakya
11
Incremental Analysis based on BC(i)
Consider three investment projects, A1, A2, and A3. Each
project has the same service life, and the present worth of
each component value (B, I, and C') is computed at 10% as
follows.
(a)
If all three projects are independent, which projects would
be selected based on BC(i) and B'C(i) criteria, respectively?
(b)
If the three projects are mutually exclusive, which project
would be the best alternative? Show the sequence of calculations
required to produce the correct results, using (1) the aggregate BIC
12
ratio, and (2) the netted BIC ratio.
Solution
(a)
Since PW(i)1, PW(i)2, and PW(i)3 are positive,
all projects would be acceptable if they were
independent.
Also, BC(i) and B'C(i) values for each project are greater
than 1, so the use of either ratio will lead to the same
accept-reject conclusion under the NPW criterion.
Lecture 11 - Dr. Shree Raj Shakya
13
(b) If these projects are mutually exclusive, we must
use the principle of incremental analysis.
1.
Remove any alternatives with a BIC ratio less than 1
2.
Arrange the remaining alternatives in the increasing
order of the denominator (I + C').
Thus, the alternative with the smallest denominator should
be first (j), the alternative with the next smallest second (k),
and so forth.
Lecture 11 - Dr. Shree Raj Shakya
14
3. Compute the incremental differences for each term (B,
I, and C') for the paired alternatives j, k) in the list.
Select A1 and A3 as they have least denominator (cost)
4. Compute the BC(i) on incremental investment by evaluating
Since the ratio is greater than 1, we prefer A3 over A1. Therefore,
A3 becomes the "current best" alternative.
15
5. Compare the selected alternative with the next one on the list by
computing the incremental benefit-cost ratio. Continue the process
until you reach the bottom of the list. The alternative selected during
the last pairing is the best.
Therefore, we need to compare A2 and A3 as follows:
The incremental B/C ratio again exceeds 1, and therefore we prefer
A2 over A3.
With no further projects to consider, A2 becomes the ultimate choice.
Lecture 11 - Dr. Shree Raj Shakya
16
Using the net B/C (B'C(i)) ratios:
If we had to use the net BIC ratio on this incremental investment
decision, we would obtain the same conclusion.
Since all B'C(i) ratios exceed 1, there will be no do-nothing alternative.
By comparing the first pair of projects on this list, we obtain
Project A3 becomes the current best.
Next, comparing A2 and A3 yields
Therefore, A2 becomes the best choice by the net BIC criterion.
Lecture 11 - Dr. Shree Raj Shakya
17
Solve 1
A city government is considering town-dump sanitary system.
The project requires an initial outlay of $400,000, with
annual operating and maintenance costs of $50,000 for
the next 15 years. Fee collections from the residents
would be $85,000 per year. The interest rate is 8%, and
there is no salvage value associated with either system.
Determine economics feasibility of the project using the
benefit-cost ratio and net benefit-cost ratio.
Lecture 11 - Dr. Shree Raj Shakya
18
Lecture 8 - Dr.Shree Raj Shakya
19
Solve 2
A city which operates automobile parking facilities is evaluating a proposal
that it erect and operate a structure for parking in a city's downtown area.
Three designs for a facility to be built on available sites have been identified.
(All dollar figures are in thousands.)
At the end of the estimated service life, whichever facility had been
constructed would be torn down and the land would be sold. It is estimated
that the proceeds from the resale of the land will be equal to the cost of
clearing the site. If the city's interest rate is known to be 10%, which
design alternative would be selected based on the benefit-cost criterion?
20
Lecture 8 - Dr.Shree Raj Shakya
21
Practice
• 16.6, 16.7, 16.8, 16.9, 16.10, 16.11, 16.12,
16.13.
22