advanced engineering economy & costing.pptx
TRANSCRIPT
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
1/85
Advanced Engineering Economy &
Costing
D.Prakash
Adminstrative block; Room no: [email protected]
0932120816
mailto:[email protected]:[email protected] -
7/27/2019 Advanced Engineering Economy & Costing.pptx
2/85
Aim
To develop economic and cost analysis models for decisions
making.
Course Description
Formulation of economic problems models. Analysis of
capital Investments, Decision analysis methods: decision tree
analysis, multi-attribute decisions, probabilistic analysis andsensitivity/risk analysis.
Stochastic techniques and risk to evaluate design
alternatives, Capital budgeting models: multi-criteria optimization,
certainty equivalence. Replacement analysis.
Costing techniques applicable in manufacturing: activitybased costing, life cycle costing, theory of constraints, cost of
quality.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
3/85
3
ENGINEERING ECONOMICS INVOLVES:FORMULATING, ESTIMATING, AND
EVALUATING ECONOMIC OUTCOMES
WHEN CHOICES ORALTERNATIVES AREAVAILABLE
What Kinds of Questions Can
Engineering Economics Answer?
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
4/85
4
How Does It Do This?
BY USING SPECIFIC
MATHEMATICAL RELATIONSHIPSTO COMPARE THE CASH FLOWS OF THE
DIFFERENT ALTERNATIVES
(typically using spreadsheets)
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
5/85
5
Where Does Engineering
Economics Fit?Here is an approach to problem-solving:
Understand the problem
Collect all relevant data/information
Define the feasible alternatives
Evaluate each alternative
Select the best alternative
Implement and monitor the decision
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
6/85
6
Where Does Engineering
Economics Fit?1. Understand the Problem
2. Collect all relevant data/information (difficult!)
3. Define the feasible alternatives
4. Evaluate each alternative
5. Select the best alternative
6. Implement and monitorThis is the major role of
engineering economics
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
7/85
Contemporary Engineering Economics, 4thedition, 2007
7
What Makes the Engineering Economic
Decision Difficult? - Predicting the Future
Estimating a Required
investment
Forecasting a product
demand Estimating a selling price
Estimating a
manufacturing cost
Estimating a product life
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
8/85
Contemporary Engineering Economics, 4thedition, 2007
8
Create & Design
Engineering Projects
Evaluate
Expected
Profitability
Timing of
Cash Flows Degree of
Financial Risk
Analyze
Production Methods
Engineering Safety
Environmental Impacts
Market Assessment
Evaluate
Impact on
Financial Statements
Firms Market Value
Stock Price
Role of Engineers in Business
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
9/85
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.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
10/85
Contemporary Engineering Economics, 4thedition, 2007
10
Present
FuturePast
Engineering EconomyAccounting
Evaluating past performance Evaluating and predicting future events
Accounting Vs. Engineering Economics
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
11/85
Contemporary Engineering Economics, 4thedition, 2007
11
Fundamental Principles of Engineering
Economics
Principle 1: A nearby dollar is worth more
than a distant dollar
Principle 2: All it counts is the differences
among alternatives
Principle 3: Marginal revenue must exceed
marginal cost
Principle 4: Additional risk is not taken
without the expected additional return
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
12/85
Contemporary Engineering Economics, 4thedition, 2007
12
Principle 1: A nearby dollar is worth more
than a distant dollar
Today 6-month later
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
13/85
Contemporary Engineering Economics, 4thedition, 2007 13
Principle 2: All it counts is the differences
among alternatives
Option MonthlyFuelCost
MonthlyMaintenance
Cashoutlay atsigning
Monthlypayment
SalvageValue atend ofyear 3
Buy $960 $550 $6,500 $350 $9,000
Lease $960 $550 $2,400 $550 0
Irrelevant items in decision making
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
14/85
Contemporary Engineering Economics, 4thedition, 2007
14
Principle 3: Marginal revenue must exceed
marginal cost
Manufacturing cost
Sales revenueMarginal
revenue
Marginal
cost
1 unit
1 unit
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
15/85
Contemporary Engineering Economics, 4thedition, 2007 15
Principle 4: Additional risk is not taken
without the expected additional return
Investment Class Potential
Risk
Expected
Return
Savings account(cash)
Low/None 1.5%
Bond (debt) Moderate 4.8%
Stock (equity) High 11.5%
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
16/85
Simple Methods
Simple Payback Period (SPP)- The timerequired for savings to offset first costs.
Simple Return on Investment (ROI)- The
simple percent return the project pays over itslife.
These methods are simple because they do
not consider the time value of money. Simple methods are OK for investments that
are very good and pay off over short timeperiods.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
17/85
17
Time Value of Money
Money has value Money can be leased or rented
Thepayment is called interest
If you put $100 in a bank at 9% interest for one time periodyou will receive back your original $100 plus $9
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
18/85
18
Compound Interest
Interest that is computed on the original
unpaid debt and the unpaid interest Compound interest is most commonly used
in practice
Total interest earned = In = P (1+i)n - P Where,
P present sum of money
i interest rate
n number of periods (years)
I2 = $100 x (1+.09)2 - $100 = $18.81
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
19/85
Present and Future Value
Present Value is the value now of an amount
of money Freceived n years in the future.
Future Value is value n years in the future of
an amount of money Preceived now.
If we can earn interest rate i on investments,
the relationship between P and F is:
F = P(1 + i)n or P = F/(1 + i)n
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
20/85
ECONOMIC MODELS
Economic modeling is at the heart of economic theory.
Modeling provides a logical,abstract template to helporganize the analyst's thoughts.
The model helps the economist logically isolate and
sort out complicated chains of cause and effect andinfluence between the numerous interacting elementsin an economy.
Through the use of a model, the economist canexperiment, at least logically, producing differentscenarios, attempting to evaluate the effect ofalternative policy options, or weighing the logicalintegrity of arguments presented in prose.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
21/85
Types of Models
visual models,
Mathematical models,
Empirical models,
Simulation models.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
22/85
1. Visual Models - Visual models are simply pictures of anabstract economy; graphs with lines and curves that tell an economic story
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
23/85
2. Mathematical Models
The most formal and abstract of the economic models are the purely
mathematical models. These are systems of simultaneous equations with anequal or greater number of economic variables.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
24/85
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
25/85
3.Empirical Models
Empirical models are mathematical modelsdesigned to be used with data. Thefundamental
model is mathematical, exactly as describedabove. With an empirical model, however,
data is gathered for the variables, and using
accepted statistical techniques, the data areused to
provide estimates of the model's values.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
26/85
"What will happen to investment if income
rises one percent?" The purely mathematicalmodel might only allow the analyst to say,"Logically, it should rise.
The user of the empirical model, on the otherhand, using actual historical data forinvestment, income, and the other variables inthe model, might be able to say,
"By my best estimate, investment should rise byabout two percent."
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
27/85
4.Simulation Models
Simulation models, which must be used with computers,
embody the very best features of mathematical models
without requiring that the user be proficient in mathematics.
The models are fundamentally mathematical (the equations
of the model are programmed in a programming language like
Pascal or C++) but the mathematical complexity is transparentto the user.
The simulation model usually starts with initial or "default"
values assigned by the program or the user, then certain
variables are changed or initialized, then a computersimulation is done.
The simulation, of course, is a solution of the model's
equations. The user can usually alter a whole range of
variables at will.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
28/85
The computerized simulation model can show
the interaction of numerous variables all at
once, including hidden feedback and
secondary effects that are not so apparent inpurely mathematical or visual models.
Macroeconomic simulation model calledHMCMacroSim
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
29/85
Static and Dynamic Models
Most of the models used in economics are
comparative statics models. Some of the more
sophisticated models in macroeconomics and
business cycle analysis are dynamic models.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
30/85
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
31/85
The initial equilibrium (point 'a') identifiesthe price and
level of output that would obtain, given assumptions aboutsupply and demand and the level of inflationaryexpectations.
Then the model is shocked by introducing a higher level ofexpectations, demonstrating a new equilibrium at point 'b'.
Obviously this movement in equilibria and the shift in themodel's solution happened over time, but neither thevisual model nor its mathematical counterpart candemonstrate what happened in the interim. The modelshows only the starting point and the ending point.
The comparative statics approach is roughly analogous tousing snapshots from a camera to record developmentsduring a dynamic event. With each snapshot a static butinformative picture is presented.
i d l
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
32/85
Dynamic Model
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
33/85
Why Comparative Statics Models are
Usually Used?
The answer is simple - comparative statics modelsare much easier to solve.
Any student of calculus knows the difficulty of
solving systems of difference or (especially)differential equations.
The latter, as soon as they achieve any complexity,are sometimes impossible to solve.
Therefore dynamic models must be kept extremelysimple and are therefore so elementary that moreis lost than gained.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
34/85
Simple dynamic models, nonetheless, often providevaluable insights into the complex interactions betweenvariables over time.
They can capture remarkably subtle feedback effects thatare easily missed by static models.
It should be noted that dynamic models are much easier tosimulate on computers than they are to solve outright.
The user can experiment with an endless variety of valuesand assumptions to see whether results obtained are
realistic or insightful. Since computers are now powerfuland cheaper, the importance of dynamic simulation modelsshould gradually grow in importance.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
35/85
Expectations-Enhanced Models
Economic models often incorporate economicexpectations, such as inflationary expectations. Suchmodels are called expectations-enhanced models.
Generally, expectations-enhanced models include oneor more variables based upon economic expectationsabout future values.
For example, if consumers, for whatever reason,
expect the inflation rate to be much higher next yearthan this year, they are said to have formedinflationary expectations.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
36/85
There are many types of expectations found in
economics.
In addition to inflationary expectations,
economists might consider interest rate
expectations, income expectations, and
wealth expectations. This list is hardly
exhaustive.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
37/85
Adaptive Expectations
The theory of adaptive expectations presumes thatexpectations are primarily learned from experience.
For example, the theory of adaptive expectationswould say that if consumers begin to actually see
prices rising, say from three percent to five percent toseven percent, over a period of, say, two years, theywill begin to form robust expectations of inflationaryexpectations perhaps even expectations of double-digit inflation.
The same theory might claim that consumers willexpect an economic recovery to begin only after ampleevidence that the turning point has been passed.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
38/85
Rational Expectations
The theory of rational expectations presumes thatexpectations are formed when economic agents see newdevelopments in the economy and they logically deduceexpectations based upon the information they have.
For example, if the Federal Reserve System were tosuddenly increase the money supply, according to thetheory of rational expectations, consumers wouldimmediately form inflationary expectations, not becauseprices are actually rising, but because they deduce thatexcessive money supply growth is likely to cause inflation.
The theory of rational expectations emphasizes the effectsof changes in economic policy upon expectations, althoughthe theory is not restricted to policy decisions alone.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
39/85
The Limitations of Models
Improper Assumptions
Oversimplification
Mathematical Intractability
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
40/85
The Model as an "Image" of Economic Activity
Two important points are being made here:
1. This model, like most in economies, is not an applied model, where anyone actually uses it to
determine appropriate prices and levels of production. (To be more specific, it is not an
applied management model; corporations don't use these models to make pricing decisions).
Instead, the model represents a type of consistent behavior that economists see in the market
place, and it presents an image of that behavior. It allows an economist to both ask and answer
the question, "What would we expect to happen in a market where prices are too high or too
low? What kind of adjustment would take place, and why?
2. The market reactions of the economic decision-makers are not undertaken by virtue of their
use of this model or any other, but is instead motivated by their necessary response tomarket signals that tell them that they must alter their decisions.
The model, therefore, simply captures their responses to a series of market signals.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
41/85
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
42/85
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
43/85
Analysis of capital investment
Present value method
Future value technique
Annual equivalent cost method
Rate of return method
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
44/85
Present value method
Using the compound interest formulas bring allbenefits and costs to present worth
Select the alternative if its net present worth 0
Net present worth =Present worth of benefits Present worth of costs
Present Worth Analysis
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
45/85
45
Present Worth Analysis
A construction enterprise is investigating the
purchase of a new dump truck. Interest rate is 9%.The cash flow for the dump truck are as follows:
First cost = $50,000, annual operating cost = $2000,
annual income = $9,000, salvage value is $10,000, life
= 10 years. Is this investment worth undertaking? P = $50,000, A = annual net income = $9,000 - $2,000
= $7,000, S = 10,000, n = 10.
Evaluate net present worth = present worth ofbenefits present worth of costs
Present Worth Analysis
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
46/85
46
Present Worth Analysis
Present worth of benefits = $9,000(PA,9%,10) =
$9,000(6.418) = $57,762
Present worth of costs = $50,000 +
$2,000(PA,9%,10) - $10,000(PF,9%,10)= $50,000 +
$2,000(6..418) - $10,000(.4224) = $58,612
Net present worth = $57,762 - $58,612 < 0 do not
invest
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
47/85
Future value technique
The future value technique of evaluating
alternatives is almost identical to the present
value method except that all costs andrevenues are stated in terms of future value.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
48/85
Annual equivalent cost method
The annual equivalent cost method of evaluating alternativeprojects states all costs and revenues over the useful life ofthe project in terms of an equal annual payment series
1. It requires less effort and fewer calculations.
2. It eliminates the problem of alternatives with incompatibleuseful lives.
3. It allows for much more sophistication when consideringinflation, increasing equipment cost, equipment depreciationschedules, etc.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
49/85
Rate of Return (ROR)
The rate of return (ROR) method of comparing
alternatives calculates the interest rate for each
alternative and selects the highest ROR.
ROR evaluates INVESTED capital and the costs of
operation and maintenance as opposed to revenues
or benefits received from the project.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
50/85
Cost-Benefit Analysis
Project is considered acceptable ifB C 0 or B/C 1. Example (FEIM):
The initial cost of a proposed project is $40M, thecapitalized perpetual annual cost is $12M, the
capitalized benefit is $49M, and the residual value is$0. Should the project be undertaken?
B = $49M, C = $40M + $12M + $0 B C = $49M $52M = $3M < 0
The project should not be undertaken.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
51/85
51
Rational Decision-Making Process
1. Recognize a decision problem
2. Define the goals or objectives
3. Collect all the relevant
information
4. Identify a set of feasibledecision alternatives
5. Select the decision criterion to
use
6. Select the best alternative
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
52/85
52
Which Car to Lease?
Saturn vs. Honda
1. Recognize a decision problem
2. Define the goals or objectives
3. Collect all the relevant
information
4. Identify a set of feasibledecision alternatives
5. Select the decision criterion
to use
6. Select the best alternative
Need a car
Want mechanical security
Gather technical as well
as financial data
Choose between Saturn
and Honda
Want minimum total cash
outlay Select Honda
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
53/85
53
Financial Data Required to Make an Economic
Decision
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
54/85
54
Predicting the Future
Estimating a Required
investment
Forecasting a product
demand
Estimating a selling price
Estimating a
manufacturing cost
Estimating a product life
Types of Strategic Engineering
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
55/85
Contemporary Engineering Economics, 4th
edition, 2007 55
Types of Strategic EngineeringEconomic Decisions in Manufacturing
SectorService Improvement
Equipment and Process Selection
Equipment Replacement
New Product and Product Expansion
Cost Reduction
S i I t H lth
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
56/85
Contemporary Engineering Economics, 4th
edition, 2007 56
Service Improvement - Healthcare
Delivery
Which plan is moreeconomically viable?
Traditional Plan: Patients
visit each service provider.
New Plan: Each service
provider visits patients
: patient
: service provider
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
57/85
Contemporary Engineering Economics, 4th
edition, 2007 57
Equipment & Process Selection
How do you choose between the Plastic SMC
and the Steel sheet stock for an auto body
panel?
The choice of material will dictate themanufacturing process for an automotive
body panel as well as manufacturing costs.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
58/85
Contemporary Engineering Economics, 4th
edition, 2007 58
Equipment Replacement Problem
Now is the time to
replace the old machine?
If not, when is the right
time to replace the old
equipment?
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
59/85
Contemporary Engineering Economics, 4th
edition, 2007 59
New Product and Product Expansion
Shall we build or acquire
a new facility to meet the
increased demand?
Is it worth spending
money to market a new
product?
T p f Str t i E i ri E i
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
60/85
Contemporary Engineering Economics, 4th
edition, 2007 60
Types of Strategic Engineering Economic
Decisions in Service Sector
Commercial Transportation
Logistics and Distribution
Healthcare Industry
Electronic Markets and Auctions
Financial Engineering
Retails
Hospitality and Entertainment Customer Service and Maintenance
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
61/85
Which Material to Choose?
New plant design Upgrade old plant
A i
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
62/85
62
Alternative 1
Description
Cash flows oversome time period
Analysis using an
engineeringeconomy model
Evaluatedalternative 1
Noneconomic issues-environmental considerations
Alternative2
Description
Cash flows oversome time period
Analysis using anengineering
economy model
Evaluatedalternative2
Income, cost estimationsFinancing strategiesTax laws
Planning horizonInterest
Measure of worth
Calculated value ofmeasure of worth
I select alternative 2
Rate of return (Alt 2)>Rate of return (Alt 1)
Alternatives
Methods of Economic Selection
Example 7 2
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
63/85
63
Compare the following machines on the basis of their equivalent
uniform annual cost. Use an interest rate of 18% per year.
Comparison point New Machine Used Machine
Capital cost 44000 m.u. 23000 m.u.
Annual operating
cost7000 m.u. 9000 m.u.
Annual repair cost210 m.u. 350 m.u.
Overhauling 2500 m.u. every 5 years 1900 m.u. every 2 years
Salvage value 4000 m.u. after 15 years 3000 m.u. after 8 years
Example 7.2
Cash flows of the two machines.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
64/85
64
0 1 2 3 4 5 6 7 8 9 10 11 12 1513 14
i = 18%
m.u.2500 m.u.2500m.u.7210/year
m.u.44000-2500
New machine
m.u.4000
m.u.2500
EUACnew = 7,210 + (440002500) (A/P, 18%, 15) + 2500 (A/P, 18%, 5) 4000 (A/F, 18%, 15)
= 7210 + 41500 (0.18 (1.1815) / (1.18151)) + 2500 (0.18 (1.185) / (1.1851))
4000 (0.18/ (1.1815-1))
EUACnew = 16094.55 m.u. per year.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
65/85
65
EUACused = 9350 + (230001900) (A/P, 18%, 8) + 1900 (A/P, 18%, 2) 3000 (A/F, 18%, 8)
= 21100 (0.18 (1.18)8 / (1.1881)) + 9350 + 1900 (0.18 (1.18)2 / (1.1821))
3000 (0.18 / (1.1881))
= 15542.4 m.u. per year.
1 2 3 4
0
5
i= 18 %
6 7 8
m.u.23000
m.u.9350/year
m.u.1900 m.u.1900 m.u.1900
se mac ne
m.u.3000
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
66/85
66
Since we have found that: EUACused
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
67/85
67
Spreadsheet Solution for example 7.2
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
68/85
Analysis
The acceptance or rejection of a project based on the IRR criterion
is made by comparing the calculated rate with the required rate of return,or cutoff rate established by the firm. If the IRR exceeds the required ratethe project should be accepted; if not, it should be rejected.
If the required rate of return is the return investors expect the
organization to earn on new projects, then accepting a project with an IRRgreater than the required rate should result in an increase of the firmsvalue.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
69/85
Analysis
There are several reasons for the widespread popularity of the IRR as an
evaluation criterion:
Perhaps the primary advantage offered by the technique is that
it provides a single figure which can be used as a measure of
project value.
Furthermore, IRR is expressed as a percentage value. Most
managers and engineers prefer to think of economic decisions
in terms of percentages as compared with absolute values
provided by present, future, and annual value calculations.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
70/85
Analysis
Another advantage offered by the IRR method is related to the
calculation procedure itself:
As its name suggests, the IRR is determined internally for
each project and is a function of the magnitude and timing of the
cash flows.
Some evaluators find this superior to selecting a rateprior to calculation of the criterion, such as in the profitability index
and the present, future, and annual value determinations. In other
words, the IRR eliminates the need to have an external interest
rate supplied for calculation purposes.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
71/85
Multiple Roots Case
One of the disconcerting aspects associated with the internal rate of return is that
more than one interest rate may satisfy the calculation. The solution procedure for IRR
is essentially the solution for an nth degree polynomial of the form:
NPV = 0 = A0 + A1X + A2X2 + A3X
3 + .... + AnXn
whereX= 1/(1 + r)
For a polynomial of this type there may be n different real roots, or values ofr,
which satisfy the equation. Multiple positive rates of return may occur when the
annual cash flows have more than one change in sign.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
72/85
Multiple Roots Case
The following example illustrates the possibility of multiple rates which satisfy the definition of IRR:
Suppose a mining operation has a remaining life of eight years, but an investment is considered to increase the production rate. This will result indepleting the deposit in six years. Assuming the following cash flows, is the investment justified?
Because there are two sign reversals in the cash flows, Descartes Rule of Signs indicates there area maximum of two real roots to the IRR polynomial.
Solving for these roots by trial and error yields the following:
Year 0 1 2 3 4 5 6 7 8
Existing 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0
Proposed (15.0) 16.0 16.0 16.0 16.0 15.0 15.0 - -
Difference (15.0) 6.0 6.0 6.0 6.0 5.0 5.0 (10.0) (10.0)
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
73/85
Multiple Roots CaseGraphically this appears as shown in the following figure:
The rates at which NPV = 0 are, by definition, the internal rates of return. By interpolation, the two solving rates of return for this example are approximately 4.5 and 12.3%
Rate 0% 4% 5% 10% 12% 15%
NPV (1.00) (0.07) 0.06 0.05 0.03 (0.25)
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
74/85
Multiple Roots Case
Should the firm invest in the project or not?
If both rates were above the firm's required rate of return there would be no
problem and the firm would accept the project.
However, what if the required rate of return is 10%? Which of the calculated
IRR values is correct? The answers to these questions are that they are both
mathematically correct, but they are meaningless from an economic standpoint.
Neither of these rates can be considered an adequate measure of the project's
rate of return because a project can not earn more than one rate of return over
its life. Therefore, the calculation of an IRR value(s) does not always enable the
decision-maker to make accept/reject decisions on investment proposals.
M lti l R t C
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
75/85
Multiple Roots Case
How often this problem of multiple rates actually occurs?
The possibility of multiple-rate occurrences is perhaps more prevalent in thecase of new mining ventures than in most other industries. The negative cash flows
are typically the result of anticipated periods of reduced market prices, major
capital expenditures for equipment replacement, expansion programs, and/or
major environmental expenditures, particularly at the end of project life.
Because of the possibility of multiple rates and the reinvestment assumptionwhen using-the IRR to rank projects, the evaluator must carefully consider the
exclusive use of this technique for decision-making.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
76/85
Sensitivity and Breakeven Analysis:
These techniques are used to see howsensitive a decision is to estimates for thevarious parameters.
BREAKEVEN ANALYSIS is done to locateconditions under which various alternativesare equally desirable. Examples include singlevs. multi-stage construction ,hours of
equipment utilization, production volumerequired, and equipment replacement analysis
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
77/85
Break-Even Analysis
Excel using a Goal Seek function
Analytical Approach
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
78/85
Excel Using a Goal Seek Function
Goal Seek
Set cell:
To value:
By changing cell:
Ok Cancel
? X
$F$5
0
$B$6
NPW
Breakeven Value
Demand
1
2
A B C D E F G
Example 10.3 Break-Even Analysis
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
79/85
2345678
910111213141516171819
2021222324
2627
293031
32333435363738
4041
Input Data (Base): Output Analysis:
Unit Price ($) 50$ Output (NPW) $0Demand 1429.39Var. cost ($/unit) 15$Fixed cost ($) 10,000$
Salvage ($) 40,000$Tax rate (%) 40%MARR (%) 15%
0 1 2 3 4 5Income StatementRevenues:
Unit Price 50$ 50$ 50$ 50$ 50$Demand (units) 1429.39 1429.39 1429.39 1429.39 1429.39Sales Revenue 71,470$ 71,470$ 71,470$ 71,470$ 71,470$
Expenses:Unit Variable Cost 15$ 15$ 15$ 15$ 15$Variable Cost 21,441 21,441 21,441 21,441 21,441Fixed Cost 10,000 10,000 10,000 10,000 10,000Depreciation 17,863 30,613 21,863 15,613 5,581
Taxable Income 22,166$ 9,416$ 18,166$ 24,416$ 34,448$Income Taxes (40%) 8,866 3,766 7,266 9,766 13,779
Net Income 13,299$ 5,649$ 10,899$ 14,649$ 20,669$
Cash Flow Statement
Operating Activities:Net Income 13,299 5,649 10,899 14,649 20,669Depreciation 17,863 30,613 21,863 15,613 5,581
Investment Activities:Investment (125,000)Salvage 40,000Gains Tax (2,613)
Net Cash Flow (125,000)$ 31,162$ 36,262$ 32,762$ 30,262$ 63,636$
Goal Seek
FunctionParameters
Analytical Approach
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
80/85
Analytical ApproachUnknown Sales Units (X)
0 1 2 3 4 5
Cash Inflows:
Net salvage 37,389
X(1-0.4)($50) 30X 30X 30X 30X 30X
0.4 (dep) 7,145 12,245 8,745 6,245 2,230Cash outflows:
Investment -125,000
-X(1-0.4)($15) -9X -9X -9X -9X -9X
-(0.6)($10,000) -6,000 -6,000 -6,000 -6,000 -6,000
Net Cash Flow -125,000 21X+
1,145
21X+6,245
21X+
2,745
21X+
245
21X+
33,617
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
81/85
PW of cash inflows
PW(15%)Inflow= (PW of after-tax net revenue)
+ (PW of net salvage value)+ (PW of tax savings from depreciation
= 30X(P/A, 15%, 5) + $37,389(P/F, 15%, 5)
+ $7,145(P/F, 15%,1) + $12,245(P/F, 15%, 2)
+ $8,745(P/F, 15%, 3) + $6,245(P/F, 15%, 4)
+ $2,230(P/F, 15%,5)
= 30X(P/A, 15%, 5) + $44,490
= 100.5650X+ $44,490
PW of cash outflows:
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
82/85
PW of cash outflows:
PW(15%)Outflow = (PW of capital expenditure_
+ (PW) of after-tax expenses
= $125,000 + (9X+$6,000)(P/A, 15%, 5)= 30.1694X+ $145,113
The NPW:
PW (15%) = 100.5650X+ $44,490
- (30.1694X+ $145,113)
=70.3956X- $100,623.
Breakeven volume:
PW (15%) = 70.3956X- $100,623 = 0
Xb =1,430 units.
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
83/85
Demand
PW of
inflow
PW of
Outflow NPW
X
100.5650X
- $44,49030.1694
X
+ $145,11370.3956
X
-$100,623
0 $44,490 $145,113 100,623
500 94,773 160,198 65,425
1000 145,055 175,282 30,2271429 188,197 188,225 28
1430 188,298 188,255 43
1500 195,338 190,367 4,970
2000 245,620 205,452 40,168
2500 295,903 220,537 75,366
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
84/85
Outflow
Break-Even Analysis Chart
0 300 600 900 1200 1500 1800 2100 2400
$350,000
300,000
250,000
200,000
150,000
100,000
50,000
0
-50,000
-100,000
Profit
Loss
Break-even Volume
Xb=1430
Annual Sales Units (X)
PW
(15%
)
Scenario Analysis
-
7/27/2019 Advanced Engineering Economy & Costing.pptx
85/85
Scenario Analysis
Variable
Considered
Worst-
Case
Scenario
Most-Likely-
Case
Scenario
Best-Case
Scenario
Unit demand 1,600 2,000 2,400
Unit price ($) 48 50 53Variable cost ($) 17 15 12
Fixed Cost ($) 11,000 10,000 8,000
Salvage value ($) 30,000 40,000 50,000
PW (15%) -$5,856 $40,169 $104,295