Advanced Engineering Economy & Costing.pptx

download Advanced Engineering Economy & Costing.pptx

of 85

Transcript of Advanced Engineering Economy & Costing.pptx

  • 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