engineering economics - present worth analysis

Applied Software Project Management

Engineering EconomicsEngineering Economics

Present-Worth Analysis Present-Worth Analysis

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RecapRecap

PART 1 UNDERSTANDING MONEY AND ITS MANAGEMENT Chapter 1 Engineering Economic Decisions Chapter 2 Time Value of Money Chapter 3 Understanding Money Management Chapter 4 Equivalence Calculations under Inflation

PART 2 EVALUATING BUSINESS AND ENGINEERING ASSETS Chapter 5 Present-Worth Analysis Chapter 6 Annual Equivalence Analysis Chapter 7 Rate-of-Return Analysis

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Road MapRoad Map

Chapter 5 Present-Worth Analysis Loan versus Project Cash Flows Initial Project Screening Methods Present-Worth Analysis Methods to Compare Mutually Exclusive Alternatives

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Loan versus Project Cash FlowsLoan versus Project Cash Flows

Loan cash flow: future return in the form of interest plus repayment of the principal in case of bank loan

Project cash flow: future return in the form of cash generated by productive use of the fixed asset, along with the capital expenditures and annual expenses (such as wages, raw materials, operating costs, maintenance costs, and income taxes)

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Loan versus Project Cash FlowsLoan versus Project Cash Flows

Similarity between loan cash flow and project cash flow future → use the same equivalence techniques to measure economic worth

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Bank Loan

BankLoan

Repayment

Customer

Investment Project

CompanyInvestment

Return

Project


Road MapRoad Map


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Initial Project Screening MethodsInitial Project Screening Methods

One of the primary concerns of most businesspeople: whether and when the money invested can be recovered

The payback method screens projects on the basis of how long it takes for net receipts to equal investment outlays: cash inflows exactly match or pay back the cash outflows

Conventional-payback method: ignore the time-value-of-money Discounted-payback method: include the time-value-of-money A common standard used to determine whether to pursue a project: a

project does not merit consideration unless its payback period is shorter than some specified period of time

Payback screening: not an end in itself, but rather a method of screening out certain obviously unacceptable investment alternatives before progressing to an analysis of potentially acceptable ones

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Benefits & Flaws of Payback ScreeningBenefits & Flaws of Payback Screening

Benefits:– Simplicity: one of its most appealing qualities. – focusing on that time at which the firm expects to

recover the initial investment. – Eliminate some alternatives: reducing a firm's need to

make further analysis efforts on those alternatives Flaws:

– fails to measure profitability

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Benefits & Flaws of Payback ScreeningBenefits & Flaws of Payback Screening

n Project 1 Project 2

0 -10,000 -10,000

1 1,000 9,000

2 9,000 1,000

3 1,000 1,000

Payback Period 2 years 2 years

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Discounted-Payback PeriodDiscounted-Payback Period

Take into account the time value of money, i.e. the cost of funds (interest) used to support the project

Payback-Period Calculation Considering the Cost of Funds

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Period Cash Flow Cost of Funds (15%) Cumulative Cash Flow

0 -85,000 0 -85,000

1 15,000 -85,000(0.15) = -12,750 -82,750

2 25,000 -82,750(0.15) = -12,413 -70,163

3 35,000 -70,163(015) = -10,524 -45,687

4 45,000 -45,687(0.15) = -6,853 -7,540

5 45,000 -7,540(0.15) = -1,131 36,329

6 45,000 36,329(0.15) = 5,449 76,778


Road MapRoad Map


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Net-Present-Worth CriterionNet-Present-Worth Criterion

Basic procedure for applying the net-present-worth criterionEvaluation of a Single Project

Step 1: – Determine the interest rate that the firm wishes to earn on its

investments. – This interest rate is often referred to as either a required rate of

return or a minimum attractive rate of return (MARR).

Step 2: Estimate the service life of the project

Step 3: Estimate the cash inflow for each period over the service life.

Step 4: Estimate the cash outflow for each period over the service life.

Step 5: Determine the net cash flows for each period

net cash flow = cash inflow - cash outflow

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Step 6: Find the present worth of each net cash flow at the MARR. Add up these present-worth figures; their sum is defined as the project's PW:

Where: PW(i) = PW calculated at i,

An = net cash flow at the end of period n

i = MARR (or cost of capital)

n = service life of the project

An: positive if the corresponding period has a net cash inflow and negative if the period has a net cash outflow.

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N)i,(P/F,APW(i)

i)(1

APW(i)

N

0nn

N

0nn

n



Step 7: a positive PW means that the equivalent worth of the inflows is greater than the equivalent worth of the outflows, so the project makes a profit.

If PW(i) > 0, accept the investment.

If PW(i) = 0, remain indifferent.

If PW(i) < 0, reject the investment.

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Above rule for evaluation of a single project Below guidelines for evaluating and comparing more than one project:

Comparing More Than One Alternative

1. To select the best alternative– Select the one with the highest PW (all the alternative have the

same service lives)– Comparison of alternatives with unequal service lives requires

special assumptions, as will be detailed in next Section.

2. Comparison of mutually exclusive alternatives with the same revenues is performed on a cost-only basis. – accept the project that results in the smallest PW of costs, – or the least negative PW (because you are minimizing costs, rather

than maximizing profits).

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Tiger Machine Tool Company is considering the acquisition of a new metal cutting machine. The required initial investment of $75,000 and the projected cash benefits over a three-year project life are as follows:

You have been asked by the president of the company to evaluate the economic merit of the acquisition. The firm's MARR is known to be 15%.

Given: Cash flows as tabulated; MARR = 15% per year. Find: PW.

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End of Year Net Cash Flow

0 -75,000

1 24,000

2 27,340

3 55,760



To bring each flow to its equivalent at time zero as shown the table:

PW = -75,000 + 24,000(P/F,15%,1) + 27,000(P/F,15%,2) + 55,760(P/F,15%,3) = 3,553

Since the project results in a surplus of $3,553, the project is acceptable. It is returning a profit greater than 15%.

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1 2 3

75,000

24,00027,000

55,760

outflow

inflow


Guidelines for Selecting a MARR Guidelines for Selecting a MARR

Return: one way to evaluate how your investments in financial assets or projects are doing in relation to each other and to the performance of investments in general.

Conceptually, the rate of return is a function of three components:– risk-free real return– inflation factor– risk premium(s)

Zero for safe or risk free (e.g. US Treasury Bills) Increase risk premium for high risk investment (e.g. Internet

Stock?)

Expect return = risk-free real return + inflation factor + risk premium(s)

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Meaning of Net Present WorthMeaning of Net Present Worth

In present-worth analysis, two sources to fund investment:– Investment pool or– Borrowed money at the MARR from the capital

markets Investment pool:

– equivalent to a firm's treasury– left in pool to earn interest at the MARR – may withdraw funds from this investment pool for

other investment purposes

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Meaning of Net Present WorthMeaning of Net Present Worth

Consider previous example which required an investment of $75,000 If the firm did not invest in the project and instead left the $75,000 in

the investment pool for three years to grow to:

75,000(F/P,15%,3) = 114,066 If the firm invest in the project, the return cash flow would also earn

interest:

24,400(F/P,15%,2) + 27,340(F/P,15%,1) + 55,760(F/P,15%,0) = 119,470

The additional cash accumulation at the end of three years from investing in the project:

119,470 – 114,066 = 5,404 → Net Future Worth of the Project

the equivalent present worth of this net cash surplus at time zero

$5,404(P/F, 15%. 3) = $3,553 same as in previous example

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Capitalized-Equivalent MethodCapitalized-Equivalent Method

A special case of the PW criterion: when the life of a proposed project is perpetual or the planning horizon is extremely long

capitalized-equivalent [CE(i)] method for evaluating such projectsPW(i) = A(P/A,i,N) with N → ∞

Same result shown in perpetuity (share of preferred stock that pays a fixed cash dividend each period and never matures)

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i

A

i)i(1

1i)(1limAN)i,(P/A,limAPW(i)

N

N

NN


Road MapRoad Map


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Applied Software Project ManagementMethods to Compare Mutually Methods to Compare Mutually Exclusive Alternatives Exclusive Alternatives

Until now, considering situations involving only a single project or projects that were independent of each other. In both cases, the decision to reject or accept each project individually, based on whether it met the MARR requirements, evaluated using the PW.

In this section, evaluation techniques to consider multiple projects that are mutually exclusive: the selection of one alternative implies that the others will be excluded.

Consider two cases:

1. analysis period equals project lives and

2. analysis period differs from project lives.

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Doing Nothing Is a Decision OptionDoing Nothing Is a Decision Option

When considering an investment, the project either: aimed at replacing an existing asset or system

– if the existing system still adequate and none of new proposals economical replacements then do nothing

– if the existing system terminally failed, the choice among proposed alternatives is mandatory

new endeavor:– None of the proposed alternatives economically

sound, do nothing is an option

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Service Projects vs. Revenue ProjectsService Projects vs. Revenue Projects

Service projects: generate revenues that do not depend on the choice of project, but must produce the same amount of output( revenue)– PW criterion used to compare alternatives to minimize

expenditures, e.g. choose the alternative with the lower present-value production cost over the service life

Revenue projects: generate revenues that depend on the choice of alternative. – Not limiting the amount of input to the project or the

amount of output that the project would generate– Select the alternative with the largest net gains (output -

input).

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Analysis Period Equals Project LivesAnalysis Period Equals Project Lives

Ansell, Inc., a medical-device manufacturer uses compressed air in solenoids and pressure switches in its machines to control various mechanical movements. Over the years, the manufacturing floor has changed layouts numerous times. With each new layout, more piping was added to the compressed-air delivery system in order to accommodate new locations of manufacturing machines. None of the extra, unused old piping was capped or removed; thus, the current compressed-air delivery system is inefficient and fraught with leaks. Because of the leaks in the current system, the compressor is expected to run 70% of the time that the plant will be in operation during the upcoming year.

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This excessive usage will require 260 kWh of electricity at a rate of $O.O5/kWh. (The plant runs 250 days a year, 24 hours per day.) Ansell may address this issue in one of two ways:

Option 1 - Continue current operation: If AnseIl continues to operate the current air delivery system, the compressor's run time will increase by 7% per year for the next five years. because of ever-worsening leaks. (After five years, the current system will not be able to meet the plant's compressed-air requirement, so it will have to be replaced.)

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Option 2 - Replace old piping now: If Ansell decides to replace all of the old piping now, it will cost $28,570. The compressor will still run for the same number of days; however, it will run 23% less (or will incur 70%(1 - 0.23) = 53.9% usage per day) because of the reduced air-pressure loss.

If Ansell's interest rate is 12% compounded annually, is it worth fixing the air delivery system now?

Given: Current power consumption, g = 7%, i = 12%, and N = 5 years.

Find: A1 and P.

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Step 1: We need to calculate the cost of power consumption of the current piping system during the first year

Power cost = (% of day operating) x (days operating per year) x (hours per day) x (kWh) x ($/kWh)

Power cost = (70%) x (250 days/years) x (24 hours/days) x (260kWh) x ($0.05/kWh) = $54,600

PW = 54,600(P/F,12%,1) + 54,600(1+7%)(P/F,12%,2) + 54,600(1+7%)2(P/F,12%,3) + 54,600(1+7%)3(P/F,12%,4)

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Step 2: Each year, if the current piping system is left in place, the annual power cost will increase at the rate of 7% over the previous year's cost. The anticipated power cost over the five-year period is summarized in Figure 5.10. The equivalent present lump-sum cost at 12% interest for this geometric gradient series is:

POption1 = 54,600(P/A1,7%,12%,5)

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222,936P

7%12%

12%)(17%)(1154,600P

Option1

55

Option1



Step 3: If Ansell replaces the current compressed-air delivery system with the new one, the annual power cost will be 23% less during the first year and will remain at that level over the next five years. The equivalent present lump-sum cost at 12% interest is

POption2 = 54,600(1 - 23%)(P/A,12%,5)

POption2 = 54,600(1 - 23%)(P/A,12%,5)

POption2 = 151,553

Step 4: The net cost of not replacing the old system now is $71,383 (= $222,936 - $ 151,553). Since the new system costs only $28,570, the replacement should be made now.

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Analysis Period Differs from Project LivesAnalysis Period Differs from Project Lives

Present-Worth Comparison: Project Lives Longer Than the Analysis Period

Waste Management Company (WMC) has won a contract that requires the firm to remove radioactive material from government-owned property and transport it to a designated dumping site. This task requires a specially made ripper-bulldozer to dig and load the material onto a transportation vehicle. Approximately 400,000 tons of waste must be moved in a period of two years. There are two possible models of ripper-bulldozer that WMC could purchase for this job:

Model A costs $150,000 and has a life of 6.000 hours before it will require any major overhaul. Two units of model A will be required in order to remove the material within two years, and the operating cost for each unit will run $40,000 per year for 2,000 hours of operation. At this operational rate, each unit will be operable for

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Applied Software Project ManagementPresent-Worth Comparison: Project Present-Worth Comparison: Project Lives Longer Than the Analysis PeriodLives Longer Than the Analysis Period

for three years, and at the end of that time, it is estimated that the salvage value will be $25,000 for each machine.

The more efficient model B costs $240,000 each, has a life of 12,000 hours before requiring any major overhaul, and costs $22,500 to operate for 2,000 hours per year in order to complete the job within two years. The estimated salvage value of model B at the end of six years is $30,000. Once again, two units of model B will be required.

Since the lifetime of either model exceeds the required service period of two years, WMC has to assume some things about the used equipment at the end of that time. Therefore, the engineers at WMC estimate that, after two years, the model A units could be sold for $45,000 each and the model B units for $125,000 each.

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After considering all tax effects. WMC summarized the resulting cash flows (in thousands of dollars) for each project as follows:

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Period Model A (thousand) Model B (thousand)

0 -300 -480

1 -80 -45

2 -80 + 90 -45 + 250

3 -80 + 50 -45

4 -45

5 -45

6 -45 + 60


Given: Cash flows for the two alternatives as shown in table; i = 15% per year.

Find: PW for each alternative: the preferred alternative.

PW(15%)A = -$300 - $80(P/A, 15%. 2) + $90(P/F, 15%, 2) = -$362;

PW(15%)B = -$480 - $45(P/A, 15%, 2) + $250(P/F, 15%, 2) = -$364.

Model A has the least negative PW costs and thus would be preferred.

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Applied Software Project ManagementPresent-Worth Comparison: Project Present-Worth Comparison: Project Lives Shorter Than the Analysis PeriodLives Shorter Than the Analysis Period

The Smith Novelty Company, a mail-order firm, wants to install an automatic mailing system to handle product announcements and invoices. The firm has a choice between two different types of machines. The two machines are designed differently, but have identical capacities and do exactly the same job. The $12,500 semiautomatic model A will last three years, while the fully automatic model B will cost $15,000 and last four years. The expected cash flows for the two machines, including maintenance costs, salvage values, and tax effects, are as follows:

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Period Model A Model B

0 -12,500 -15,000

1 -5,000 -4,000

2 -5,500 -4,500

3 -6,000 + 2,000 -5,000

4 -5,500 + 1,500

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Once again, as business grows to a certain level, neither of the models may be able to handle the expanded volume at the end of year 5. If that happens, a fully computerized mail-order system will need to be installed to handle the increased business volume. With this scenario, which model should the firm select at MARR = 15%?

Given: Cash flows for the two alternatives as shown in Figure 5.12, analysis period of five years. and i = 15%.

Find: PW for each alternative: the preferred alternative.

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Since both models have a shorter life than the required service period (five years), we need to make an explicit assumption of how the service requirement is to be met. Suppose that the company considers leasing comparable equipment (Model A) that has an annual lease payment of $5,000 (after taxes). With an annual operating cost of $6,500 for the remaining required service period. The anticipated cash flows for both models under this scenario are as follows:

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Period Model A Model B

0 -12,500 -15,000

1 -5,000 -4,000

2 -5,500 -4,500

3 -6,000 + 2,000 -5,000

4 -6,500 – 5,000 -5,500 + 1,500

5 -6,500 – 5,000 -6,500 – 5,000


PW(15%)A = -12,500 – 5,000(P/F,15%,1) - 5,500(P/F,15%,2) - 6,000(P/F,15%,3) - 11,500(P/A,15%,2)(P/F,15%,3) + 2,000(P/F,15%,3)

PW(15%)B = -15,000 – 4,000(P/A,15%,4) – 500[(1+15%)3 –i3 – 1]/[(15%)2(1+15%)3] - 11,500(P/F,15%,5) + 1,500(P/F,15%,4)

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TutorialTutorial

Do end chapter problems: 5.1, 5.3, 5.7, 5.8, 5.16, 5.18, 5.30, 5.33, 5.47, 5.49

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engineering economics - present worth analysis

Documents

air delivery

worth analysis

project lives

annual power

project lives

equivalent

net present

current piping