time value analysis

8/4/2019 Time Value Analysis

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February 17, 2009

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Engineering Economics


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Engineering Economy

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It deals with the concepts and techniques of analysisuseful in evaluating the worth of systems, products,and services in relation to their costs


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Engineering Economy

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It is used to answer many different questions

Which engineering projects are worthwhile? Has the mining or petroleum engineer shown that the mineral or

oil deposits is worth developing?

Which engineering projects should have a higherpriority? Has the industrial engineer shown which factory improvement

projects should be funded with the available dollars?

How should the engineering project be designed? Has civil or mechanical engineer chosen the best thickness for

insulation?


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Basic Concepts

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Cash flow

Interest Rate and Time value of money Equivalence technique


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Cash Flow

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Engineering projects generally have economic

consequences that occur over an extendedperiod of time For example, if an expensive piece of machinery is

installed in a plant were brought on credit, the simple

process of paying for it may take several years The resulting favorable consequences may last as long

as the equipment performs its useful function

Each project is described as cash receipts or

disbursements (expenses) at different points intime


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Categories of Cash Flows

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The expenses and receipts due to

engineering projects usually fall into one ofthe following categories: First cost: expense to build or to buy and install

Operations and maintenance (O&M): annualexpense, such as electricity, labor, and minorrepairs

Salvage value: receipt at project termination for saleor transfer of the equipment (can be a salvage cost)

Revenues: annual receipts due to sale of products

or services Overhaul: major capital expenditure that occurs

during the assets life


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Cash Flow diagrams

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The costs and benefits of engineering projectsover time are summarized on a cash flowdiagram (CFD). Specifically, CFD illustratesthe size, sign, and timing of individual cashflows, and forms the basis for engineering

economic analysis A CFD is created by first drawing a

segmented time-based horizontal line, dividedinto appropriate time unit. Each time when

there is a cash flow, a vertical arrow is added pointing down for costs and up for revenuesor benefits. The cost flows are drawn torelative scale


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Drawing a Cash Flow Diagram

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In a cash flow diagram (CFD) the end of period t isthe same as the beginning of period (t+1) Beginning of period cash flows are: rent, lease, and

insurance payments End-of-period cash flows are: O&M, salvages,

revenues, overhauls The choice of time 0 is arbitrary. It can be when a

project is analyzed, when funding is approved, orwhen construction begins

One persons cash outflow (represented as a negative

value) is another persons inflow (represented as apositive value) It is better to show two or more cash flows occurring

in the same year individually so that there is a clearconnection from the problem statement to each cashflow in the diagram


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An Example of Cash Flow Diagram

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A man borrowed $1,000 from a bank at 8%

interest. Two end-of-year payments: at theend of the first year, he will repay half of the$1000 principal plus the interest that is due. Atthe end of the second year, he will repay the

remaining half plus the interest for the secondyear.

Cash flow for this problem is:End of year Cash flow

0 +$1000

1 -$580 (-$500 - $80)

2 -$540 (-$500 - $40)


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Cash Flow Diagram

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$1,000

0

1 2

$580 $540


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Time Value of Money

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Money has value

Money can be leased or rented

The payment is called interest

If you put $100 in a bank at 9% interest for one timeperiod you will receive back your original $100 plus $9

Original amount to be returned = $100

Interest to be returned = $100 x .09 = $9


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Compound Interest

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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


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Future Value of a Loan WithCompound Interest

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Amount of money due at the end of a loan F = P(1+i)1(1+i)2..(1+i)n or F = P (1 + i)

n

Where, F = future value and P = present value

Referring to slide #10, i = 9%, P = $100 and say n= 2.Determine the value of F.

F = $100 (1 + .09)2 = $118.81


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Notation forCalculating a Future Value

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Formula:

F=P(1+i)n is the

single payment compound amount factor. Functional notation:

F=P(F/P,i,n) F=5000(F/P,6%,10)

F =P(F/P) which is dimensionally correct.


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Notation forCalculating a Present Value

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P=F(1/(1+i))n=F(1+i)-n is the

single payment present worth factor.

Functional notation:

P=F(P/F,i,n) P=5000(P/F,6%,10)Interpretation of (P/F, i, n): a present sum P, given a

future sum, F, n interest periods hence at aninterest rate i per interest period


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Equivalence

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Relative attractiveness of different alternatives can

be judged by using the technique of equivalence We use comparable equivalent values of

alternatives to judge the relative attractiveness of thegiven alternatives

Equivalence is dependent on interest rate Compound Interest formulas can be used to

facilitate equivalence computations


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Technique of Equivalence

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Determine a single equivalent value at a point intime for plan 1.

Determine a single equivalent value at a point intime for plan 2.

Both at the same interest rate and at the same time point.

Judge the relative attractiveness of thetwo alternatives from the comparable

equivalent values.


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Engineering Economic Analysis Calculation

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Generally involves compound interest formulas(factors)

Compound interest formulas (factors) can beevaluated by using one of the three methods

Interest factor tables Calculator

Spreadsheet


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Given the choice of these two plans whichwould you choose?

Year Plan 1 Plan 2

0 $5,000

1 $1,000

2 $1,0003 $1,000

4 $1,000

5 $1,000Total $5,000 $5,000

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To make a choice the cash flows must be altered

so a comparison may be made.

eso v ng as ows to qu va ent resent


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eso v ng as ows to qu va ent resentValues

P = $1,000(PA,10%,5)

P = $1,000(3.791) =$3,791

P = $5,000

Alternative 2 is better than

alternative 1 sincealternative 2 has a greaterpresent value

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A E l f F V l


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An Example of Future Value

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Example: If $500 were deposited in a banksavings account, how much would be in theaccount three years hence if the bank paid 6%interest compounded annually?

Given P = 500, i = 6%, n = 3, use F =

FV(6%,3,,500,0) = -595.91 Note that the spreadsheet gives a negative

number to find equivalent of P. If we find P usingF = -$595.91, we get P = 500.

A E l f P V l


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An Example of Present Value

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Example 3-5: If you wished to have $800 in asavings account at the end of four years, and 5%interest we paid annually, how much should youput into the savings account?

n = 4, F = $800, i = 5%, P = ?

P = PV(5%,4,,800,0) = -$658.16 You should use P = $658.16


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Economic Analysis Methods

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Three commonly used economic analysis methods

are Present Worth Analysis

Annual Worth Analysis

Rate of Return Analysis


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Present Worth Analysis

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Steps to do present worth analysis for a single

alternative (investment) Select a desired value of the return on investment (i)

Using the compound interest formulas bring all benefitsand costs to present worth

Select the alternative if its net present worth (Presentworth of benefits Present worth of costs) 0


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Steps to do present worth analysis for selecting a

single alternative (investment) from among multiplealternatives

Step 1: Select a desired value of the return oninvestment (i)

Step 2: Using the compound interest formulas bring allbenefits and costs to present worth for each alternative

Step 3: Select the alternative with the largest netpresent worth (Present worth of benefits Presentworth of costs)


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A construction enterprise is investigating the

purchase of a new dump truck. Interest rate is9%. The cash flow for the dump truck are asfollows:

First cost = $50,000, annual operating cost =

$2000, annual income = $9,000, salvage value is$10,000, life = 10 years. Is this investment worthundertaking?

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


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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 donot invest

What should be the minimum annual benefit formaking it a worthy of investment at 9% rate ofreturn?


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Present worth of benefits = A(PA,9%,10) = A(6.418)

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

A(6.418) = $58,612 A = $58,612/6.418 =

$9,312.44


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Cost and Benefit Estimates

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Present and future benefits (income) and costs need

to be estimated to determine the attractiveness(worthiness) of a new product investment alternative


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Annual costs and Income for a Product

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Annual product total cost is the sum of annual

material, labor, and overhead (salaries, taxes,marketing expenses, office costs, and related costs),annual operating costs (power, maintenance,repairs, space costs, and related expenses), andannual first cost minus the annual salvage value.

Annual income generated through the sales of aproduct = number of units sold annuallyxunit price


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Single alternative case

In this method all revenues and costs of thealternative are reduced to a single percentagenumber

This percentage number can be compared to other

investment returns and interest rates inside andoutside the organization


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Steps to determine rate of return for a single stand-

alone investment Step 1: Take the dollar amounts to the same point in

time using the compound interest formulas

Step 2: Equate the sum of the revenues to the sum ofthe costs at that point in time and solve for i

R t f R t A l i


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An initial investment of $500 is being considered.

The revenues from this investment are $300 at theend of the first year, $300 at the end of the second,and $200 at the end of the third. If the desired returnon investment is 15%, is the project acceptable?

In this example we will take benefits and costs to thepresent time and their present values are thenequated

R t f R t A l i


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$500 = $300(PF, i, n=1) + 300(PF, i, n=2) +$200(PF, i, n=3)

Now solve for i using trial and error method

Try 10%: $500 = ? $272 + $247 + $156 = $669(not equal)

Try 20%: $500 = ? $250 + $208 + $116 = $574(not equal)

Try 30%: $500 = ? $231 + $178 + $91 = $500(equal) i = 30%

The desired return on investment is 15%, theproject returns 30%, so it should be implemented

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