1 Engineering Economy: Eide (chapter 13) & Chase (pages 703-719) u Definition of terms u Concept of Equivalence in Engineering Economy u Simple and Compound.

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Engineering Economy: Eide (chapter 13) & Chase (pages 703-719)

Definition of terms Concept of Equivalence in Engineering Economy Simple and Compound Interest Calculation Engineering Economy Symbols (P, F. A, n, and i) Cash-flow and End-of-period conventions Constructing cash-flow diagrams

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Definitions from Chase, pp 703 - 719 Assets Fixed and Variable Costs: Sunk Costs: Economic Life & Obsolescence: Book Value: Investment cost - Depreciation Depreciation & Amortization:

– depreciation is allocation of costs to tangible assets (land,equipment, buildings)

– amortization is allocation of intangible assets (patents, licenses, franchises, goodwill)

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Definitions Fixed Costs - expense that remain constant

regardless of varying levels of production output e.g. rent, property tax, depreciation– no costs are truly fixed, WHY ?

Variable Costs - expense that vary (fluctuate) with changes in output levels e.g. materials, labor, energy usage

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Definitions continued. Sunk Costs - past expenses that have no salvage

value. They are therefore not considered in evaluating investment alternatives. They could be current but fixed costs e.g. rent on building.

Opportunity Costs - benefit or advantage forgone that results from choosing one alternative over another– IBM not getting into pc market sooner– Investment A has $25,000 profit and Investment B has

$20,000 profit, opportunity cost for choosing B is $5000.

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Definitions continued. Economic Life and Obsolescence

– Economic Life - for a piece of equipment, is the period over which it provides the best method for performing its task. When a superior method is developed, the machine becomes obsolete - thus book value becomes a meaningless figure in the accounting records.

– Book Value - used in accounting to reflect the value at a period after deducting depreciation from prior periods.

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Engineering Economy is a collection of mathematical techniques that

simplify economic comparisons. it provides a rational and systematic approach for

evaluating different economic decision E.G.

– purchase of a new manufacturing equipment– evaluating different manufacturing methods in terms of

economic value to the company– replacing existing manufacturing equipment or method

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Alternatives Alternatives are always present for any economic

decision Identifying appropriate alternatives is as important as

-if not more important than - evaluating the alternatives

Alternatives evaluation variables:– initial cost; Interest rate (rate of return)– anticipated life of equipment (useful/economic life)– annual maintenance or operating cost– resale or salvage value

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Basis of Alternative Evaluation $$$$ is generally the basis for comparison method with lowest overall cost is usually selected intangible factors (e.g. effect of process changes,

human factors, social and others) must be considered – Example: impact on employee morale; relevance to

strategic direction, etc.

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Time Value of Money change in the amount of money over time is called

time value of money Impact of Interest:

– Interest = total $ accumulated - original investment– OR =amount owed - original loan– % Int. Rate = [$ accrued/original amt] X 100%

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Example - Interest You borrowed $10,000 for 1 yr. at 15% interest.

Calculate (a) the interest and (b) total amount due after one year– a) Interest = $10,000(0.15) = $1,500– b) Total Due = $10,000 + $1,500 = $11,500– Total Amount Due = P(1 + interest rate)

What is :– Principal amount (P); Period (n)

Multiple periods - simple or compound interest considerations

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Equivalence basis for engineering economy analysis combination of time value of money and interest

rate generate the concept of equivalence it means that different sums of money at different

times can be equal in economic value

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Example - Equivalence You will be indifferent (i.e.. in economic terms) to a

gift of $100 today or a gift of $112 a year from now if interest rate is 12% per annum

Because: – Amount after 1 yr.. = 100 + (100x0.12) = $112

If interest rate were different, your decision will be different

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Equivalence continued Example: $5,000 loan that has to be paid back in a

five-year period at 15% interest per year– Plan 1: no int or principle recovered until year 5. Int

however accumulates each year and is added to principal– Plan 2: pay annual interest each year and principal is

recovered at end of year 5– Plan 3: accrued int. and 20% of principal is paid each yr– Plan 4: Equal payments are made annually with some

portion going towards principal recovery Which plan is better? ...

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Example: Relationship between time-value-of-money and

equivalence

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Simple and Compound Interest simple interest is calculated using the principal

only-i.e. ignoring any int. that was accrued in preceding interest periods

Simple int. = P x n x i E.G. $1000 borrowed for 3 yrs. at 14% per year

simple int.– Int. per yr. = 1000(.14) = $140– Total Int. for 3 yrs. = (1000)(3)(.14) = $420– Amt due in 3 yrs. =1000 + 420 = $1420

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Compound Interest in this case, the int. for a period is calculated on the

principal plus the total int accumulated in previous periods

$1000 @ 14% compound int. for 3 yrs.

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Symbols of Engineering Economy P = sum/value at time denoted as the present F = sum at some future time A = equal end-of-yr amounts of money n = # of int. periods - months, yrs., weeks, etc. i = int. rate per interest period - % per yr., % per

month, etc.

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Cash- Flow Diagrams receipts and disbursements over a span of time is

called cash flow - companies and individuals have these.

positive cash flows usually represent receipts negative cash flow represent disbursements it is assumed in engineering economy that cash flow

occurs at the end of the interest period. This is called the “end-of-period” convention .....

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Cash flows continued simplifying assumption for multiple receipts within

a period ?

present is zero time

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Example You make 5 deposits of $1000 per yr in a 17% per

annum account, how much money will be accumulated immediately after you made the last deposit

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In-Class Exercise A company invested $2500 in a new compressor 7

yrs ago, Annual income from the compressor was $750. During the first yr, $100 was spent on maintenance, this cost increased each yr. by $25. The company wants to sell the compressor for salvage at the end of next yr. for $150. Construct the cash flow diagram for the piece of equipment.

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Example: (1.20) If you buy a new television set in 1996 for $900,

maintain it for three yrs at a cost of $50 per yr, and then sell it for $200 at the end of the 3rd year, show a diagram of your cash flows and label each arrow as P, F, or A with its respective dollar value so that you can find the single amount in 1995 that would be equivalent to all of the cash flows shown. Assume an interest rate of 12% per yr.

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Derivation of Single-Payment Formulas: Given:

– P = an amt invested at t = 0– F1 = amt accumulated at period = 1– i = interest rate per period– then, at end of period 1

» F1 = P + Pi

» F1 = P(1+ i)

– at the end of period 2,» F2 = F1 + F1x i = P(1+ i) + P(1 + i) = P(1 + 2i + i2)

» F2 = P(1 + i)2 ......

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Derivation contd. at end of period 3.

– F3 = F2 + F2 x i = – = P(1+i)(1+i)2 = P (1 + i)3

In general,– F = P (1 + i)n

– P = F [ 1/(1+i)n ]– (1 + i)n = single payment compound amount

factor(SPCAF)– [ 1/(1+i)n ] = single payment present-worth factor

(SPPWF)

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Single Payment formulas continued. Note that single payment formulas are used to find

the present or future amount when payment or receipt is involved.

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Derivation of Uniform-series present-worth series

P = A

– uniform-series present-worth factor (USPWF) A = P

– capital-recovery factor (CRF)– A = uniform annual worth over n periods (e.g. years) of a

given investment P when interest rate is i

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Derivation contd. (pages 27 and 28)

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Derivation of uniform-series compound-amount factor and the Sinking-Fund factor

A =

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Standard Notation and Use of Int Table ( X/Y, i%, n )

– X - represents what you want– Y - represents what is given– i - is interest rate in percent– n - periods

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Using the Interest Tables If a woman deposits $600 now, $300 2 yrs. from

now and $400 5 yrs from now, How much will she have in her account in 10 yrs. if the interest rate is 5% per annum.

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Using the tables - interpolating determine the value of the A/P factor for an interest

rate of 7.3% and n of 10 yrs. - (A/P, 7.3%, 10)

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Calculating unknown periods (yrs) important in breakeven or pay-back period

economic analysis Example: How long will it take for $1000 invested

today to double in value if interest is 5% per annum