engineering economics. john ayers september 17, 2004

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Engineering Economics . John Ayers September 17, 2004

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Page 1: Engineering Economics. John Ayers September 17, 2004

Engineering Economics

.

John AyersSeptember 17, 2004

Page 2: Engineering Economics. John Ayers September 17, 2004

Engineering Economics

• Why is it important?• Value and Interest• Cash Flow Diagrams and Patterns• Equivalence of Cash Flow Patterns• Evaluating Alternatives• Break-Even Analysis• Income Tax and Depreciation• Inflation• Conclusion

Page 3: Engineering Economics. John Ayers September 17, 2004

Why do we care about Engineering Economics?

• Engineering designs are intended to produce good results.

• They are accompanied by undesirables (costs).• If outcomes are evaluated in dollars, and “good” is

defined as profit, then decisions will be guided by engineering economics.

• This process maximizes goodness only if all outcomes are anticipated and can be monetized. $

Page 4: Engineering Economics. John Ayers September 17, 2004

Value and Interest

• The “value” of money depends on the amount and when it is received or spent.

1 2

$1000

$1166

Example: What amount must be paid to settle a current debt of $1000 in two years at an interest rate of 8% ?

Solution: $1000 (1 + 0.08) (1 + 0.08) = $1166

Page 5: Engineering Economics. John Ayers September 17, 2004

Cash Flow Diagrams

P-Pattern1 2 3 n

“present”

F-Pattern1 2 3 n

“future”

A-Pattern1 2 3 n

“annual”

G-Pattern1 2 3 n

“gradient”

Page 6: Engineering Economics. John Ayers September 17, 2004

Equivalence of Cash Flow Patterns

To Find Given Multiply By Formula

F P

P F

A P

A G

niPF )/(

niFP )/(

niPA )/(

niGA )/(

ni)1(

ni)1(

1

1)1(

)1(

n

n

i

ii

1)1(

1

ni

ni

Page 7: Engineering Economics. John Ayers September 17, 2004

Example: A new circuit board component insertion tool will save $50,000 in production costs each year and will have a life of seven years. What is the highest price that can be justified for the tool using a 12% interest rate?

Solution:1 2 3 74 5 6

P

50k 50k 50k 50k 50k 50k 50k

k

Aii

iAAPP n

n

228$000,50$56.4

000,50$)12.01(12.0

1)12.01(

)1(

1)1()/(

7

7

%127

Page 8: Engineering Economics. John Ayers September 17, 2004

Evaluating Alternatives

• Annual Equivalent Cost Comparisons

• Present Equivalent Cost Comparisons

• Incremental Approach

• Rate of Return Comparisons

• Benefit/Cost Comparisons

Minimum Attractive Rate of Return (MARR): The lowest rate of return that the organization will accept.

Page 9: Engineering Economics. John Ayers September 17, 2004

Annual Equivalent Cost Comparison

• Incomes are converted to an A-pattern.• Costs are converted to an A-pattern.• The costs are subtracted from the incomes to

determine the ANEV.• Mutually Exclusive Alternatives – choose the one

with highest ANEV• Independent Alternatives – choose all with

positive ANEV

ANEV: Annual Net Equivalent Value

Page 10: Engineering Economics. John Ayers September 17, 2004

Example: A new circuit board component insertion tool is needed. Which should you buy?

Solution: The ANEV is calculated for each:

kkkk

kFAkkPAANEV

9.539.1208.35

30)/(20220)/( %1010

%1010

Model Price Annual Maintenance Salvage Value Life

JACO $220k $20k $30k 10 years

Cheepo $100k $35k 0 5 years

JACO:

k

kkPAANEV

4.61$

35100)/( %105

Cheepo:

JACO

Page 11: Engineering Economics. John Ayers September 17, 2004

Present Equivalent Cost Comparison

• Incomes and costs are converted to P-patterns.• The costs are subtracted from the incomes to

determine the PNEV.• Mutually Exclusive Alternatives – choose the one

with highest PNEV• Independent Alternatives – choose all with

positive PNEV

PNEV: Present Net Equivalent Value, also called “life cycle cost,” “present worth,” “capital cost,” and “venture worth.”

Page 12: Engineering Economics. John Ayers September 17, 2004

Incremental Approach

• For a set of mutually exclusive alternatives, only the differences in amounts need to be considered.

kkkkk

kFPkFPkAPkPNEV

9.45$6.111.622.92120

30)/(100)/(15)/(120 %1010

%105

%1010

Model Price Annual Maintenance Salvage Value Life

JACO $220k $20k $30k 10 years

Cheepo $100k $35k 0 5 years

JACO- Cheepo:

JACO

Page 13: Engineering Economics. John Ayers September 17, 2004

Rate of Return Method

• ANEV or PNEV is formulated• From this, we solve for the interest rate that will

give zero ANEV or PNEV• This interest rate is the ROR of the alternative• For mutually exclusive alternatives, the one with

the highest ROR is chosen• For independent alternatives, all with a ROR

greater than MARR are accepted

ROR: Rate of Return on Investment

Page 14: Engineering Economics. John Ayers September 17, 2004

Benefit/Cost Comparisons

• The benefit/cost ratio is determined from

• For mutually exclusive alternatives, the one with the highest B/C is chosen.

• For independent alternatives, all with B/C > 1 are accepted.

The MARR is used to determine the numerator (benefits).

cost initial of equivalent annualbenefits annualnet uniform

CB

Page 15: Engineering Economics. John Ayers September 17, 2004

Break-Even Analysis

• Break-even point: the value of an independent variable such that two alternatives are equally attractive.

• For values above the break-even point, one alternative is preferred.

• For values below the break-even point, the other is preferred.

• Break-even analysis is useful when dealing with a changing variable (such as MARR).

Page 16: Engineering Economics. John Ayers September 17, 2004

Income Tax and Depreciation

• Businesses pay the IRS a tax:

• Depreciation: method of charging the initial cost of an asset against more than one year.

• An asset is depreciable if :– It is used to produce income,– Has a life greater than one year, but– Decays, wears out, becomes obsolete, or gets used up.

ondepreciati-paidinterest -

costs operating-revenue grossRTAX

ACRS: Accelerated Cost Recovery System, used by IRS since 1980.

Page 17: Engineering Economics. John Ayers September 17, 2004

Inflation

• The buying power of money changes with time.• Inflation, if anticipated, can be put to good use by

fixing costs and allowing income to rise by– Entering long-term contracts for materials or wages

– Purchasing materials long before they are needed

– Stockpiling product for sale later.

Page 18: Engineering Economics. John Ayers September 17, 2004

Conclusion

• For-profit enterprises exist to make money.

• Non-profit entities also make decisions to maximize the goodness of outcomes by assigning dollar values.

• Your engineering decisions will be shaped by economics.

$