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

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.

$

Value and Interest

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

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 $1000 1 2 $1166

Cash Flow Diagrams

P-Pattern 1 2 3 F-Pattern 1 2 3 A-Pattern 1 2 3 G-Pattern 1 2 3 n n n n “present” “future” “annual” “gradient”

Equivalence of Cash Flow Patterns

To Find Given Multiply By Formula F P (

F

/

P

)

i n

P A A F P G (

P

/

F

)

i n

(

A

/

P

)

i n

(

A

/

G

)

i n

( 1 

i

)

n

1 ( 1 

i

)

n i

( 1 

i

)

n

( 1 

i

)

n

 1

i

1 

n

( 1 

i

)

n

 1

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?

50k 50k 50k 50k 50k 50k 50k Solution: 1 2 3 4 5 6 7 P

P

 (

P

/

A

) 12 7 %

A

 ( 1 

i i

( 1  )

n

 1

A i

)

n

 ( 1  0 .

12 ) 7  1 0 .

12 ( 1  0 .

12 ) 7 $ 50 , 000  4 .

56  $ 50 , 000  $ 228

k

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.

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

Example: A new circuit board component insertion tool is needed. Which should you buy? Model Price Annual Maintenance Salvage Value Life JACO Cheepo $220k $100k $20k $35k $30k 0 10 years 5 years Solution: The ANEV is calculated for each: JACO:

ANEV

   35 .

8

k

 ( 

A

/ 20

k P

) 10 10 %  1 .

9

k

220

k

   53 20 .

9

k k

 (

A

/

F

) 10 10 % 30

k

Cheepo:

ANEV

   $ 61 .

4

k

 (

A

/

P

) 10 5 % 100

k

 35

k

JACO

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.”

Incremental Approach

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

Model JACO Cheepo Price $220k $100k Annual Maintenance $20k $35k Salvage Value $30k 0 Life 10 years 5 years JACO- Cheepo: 

PNEV

 120

k

   120

k

92 .

2

k

  (

P

62 .

/ 1

A

) 10 10 % 15

k k

 11 .

6

k

  (

P

$ / 45

F

) 10 5 % .

9

k

100

k

 (

P

/

F

) 10 10 % 30

k

JACO

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

Benefit/Cost Comparisons

• The benefit/cost ratio is determined from

B C

 uniform net annual benefits annual equivalent of initial cost • 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).

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).

Income Tax and Depreciation

• Businesses pay the IRS a tax:

TAX



R

   gross revenue interest paid operating costs depreciati on   • 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.

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

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.

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.

$