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Engineering Orientation

Engineering Orientation. Engineering Economics. Engineering Economics. Value and Interest Cost of Money Simple and Compound Interest Cash Flow Diagrams Cash Flow Patterns Equivalence of Cash Flow Patterns. Value and Interest.

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Engineering Orientation

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  1. Engineering Orientation Engineering Economics

  2. Engineering Economics • Value and Interest • Cost of Money • Simple and Compound Interest • Cash Flow Diagrams • Cash Flow Patterns • Equivalence of Cash Flow Patterns

  3. Value and Interest • “Value” is not synonymous with “amount”. For a fixed amount of money, the value may change over time. • The difference between the anticipated amount in the future and its current value is called interest. • At an annual interest rate of 10% what is the value now of the expectation of receiving $1 in one year?

  4. Terms and formulae • P or PV Principal is the amount borrowed • N # of pay periods • r Interest rateper period • F or FV, Future worth, value in the future of what you have to payback • Formulae: • Simple interest = P(1 + Nr) ( = $137,500) • Compound interest = P(1 + r)N ( = $143,563)

  5. Pay periods • Calculate FV • Assume your loan is compounded quarterly, monthly or daily instead of yearly. • Student loan of $25,000 at 8% for • Annually for two years, • Quarterly for two years and • Daily for two years

  6. Pay periods

  7. Study Examples • Compute the effective annual interest rate ie equivalent to 8% nominal annual interest compounded continuously. • Calculate the FV • In the limit

  8. Example • What amount must be paid in two years to settle a current debt of $1,000 if the interest rate is 6% Annually?

  9. Cash Flow Diagrams

  10. Cash Flow Patterns

  11. Example • A new widget twister, with a life of six years, would save $2,000 in production costs each year. Using a 12% interest rate, determine the highest price that could be justified for the machine. Although the savings occur continuously throughout each year, follow the usual practice of lumping all amounts at the ends of years.

  12. Example • A new widget twister, with a life of six years, would save $2,000 in production costs each year. Using a 12% interest rate, determine the highest price that could be justified for the machine. Although the savings occur continuously throughout each year, follow the usual practice of lumping all amounts at the ends of years.

  13. Example • How soon does money double if it is invested at 8% interest?

  14. Example • Compute the annual equivalent maintenance costs over a 5-year life of a laser printer that is warranted for two years and has estimated maintenance costs of $100 annually. Use i = 10%.

  15. Engineering Economics • Value and Interest • Cost of Money • Simple and Compound Interest • Cash Flow Diagrams • Cash Flow Patterns • Equivalence of Cash Flow Patterns

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