Engineering orientation
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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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Engineering orientation

Engineering Orientation

Engineering Economics


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

Value and Interest

  • “Value” is not synonymous with “amount”. The value of an amount of money depends on when the amount is received or spent.

  • First Cost is what you pay for an item when you buy it


Value and interest1

Value and Interest

  • The difference between the anticipated amount in the future and its current value is called interest.

  • At an interest rate of 10% what is the value now of the expectation of receiving $1 in one year?


Cost of money

Cost of Money

  • Interest that could be earned if the amount invested in a business or security was instead invested in government bonds or in time deposit.


Cost of money1

Cost of Money

  • Buy a car for $20,000 of your own cash vs. US bonds returning 5%/yr ($1,000 forever)

  • In effect you are paying $1,000 for ever (even after the car is a certifiable clunker destined for destruction)


Simple and compound interest

Simple and Compound Interest

  • You have a business project costing $100,000

    • You get a loan for 7.5% yearly for 5 years at simple interest payable at the end of the loan

      • The loan costs $7,500 for each of five years for a total interest of $37,500

  • Total cost over 5 years = $137,500


Simple and compound interest1

Simple and Compound Interest

  • Is the banker really willing to lend you money for 5 years? Isn’t he also lending you

    • $7,500 for four years,

    • $7,500 for three years,

    • $7,500 for two years,

    • $7,500 for one year


Terms and formulae

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)


Pay periods

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


Pay periods1

Pay periods


Study examples

Study Examples

  • Compute the effective annual interest rate ie equivalent to 8% nominal annual interest compounded continuously.

  • Calculate the PV


Example

Example

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


Cash flow diagrams

Cash Flow Diagrams


Cash flow patterns

Cash Flow Patterns


Example1

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.


Example2

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.


Example3

Example

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


Example 2

Example 2

  • Find the value in 2002 of a bond described as “Acme 8% of 2015” if the rate of return set by the market for similar bonds is 10%.


Example4

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%.


Return on investment

Return on Investment

  • ROI = The ratio of annual return to the cost of the investment

  • If an investment of $500,000 produces an income of $40,000 per year, its ROI = $40,000/$500,000 = 0.08 = 8%.

  • Many successful large companies operate with ROI’s of 15% or more


Return on investment1

Return on Investment


Engineering orientation

  • Unusual Cash Flows and Interest Periods


Payments at beginnings of years

PAYMENTS AT BEGINNINGS OF YEARS

  • Using a 10% interest rate, find the future equivalent of:


Cost of losing one semester

Cost of losing one semester

  • Two students, Frank and Mary start they Engineering Studies on the same date and they make the commitment of retiring thirty years after their forecasted graduation date (the date they would graduate if no delays are introduced). This date will not change if any delays make any of them graduate later.

  • Calculate the difference in their earnings if for some reasons Frank is required to graduate one semester later than what was intended.


The model

The Model

  • Assumptions:

    • We have a constant inflation

    • You have a yearly Salary Increase greater than what you loose because of inflation

    • Salary increases and inflation are constant

    • They work for the same company their entire carrier


Study examples1

Study Examples

  • Your perfectly reliable friend, Frank, asks for a loan and promises to pay back $150 two years from now. If the minimum interest rate you will accept is 8%, what is the maximum amount you will loan him?

    • a) $119 b) $126 c) $129 d) $139

  • The annual amount of a series of payments to be made at the end of each of the next twelve years is $500. What is the present worth of the payments at 8% interest compounded annually?

    • a) $500 b) $3,768 c) $6,000 d) $6,480


Study examples2

Study Examples

  • Maintenance expenditures for a structure with a twenty-year life will come as periodic outlays of $1,000 at the end of the fifth year, $2,000 at the end of the tenth year, and $3,500 at the end of the fifteenth year. With interest at 10%, what is the equivalent uniform annual cost of maintenance for the twenty-year period?

    • a) $200 b) $262 c) $300 d) $325

  • The purchase price of an instrument is $1 2,000 and its estimated maintenance costs are $500 for the first year, $1 500 for the second and $2500 for the third year. After three years of use the instrument is replaced; it has no salvage value. Compute the present equivalent cost of the instrument using 1 0% interest.

    • a) $14,070 b) $15,570 c) $15,730 d) $16,500


Study examples3

Study Examples

  • If $10,000 is borrowed now at 6% interest, how much will remain to be paid after a $3,000 payment is made four years from now?

    • a) $7,000 b) $9,400 c) $9,625 d) $9,725


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