Engineering Economics Exploring Engineering

2 Engineering economics  How much will an engineering project cost?  Simple and compound interest  Cost of borrowing money for an engineering project  Mathematical and Excel formulae  Breakeven point  Return on Investment

3 Engineering Economics  “Money makes the world go around …”  6u7Q 6u7Q  True in engineering too!  “Cost of Money”: Interest that could be earned if the amount invested in a business or security was instead invested in government or in time deposit.  In other words, the business investment vs. a guaranteed return

4 Cost of money  Buy a car for $20,000 of your own cash vs. US bonds returning 5%/yr ($600 … forever)  In effect you are paying $1,000 for ever (even after the car is a certifiable clunker destined for destruction)  Likewise, engineering economics looks beyond the first cost and adds the interest you have to pay to get the money to invest

5 Simple and Compound Interest  You have a business project costing $100,000  You get a loan for 7.5% 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  Is the banker really willing to lend you money for 5 years? Isn’t he also lending you $100, % of $7,500 for four years, $15,000 for three years, $22,500 for two years, $30,000 for four years and $37,500 for five years?

6 Simple and Compound Interest  Guess what? The banker does think you owe him interest on the interest (known as compound interest)  He will charge you about $375 after year 1, $750 after year 2. $1,125 after year 3, $1,500 after year 4 and $1,875 after year 5  The cost of the 5 year project is thus about $142,125  Compound interest can be a significant part of an engineering project

7 Terms and formulae  Principal P is the amount borrowed  # of pay periods, N  Interest rate r per period  Future worth, F, total of how much you have to payback  Formulae:  Simple interest = P(1 + Nr) ( = $137,500)  Compound interest = P(1 + r)N ( = $143,563)

8 SI and CI formulae

9 Pay periods  Suppose your load is compounded quarterly, monthly or daily instead of yearly.  Student loan of $25,000 at 8% for 1) annually for two years, 2) quarterly and 3) daily  1) r = 0.08 per yr, N = 2 and  F = $29,160  2) r = 0.02 per qtr N = 8 and  F = $29,291  r = 2.19 x per day, N = 730 and  F = $29,337  Morale: Watch the effect of increased compounding!

10 Excel rides to the rescue …

11 Example  A nuclear reactor has cost $5 Billion when test trials start that take an additional 4 years to complete. If interest rates are 12% annually (payable quarterly), what’s the final reactor cost?

12 Example in Excel

13 Example  The $5B reactor ends up costing a cool $8 B  Nuclear reactors are only economical if they are built during times of low cost of money!

14 When Is An Investment Worth It?  ‘Break Even Point’ (BEP) has a simple definition:  BEP occurs when the project has earned back the cost it took to make it.

15 Example  Cost of producing new widget is $1,000,000. If profit per widget is $1.00 and we’re selling 1,000/day when is the BEP?.  Need: BEP = _____ years  Know - How: Equate cost to total money stream.  Solve: 1,000 [widgets/day]  1.00 [$/widget]  D [days] = $1,000,000. Solving for D gives:  D = 1,000 days = 2.74 years.  Most companies require BEP of months to fund a new widget

16 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

17 Return on Investment

18 Summary  Manufacturing businesses add to their costs the cost of borrowing  Compound interest is the only way money is lent  More payment periods is a more expensive loan  Breakeven Point and Return on Investment are the principal business criteria for a successful investment  BEP needs to be about months and ROI needs to be about 15% for a sturdy investment.