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Why do engineers care about finance?  Projects often require an investment of money up front.  Often receive money back in later years after project.

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Presentation on theme: "Why do engineers care about finance?  Projects often require an investment of money up front.  Often receive money back in later years after project."— Presentation transcript:

1 Why do engineers care about finance?  Projects often require an investment of money up front.  Often receive money back in later years after project completion.  Need to determine if project will be profitable.  Must compare monies spent today to monies received in the future.

2 $$ Time Value of Money $$  Why is "time value of money" important?  Money grows through compound interest.  A savings account with 5% interest paid annually.  Put $100 in account now, how much do you have in a year?  The money grows every year by 5%. One year 100 (1+ 0.05) = $105 2 years: $100 (1+ 0.05) (1+ 0.05) = 100 ( 1+ 0.05) 2 = $110.25 10 years: $100 ( 1+ 0.05) 10 = $162.90  Basic Formula:  FV=Future Value, PV=Present Value for a fixed interest rate in decimals (i) for n periods

3 What if I compounded more than annually (once a year)?  So we can compare, we’ll use the same numbers: A savings account starting with $100 with 5% interest PAID MONTHLY. Now how much do you have in a year? New formula: A=Future Value and P=Present Value This time we get $105.12, instead of just $105.00....

4 What if I compounded “continuously”?  We’ll use the same numbers:  A savings account starting with $100 with 5% interest compounded continuously. Now how much do you have in a year?  New formula using Euler’s constant, e ≈ 2.7128: Future Value (FV)=Present Value (PV) * e rt This time we get $105.13, instead of just $105.12 (compounded monthly) or $105.00 (annual interest rate). This is from only $100.00 initial investment. A ten million dollar investment yields a $13,000 difference between continuous compound interest and annual interest.

5 Handy formula for future value. where : N = annual amount received r = interest rate n = total no. of payment periods.  Important: This formula is applicable when cash amount N is the same every payment.

6 You borrow $500 from your dad and have to pay it back in 1 year.  How much would you pay per month if the saving account interest is 0.4% monthly?

7 If you won the $100M lottery, would you take 50% now OR even monthly payments worth the $100M for 25 years?  If the full payment is based on an annual 5% return over 25 years, what would you get each month?  That’s $52.5M at the end, $2.5M more than the $50M you’d get if you took the 50% offer now.

8 Would you rather have $100 today or $162.89 in 10 years?  Compound interest says they are the same !!  Assumes future cash flows have no risk - Is that always true?  $162.89 in 10 years is equal to $100 today  Assumes 5% interest rate for every year - Is that going to be true?  We say $162.89 in 10 years has a “present value” of $100  We also say $100 has a “future value” of $162.89 in 10 years.  Savings accounts are a way of making sure your money maintains its “present value”.

9 Annual Compound interest in Reverse...  $100 you get next year is worth $95.24 today.  The $162.50 received in 10 years has a value today of:  This is called “discounted cash flow”.  It gives a way to compare future money with present money.

10 Would you rather have $20 a year for 5 years OR receive what $25 is worth at 6% interest each year for 5 years?  Cash by 5 annual payments of $20 = $100  Cash back must be discounted, assume an annual interest rate of 6%. $25 in 1 year: $25/(1.06) = $23.58 $25 in 2 years: $25/(1.06) 2 = $22.25 $25 in 3 years: $25/(1.06) 3 = $20.99 $25 in 4 years: $25/(1.06) 4 = $19.80 $25 in 5 years: $25/(1.06) 5 = $18.68 Present Value of cash received= $105.30  This will generate $105.30 - $100.00 = $5.30 in today’s dollars.  Multiply by a million for an engineering project.

11 Discount everything to the present time. where N = annual amount received, r = interest rate, n = total no. of years  This formula is applicable when the cash amount N is the same every year.  For our example: N=$25, r=.06 n= 5:

12 Engineering Economics  Answers the question: Will this project be profitable? How long will it take to get to get my money back? How much money do I have at risk?  Requires assumptions about future costs and revenue.  Considers the “time value of money”.  Allows alternative implementation scenarios. Can I reduce risk by implementing the project slower (or faster)?  Allows a comparison of alternative projects to determine which is most profitable.

13 Tonight’s Homework (from ‘07 Final) Problem 1: Your company assigns you the task of evaluating the costs of obtaining and running a $19K company car for 10 years. Three options are possible: 1)Placing $500/month in a savings account at 5% annual interest, compounded monthly until you’ve accumulated $19K, then buying the car for cash 2)Buying the car with no money down and $500/month payments 3)Leasing a car, 39 months at a time, for $200/month plus a down payment of $4K.  What are the total costs for each option?  Why can’t these costs can’t all be legitimately compared? Problem 2: It’s never too early to begin planning for retirement, so you begin socking away $200/month in an account that pays 6% annually, compounded monthly. After some number of months you’ll have enough money in the account so that you can stop making payments and start drawing out $200/month—forever. How many months will it take?

14 Request for Proposal (RFP)  Teams of 3  Respond to the City Councils of two separate towns who are looking for a cost effective and dependable engineering solution to individual problems facing their communities.  Each team will deliver a 12-minute formal PowerPoint presentation to City Council Members which will include:  Detailed Graphic representing solution  Specific timeline of development, testing and implementation  Present and Future Cost Analysis including: 1) upfront monies required, 2) production costs (manpower, materials, etc.), 3) reoccurring costs 4) short and long term maintenance costs  Safety considerations

15 RFP Teams 2010 Team 1 Colin Jessica Jesus Team 2 Flaviu Eduardo Nick Team 3 Jim Jose Sam Team 4 Sami Joseph Kelly Team 5 Arjun Michelle Craig Team 6 Mike Ivan Ryan Team 7 Alejandra Niko Chris

16 What data do you need?  One time costs (or capital investments) Buildings, equipment, research etc.  Recurring costs (non-revenue producing) Salaries, Rent, utilities, insurance etc.  Production costs Materials, packaging, shipping etc.  Revenue projections How much can I sell it for? When will I get paid?


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