Lecture 3

Lecture 3 - ISE460:Engineering Economy Fall2010 JosephChow...

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ISE 460: Engineering  Economy University of Southern California Fall 2010 Joseph Chow Week 2: Lecture 3
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Review Role of Engineering Economics Accounting Basics Time Value of Money Compound Interest Today…  Determining economic equivalence between  alternative cash flows/investments Thursday… Nominal and effective interest rates 10/12/10 2
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Clarification: End of Period  Convention 10/12/10 3 0 1 0 1 Is assumed  equivalent to… $20 $20 $30 $30 $100 = $20 + $30 + $20 + $30
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Clarification: Accounting Terms Working Capital = Current Assets – Current  Liabilities Current Assets = Cash + Cash Equivalents +  Accounts Receivable + Inventory Current Liabilities = Accounts Payable + Notes  Payable + Accrued Expenses Taxable/Operating Income = Net Revenue – Cost  of Revenue – Operating Expenses (which includes  Depreciation and Amortization) Net Income = Taxable Income – Income Taxes Net Cash Flow = Net Income + Depreciation (if  any) + Amortization (if any) 10/12/10 4
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Recall: “ All that counts are the  differences among alternatives 10/12/10 5
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Economic  Equivalence From PARK: “the process of comparing  two different cash amounts at different  points in time”    extends to comparing  alternatives Given the following for each of  k   alternatives… Magnitude of payment Receipt vs disbursement Timing of payment Interest rate in effect Which alternative should you choose? 10/12/10 6
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Economic  Equivalence  of  Alternatives $20,000 today and $50,000 ten years from  now, or $8,000 each year for next ten years? Which alternative should you choose?  10/12/10 7
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Four Principles for Determining  Economic Equivalence 10/12/10 8
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Principle 1: Common Base  Period Ex: How do you use compound interest formula to  compare the following alternatives: Deposit $1000 in a savings account at 2% for 5 year  compounded annually (ALT 1 (base)) Invest in a friend who promises to return $1100 at the  end of the 5 th  year (ALT 2) 10/12/10 9 0 5 $1000 1 2 3 4 0 5 $1100 1 2 3 4 0 5 $1000 1 2 3 4 0 5 $1100 1 2 3 4 ALT  1 ALT  2 ALT  1 ALT  2
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Different Period $1000 at Year 3, 5% annual interest rate What is it worth at Year 5? F 5  = P 3  (1 +i) N  = 1000 (1 + 0.05) 5-3   =  $1102.50 What is it worth at Year 0? F = P (1 +i)
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Lecture 3 - ISE460:Engineering Economy Fall2010 JosephChow...

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