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Present Worth Analysis Lecture No.15 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005

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Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. Decision Rule: Accept the project if the net surplus is positive. 2 3 4 5 0 1 Inflow Outflow 0 PW(i) inflow PW(i) outflow Net surplus PW(i) > 0

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Example 5.3 - Tiger Machine Tool Company $75,000 $24,400 $27,340 $55,760 0 12 3 outflow inflow

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ABC 1PeriodCash Flow 20($75,000) 31$24,400 42$27,340 53$55,760 6 7PW(15%)$3,553.46 =NPV(15%,B3:B5)+B2 Excel Solution

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Present Worth Amounts at Varying Interest Rates i (%)PW(i)i(%)PW(i) 0$32,50020-$3,412 227,74322-5,924 423,30924-8,296 619,16926-10,539 815,29628-12,662 1011,67030-14,673 128,27032-16,580 145,07734-18,360 162,07636-20,110 17.45*038-21,745 18-75140-23,302 *Break even interest rate

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-30 -20 -10 0 10 20 30 40 0 510152025303540 PW (i) ($ thousands) i = MARR (%) $3553 17.45% Break even interest rate (or rate of return) Accept Reject Present Worth Profile

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Future Worth Criterion Given: Cash flows and MARR (i) Find: The net equivalent worth at the end of project life $75,000 $24,400 $27,340 $55,760 0 12 3 Project life

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Future Worth Criterion

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ABC 1PeriodCash Flow 20($75,000) 31$24,400 42$27,340 53$55,760 6PW(15%)$3553.46 7FW(15%)$5,404.38 =FV(15%,3,0,-B6) Excel Solution

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How To Use Cash Flow Analyzer Cash Flow Input Fields Output or Analysis results Graphical Plots

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Solving Example 5.3 with Cash Flow Analyzer Net Present Worth Net Future Worth Payback period Project Cash Flows

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Obtaining a NPW Plot Specify the Range of Interest Rate to plot NPW plot Between 0% and 100%

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Can you explain what $3,553 really means? 1. Project Balance Concept 2. Investment Pool Concept

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Project Balance Concept N0123N0123N0123N0123 Beginning Balance Interest Payment Project Balance -$75,000 -$11,250 +$24,400 -$61,850 -$9,278 +$27,340 -$43,788 -$6,568 +$55,760 +$5,404 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

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Project Balance Diagram 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -80,000 -100,000 -120,000 01230123 -$75,000 -$61,850 -$43,788 $5,404 Year(n) Terminal project balance (net future worth, or project surplus) Discounted payback period Project balance ($)

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Investment Pool Concept Suppose the company has $75,000. It has two options. (1)Take the money out and invest it in the project or (2) leave the money in the company. Let’s see what the consequences are for each option.

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$75,000 0 1 2 3 $24,400 $27,340 $55,760 Investment pool How much would you have if the Investment is made? $24,400(F/P,15%,2) = $32,269 $27,340(F/P,15%,1) = $31,441 $55,760(F/P,15%,0) = $55,760 $119,470 How much would you have if the investment was not made? $75,000(F/P,15%,3) = $114,066 What is the net gain from the investment? $119,470 - $114,066 = $5,404 Project Return to investment pool N = 3 Meaning of Net Present Worth PW(15%) = $5,404(P/F,15%,3) = $3,553

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What Factors Should the Company Consider in Selecting a MARR in Project Evaluation? Risk-free return Inflation factor Risk premiums

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Guideline for Selecting a MARR Real Return2% Inflation4% Risk premium0% Total expected return 6% Real Return2% Inflation4% Risk premium20 % Total expected return 26 % Risk-free real return Inflation Risk premium U.S. Treasury Bills Amazon.com Very safe Very risky

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Capitalized Equivalent Worth Principle: PW for a project with an annual receipt of A over infinite service life Equation: CE(i) = A(P/A, i, ) = A/i A 0 P = CE(i)

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Practice Problem 10 $1,000 $2,000 P = CE (10%) = ? 0 Given: i = 10%, N = ∞ Find: P or CE (10%) ∞

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Solution 10 $1,000 $2,000 P = CE (10%) = ? 0 ∞

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A Bridge Construction Project Construction cost = $2,000,000 Annual Maintenance cost = $50,000 Renovation cost = $500,000 every 15 years Planning horizon = infinite period Interest rate = 5%

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$500,000 $2,000,000 $50,000 0 15 30 4560 Years

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Solution: Construction Cost P 1 = $2,000,000 Maintenance Costs P 2 = $50,000/0.05 = $1,000,000 Renovation Costs P 3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60). = {$500,000(A/F, 5%, 15)}/0.05 = $463,423 Total Present Worth P = P 1 + P 2 + P 3 = $3,463,423

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Alternate way to calculate P 3 Concept: Find the effective interest rate per payment period Effective interest rate for a 15-year cycle i = (1 + 0.05) 15 - 1 = 107.893% Capitalized equivalent worth P 3 = $500,000/1.07893 = $463,423 15 304560 0 $500,000

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Practice Problem Want to purchase an electrical motor rated at 15HP for $1,000. The service life of the motor is known to be 10 years with negligible salvage value. Its full load efficiency is 85%. The cost of energy is $0.08 per kwh. The intended use of the motor is 4,000 hours per year. Find the total cost of owning and operating the motor at 10% interest.

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Solution

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0 1 2 3 4 5 6 7 8 9 10 $4,211 $1,000

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