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L15: Present Worth Analysis

L15: Present Worth Analysis. ECON 320 Engineering Economics Mahmut Ali GOKCE Industrial Systems Engineering Computer Sciences. PW( i ) > 0. Net Present Worth Measure. Principle : Compute the equivalent net surplus at n = 0 for a given interest rate of i .

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L15: Present Worth Analysis

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  1. L15: Present Worth Analysis ECON 320 Engineering Economics Mahmut Ali GOKCE Industrial Systems Engineering Computer Sciences

  2. PW(i) > 0 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. Inflow 0 1 2 3 4 5 Outflow Net surplus PW(i) inflow 0 PW(i) outflow

  3. Example 5.3 - Tiger Machine Tool Company inflow $55,760 $27,340 $24,400 0 3 1 2 outflow $75,000

  4. Excel Solution =NPV(15%,B3:B5)+B2

  5. Present Worth Amounts at Varying Interest Rates *Break even interest rate

  6. Present Worth Profile 40 Reject Accept 30 20 Break even interest rate (or rate of return) 10 $3553 17.45% 0 PW (i) ($ thousands) -10 -20 -30 0 5 10 15 20 25 30 35 40 i = MARR (%)

  7. Future Worth Criterion • Given: Cash flows and MARR (i) • Find: The net equivalent worth at the end of project life $55,760 $27,340 $24,400 0 3 1 2 $75,000 Project life

  8. Future Worth Criterion

  9. Excel Solution =FV(15%,3,0,-B6)

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

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

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

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

  14. Project Balance Concept N 0 1 2 3 Beginning Balance Interest Payment Project Balance -$75,000 -$11,250 +$24,400 -$61,850 -$61,850 -$9,278 +$27,340 -$43,788 -$43,788 -$6,568 +$55,760 +$5,404 -$75,000 -$75,000 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

  15. Project Balance Diagram 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -80,000 -100,000 -120,000 Terminal project balance (net future worth, or project surplus) $5,404 Discounted payback period Project balance ($) -$43,788 -$61,850 -$75,000 0 1 2 3 Year(n)

  16. 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.

  17. Meaning of Net Present Worth N = 3 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 Investment pool $75,000 Return to investment pool $55,760 $27,340 $24,400 Project 0 1 2 3 PW(15%) = $5,404(P/F,15%,3) = $3,553

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

  19. Guideline for Selecting a MARR U.S. Treasury Bills Risk-free real return Very safe Risk premium Inflation Very risky Amazon.com

  20. 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)

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

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

  23. 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%

  24. Years 15 30 45 60 0 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000

  25. Solution: • Construction Cost • P1 = $2,000,000 • Maintenance Costs • P2 = $50,000/0.05 = $1,000,000 • Renovation Costs • P3 = $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 = P1 + P2 + P3 = $3,463,423

  26. 30 45 60 0 15 Alternate way to calculate P3 • Concept: Find the effective interest rate per payment period $500,000 $500,000 $500,000 $500,000 • Effective interest rate for a 15-year cycle • i = (1 + 0.05)15 - 1 = 107.893% • Capitalized equivalent worth • P3 = $500,000/1.07893 • = $463,423

  27. 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.

  28. Solution

  29. 0 1 2 3 4 5 6 7 8 9 10 $1,000 $4,211

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