1 / 24

ENGR 112

ENGR 112. Economic Analysis II. Engineering Economic Analysis. Time Value of Money $1 today is more valuable than $1 a year later Engineering economy adjusts for the time value of money to balance current and future revenues and cost. 10. 0. 5. Years. Recap. Capital vs. Interest

gelsey
Download Presentation

ENGR 112

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ENGR 112 Economic Analysis II

  2. Engineering Economic Analysis • Time Value of Money • $1 today is more valuable than $1 a year later • Engineering economy adjusts for the time value of money to balance current and future revenues and cost 10 0 5 Years

  3. Recap • Capital vs. Interest • Types of Interest • Simple vs. Compound • Cash Flow Diagrams • Calculate a Future given a Principal and vice versa • P/F and F/P  F = P(1 + i)N • Frequency of Compounding • Quarterly, monthly, daily • n = m x N  F = P(1 + i/m)m*N

  4. Annuity • A series of uniform cash flows • Begins at end of period one • Continues through the end of period N • Annuity given a Principal (A/P) • Annuity given a Future (A/F)

  5. IDDI Company Case The I Din’t Do It (IDDI) Company plans to borrow $150,000 from the Trust Me Bank (TMB) at an interest rate of 9% per year, compounded quarterly, in order to finance the installation of pollution reduction equipment. The TMB will agree to the loan only if IDDI makes quarterly payments during the next 4 years. Use the A/P formula to determine the quarterly payment. P = $150,000 i = 9% per year, compounded quarterly m = 4 N = 4

  6. Modifying formula  = $11,267.49 IDDI Company Case Solution Quarterly compounding  i/4 = 0.09/4 = 0.0225

  7. $150,000 …….. 0 1 2 14 15 16 $11,267.49 IDDI Company Cash Flow Diagram

  8. In Class Problem What is the total amount of interest paid over the 4 years of the loan period?

  9. Analyzing a Project • Present Worth • Internal Rate of Return (IRR)

  10. Analyzing a Project Present Worth • Value at time zero (e.g., today) that is equivalent to the cash flow series of a proposed project or alternative • Easy to understand and use • Matches our intuitive understanding of money • PW > 0  Desired Value!! • PW = 0  Economic indifference • PW < 0  Should be avoided, if possible

  11. Present Worth Useful when • Setting a price to buy or sell a project • Evaluating an investment or project when the price to invest or the first cost is given; • Calculating and equivalent value for an irregular series of cash flows

  12. Present Worth Assumptions 1. Cash flows occur at the end of the period, except for first costs and pre-payments • i.e., insurance, leases 2. Cash flows are known, certain values 3. The interest rate (i) is given 4. The problem’s horizon (N) is given

  13. Voice Recognition for Construction Inspection A large construction job will take 4 years to complete. The costs of the many required inspections can be reduced by purchasing a voice recognition system to act as the “front end” for a word processor. Then inspectors can be trained to record their comments and revise the written output of an automated form. The firm interest rate is 12%. What is the PW of the cash flows for the voice recognition system? The first cost for purchase and training is 100K. Savings are estimated at $30K, $40K, $65K, and $35K for the 4 years of construction.

  14. $65K $40K $35K $30K 0 1 2 3 4 $100K VR Project Cash Flow Diagram

  15. i PW 5% $49,796 8% $39,396 20% $ 7,272 30% $11,414 VR Project Solution PW = -100K + 30K(P/F,0.12,1) +40K(P/F,0.12,2) + 65K(P/F,0.12,3) +35K(P/F,0.12,4) PW =$27,182

  16. In Class Problem Find the present worth of the 4 different years using formulas in Excel. Find the present worth of the 4 different years using Excel financial functions.

  17. Analyzing a Project Internal Rate of Return (IRR) • Interest rate that makes the PW equal zero • Firms, government agencies and individuals must compute the cost of financial projects • Solution • Calculate the IRR for a project or for a financial source • Then, compare the resulting IRR with an interest rate for the time value of money

  18. Golden Valley Manufacturing By replacing its assembly line conveyor with an asynchronous conveyor, Golden Valley Manufacturing will save $50,000 per year in rework, inspection, and labor costs. The asynchronous conveyor will cost $275,000, and its life is 15 years (when its salvage value equals 0). If Golden Valley uses a 10% interest rate, should the asynchronous conveyor be installed?

  19. $50,000 …….. 0 1 2 13 14 15 $275,000 GVM Cash Flow Diagram

  20. - - - - GVM Solution PW = 0 = -275K + 50K(P/A, IRR, 15) (P/A, IRR, 15) = 275K / 50K = 5.500 From tables: (P/A, 0.16, 15) = 5.575 (P/A, IRR, 15) = 5.500 (P/A, 0.17, 15) = 5.324 ¸ ¸

  21. In Class Problem #4 Use Excel to find the value of IRR of return using interpolation

  22. 0.4261 GVM Solution 16.30% > 10% GOOD INVESTMENT!!

  23. In Class Problem #5 Use Excel to find IRR

  24. IRR = 16.3% Internal Rate of Return +PW Present Worth PW 0 -PW i % Interest rate i

More Related