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 05 10 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 m = 4 i = 9% per year, compounded quarterly N = 4

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

7. IDDI Company Cash Flow Diagram 012141516 $150,000 $11,267.49 ……..

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 (1) Setting a price to buy or sell a project (2) Evaluating an investment or project when the price to invest or the first cost is given; (3) 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. VR Project Cash Flow Diagram 01234 $100K $30K $40K $65K $35K

15. VR Project Solution i PW 5%$49,796 8%$39,396 20%$ 7,272 30%$11,414 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. GVM Cash Flow Diagram 012131415 $275,000 $50,000 ……..

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

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

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