Engineering Economic Analysis Chapter 16 Economic Analysis in

Engineering Economic Analysis Chapter 16 Economic Analysis in the Public Sector 12/5/2020 rd 1

Investment Objective Promote the General Welfare Dam – loss of land upstream due to backed up water loss of arable farm land concerns of environmentalists downstream consequences Town rose garden – parking, traffic flow consequences, noise abatement, aphids Mall COMPROMISE Eminent domain … the benefits to whosoever they accrue 12/5/2020 rd 2

View Point External consequences – ignored during the past Town vs. State or Federal governments Example: A municipal project costs $1 M with the federal government putting up 50% of the cost. Accrued benefits to the town are $750 K for $500 K cost. Overall cost benefit < 1 => should not fund. Viewpoint should include those who pay and those who benefit. 12/5/2020 rd 3

Interest Rate No time value of money, Money spend as soon as printed or collected => 0%. General obligation municipal bonds – government asssted Revenue bonds – not supported by tax authority. Rate should be the highest of cost of capital, government opportunity cost, or taxpayer's opportunity cost. 12/5/2020 rd 4

Project Politics Read text page 508 -509. 12/5/2020 rd 5

Benefit/Cost Ratios Conventional = benefits / costs vs. Modified = (benefits – disbenefits)/ first-cost Interest rates Project duration Politics 12/5/2020 rd 6

16 -10 Initial cost = $100 K end of year 1 $150 K, years 2 -10 20 K B/C = [20 K(P/A, 7%, 9) (P/F, 7%, 1)]/(100 K + 150 K(1. 07)-1 121780. 04 / 240186. 92 = 0. 5070 12/5/2020 rd 7

16 -11 (PWB)2 – 22(PWC) + 44 = 0 Find PWC for optimal size project. Let x = PW of costs and y = PW of benefits y 2 -22 x + 44 = 0 or implicitly differentiating 2 yy' – 22 = 0 => y' = 11/y = 1 => y = 11 => x = 7. 5 22 x = 165 => x = 7. 5 12/5/2020 rd 8

16 -15 Benefits = Costs = (550 K - 35 K)(P/A, 8%, 20) (750 K + 2, 750 K) + 185 K(P/A, 8%, 20) = 0. 9510 Try at least 25 years in calculation to get B/C = 1. 004 Mayor is playing politics. 12/5/2020 rd 9

16 -26 Life = 15 First Cost Ann Savings Ann costs Salvage A 9. 5 K 3. 2 K 1 K 6 K B 18. 5 K 5 K 2. 75 K 4. 2 K C MARR = 12% 22 K 9. 8 K 6. 4 K 14 K a) Conventional B/C Eliminate B A B C PW Numerator 21. 7947 34. 054 66. 746 PW Denominator 15. 215 36. 463 63. 032 1. 43 0. 93 1. 06 B/C Ratio B/CC-A = 44. 95/ = 0. 94 => A is best. 12/5/2020 rd … continued 10

16 -26 continued Modified B/C Life = 15 First Cost Ann Savings Ann costs Salvage A 9. 5 K 3. 2 K 1 K 6 K B 18. 5 K 5 K 2. 75 K 4. 2 K C MARR = 12% 22 K 9. 8 K 6. 4 K 14 K A B C 14. 984 15. 324 23. 157 PW Denominator 8. 404 17. 733 19. 142 B/C Ratio 1. 78 0. 86 1. 19 PW Numerator B/CC-A = 8. 173/10. 738/ = 0. 76 => A is best. 12/5/2020 rd 11