ME 443

1. ME 443 ECONOMIC ANALYSIS of PROJECTS in the PUBLIC SECTOR Prof. Dr. Mustafa Gökler

2. INTRODUCTION There are many types of government projects and many agencies involved. Four classes reasonably cover the spectrum of projects entered into by government. They include • cultural development, • protection, • economic services, and • natural resources.

3. INTRODUCTION Cultural developmentis enhanced through education, recreation, historic, and similar institutions or preservations. Protection is achieved through military services, police and fire protection, and the judicial system. Economic services include transportation, power generation, and housing loan programs.

4. INTRODUCTION Natural resourceprojects mightentail wild land management, pollution control, and flood control. Although these are obviously incomplete project lists in each class, it is not so obvious that some projects belong in more than one area. For example, flood control is certainly a form of protection for some, as well as being related to natural resources.

5. INTRODUCTION Fortunately, analysis methods for public and private projects are very similar, even though there are some basic, significant differences between the two. Those most frequently used method in evaluating government (national, state, or local) projects i s the benefit-cost method

6. BENEFIT/COST Benefit/Cost Analysis (Savings/Investment) Analysis Bjt :public benefits associated with project jduring year t Cjt : governmental costs associated with project j during year t

7. BENEFIT/COST When two or more project alternatives are being compared using B/C ratio, the analysis should be done on an incremental basis If it is greater than 1.0, Alternative 2 is justified.

8. BENEFIT - COST If B-C Criterion is used, it is similar to the present worth method

9. BENEFIT - COST When two or more project alternatives are being compared using B-C criterion, the analysis should be done onan incremental basis Δ(B-C)2-1 = ΔB2-1 -ΔC2-1 Δ(B-C)2-1 > 0 => Alt. 2 is justified.

10. EXAMPLE We are given the task of deciding between three highway alternatives to replace a winding, old, dangerous road. The length of the current route is 26 miles. Alternative A is to overhaul and resurface the old road at a cost of $3 million. Resurfacing will then be required at a cost of $2.5 million at the end of each 10-year period. Annual maintenance for Alternative A will cost $10,000/mile.

11. EXAMPLE Alternative B is to cut a new road following the terrain; it will be only 22 miles long. Its first cost will be $10 million, and surface renovation will be required every 10 years at a total cost of $2,250,000. Annual maintenance will be $10,000/mile.

12. EXAMPLE Alternative C also involves a new highway which, for practical considerations, will be built along a 20.5-mile straight line. Its first cost, however, will be $18 million, because of the extensive additional excavating necessary along this route. It, too, will require resurfacing every 10 years at a cost of $2,250,000. Annual maintenance will be $18,000/mile. This increase over route B is due to the additional roadside bank retention efforts that will be required.

13. EXAMPLE Our task is to select one of these alternatives, considering a planning horizon of 30 years with negligible residual value for each of the highways at that time. One of these alternatives is required, since the old road has deteriorated below acceptable standards. We can calculate the annual equivalent first cost and maintenance cost of each alternative using an interest rate of 8%.

14. EXAMPLE Cost of Road : Route A: [$3,000,000 + $2,500,000(P|F 8,10) + $2,500,000(P|F 8,20)](A|P 8,30) = [$3,000,000 + $2,500,000(0.4632) + $2,500,000(0.2145)]x (0.0888) = $416,849/year

15. EXAMPLE Route B: [$10,000,000+ $2,250,000(P|F 8,10) + $2 ,250 ,000( P | F 8 ,20)](A | P 8, 30) = [$10,000,000 + $2,250,000(0.4632) + $2,250,000(0.2145)]x (0.0888) = $1,023,404/year

16. EXAMPLE Route C: [$18,000,000+ $2,250,000(P|F 8,10) + $2,250,000(P| F 8 ,20)](A | P 8 ,30) = [$18,000,000 + $2,250,000(0.4632) + $2,250,000(0.2145)]x (0.0888) =$ 1733 .804/year

17. EXAMPLE

18. EXAMPLE Maintenance cost: Route A : (10 000 $/ mile - year) (26 miles) = $260,000/year Route B: ($10,000)(22) = $220,000/year Route C: ($18,000)(20.5) = $369,000/year

19. EXAMPLE

20. EXAMPLE Traffic density along each of the three routes will fluctuate widely from day to day, but will average 4000 vehicles/day throughout the year. This volume is composed of 350 light commercial trucks, 250 heavy trucks, 80 motorcycles, and the remainder are automobiles. The average cost per mile of operation for these vehicles is $0.35, $0.50, $0.10, and $0.20, respectively.

21. EXAMPLE Operational cost: Route A : (350 light trucks/day)(26miles) (0.35$/mile)(365days/year)+(250heavytrucks/day) (26miles)(0.50$/mile)(365days/year) +(80motorcycles/day)(26miles)(0.10$/mile) (365 days/year)+ (3320automobiles/day)(26miles) (0. 20 $/mile)(365 days/year) = $ 8726055 / year

22. EXAMPLE Operational cost: Route B : (350 light trucks/day)(22miles) (0.35$/mile)(365days/year)+(250heavytrucks/day) (22miles)(0.50$/mile)(365days/year) +(80motorcycles/day)(22miles)(0.10$/mile) (365 days/year)+ (3320automobiles/day)(22miles) (0. 20 $/mile)(365 days/year) = $7383585 / year

23. EXAMPLE Operational cost: Route C : (350 light trucks/day)(20.5miles) (0.35$/mile)(365days/year)+(250heavytrucks/day) (20.5miles)(0.50$/mile)(365days/year) +(80motorcycles/day)(20.5miles)(0.10$/mile) (365 days/year)+ (3320automobiles/day)(20.5miles) (0. 20 $/mile)(365 days/year) = $6880159 / year

24. EXAMPLE

25. EXAMPLE There will be a time savings because of the different distances along each of the routes, as well as the different speeds which each of the routes will sustain. Route A will allow heavy trucks to average 35 miles/hour, while other traffic can maintain an average speed of 45 miles/hour.

26. EXAMPLE Routes B and C will allow heavy trucks to average 40 miles/hour, and the rest of the vehicles can average 50 miles/hour. The cost of time for all commercial traffic is valued at $15/vehicle/hour, and for noncommercial traffic, $5/vehicle/hour. Twenty-five percent of the automobiles and all of the trucks are considered commercial.

27. EXAMPLE Time cost:(Route A) : (350lighttrucks/day)(26miles/lighttrucks)(1hour/45miles) (365days/year)(15$/hour)+(250heavytrucks/day) (26miles/heavytrucks)(1hour/35miles)(365days/year)(15$/hr) +(80motorcycles/day)(26miles/motorcycles)(1 hour/45 miles) (365 days/year) (5$/hour) + (3320automobiles/day) (26miles/automobile)( 1 hour/45 miles) (365 days/year) (0.25X 15$/hour+0.75X 5$/hour) = $7459441 / year

28. EXAMPLE Time cost:(Route B) : (350lighttrucks/day)(22miles/lighttrucks)(1hour/50miles) (365days/year)(15$/hour)+(250heavytrucks/day) (22miles/heavytrucks)(1hour/40miles)(365days/year)(15$/hr) +(80motorcycles/day)(22miles/motorcycles)(1 hour/50 miles) (365 days/year) (5$/hour) + (3320automobiles/day) (22miles/automobile)( 1 hour/50 miles) (365 days/year) (0.25X 15$/hour+0.75X 5$/hour) = $5659143 / year

29. EXAMPLE Time cost:(Route C) : (350lighttrucks/day)(20.5miles/lighttrucks)(1hour/50miles) (365days/year)(15$/hour)+(250heavytrucks/day) (20.5miles/heavytrucks)(1hour/40miles)(365days/year) (15$/hr) +(80motorcycles/day)(20.5miles/motorcycles)(1 hour/50 miles) (365 days/year) (5$/hour) + (3320automobiles/day) (20.5miles/automobile)( 1 hour/50 miles) (365 days/year) (0.25X 15$/hour+0.75X 5$/hour) = $ 5273292 / year

30. EXAMPLE

31. EXAMPLE Finally, there is a significant safety factor that should be included. Along the old winding road, there has been an excessive number of accidents per year. Route A will reduce the number of vehicles involved in accidents to 105, and routes B and C are expected to involve only 75 and 70 vehicles in accidents, respectively, per year.

32. EXAMPLE The average cost per vehicle in an accident is estimated to be $7500, considering actual physical property damages, lost wages because of injury, medical expenses, and other relevant costs.

33. EXAMPLE Safety cost: Route A : (105 vehicles/year)(7500 $/ vehicle)= $787500/year Route B: (75) (7500 $) = $562500/year Route C: (70) (7500 $) = $525500/year

34. EXAMPLE

35. EXP.(7.1) B/C Public Benefits = - Public Costs ΔBB-A(8%)= Public CostsA-Public CostsB = $16972996-$13605228=$3367768/year ΔCB-A(8%)=Government CostsB-Government CostsA = $1243404-$676849 = $566555/year ΔB/CB-A = ΔBB-A /ΔCB-A= $3367768/$566555 = 5.94 > 1.00 => B is justified.

36. EXP.(7.1) B/C ΔBC-B(8%)= Public CostsB-Public CostsC = $13605228-$12678451=$926777/year ΔCC-B(8%)=Government CostsB-Government CostsA = $2102804-$1243404 = $859400/year ΔB/CC-B = ΔBC-B /ΔCC-B= $926777/$859400 = 1.08 > 1.00 => C is justified and prefered.

37. EXAMPLE(7.2) B-C ΔBB-A(8%)= $3367768/year ΔCB-A(8%) = $566555/year Δ(B-C)B-A = ΔBB-A - ΔCB-A= $3367768-$566555 = $2801213/year > 0 => B is justified. ΔBC-B(8%)= $926777/year ΔCC-B(8%)= $859400/year Δ(B-C)C-B = ΔBC-B -ΔCC-B= $926777-$859400 = $67377/year > 0 => C is justified and prefered.

38. EXP.(7.3) PW The present worth of all costs for each highway alternative is given by the following formula: PW(total)= PWtotal government costs (i) + PWtotal public costs (i) Route A: PWA(8%) = $676,849(P|A 8,30) + $16,972,996(P|A 8,30) = $676,849(11.2578) + $16,972,996(11.2578) = $ 198,698,425

39. EXP.(7.3) PW Route B : PWB(8%) = $1,243,404(P|A 8,30) + $13,605,228(P|A 8,30) =$1,243,404(11.2578) + $13,605,228(11.2578) = $167,162,930 RouteC: PWC(8%) = $2,102,804(P|A 8,30) + $12,678,451(P|A 8,30) = $2,102,804(11.2578) + $12,678,451(11.2578) = $166,404,413 Since the present worth of all costs for route C is smallest, we again see that this alternative is preferred.

40. EXP.(7.4) AW The annualworth of all costs for each highway alternative is given by the following formula: AW(total)=AWtotal government costs (i) + AWtotal public costs (i) Route A: AWA(8%) = $676,849 + $16,972,996 = $ 17649845/year

41. EXP.(7.4) AW Route B : AWB(8%) = $1,243,404 + $13,605,228 = $14848632/year RouteC: AWC(8%) = $2,102,804 + $12,678,451 = $ 14781255/year Since the annual worth of all costs for route C is smallest, we again see that this alternative is preferred. However, PW and AW are seldom used in government projects.