1. 1/26/2016rdcw2a-1 Engineering Economic Analysis Chapter 2  Engineering Costs and Cost Estimating

2. 1/26/2016rdcw2a-2 Cost Explicit vs. Implicit Cash vs. Book Fixed vs. Variable Marginal vs. Average Sunk, Opportunity Recurring, Nonrecurring Incremental, Cash (out of pocket) vs. Book Life Cycle

3. 1/26/2016rdcw2a-3 Costs Life cycle First Operating & Maintenance Disposal Sunk Future Opportunity Direct Indirect Overhead Fixed Variable 1/26/2016rd3

4. 1/26/2016rdcw2a-4 Cost Estimation What cost component must be estimated? What approach to cost estimation will be applied? How accurate should the estimates be? What estimation techniques will be utilized?

5. 1/26/2016rdcw2a-5 Cost Estimating What cost components must be estimated? First component ~ equipment, delivery, installation insurance, training Elements – Direct labor and materials, maintenance What approach to cost estimation will be applied? Bottom-up, design-to-cost or top-down How accurate should estimates be? ROMs ~  20% of actual costs, Detail  5% What estimation techniques will be used? Unit, cost indices, learning curves, capacity, analogy 1/26/2016rd5

6. 1/26/2016rdcw2a-6 Cost Estimation Cost component estimates Cost Estimates 1/26/2016rd6 Required price Bottom-Up Top-Down DTC Required price

7. 1/26/2016rdcw2a-7 First Cost Component Equipment Cost Delivery charges Installation costs Insurance coverage Initial training of personnel for equipment use

8. 1/26/2016rdcw2a-8 Unit Method C T = U x N Cost of operating a car (43 cents / mile) Cost to bury fiber cable ($30K / mile) Cost to construct a parking space in parking garage $4500 Cost to construct Interstate highway ($6.2 M /mile) Cost of house construction ($225 /ft 2 ) 1/26/2016rd8

9. 1/26/2016rdcw2a-9 Annual Operating Costs Direct labor cost for operating personnel Direct materials Maintenance (daily, periodic, repairs, etc.) Rework and rebuild Can use canned cost estimating packages for bridges, pavement, etc.

10. 1/26/2016rdcw2a-10 Cost Estimates Rough Order of Estimates (ROMs) Semi-detailed Detailed (WBS) Segmented (divide and conquer) Analogy Parametric (e.g., by foot or mile) Cost Index Triangulation (Delphi) Learning Curve

11. 1/26/2016rdcw2a-11 Cost Index An engineer finds a project of similar complexity was done 5 years ago at a skilled labor cost of $360K. The skilled labor index was 3496 and is now 4038. Estimate cost today for the skilled labor. C t = C 0 (I t /I 0 ) = 360K * 4038 3496 = $415,812.

12. 1/26/2016rdcw2a-12 Cost Capacity Equation C 2 = C 1 (Q 2 /Q 1 ) x Cost for flow rate of 0.5 MGD is $1.7M in year 2000. Estimate today's cost for 2 MGD flow rate. Cost index for 2000 is 131 and today is 225 and C 2 = 1.7M * 2/0.5 0.14 * 225 131 =$3,545,264

13. 1/26/2016rdcw2a-13 Cost Capacity C 2 = C 1 (Q 2 / Q 1 ) x (I t /I 0 ) The total cost for a digester with a flow rate of ½ MGD was $1.7M in 2000. Estimate the cost for a flow rate of 2 MGD. The cost index in 2000 of 131 is now 225 and the exponent x = 0.14 for the range flow 0.2 to 40 MGD. C 2 = 1,700,000 * (2/0.5) 0.14 (225/131) = $3,546,178. ComponentSize RangeExponent Activated sludge plant1-100 MGD 0.84 Aerobic digester0.2-40 MGD 0.14 …. … … 1/26/2016rd13

14. 1/26/2016rdcw2a-14 DK Enterprise Fixed Variable Bus $80 Event ticket $12.50 Gas 75 Treats 7.50 Oil etc 20 $20.00 Driver 50 $225 Ticket price $35, Let X be number of people Profit = (35 – 20)X – 225 = 0 => Breakeven X = 15

15. 1/26/2016rdcw2a-15 Production Costs (1) Q (2) TFC (3) TVC (4) TC (5) AFC (6) AVC (7) AC (8) MC 160309060309010 26040100302050 5 3604510520153510 460551151513.7528.7520 5607513512152745 660120180102030

16. 1/26/2016rdcw2a-16 Incremental Costs Cost Item Model A Model B Difference First cost $10K $17.5K-$7.5K Installation 3.5K 5K -1.5K AnnMaintCost 2.5K 0.75K 1750 Annual utility 1.2K 2K -800 Salvage value 700 500 -200

17. 1/26/2016rdcw2a-17 Problem 2-5 C = 3M – 18K Q + 75Q 2 cost/production dC/dQ = -18K + 150 Q = 0 when Q = 120 units dC 2 /dQ 2 = 150 > 0 => relative and absolute minimum as C(0) = $3M and C(  ) = . C(120) = 3M – 18K * 120 + 75 * 120 2 = $1.92M for 120 units => 1.92M/120 = $16K average unit cost C(110) = 3M – 18K * 110 + 75 * 110 2 = $1,927,500 for 110 units => $17,522.73/unit Save 1.9275M – 1.92M = $75K

18. 1/26/2016rdcw2a-18 Problem 2-10 C A = $100K + 20.5x C B = $350K + 10.5x 0 < x < 150K production C C = $600K + 8x Breakeven BE AB => 250K = 10x => x = 25K BE BC => 250K = 2.5x => x = 10K BE AC => 500K = 12.5x => x = 40K x 0 50 100 150 A1001125 2150 3175 25K A B 350 875 1400 1925 25K B C 600 1000 1400 1800 80K C x CACA CBCB C

19. 1/26/2016rdcw2a-19 Problem 2-14 P = 100 – S; C = 1000 + 10S where P is unit selling price and S is Sales volume. C is the total cost of production. Profit = (100 - S)S – 1000 – 10S = -S 2 + 90 S - 1000 Profit' = -2S + 90 = 0 when S = 45 Profit"(45) = 0 Use first derivative test showing 45 is max. Breakeven equation set Profit = 0 or S 2 – 90S + 1000= 0 (quadratic 1 -90 1000)  (77.02 12.98) or S = 77, 13

20. 1/26/2016rdcw2a-20 Learning Curve (80%) n Time (hrs) 120 216 = 0.8 * 20 (doubled) 314 412.8 = 0.8 * 16 (doubled) 6 ? T n = T 1 * n b ; T 1 = 20, b is slope of curve 12.8 = 20 * 4 b => b = -0.321928 (log 0.8 2) = -0.321928  2 x = 0.8

21. 1/26/2016rdcw2a-21 Learning Curve 80% (Sim-LC 20 4 4/5)  Unit Cost Cumulative 1 20.00 20.00 2 16.00 36.00 3 14.04 50.04 4 12.80 62.84 5 11.91 74.75 6 11.23 85.99 7 10.69 96.68 8 10.24 106.92 The slope of 80% learning curve is -6.9658

22. 1/26/2016rdcw2a-22 Example 2-10 (sim-lc 9.6 20 85)  Unit Hours Cumulative 1 9.60 9.60 2 8.16 17.76 3 7.42 25.18 4 6.94 32.12 5 6.58 38.70 6 6.31 45.01 7 6.08 51.09 8 5.90 56.98 9 5.74 62.72 10 5.60 68.31 11 5.47 73.79 12 5.36 79.15 13 5.26 84.41 14 5.17 89.58 15 5.09 94.67 16 5.01 99.68 17 4.94 104.62 18 4.87 109.49 19 4.81 114.31 20 4.76 119.06

23. 1/26/2016rdcw2a-23 Learning Curve The time to make the first unit is $1000 and the learning curve is 90%. What is the cost to make the 2 nd, 4 th and 8 th units? T n = T 1 * n b T 2 = 1000 * 2(log 0.90 2) = 1000 * 2 -0.1520 = $900 T 4 = 0.90 * T 2 = 0.90 * $900 = $810 T 8 = 0.90 * T 4 = 0.90 * $810 = $729. 2 b = 0.90 => b Ln 2 = Ln 0.90 => b = -0.1520031 (Tn% 1000 8 0.90)  729

24. 1/26/2016rdcw2a-24 Learning Curve 2-31 200 labor hours produced unit 1; 60 hours produced unit 7. Find learning curve rate in percent. T n = T 1 * n b 60 = 200 * 7 b => b = Ln 0.3/ Ln 7 = -0.6187196 LC% = 2 b = 2 -0.62 = 65%. (log 0.65125 2)  -0.61872 = b

25. 1/26/2016rdcw2a-25 Life Cycle Costs Concurrent Engineering Later the design changes, the more expensive the costs Early decisions lock in later costs. More accurate estimates Benefits usually overestimated Costs usually underestimated

26. 1/26/2016rdcw2a-26 Benefits vs. Costs Estimating Benefits – tend to be overestimated and accumulate later in project. Costs – tend to be underestimated and mostly occur during beginning of project.

27. Earned Value Analysis (EVA) Budgeted cost of work scheduled (BCWS) Budgeted cost of work performed (BCWP) Actual cost of work performed (ACWP) Cost Variance (CV) = BCWP – ACWP Schedule variance (SV) = BCWP – BCWS Estimated cost at completion (ECAC) = ACWP * BAC/BCWP Estimated time at completion (ETAC) = BCWS * TAC/BCWP 1/26/2016rdcw2a-27

28. Example of EVA You are PM at the 18 th month of a 24 month project with $400K budget. You find that BCWS = $300K, ACWP = $310K and BCWP = $280K. Find the current estimates of CV and SV and ECAC and ETAC. CV = BCWP – ACWP = $280K - $310K = -$30K SV = BCWP – BCWS = $280K - $300K = -$20K ECAC = ACWP * BAC / BCWP = 320K * 400K/280K = $442,857 ETAC = BCWS * TAC / BCWP = 300K * 24 /280K = 25.7 months From $400K to $442,857 and from 24 months to 25.7 months. Overruns in cost and schedule. 1/26/2016rdcw2a-28

29. 1/26/2016rdcw2a-29 BCWS ACWP BCWP Cost variance Schedule Variance BAC Cost Current review time TAC Time