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Pramen P. Shrestha, Ph.D., P.E. February 6, 2012.

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Presentation on theme: "Pramen P. Shrestha, Ph.D., P.E. February 6, 2012."— Presentation transcript:

1 Pramen P. Shrestha, Ph.D., P.E. February 6, 2012

2 Topics to be Covered Construction Scheduling Construction Estimating Project Controls

3 Construction Scheduling Project Scheduling Methods to Calculate Total Project Duration Critical Path Method (CPM) Precedence Diagram Method (PDM) Float Calculation Program Evaluation & Review Technique (PERT)

4 Project Scheduling To arrange the project activities in order to get the total project completion duration Predecessor and Successor Predecessor Activity Successor Activity Activity Relationship = Finish to Start (FS) Excavate Earthwork Place Formwork

5 Activity Relationships Finish to Start Relationship Predecessor Activity Successor Activity Activity Relationship = Finish to Start (FS) Excavate Earthwork Place Formwork

6 Activity Relationships Finish to Finish Relationship Predecessor Activity Successor Activity Activity Relationship = Finish to Finish (FF) Excavate Earthwork Place Formwork

7 Activity Relationships Start to Start Relationship Predecessor Activity Successor Activity Activity Relationship = Start to Start (SS) Excavate Earthwork Place Formwork

8 Activity Relationships Start to Finish Relationship with Lead Predecessor Activity Successor Activity Activity Relationship = Start to Finish (FS) Order Concrete from Supplier Place Concrete in Formworks SF/5

9 Activity Relationships with Lag Finish to Start Relationship with Lag Lag means delayed Predecessor Activity Successor Activity Activity Relationship = Finish to Start (FS) with Lag Excavate Earthwork Place Formwork FS, Lag =3

10 Methods to Calculate Total Project Duration Bar Chart Critical Path Diagram (CPM) Precedence Diagram Method (PDM) Program Evaluation and Review Technique (PERT)

11 Bar Chart

12 Critical Path Method (CPM) Activity on Arrow (AOA)

13 Early Start Date Calculation Forward Pass ESLF ES= Early Start LS= Late Finish

14 Early Start Date Calculation Forward Pass ESLF ES= Early Start LF= Late Finish 14 22

15 Late Start Date Calculation Backward Pass ESLF ES= Early Start LF= Late Finish 14 22

16 Critical Path ESLF ES= Early Start LS= Late Finish Critical Path

17 Precedence Diagram Method Activity on Node (AON)

18 PDM (Forward Pass)

19 PDM (Backward Pass)

20 PDM (Critical Path)

21 Total and Free Float Total Float The total number of days that the activity can be delayed without delaying the total project Free Float The total number of days that the activity can be delayed without delaying the successor activity Total Float and Free Float will be zero in critical path of the schedule

22 Total Float Calculation Total Float (TF) = LS- ES = LF-EF TF = 0 TF = 5-3=2 TF = 7-5=2 TF = 0

23 Free Float Calculation Free Float (FF) = ES J -EF I where I is the predecessor and J is successor activity. FF = 3-3=0 FF = 5-5=0 FF = 14-12=2 TF = 7-7=0FF = 14-14=0

24 Sample Question from NCEES

25 Question An activity-on-node network for a project is shown in the following figure. All relationships are finish-to-start with no lag unless otherwise noted. If all activities begin at their early start except Activity E, which is delayed by 2 days from its early start, which of the following statements is true? A. Activity E will have no impact on the start time of any other activity B. Activity E will delay the start of Activity G by 1 day but will not delay project completion. C. Activity E will delay the start of Activity G by 2 days but will not delay project completion. D. Activity E will delay the completion of the project by 2 days

26 Forward Pass to calculate ES & EF

27 Backward Pass to calculate LS & LF

28 Critical Path

29 Total and Free Float Calculation

30 TF and FF Calculation

31 Answer Total Float of Activity E = 3 days Free Float of Activity E = 2 days By starting Activity E, 2 days late will not delay the project as well as not delay its successor activity (Activity G). Choice A is correct.

32 PERT Program Evaluation and Review Technique Probability method Most Likely Duration - m Pessimistic Duration (Longer duration) -b Optimistic Duration (Shorter duration) -a Weighted most likely duration = (a+4m+b)/6 Variance = [(b-a)/6] 2 Standard Deviation = Square Root of Variance

33 PERT Problem- Critical Activities Find out the probability of completing the project at 90 days? Activity ES Duration (Days) LS EFW. DurationLF ABC 12/18/42 18/33/60 8/11/20

34 Calculated Weighted Duration

35 Forward Pass

36 Backward Pass

37 Variance Calculation

38 Probability Calculation Total Variance of Critical Path = = 78 Standard Deviation = = 8.83 days Total Critical Path Duration = 68 days Probability of completing project in 90 days Z = (90-68)/8.83= 2.49 standard deviation Referring to Standard Normal Curve, Probability = = 99.4%

39 Standard Normal Curve

40 Recommended Book for Construction Scheduling Project Management for Construction & Engineering Garold D. Oberlender Construction Planning and Scheduling Jimmie W. Hinze Computer-based Construction Project Management Tarek Hegazy

41 Construction Estimating Construction Cost consists of Direct Cost Labor, material, equipment, and sub-contractor cost Indirect Cost Overhead, taxes, bonds, insurance cost Contingency Cost Potential unforeseen work based on the amount of risk Profit Compensation costs for performing the work

42 Steps for Preparing an Estimate Review the scope of the project Consider effect of location, security, available storage, traffic on costs Determine quantities Material quantity takeoff Price material Material cost = Quantity x Unit price of material

43 Steps for Preparing an Estimate Price labor Based on labor production rates and crew sizes Labor cost = [ (quantity)/(labor production rates)] x [labor rate] Price equipment Based on equipment production rates and equipment spreads Equipment cost = [ (quantity)/(equip. production rates)] x [equip. rate]

44 Steps for Preparing an Estimate Obtain specialty sub-contractors bid Obtain suppliers bid Calculate taxes, bonds, insurance, and overhead Contingency Potential unforeseen work based on the amount of risk Profit Compensation costs for performing the work

45 Types of Estimate Conceptual Cost Estimate Preliminary, feasibility, budget estimate etc. Conducted before detail design Conducted in planning or feasibility stage Detailed Cost Estimate Conducted after the detail design is complete Basis for bid

46 Conceptual Estimates Prepared from completed similar projects Size of project No. of unit No. of SF No. of cars in a parking garage Developed from unit cost Weighting of average, maximum and minimum value

47 Estimating Equation Weighted Unit Cost Estimating Equation to forecast unit cost UC = (A + 4B + C) / 6 Where UC = Unit Cost A = Minimum unit cost of previous projects B = Average unit cost of previous projects C = Maximum unit cost of previous projects

48 Adjustments Time Location Size Complexity Need appropriate contingency

49 Weighted Unit Cost Estimate 1. Weighted Unit Cost Estimating Problem: Cost information from 6 previously completed housing projects are shown in the following table. These projects were completed in Las Vegas in Now a contractor has to build a house (2000 SF) in New Orleans, in Estimate the cost of that house using conceptual estimating method. ProjectsCostSquare Foot Cost/ SF 1$500,000 2,000 $250 2$351,000 1,300 $270 3$371,000 1,400 $265 4$550,000 2,500 $220 5$600,000 3,000 $200 6$200,000 1,100 $182 Cost Indices YearsIndices Location Indices LocationIndex Las Vegas1205 Austin1000 Los Angeles1665 New Orleans1050

50 Weighted Unit Cost Estimate Solution: From historical data: Average cost of building per SF = ($250 + $270 + $265 + $220 + $200 + $182)/6 = $ Minimum SF Rate = $182 Maximum SF Rate = $270 Weighted Unit Cost = ($ x $ $270) / 6 = $ / SF Conceptual cost estimate for 2,000 SF of building in Las Vegas, in 2004 = 2,000 SF x $229.45/ SF = $458,900 Adjustment for time Find out the average yearly interest rate {4877 / 3980} = (1+i) n Where is i = average yearly interest rate n = number of years = 3 Substituting the n value = (1+i) 3 i = 7% Time Adjustment factor building in the year 2009 (n = 5 years) for Las Vegas = (1+.07) 5 = Adjustment Factor for Location = (1050 / 1205) = Adjusted Cost for the building = x x $458,900 = $560,382

51 Detailed Estimate – Labor Cost Straight time wage rate Overtime wage rate Workdays – more than 8 hour per day should be paid 1.5 times straight time wage rate Weekends – all hours should be paid 2 times straight time wage rate

52 Cost of Labor Straight time or overtime wage Social security tax (FICA) Unemployment compensation tax Workers compensation insurance Public liability and property damage insurance Fringe Benefits

53 Labor Production Rates Number of units of work produced by a person in a specified time (hour or day) For example 1,000 bricks laid in 12 hours Production rate should be calculated considering person will not work 60 min/hr Production rates depend upon Climatic condition Job supervision Complexity of the job

54 Labor Cost Calculation Brick layers Cost Base wage = $22.50/hr FICA Tax = 7.65% Unemployment Tax = 3.00% Workers Compensation Insur. = $15 per $100 PL & PD Insurance = $2.50 per $100 Fringe Benefits = $3.50 per hour Work 10 hours and 6 days per week (Mon – Saturday)

55 Labor Cost Calculation Actual Hours and Pay Hours Calculation: Actual hours = 10 hours/day X 6 days = 60 hours Pay hours: Normal hours = 8 hours/day X 5 days = 40 hours Overtime hours in weekdays = 1.5 X 2 hours/day X 5 days = 15 hours Overtime hours in weekend = 2 X 10 hours/day x 1 day = 20 hours Total Pay hours = 40 hours + 15 hours + 20 hours = 75 hours

56 Labor Cost Calculation Average Hourly Pay Rate Calculation: Average actual hourly pay = $22.50/ hour X (75hr / 60 hr) = $28.12/ hour 3. FICA Tax Calculation: FICA tax per hour = 7.65% of actual hourly pay = X $28.12/hour = $2.15 /hour 4. Unemployment Tax Calculation: Unempl. tax per hour = 3.00% of actual hourly pay = 0.03 X $28.12/hour = $0.84 /hour 5. Workers Compensation Insurance Calculation: Work Comp. Ins. per hour = $15 per $100 of base hourly pay = ($15/$100) X $22.50/hour = $3.37 /hour 6. PL & PD Insurance Calculation: PL & PD Ins. per hour = $2.50 per $100 of base hourly pay = ($2.50/$100) X $22.50/hour = $0.56 /hour 7. Fringe Benefits:= $3.50 /hour

57 Labor Cost Calculation Total cost per hour = $ $ $ $ $ $3.50 = $38.54 Total cost per day= 10 hours x $38.54 /hours = $ Total cost per week= 60 hours x $38.54 /hours = $2, Total cost per month= $2, /week x (52 weeks / 12 months) = $10, Total cost per year= $10, /month x 12 months = $120,244.80

58 Equipment Costs If Equipment is rented: Rental Cost Monthly rent fixed by the rental company Rate obtained from Rental Rate Blue Book by Dataquest Inc., DOT etc. Does not include operating cost Include fuel, oil and lubricants If Equipment is purchased Ownership Costs Operating Costs

59 Ownership Costs Cost associated with equipment whether it is used or not Useful life – expected duration the equipment can be used It includes Investment cost Depreciation Taxes and insurance

60 Investment Costs Cost of purchasing equipment Money borrowed from bank or lender Interest cost associated with borrowed money Interest cost will depend upon economic condition Most contractor will use current interest rate and add interest for risk in buying equipment

61 Depreciation Costs Loss of value due to use and age Salvage value – equipment value at the end of useful life If P is purchase amount and F is salvage value Depreciation = P - F

62 Depreciation Methods Straight line depreciation method Yearly depreciation = (P – F) / No. of years Depreciation cost is determine for Estimate equipment cost for the project Depreciate cost for tax purpose Straight line depreciation is used for estimate Double-declining-balance and sum of years digit method are used for tax purpose

63 Hourly Ownership Cost Ownership cost is calculated on hourly basis method This method used time-value-of-money Generally provided in Engineering Economic analysis book This method used two equations Capital Recovery Equation Sinking Fund Equation

64 Capital Recovery Equation A = Equivalent Annual Value P = Purchase Price i = Annual Interest Rate n = Useful Life in Years Economic Analysis Table A = P x (A/P, i%, n)

65 Sinking Fund Equation A = Equivalent Annual Value F = Future Salvage Value i = Annual Interest Rate n = Useful Life in Years Economic Analysis Table A = F x (A/F, i%, n)

66 Engineering Economic Table

67 Steps for Determining Annual Ownership Cost Obtain Purchase Price (P) Estimate Salvage Value (F) Estimate Useful Life (n) Estimate Interest Rate First estimate interest rate for borrowing money Estimate taxes, insurance, and storage and convert these in equivalent interest rate Add these two rates to get Minimum Attractive Rate of Return (MARR) Use Capital Recovery and Sinking Fund equations to find out annual ownership cost

68 Equipment Annual Ownership Cost Net equipment value = $90,000 Expected salvage value = $20,000 Useful life = 5 years Interest Rate = 17%

69 Operating Costs Maintenance and Repair Costs Replacement cost, labor, oil & lubricating services Varies depending upon the equipment Cost is expressed in % of purchase cost or depreciation cost Fuel Costs Consumption per hour calculated from the equation Lubricating Oil Costs Consumption per hour Cost of Rubber Tires % of depreciation cost

70 Fuel Consumption Cost Fuel Consumed per hour = OF x Engine HP x [0.04 gal/(hp-hr)] for Diesel engine = OF x Engine HP x [0.06 gal/(hp-hr)] for Gasoline engine OF = Operating Factor = Time Factor x Engine Factor Time Factor = % time the engine runs in one hour Engine Factor = % of time the engine operated at a full power

71 Lubricating Oil Consumption Cost Oil Consumed per hour = [HP x OF x lb/(hp-hr) / 7.4 lb/gal] + c/t Where, OF = Operating Factor = 0.6 c = capacity of crankcase in gallons t = hours between oil change For example Engine 100 hp, crankcase capacity 4 gal and oil change every 100 hrs Oil consumed =(100 x 0.6 x 0.006/7.4) + (4/100) = gal/hr

72 Handling and Transporting Material Loading Material being loaded in truck Haul (Loaded) Material being transported to site Unload Material being dumped on site Return (Empty) Truck returns back to loading site

73 Cycle Time Total time to complete these four activities is called cycle time Hauling and Returning time can be combined if they require same time Distance between material site and jobsite Effective speed of the vehicle Vehicle Traffic congestion Condition of road Other factors

74 Production Rate Total quantity of work done per hour Labor and Equipment Production rate Labor and equipment can be involved together Only equipment can be involved Only labor can be involved Analysis process is very important Estimate must be done with different alternatives

75 Earthwork Measurement Excavated: Cut or bank measure ( Unit weight 95 lb/cf to 105 lb/cf) Hauled: Loose measure ( Volume increases due to swell but unit weight reduces- 80 to 95 lb/cf) Compacted: Fill or compacted measure (Volume decreases but unit weight increases – 110 to 120 lb/cf)

76 Soil Report Soil classification Unit weight Moisture content Swell factor ( % of swell) - % gain in volume compared to original volume Shrinkage factor- When soil is compacted, volume decreases

77 Correlation between Volume, Swell, and Shrinkage w L = (1 + S w /100) B C = (1 – S h /100) B Where L = Volume of loose soil (Loose measure) B = Volume of undisturbed soil (Bank measure) C = Volume of compacted soil (Fill or Compacted measure) S w = % of swell S h = % of shrinkage

78 Correlation between Weight, Swell, and Shrinkage w L = B / (1 + S w /100) C = B / (1 – S h /100) Where L = Unit weight of loose soil (Loose measure) B = Unit weight of undisturbed soil (Bank measure ) C = Unit weight of compacted soil (Compacted measure) S w = % of swell S h = % of shrinkage

79 Earthwork Excavation Estimation Example Volume of earthwork = 60,000 cy bank measure Production rate of backhoe = 300 cy / hr bank measure Swell factor = 25% Capacity of truck = 20 cy loose measure = (20/1.25) bank cy = 16 bank cy Hauling distance = 4 miles Average speed = 30 mph Dump time = 4 minutes Truck wait time = 3 minutes Efficiency 45 minute per hour Cost data: Cost of backhoe = $75.00/ hr Cost of operator = $35.00 /hr Cost of trucks = $45.00 /hr Cost of drivers = $30.00 /hr Cost of supervisor = $45.00 /hr Find out production rate in bank cubic yard, cost per cubic yard of bank measure, and total cost.

80 Production Rate of Hauling Solution: Cycle Time Loading time: (16 cy / 300 cy)= 0.05 hrs Hauling time and return ( 4+4) mi / 30 mph= 0.27 hrs Dump Time = 4 min = 4 min / 60 min/hr= 0.07 hr Waiting to load= 3 min= 3 min/ 60 min/hr= 0.05 hr Total cycle time = 0.05 hr hr hr hr = 0.44 hrs Production Rate Number of trips per hour = 1/ 0.44 hrs / trip = 2.3 trips Quantity hauled per hour = 16 cy/ trip x 2.3 trip/hr bank measure = 36.8 cy/ hr bank measure Assuming operating factor as 45 min/ hr Production rate= (45 min / 60 min/hr) x 36.8 bcy/hr = 27.6 bcy/hr

81 Cost Estimation of Hauling Let us find out number of trucks to balance the production rate. Loading time = 0.05 hr Total cycle time= 0.44 hr Number of trucks required= 0.44 hr / 0.05 hr / truck = 8.8 trucks Consider using 8 trucks: If 8 trucks are used, then there are less trucks used than required. Therefore, the production rate will be governed by the production rate of the trucks. Production rate of backhoe= 300 cy/hr x (45 min / 60 min/hr) = 225 cy/ hr bank measures Production rate of 1 truck= 27.6 cy/ hr bank measures Production rate of 8 trucks= 8 x 27.6 bcy / hr = bcy/ hr Use production rate of cy/ hr Cost per hour: = $ $ ($ $30.00) + $45.00 = $755/ hr Cost per cubic yard bank measure = $ /hr / bcy/hr = $3.42 per cy bank measure

82 Cost Estimation of Hauling Consider using 9 trucks: If 9 trucks are used, then there are more trucks used than required. Therefore, the production rate will be governed by the production rate of the loader. Production rate of backhoe= 300 cy/hr x (45 min / 60 min/hr) = 225 cy/ hr bank measures Production rate of 1 truck= 27.6 cy/ hr bank measures Production rate of 9 trucks= 9 x 27.6 bcy / hr = bcy/ hr Use production rate of 225 cy/ hr Cost per hour: = $ $ ($ $30.00) + $45.00 = $830/ hr Cost per cubic yard bank measure = $ /hr / 225cy/hr = $3.34 per bcy bank measure Therefore using 9 trucks is more economical Total Cost = 60,000 bcy x $3.34/bcy = $200,400

83 Sample Question from NCEES Find the productivity of truck in bank measure: Maximum vehicle weight = 37,800 lb Empty vehicle weight = 10,800 lb Heap capacity = 12 yd 3 Struck capacity = 10 yd 3 Load + haul + return + dump times = 17 min Delay time= 5 min/ hr Bulk density = 110 pcf Loose density = 100 pcf

84 Solution for productivity Find the volume of truck in bank cubic yard Weight of earth in each truck = 37,800–10,800=27,000 lbs Volume (bank) = Wt/ Bank Density = 27,000 / 110 x 27 = 9.09 yd 3 Find no. of trips per hour Trips per hour (considering 55 min.) = 55/ 17 = 3.24 trips Multiply no. of trips per hour and volume measured in bank cubic yard Productivity = 3.24 trips/ hr x 9.09 yd 3 / trip = yd 3

85 Sample Question from NCEES Size of concrete wall = 72 ft (L) x 12 ft (B) x 1 ft (thick) Build in three equal pours. Include 10% waste on concrete Labor rate: Carpenter= $32.73/hr, Laborer = $26.08/hr Supervisor = $35.37/hr Productivity: Erect forms = 5.5 ft 2 /LH, Strip forms = 15 ft 2 /LH, Place concrete = 2.2 yd 3 /LH Crews: Erect and strip : 4 Carpenters, 2 Laborers, 1 Supert. Place concrete: 3 Laborers, 1 Carpenters, 1 Supervisor Materials: Formwork Initial erection: $2.66/ft 2, Reuse (2 times) = $0.34/ft 2, Concrete = $97.20/yd 3, Reinforcing sub contract = $120/ yd 3

86 Cost Estimation Volume of concrete = 1.10(72*12*1 /27) = 35.2 cubic yard Area of Wall = 2 sides ( 72*12) = 1728 ft 2 Material Cost Form work (initial) = (1728*2.66)(1/3) = $1,532 Reuse formwork = (1728*0.34)(2/3) = $392 Concrete = 35.2* 97.20= $3,421 Reinforcing subcontract = (35.2/1.1) x $120 = $3,840 Total material cost = $9,185

87 Cost Estimation (contd.) Labor Cost: Erect hours = 1728/5.5 = hrs Strip hours = 1728/15 = hrs Total erect and strip hours = hrs Concrete placing hours = 35.2/2.2 = 16 hrs Erect and strip cost / LH = (4* * )/7 =$31.21/LH Concrete place cost/hr = (3* )/5 = $29.27/ LH

88 Cost Estimation (contd.) Labor Cost Erect and strip cost = LH x $31.21 = $13,402 Concreting placing cost = 16 LH x $29.27 = $468 Total Cost = $13,870 Total Cost Material cost = $9,185 Labor cost = $13,870 Total cost = $23,055

89 Painting Cost Estimation A 200 ft x 100 ft room has been prepared for painting. The walls are 7 ft high and will require two coats of paint on the previously painted surface for proper coverage. If 1 gal of paint covers 300 ft 2, the number of gallons of paint needed to paint the walls is mostly nearly? Area of painting = 2 sides ( 200* *7) = 4,200 ft2 Paint required for one coat = 4,200 / 300 = 14 gallons Paint required for two coats = 2 x 14 gal. = 28 gallons

90 Recommended Books for Estimating Estimating Construction Cost Robert L. Peurifoy & Garold D. Oberlender Walker Building Estimators Reference RS Means Cost Guide Engineering News Record Cost Book

91 Project Control Budgeted Cost Work Schedule (BCWS) = Measures What is Planned in terms of budget cost of the work (according to baseline schedule of work)= Budgeted Cost of Work Performed (BCWP) = Earned Value: measures What is Done in terms of the budget cost of work Actual Cost Work Performed (ACWP)= Measures What is Paid in terms of the actual cost of work that has been accomplished to date.

92 Project Control Cost Variance Schedule Variance Actual Cost Work Performed Budgeted Cost Work Schedule Budgeted Cost Work Performed

93 Cost and Schedule Variance Cost Variance: Difference between BCWP and ACWP CV = BCWP - ACWP If CV > 0, indicates cost saving Schedule Variance: Difference between the BCWP and BCWS. SV = BCWP - BCWS If SV > 0, indicates schedule advantage.

94 Sample Question from NCEES A formal CPM analysis for a project shows the planned costs to date are $85,000, and the accounting department reports charges to the job of $90,000. If the reported earned value to date is $70,000, the cost and schedule status of the project are most nearly: Solution: BCWS = $85,000 ACWP = $90,000 BCWP = $70,000

95 Solution of Project Control Problem Cost Variance = BCWP – ACWP = $70,000 - $90,000 = -$20,000 Therefore, CV < 0, means the project is over budget Schedule Variance = BCWP – BCWS = $70,000 - $85,000 = -$15,000 Therefore, SV< 0, means the project is behind schedule. Answer: The project is over budget and behind schedule.

96 Thank You


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