PE Review Course Construction Engineering

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# PE Review Course Construction Engineering - PowerPoint PPT Presentation

PE Review Course Construction Engineering. Pramen P. Shrestha, Ph.D., P.E. October 5, 2009. Topics to be Covered. Construction Estimating Construction Scheduling Project Controls Engineering Economics. Construction Estimating. Construction Cost consists of Direct Cost

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### PE Review CourseConstruction Engineering

Pramen P. Shrestha, Ph.D., P.E.

October 5, 2009

Topics to be Covered
• Construction Estimating
• Construction Scheduling
• Project Controls
• Engineering Economics
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
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
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]
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
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
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
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
• Time
• Location
• Size
• Complexity
• Need appropriate contingency
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 2004. Now a contractor has to build a house (2000 SF) in New Orleans, in 2009. Estimate the cost of that house using conceptual estimating method.
• Projects Cost Square 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
• Years Indices
• 2004 3980
• 2005 4339
• 2006 4614
• 2007 4877
• Location Indices
• Location Index
• Las Vegas 1205
• Austin 1000
• Los Angeles 1665
• New Orleans 1050
Weighted Unit Cost Estimate
• Solution:
• From historical data:
• Average cost of building per SF = (\$250 + \$270 + \$265 + \$220 + \$200 + \$182)/6 = \$231.17
• Minimum SF Rate = \$182
• Maximum SF Rate = \$270
• Weighted Unit Cost
• = (\$182 + 4 x \$231.17 + \$270) / 6
• = \$229.45 / SF
• Conceptual cost estimate for 2,000 SF of building in Las Vegas, in 2004
• = 2,000 SF x \$229.45/ SF
• = \$458,900
• 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
• 1.225 = (1+i)3
• i = 7%
• Time Adjustment factor building in the year 2009 (n = 5 years) for Las Vegas
• = (1+.07)5
• = 1.402
• = (1050 / 1205) = 0.871
• Adjusted Cost for the building
• = 1.402 x 0.871 x \$458,900
• = \$560,382
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
Cost of Labor
• Straight time or overtime wage
• Social security tax (FICA)
• Unemployment compensation tax
• Worker’s compensation insurance
• Public liability and property damage insurance
• Fringe Benefits
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
Labor Cost Calculation
• Brick layers Cost
• Base wage = \$22.50/hr
• FICA Tax = 7.65%
• Unemployment Tax = 3.00%
• Worker’s 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)

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
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
• = 0.0765 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. Worker’s 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
Labor Cost Calculation
• Total cost per hour = \$28.12 + \$2.15 + \$0.84 + \$3.37 + \$0.563 + \$3.50

= \$38.54

• Total cost per day = 10 hours x \$38.54 /hours

= \$385.40

• Total cost per week = 60 hours x \$38.54 /hours

= \$2,312.40

• Total cost per month = \$2,312.40 /week x (52 weeks / 12 months)

= \$10,020.40

• Total cost per year = \$10,020.40 /month x 12 months

= \$120,244.80

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
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
Investment Costs
• 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
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

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
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
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)

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)

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

Lubricating Oil Consumption Cost
• Oil Consumed per hour

= [HP x OF x 0.006 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)

= 0.089 gal/hr

Handling and Transporting Material
• Material being loaded in truck
• Material being transported to site
• Material being dumped on site
• Return (Empty)
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
• Other factors
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
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)
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
Correlation between Volume, Swell, and Shrinkage
• L = (1 + Sw /100) B
• C = (1 – Sh /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)

Sw = % of swell

Sh = % of shrinkage

Correlation between Weight, Swell, and Shrinkage
• L = B / (1 + Sw /100)
• C = B / (1 – Sh /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)

Sw = % of swell

Sh = % of shrinkage

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.

Production Rate of Hauling
• Solution:
• Cycle Time
• 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 + 0.27 hr + 0.07 hr + 0.05 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
Cost Estimation of Hauling
• Let us find out number of trucks to balance the production rate.

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

= 220.8 bcy/ hr

Use production rate of 225 cy/ hr

Cost per hour:

= \$75.00 + \$35.00 + 8 (\$45.00 + \$30.00) + \$45.00

= \$755/ hr

Cost per cubic yard bank measure = \$755.00 /hr / 220.8 bcy/hr

= \$3.42 per cy bank measure

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

= 248.4 bcy/ hr

Use production rate of 225 cy/ hr

Cost per hour:

= \$75.00 + \$35.00 + 9 (\$45.00 + \$30.00) + \$45.00

= \$830/ hr

Cost per cubic yard bank measure = \$830.00 /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
Sample Question from NCCES
• Find the productivity of truck in bank measure:
• Maximum vehicle weight = 37,800 lb
• Empty vehicle weight = 10,800 lb
• Heap capacity = 12 yd3
• Struck capacity = 10 yd3
• Load + haul + return + dump times = 17 min
• Delay time = 5 min/ hr
• Bulk density = 110 pcf
• Loose density = 100 pcf
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 yd3

• 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 yd3/ trip = 29.45 yd3
Sample Question from NCCES
• 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 ft2/LH, Strip forms = 15 ft2/LH, Place concrete = 2.2 yd3/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/ft2, Reuse (2 times) = \$0.34/ft2, Concrete = \$97.20/yd3, Reinforcing sub contract = \$120/ yd3
Cost Estimation
• Volume of concrete = 1.10(72*12*1 /27) = 35.2 cubic yard
• Area of Wall = 2 sides ( 72*12) = 1728 ft2
• 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

Cost Estimation (contd.)
• Labor Cost:
• Erect hours = 1728/5.5 = 314.2 hrs
• Strip hours = 1728/15 = 115.2 hrs
• Total erect and strip hours = 429.4 hrs
• Concrete placing hours = 35.2/2.2 = 16 hrs
• Erect and strip cost / LH = (4*32.73 + 2*26.08 + 35.37)/7

=\$31.21/LH

• Concrete place cost/hr = (3*26.08 + 32.73+35.37)/5

= \$29.27/ LH

Cost Estimation (contd.)
• Labor Cost
• Erect and strip cost = 429.4 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

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 ft2, the number of gallons of paint needed to paint the walls is mostly nearly?
• Area of painting = 2 sides ( 200*7 + 100*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
Recommended Books for Estimating
• Estimating Construction Cost

Robert L. Peurifoy & Garold D. Oberlender

• Walker’ Building Estimator’s Reference
• RS Means Cost Guide
• Engineering News Record Cost Book
Construction Scheduling
• Project Scheduling
• Methods to Calculate Total Project Duration
• Critical Path Method (CPM)
• Precedence Diagram Method (PDM)
• Float Calculation
Project Scheduling
• To arrange the project activities in order to get the total project completion duration
• Predecessor and Successor

Predecessor Activity

Successor Activity

Excavate Earthwork

Place Formwork

Activity Relationship = Finish to Start (FS)

Activity Relationships
• Finish to Start Relationship

Predecessor Activity

Successor Activity

Excavate Earthwork

Place Formwork

Activity Relationship = Finish to Start (FS)

Activity Relationships
• Finish to Finish Relationship

Predecessor Activity

Successor Activity

Excavate Earthwork

Place Formwork

Activity Relationship = Finish to Finish (FF)

Activity Relationships
• Start to Start Relationship

Predecessor Activity

Successor Activity

Excavate Earthwork

Place Formwork

Activity Relationship = Start to Start (SS)

Activity Relationships
• Start to Finish Relationship

Predecessor Activity

Successor Activity

SF/5

Order Concrete from Supplier

Place Concrete in Formworks

Activity Relationship = Start to Finish (FS)

Activity Relationships with Lag
• Finish to Start Relationship with Lag
• Lag means delayed

Predecessor Activity

Successor Activity

FS, Lag =3

Excavate Earthwork

Place Formwork

Activity Relationship = Finish to Start (FS) with Lag

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

Activity on Arrow (AOA)

Early Start Date Calculation

ES

LF

ES= Early Start

LS= Late Finish

Forward Pass

Early Start Date Calculation

ES

LF

ES= Early Start

LF= Late Finish

5

0

14

3

22

7

Forward Pass

Late Start Date Calculation

ES

LF

ES= Early Start

LF= Late Finish

5

7

0

0

14

14

3

3

22

22

7

7

Backward Pass

Critical Path

ES

LF

ES= Early Start

LS= Late Finish

5

7

0

0

14

14

3

3

22

22

7

7

Critical Path

Precedence Diagram Method

Activity on Node (AON)

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
Total Float Calculation

Total Float (TF) = LS- ES = LF-EF

TF = 7-5=2

TF = 5-3=2

TF = 0

TF = 0

TF = 0

Free Float Calculation

Free Float (FF) = ESJ-EFI

FF = 14-12=2

FF = 5-5=0

FF = 3-3=0

TF = 7-7=0

FF = 14-14=0

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

• 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.
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
PERT Problem- Critical Activities

Find out the probability of completing the project at 90 days?

Probability Calculation
• Total Variance of Critical Path = 25+49+4 = 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 = 0.9936 = 99.4%

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.
Project Control

Budgeted Cost Work Schedule

Actual Cost Work Performed

Schedule Variance

Cost Variance

Budgeted Cost Work Performed

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

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.

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

Engineering Economic Analysis
• Time Value of Money
• Simple interest
• Compound interest
• Depreciation
• Methods of depreciation
• Economic evaluation of construction project
• Absolute measures of worth
• Relative measures of worth
Time Value of Money
• A dollar today is not worth a dollar tomorrow
• Types of interest
• Simple interest:
• Interest is paid only on principal, not on the earned interest
• If P is Principal, i is interest rate per year, n is number of years,
• Interest = PiN
• Future value of money (F) = P+i = P(1+iN)
Simple Interest Problem
• Principal (P) = \$10,000
• Interest Rate (i) = 5% per year
• Number of years (N) = 5 years

Future Value of Money

F = P( 1+iN)

= \$10,000 (1+.05x5)

= \$10,000 x 1.25

= \$12,500

Compound Interest
• Compound interest is paid on the principal amount and on any interest that has preciously accrued over time.
• If P is principal, i is interest rate, N is number of years
• Then Future value of money

F = P (1+i) N

Compound Interest Problem
• Principal (P) = \$10,000
• Interest Rate (i) = 5% per year
• Number of years (N) = 5 years

Future Value of Money

F = P( 1+i)N

= \$10,000 (1+.05)5

= \$10,000 x (1.05)5

= \$10,000 x 1.276

= \$12,760

Depreciation
• Whenever equipment is purchased, it is expensed over time according to pre-defined set of rules.
• It is required for tax purpose
• Purchase cost is P
• If the equipment can be sold at the end of its useful life, the cost is called Salvage Value (S)
• Total depreciable amount = (P-S)
Methods of Depreciation
• Straight line method
• The asset is expensed an equal amount in each year over its useful life
• If D is depreciation per year, and N is useful life in years

D = (P-S) / N

Economic Evaluation of Projects
• Minimum Attractive Rate of Return (MARR)
• MARR = Borrowing rate + Equivalent interest rate (Risk, Bonds etc.)
• Economic Evaluation is conducted to make “go” or “No go” decision
• Present Worth
• Future Worth
• Annual Equivalent Worth
• Internal Rate of Return (IRR)
• Benefit/ Cost Ratio (B/C Ratio)
Present Worth Analysis
• Single Payment
• P= Present value, F= Future value, i = interest rate
• n = number of years
• Uniform Series Present Worth
Future Worth Analysis
• Single Payment Compound Amount
• Uniform Series Compound Amount
Internal Rate of Return
• Analyzing the investment according to its return
• Calculate the interest rate from the known P, A, and n value
• The interest calculated is called IRR
• If IRR is greater than MARR, accept the project
• If IRR is less than MARR, reject the project
• If IRR is equal to MARR, indifferent about the project
Benefit Cost Ratio
• Way of evaluating public projects
• Convert the benefits for life cycle of project to present worth value
• Calculate the cost of project in present worth
• Calculate the Benefit/ Cost ratio
• If the project B/C ratio is greater than 1, accept it
• If it is less than 1, reject it
• If it is equal to 1, indifferent about the project
Sample Question from NCEES
• After purchasing a quarry and basic crushing equipment, the contractor is considering an alternative plan to improve the operation of the quarry. The alternative plan will produce an equal amount of crushed rock and equal revenue.
Solution
• Benefit from alternative plan = \$250,000 - \$248,000

= \$2,000/ yr

• Convert all the extra cost of Alternative Plan to Present value

= 10,000*0.2638 – 1,000*0.1638 = \$2,474/yr

Benefit / Cost = 2,000 / 2,474 = 0.8