PE Review Course Construction Engineering

1
PE Review CourseConstruction Engineering

• Pramen P. Shrestha, Ph.D., P.E.
• February 2, 2009

2
Topics to be Covered

• Construction Estimating
• Construction Scheduling
• Project Controls
• Engineering Economics

3
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

4
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

5
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

6
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

7
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

8
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

9
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

10
Adjustments

• Time
• Location
• Size
• Complexity
• Need appropriate contingency

11
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

12
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
• Adjustment for time
• Find out the average yearly interest rate
• 4877 / 3980 (1i)n
• Where is i average yearly interest rate
• n number of years 3
• Substituting the n value
• 1.225 (1i)3
• i 7

13
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

14
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

15
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

16
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)

17
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

18
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. 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

19
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

20
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

21
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

22
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

23
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

24
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

25
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

26
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)
27
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)
28
Engineering Economic Table
29
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

30
Equipment Annual Ownership Cost

• Net equipment value 90,000
• Expected salvage value 20,000
• Useful life 5 years
• Interest Rate 17

31
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

32
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

33
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

34
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

35
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

36
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

37
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)

38
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

39
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

40
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

41
Earthwork Excavation Estimation

• Example
• Volume of earthwork 60,000 cy bank measure
• Production rate of backhoe 300 cy / hr bank
  measure
• Hauling distance 4 miles
• Capacity of truck 20 cy loose measure
• 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 and
  cost per cubic yard of bank measure.

42
Production Rate of Hauling

• Solution
• Cycle Time
• Loading time (16 cy / 300 cy) 0.05 hrs
• Hauling time and return ( 44) 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
  cy/hr
• 27.6 cy/hr

43
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 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 8 trucks 9 x 27.6 cy / hr
• 248.4 cy/ hr
• Use production rate of 220.8 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
  / 248.4cy/hr
• 3.34 per cy bank measure

44
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
• Bank density 110 pcf
• Loose density 100 pcf

45
Solution for productivity

• Find the volume of truck in bank cubic yard
• Weight of earth in each truck
  37,80010,80027,000 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

46
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 Sup
• Place concrete 3 Laborers, 1 Carpenters, 1
  Supervisor
• Materials Formwork Initial erection 2.66/ft2,
  Reuse 0.34/ft2, Concrete 97.20/yd3,
  Reinforcing sub contract 120/ yd3

47
Cost Estimation

• Volume of concrete 1.10(72121 /27) 35.2
  cubic yard
• Area of Wall 2 sides ( 7212) 1728 ft2
• Material Cost
• Form work (initial) (17282.66)(1/3) 1,532
• Reuse formwork (17280.34)(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

48
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 (432.73 226.08
  35.37)/7
• 31.21/LH
• Concrete place cost/hr (326.08
  32.7335.37)/5
• 29.27/ LH

49
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

50
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 ( 2007 1007)
  4,200 ft2
• Paint required for one coat 4,200 / 300 14
  gallons
• Paint required for two coats 2 x 14 gal. 28
  gallons

51
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

52
Construction Scheduling

• Project Scheduling
• Methods to Calculate Total Project Duration
• Critical Path Method (CPM)
• Precedence Diagram Method (PDM)
• Float Calculation

53
Project Scheduling

• To arrange the project activities in order to get
  the total project completion duration
• Predecessor and Successor

54
Activity Relationships

• Finish to Start Relationship

55
Activity Relationships

• Finish to Finish Relationship

56
Activity Relationships

• Start to Start Relationship

57
Activity Relationships

• Start to Finish Relationship

SF/5
58
Activity Relationships with Lag

• Finish to Start Relationship with Lag
• Lag means delayed

FS, Lag 3
59
Methods to Calculate Total Project Duration

• Bar Chart
• Critical Path Diagram (CPM)
• Precedence Diagram Method (PDM)
• Program Evaluation and Review Technique (PERT)

60
Bar Chart
61
Critical Path Method (CPM)
Activity on Arrow (AOA)
62
Early Start Date Calculation
ES
LF
ES Early Start LS Late Finish
Forward Pass
63
Early Start Date Calculation
ES
LF
ES Early Start LF Late Finish
5
0
14
3
22
7
Forward Pass
64
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
65
Critical Path
ES
LF
ES Early Start LS Late Finish
5
7
0
0
14
14
3
3
22
22
7
7
Critical Path
66
Precedence Diagram Method
Activity on Node (AON)
67
PDM (Forward Pass)
68
PDM (Backward Pass)
69
PDM (Critical Path)
70
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

71
Total Float Calculation
Total Float (TF) LS- ES LF-EF
TF 7-52
TF 5-32
TF 0
TF 0
TF 0
72
Free Float Calculation
Free Float (FF) ESj-EFi
FF 14-122
FF 5-50
FF 3-30
FF 7-70
FF 14-140
73
Sample Question from NCCES
74
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

75
Forward and Backward Pass
76
Critical Path
77
Total and Free Float Calculation
78
TF and FF Calculation
79
Answer

• Total Float of Activity E 3 days
• Free Float of Activity E 3 day
• 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.

80
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 (a4mb)/6
• Variance (b-a)/62
• Standard Deviation Square Root of Variance

81
PERT Problem
Find out the probability of completing the
project at 90 days?
82
Calculating Weighted Duration
83
Variance Calculation
84
Probability Calculation

• Total Variance of Critical Path 25494 78
• Standard Deviation 781/2 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

85
Standard Normal Curve
86
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.

87
Project Control
Budgeted Cost Work Schedule
Actual Cost Work Performed
Schedule Variance
Cost Variance
Budgeted Cost Work Performed
88
Cost and Schedule Variance

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

89
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

90
Solution of Project Control Problem

• Cost Variance BCWP ACWP
• 70,000 - 90,000 -20,000
• Therefore, CV gt 0, means the project is over
  budget
• Schedule Variance BCWP BCWS
• 70,000 - 85,000 -15,000
• Therefore, SVgt 0, means the project is behind
  schedule.
• Answer The project is over budget and behind
  schedule.

91
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

92
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

93
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) Pi P(1iN)

94
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( 1iN)
• 10,000 (1.05x5)
• 10,000 x 1.25
• 12,500

95
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 (1i) N

96
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( 1i)N
• 10,000 (1.05)5
• 10,000 x (1.05)5
• 10,000 x 1.276
• 12,760

97
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)

98
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

99
Declining-Balance Method

• An equal percentage of the value of an asset is
  expensed in each year
• If purchase price (P) 100,000
• Number of years (N) 5
• Fixed percentage 100/ 5 20
• First year depreciation 20 of 100,000
• 20,000
• Book value of an asset 100,000 - 20,000
• 80,000
• Second year of depreciation 20 of 80,000
• 16,000

100
Double-Declining-Balance

• Similar to declining-balance method
• First the percentage of declining is calculated
  by dividing 100 percent with number of useful
  life
• Then this percentage is double
• If useful life (N) 10 years
• Percentage of declining 100/ 10 10
• Double declining percentage 2 x 10 20
• Perform the calculation as in declining-balance
  method

101
MACRS Method

• Modified Accelerated Cost Recovery System (MACRS)
• U.S. Government has provided guidelines to
  calculate depreciation charges
• Table gives to depreciate in each year
  depending upon useful life of equipment

102
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)

103
Present Worth Analysis

• Single Payment
• P Present value, F Future value, i interest
  rate
• n number of years
• Uniform Series Present Worth

104
Future Worth Analysis

• Single Payment Compound Amount
• Uniform Series Compound Amount

105
Engineering Economic Table
106
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

107
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

108
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.

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Solution

• Benefit from alternative plan 250,000 -
  248,000
• 2,000/ yr
• Convert all the extra cost of Alternative Plan to
  Present value
• 10,0000.2638 1,0000.1638 2,474/yr
• Benefit / Cost 2,000 / 2,474 0.8

110

• Thank You