1. PE Review Course Construction Engineering
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 CalculationES LF
ES= Early Start
LS= Late Finish
Forward Pass

14. Early Start Date CalculationES LF
ES= Early Start
LF= Late Finish 5 14 3 22 7
Forward Pass

15. Late Start Date CalculationLF
ES= Early Start
LF= Late Finish 5 7 14 14 3 3 22 22 7 7
Backward Pass

16. Critical Path ES LF ES= Early Start LS= Late Finish 5 7 14 14 3 3 2214 14 3 3 22 22 7 7
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 Free 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 = 7-5=2
TF = 5-3=2
TF = 0
TF = 0
TF = 0

23. Free Float Calculation
Free Float (FF) = ESJ-EFI where I is the predecessor and J is successor activity.
FF = 14-12=2
FF = 5-5=0
FF = 3-3=0
TF = 7-7=0
FF = 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
Activity ES
Duration (Days) LS EF
W. Duration LF A B C
12/18/42
18/33/60
8/11/20
Find out the probability of completing the project at 90 days?

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 Detailed 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.
Projects Cost Square Foot Cost/ SF
1 $500, , $250
2 $351, , $270
3 $371, , $265
4 $550, , $220
5 $600, , $200
6 $200, , $182
Cost Indices
Years Indices
Location Indices
Location Index
Las Vegas
Austin
Los Angeles
New Orleans

50. 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 = $270
Weighted Unit Cost
= ($ 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
1.225 = (1+i)3
i = 7%
Time Adjustment factor building in the year 2009 (n = 5 years) for Las Vegas
= (1+.07)5
= 1.402
Adjustment Factor for Location
= (1050 / 1205) = 0.871
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
Worker’s 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%
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)

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. 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

57. Labor Cost Calculation
Total cost per hour = $ $ $ $ $ $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, /week x (52 weeks / 12 months)
= $10,020.40
Total cost per year = $10, /month x 12 months
= $120,244.80

58. Equipment Costs If Equipment is rented: If Equipment is purchased
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 Economic Analysis Table A = F x (A/F, i%, n)
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 Fuel 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
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

78. 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

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 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

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 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 = yd3

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 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

86. 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* = $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 ft2, 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 Estimator’s 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 Budgeted Cost Work Schedule Actual Cost Work Performed
Schedule Variance
Cost Variance
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