Time Value of Money Compound Interest Internal Rate of Return NCEES – FE Civil Engineering Topics
Construction Management 10% = 6/60 A. Procurement methods (e.g., build, design-bid-build, qualifications based)
B. Allocation of resources (e.g., labor, equipment, materials, money, time)
C. Contracts/contract law
D. Project scheduling (e.g., CPM, PERT) E. Engineering economics
F. Project management (e.g., owner/contractor/client relations, safety)
G. Construction estimating
fast facts
I. Estimating is a complex process involving collection of available and pertinent information relating to the scope of a project, expected resource consumption, and future changes in resource costs.
II. The estimating process involves a combination of evaluating information through a mental process of visualization of the constructing process for the project. This visualization is mentally translated into an approximation of the final cost.
III. At the outset of a project, the estimate cannot be expected to carry a high degree of accuracy, because little information is known. As the design progresses, more information is known, and accuracy should improve.
IV. Estimating at any stage of the project cycle involves considerable effort to gather information. The estimator must collect and review all of the detailed plans, specifications, available site data, available resource data (labor, materials, and equipment), contract
documents, resource cost information, pertinent government regulations, and applicable owner requirements. Information gathering is a continual process by estimators due to the
uniqueness of each project and constant changes in the industry environment.
V. Unlike the production from a manufacturing facility, each product of a construction firm represents a prototype. Considerable effort in planning is required before a cost estimate can be established.
Most of the effort in establishing the estimate revolves around determining the approximation of the cost to produce the one-time product.
CONSTRUCTION MANAGEMENT - PROCUREMENT METHODS
24. - Question Which of the following statements about the construction contract process are true:
I. Surety bond assures the project owner a guarantee of funds equivalent to a promissory note. The surety bond is a promise to pay the owner a certain amount if the contractor fails in fulfilling the terms of a contract.
II. Performance bond is a surety bond issued by an insurance company or a bank to guarantee satisfactory completion of a project by a contractor.
Performance bonds are issued upon Contract award and cost approximately 0.50% to 1.25% of the total contract value.
III. Builder’s risk insurance is a special type of property insurance which
indemnifies against damage to buildings while they are under construction.
Builder's risk insurance is coverage that protects a person's or an
organization's insurable interest in materials, fixtures and/or equipment being used in the construction or renovation of a building or structure should those items sustain physical loss or damage from a covered cause
IV. A bid bond is issued as part of a bidding process by the surety to the project owner, to guarantee that the winning bidder will undertake the contract under the terms at which they bid. The cash deposit is subject to full or partial forfeiture if the winning contractor fails to either execute the contract or provide the required performance and/or payment bonds. The bid bond assures and guarantees that should the bidder be successful, the bidder will execute the contract and provide the required surety bonds.
V. Bonds are not insurance. Bonds are a guarantee to pay made by a cosigner who is liable only if the principal fails to discharge the obligations under the Contract.
a. I & II b. I, II, & III c. I, II, III, & IV d. I, II, III, IV, & V
Solution: All are true.
COST ESTIMATING
25. - Question A capital improvement project requires the installation of a property line fence along the 250-ft Northern
boundary line. The decorative aluminum fence is constructed of posts spaced at 10-ft centers and an ornate picket infill panel. The material costs for this scope of work is most nearly:
a. $65,771.25 b. $66,416.60 c. $68,402.10 d. $71,065.58
Solution: Develop a Bill of Materials and multiply quantities by the costs:
Aluminum Posts: 26 x $645.35 = $16,779.10 Picket Infill Panel 25 x $1,985.50 = $49,637.50
Grand Total $66,416.60 (answer is b)
Material Costs:
Aluminum Posts $645.35 -each Picket Infill Panel $1,985.50- each Placed Concrete $498.00/CY Ironworker $78.00/hr State Sales Tax 7%
fast facts
The most common blunder during quantity take–off estimating is to omit the
“zero” position during the count. To help with the analysis, sketch the work so as to better visualize the quantity take-off. Also, material cost on a capital improvement project is non-taxable. Remember that “distracters” are included in questions to test your engineering judgment.
ESTIMATING TAKEOFF QUANTITIES
26. - Question A 450-ft length canal is to be lined with concrete for erosion control. It is estimated that there will be 10% waste. Unit material costs of concrete and curing compound based on other recent projects in the area with similar volume are $98/ yd3 and $40/5-gal, respectively.
Project specifications require an application rate of curing compound at 1-gal per 300-ft2.
[not to scale]
Determine the following:
a. The total material cost for delivered concrete is most nearly:
a. $85,000.00 b. $90,160.00 c. $99,176.00 d. $109,094.00
b. The total material cost for the concrete curing compound is most nearly:
a. $1,120.00 b. $1,160.00 c. $1,200.00 d. $1,240.00
20-ft
19-ft
2 3
8-in - concrete
Solution:
a. Horizontal length of side slope = (20 / 2) x 3 = 30 ft Slope length = √ [(20)2 + (30)2] = 36.06 ft
Cross-sectional area of lining = [(2 x 36.06) + 19] 8/12 = 60.74 ft2 Volume of lining = (60.74 x 450) / 27 = 1,012- yd3
Delivered volume (add waste) = 1,012-yd3 x 1.10 = 1,113- yd3 Material Cost = $98.00/yd3 x 1,113-yd3 = $109,094.
(answer is d)
b. Surface Area of Canal = (19-ft + 36.06-ft x 2) x 450-ft = 41,004 ft2 Quantity of Curing Compound = 41,004 ft2 / (300 ft2/gal.) = 136.68gal.
Calculate waste: 136.68 gal. x 1.10 = 150.35-gal
Convert to purchase within 5-gal containers: 150.35-gal / 5 = 30.07 containers
Material Cost = 31-containers x $40.00/container = $1,240.00 (answer is d)
fast facts
The manufacturer’s specification cannot be deviated from. This question illustrates the importance of rounding up to meet the product specifications.
The seven-hundredths of a 5-gallon container in the example is enough to support the manufacturer’s position that the coverage rate was not met. Always round up in this situation.
27. - Question A A concrete crew will use the available steel form panels measuring 2’-6” wide x 4’-0” high to construct a 40’-6” long by 3’-2”
high by 1’-6” wide concrete knee wall. The square foot of contact area for the formwork is most nearly:
a. 105-ft2 b. 134-ft2 c. 266-ft2 d. 368-ft2 Solution:
The square foot of contact area consists of the surface of the formwork
“touched” by the concrete. Therefore, apply the equation to calculate the contact area:
(40.5-ft + 1.5-ft + 40.5-ft + 1.5-ft) x 3.167-ft = 266.03-ft2 (answer is c)
Plan View
Section View 1’-6”
40’-6”
1’-6”
3’-2”
Concrete Knee Wall
Not to scale
fast facts
Cost per Square Foot of Contact Area
The cost of concrete formwork is influenced by three factors:
1. Initial cost or fabrication cost, which includes the cost of transportation, materials, assembly, and erection.
2. Potential reuse which decreases the final total cost per square foot of contact area. The data in Table indicates that the maximum economy can be achieved by maximizing the number of reuses.
3. Stripping costs, this also includes the cost of cleaning and repair. This item tends to remain constant for each reuse up to a certain point at which the total cost of repairing and cleaning start rising rapidly.
In deciding to use a specific formwork system, the initial cost should be evaluated versus the available budget for formwork cost. Some
formwork systems tend to have a high initial cost, but through repetitive reuse, they become economical.
For example, slip forms have a high initial cost, but the average potential reuse (usually over 100 times) reduces the final cost per square foot of contact area for the type of formwork.
In the case of rented formwork systems, the period of time the formwork is in use has a great effect on the cost of the formwork.
28. - Question A contractor will place 90-cubic yards of concrete for a housekeeping pad on the rooftop of a 36-ft tall building. Site conditions dictate that the safest and best method of placement is to use a crane and a 2-cubic-yard bucket. To perform the task efficiently, five Union laborers are needed — one at the concrete truck, three at the point of placement, and one on the portable internal vibrator. The wage rate for laborers is $22.00/hr (Union overtime rate is 1.5 times the wage rate after 8-hour day). Time needed for the operation is: Setup: 15-min;
Cycle time consists of Load = 3-min., plus Swing, dump and return = 6-min.
which allows a total cycle time = 9-min; Demobilize operation, 10-min.
Supervision is done by the superintendent. Allow a 10% factor for
inefficiencies during the cycle time. Crane rental cost is $1,800 per 8-hr day. The total labor cost per cubic yard for the concrete crane placement of the 90 cubic yards is most nearly:
a. $8.75 b. $9.78 c. $10.12 d. $10.65
Solution:
Identify (by underlining) relevant cost items and calculate summary quantities.
No. of cycles 90-CY/2-CY/Bucket = 45 cycles Total cycle time 45-cycles x9-min/cycle = 405 min Inefficiency(labor,delays,etc.)10%of cycle time = 41-min Setup and demobilize: Sub-total = 25 min
Total operation time: 405 + 41 + 25 = 471-min or 7.85-hrs Amount of time needed (adjusted to workday) = 8 hr
Laborers — five for 8 hours at $22.00/hr = $880.00 Total labor cost per 90-yd3 = $880.00 Cost per cubic yard $880/90-yd3 = $9.78/yd3
(answer is b)
29. - Question The Owner’s representative requested a cost proposal for the Architect’s design change. The revised Scope of Work (SOW) is to provide a credit for the installation of one layer of ½” Gypsum Wallboard (GWB) and provide a new
scope of work consisting of: (1) install two (2) layers of 5/8” Fire Code (FC) GWB on the inside face of the existing framing system, (2) as part of the building code requirement mud coat the first layer of the GWB, (3) mud coat and finish coat the second layer of GWB, (4) and provide sound-batt insulation within the existing metal stud wall cavity. The room is located within a warehouse with dimensions: 180-ft x 200-ft; 10-ft high floor to underside of deck dimension; and, eight (8) 6’-0”wide x 10’-0” high openings. The total cost for the new scope of work is most nearly:
a. $19,500 b. $20,500 c. $21,000 d. $22,500
Use the following excerpt from the Company’s cost standards:
Work Crew 4 Carpenters 2 Laborers
Working Foreman Labor Rates per hour
Carpenter Foreman (working) $46.35 fully burdened Carpenter (journeyman) $38.85 fully burdened
Laborer $28.45 fully burdened
Work Crew Productivity (based on 8-hr/day)
GWB 960 ft2/Work Crew Hr
Insulation 1,920 ft2/ Work Crew Hr Tape &Spackle Mud Coat 2,800 ft2/ Work Crew Hr Tape &Spackle Finish Coat 1,800 ft2/ Work Crew Hr
Material Costs price includes all taxes and delivery; add a 10% waste factor to all materials.
4’-0” x 10’-0” x ½” GWB $0.185/ ft2
4’-0” x 10’-0” x 5/8” GWB (FC) $0.285 / ft2 Sound Batt Insulation (65-ft2/Bag) $0.45/ ft2 Mud Coat (coverage 150- ft2/gal) $0.125/ft2 Finish Coat(coverage 300- ft2 /gal) $0.08/ft2 Contractor Change Order Pricing
Contactor’s Overhead 10%
Contractor’s Profit 5%
Solution: Step 1; Sketch the project,
[not to scale]
Step 2 - Determine GWB surface Area Walls
7,600-ft - 480-ft = 7,120-ft2
200-ft
OH&P are cumulative not additive
Round Up material quantities
Follow mfg’s application rate
Include the given waste amount
Use product coverage Qtys’
Follow bid document info
Calculate work crew rates
Use burdened labor rates
Sketch the work
Step 3 - Calculate Fully Burdened Cost Rate for Crew Day
Trade QTY U/M Rate/hr Total
Foreman ea1 $ 46.35 $ 46.35 Journeyman ea4 $ 38.85 $ 155.40 Laborer ea2 $ 28.45 $ 56.90
TOTAL $ 258.65 per crew hour
TOTAL $ 2,069.20 per crew day (8-hrs)
Step 5 - Calculate Change Order amount for the new SOW
Description QTY U/M Production U/M Unit Cost U/M Crew Hrs
Total Labor
Step 4 - Determine the Material List for the new Scope of Work
5/8" GWB 7,120 x 2-layers x 1.10 (waste) / 40-SF/BD = 392 boards Sound Batt 7,120 / 65ft^2/bag x 1.10 (waste) = 121 bags Mud Coat 14,240 /150-ft^2/gal x 1.10 (waste) = 105 gallons Finish 7,120 /300-ft^2/gal x 1.10 (waste) = 27 gallons
fast facts
Construction Estimating calculates the total fully burden cost for labor, material and equipment. Once the “raw” costs are determined, the contractor’s Over Head and Profit (plus Bond) are added to the bottom line costs.
COST ESTIMATING – BOARD FEET
Sample calculations provided in Table below for 2” thick stock.
fast facts
One board foot equals 144 cubic inches. The thickness in inches, times the width in inches, times the length in inches, divided by 144 cubic inches, equals total board feet in a piece of stock. For instance, a piece two inch thick by twelve inches wide by twelve inches long would be 2" x 12" = 24" x 12" = 288 cubic inches ÷ 144 cubic inches = 2 board feet.
30. - Question A project engineer has ordered 100 hardwood boards, 1-1/4-in thick by 8-in wide and 8-ft long. The total board feet ordered is most nearly:
a. 533 b. 600 c. 667 d. 1,067 Solution:
1.25” x 8” x 8’ x 100-pcs = 666.67 board feet (answer)
12
Or, 1.25” x 8” x (8’x 12”/1’) x 100-pcs = 666.67 board feet (answer is c) 144
fast facts
The questions on the NCEES exam generally do not include units with the associated answer choices. You are more likely to see answers such as 20, 40, 60, and 80, rather than 20-PSF, 40-PSF, etc.
Read the problem statement carefully to ensure that you know what units to solve for. Some of the answers are logical distracters and are only included to test your “engineering judgment”.
Four minutes per question requires focusing your attention to the time variable of the exam.
METHODS OF BUDGETING
31. - Question The total equipment cost for an addition to a pharmaceutical plant is estimated to be $5,000,000. The percentages provided in the Table represent the average expenditures in other cost phases within the project budget. The total cost for the project is most nearly:
a. $5,000,000 b. $12,500,000 c. $14,000,000 d. $16,700,000
Solution:
Calculate the total percentage of the items provided:
Total percentage = 70%
Apply the equation:
Total Project Cost = $5,000,000 = $16,666,667 (answer is d)
(1 – .70)
Description Percent
Engineering, overhead, and fees 22%
Equipment Storage 5%
Services 2%
Utilities 6%
Piping 20%
Instrumentation 5%
Electrical 6%
Buildings 4%
Estimated Equipment Cost = Total Project Cost (1 – Percent of Other Costs)
CONSTRUCTIONHISTORIC DATA
32. - Question A concrete specialty company’s uses productivity standards calculated as averages from historical data. On average 100-square feet of formwork requires 6 hours of carpenter time and 5
hours of common laborer time. The wage rate for a carpenter is $38.97/hr plus a 54% burden rate. The wage rate for common laborers is $14.87/hr plus a 48% burden rate. The total square foot cost is most nearly:
a. $3.08/ft2 b. $3.25/ft2 c. $4.25/ft2 d. $4.70/ft2
Solution: Burden Rates are amounts charged over and above the actual costs for labor, materials and/or taxes. Add the allotted burden rate to the trade labor rate to determine the total SF cost.
The unit cost may be calculated as follows:
Carpenter — 6 hr at $60.00/hr = $360.00 Laborer — 5 hr at $22.00/hr = $110.00 Total labor cost for 100 ft2 = $470.00
Labor cost per ft2 = $470.00 ÷ 100-ft2 = $4.70-ft2 (answer is d)
fast facts
Labor burden is the cost to a company to carry their labor force aside from salary actually paid to them. Simply stated, burden is the benefits and taxes that a company must pay on their payroll. It is important to stress that burden typically should not include any profit, markup or expenses unrelated to employee compensation, but should be the actual cost to carry the labor. These can include, but are not limited to, all of the following:
Payroll Taxes – both Federal and State (Statutory)
When applicable, Union Fringe Benefits Package
Vacation Pay allocation
Retirement/Pension Costs
Health Care
Life/Accidental Death & Dismemberment Insurance (AD&D)
Worker’s Compensation Costs
Long-Term Disability Insurance
Short-Term Disability Insurance
F ENGINEERING ECONOMICS
P – Present worth of money F – Future worth of money
A – An end-of-period cash receipt or disbursement in a uniform series i – Interest rate per interest period
n – Number of interest periods Single Payment Present Worth P = F(P/F, i, n) = n
i) (1
F
Interest Formulas: Payments
Single Payment Compound Amount Factor (F/P, i%, n) = (1 + %, i )n
Single Payment Present Worth Factor
(P/F, i%, n) = 1/ (1 + %, i )n= 1/ (F/P, i%, n) Uniform Series Compound amount Factor
(F/A, i%, n) = (1 + %, i )n - 1 / i Uniform Series Sinking Fund Factor
(A/F, i%, n) = i / (1 + i)n - 1 = 1 / (F/A, i%, n)
P
fast facts
Factor Tables are derived from the equations shown in the introductory pages of this section. The Factor Tables are a convenience, however, not all % factors are available and although interpolation can be used, it is strongly recommended to become as familiar with the equations. FE – Ref Hb. provides a summary of the varying equations used in engineering economics.
See pages 114 through 120 in the NCEES Supplied Reference Handbook for Factor Tables
For example, where
F/P
is the column selector for the Factor Tables, the interpretation is:Calculate “F” / Given “P”
Where: A = Annual Amount F = Future Worth P = Present Worth
G = Uniform Gradient Amount
n = number of compounding periods or life of asset i = effective rate per period
p
FACTOR TABLE QUICK VIEW EXERCISE
The following factor table quick exercise uses the value of $1.00 using 10% for 10-yrs.
Using the appropriate Factor Table found in the NCEES Supplied Reference Handbook, find the appropriate answer. Use the Calculate “x” / Given “y” study aid.
a. If you want $1.00, 10-yrs from now, deposit $_____ now.
a. 0.3855
b. If you deposit $1.00 at the end of every year for 10-yrs the present value is?
a. 0.3855
c. $1.00 today is worth $_______10-yrs from now in an account yielding 10%
a. 0.3855
d. The annual amount of interest deposited in the account with a starting balance of $1.00 at 10% for 10-yrs is?
e. If you want to have $1.00 in the bank, deposit $_____every year for 10–years at 10% an annual yielding account.
a. 0.3855 b. 6.1446
c. 2.5937 d. 0.1627
e. 0.0627 f. 15.9374 g. 22.8913
f. If you deposit $1.00 every year in an account yielding 10%, you will have this amount in 10-yrs $__________.
a. 0.3855 b. 6.1446
c. 2.5937 d. 0.1627
e. 0.0627 f. 15.9374 g. 22.8913
g. If you deposit $1 in yr-1; $2-at the beginning of yr-2; $3 at the beginning of yr-3, and so on to the 10th year, the present worth of the deposits is $__________.
a. 0.3855 b. 6.1446
c. 2.5937 d. 0.1627
e. 0.0627 f. 15.9374 g. 22.8913
TIME VALUE OF MONEY
33. - Question If you want to have $60,000 in 10 years, the amount that should be put into a 6.0% (effective annual rate) savings account now is most nearly:
a. $33,503.69 b. $43,000.39 c. $48,475.09 d. $53,500.60 Solution:
This problem could also be stated: What is the equivalent present worth of $60,000 ten years from now if monthly money is worth 6% per year?
P = F(P/F, I, n) = $60,000(P/F, 6%, 10) = $60,000 * 0.5584 = $33,503.69 (answer is a)
34. - Question The cost of utilities, taxes and maintenance on a home is $3,000 per year. The amount of money that would have to be invested now at 8% to cover these expenses for the next 5 years is most likely: (Assume no inflation or tax increase).
a. $10,980 b. $11,980 c. $12,980 d. $13,980 Solution:
Referencing Appendix 86-A (pg. A-136) in CERM-11 and using the Factor Tables
$3,000 (P/A, 8%, 5) = $3,000 x 3.9927 =
$11,978.10 (answer is b)
p
35. - Question Your home mortgage is $300,000 for 30 years with a nominal annual rate of 7%. The monthly payment is most nearly:
a. $1,899.00 b. $1,900.00 c. $1,995.10 d. $2,015.00 Solution:
n = 360 months interest = 7%-annual ÷ 12-months/year = 0.583% per month
$300,000(A/P, 0.00583, 360) = Apply the equation:
= 0.006650339
A = 300,000 x 0.006650339 = $1,995.10 per month (answer is c) i(1+i)n (1+i)n - 1
fast facts
This question illustrates the importance of interpreting the information provided before running through the computations. Although the nominal annual rate is given as 7%, the monthly rate needs to be computed. A common approach would be to use the CERM Appendix Factor Tables to find the monthly payments which would yield:
(A/P, 7%, 30) = $300,000(A/P, 0.0806, 30) = $24,180/12 = $2,015. / month or a 1%
error.
Conclusion: Be familiar and comfortable with both the Factor Tables and the Equations which comprise the results in the Tables.
p
36. - Question You wish to buy a house and you can afford to make a down payment of $50,000. Your monthly mortgage payment cannot exceed $2,000. If 30-year loans are available at 7.5% annual interest rate which is compounded monthly, the highest price that you may consider is most nearly:
a. $332,000 b. $334,000 c. $336,000 d. $338,000 Solution:
n = 360 months interest = 7.5%-annual ÷ 12-months/year = 0.625% per month Apply the equation:
This yields (A/P, 0.00625, 360) = 0.00699
Apply the following equation to find the highest price to consider:
P x 0.00699 = (Highest $ - $50,000) x 0.00699 ≤ $2,000
Highest $ ≤ ($2,000/0.00699) + $50,000 = $336,123 (answer is c) i(1+i)n
(1+i)n - 1
fast facts
Alternate method - Use one year as the time period.
Then, n = 30 years, and i = (1 + 0.00625)12 – 1 = 7.763%
Then, (A*/P, 0.07763, 30) = 0.0867 [A* = 0.0867 P per year]
Your effective payment per year is:
A* = $2,000 x (F/A, 0.00625, 12) = $2,000 x12.4212 = $24,842 P ≤ (24,842 / 0.0867) + $50,000 = $336,533 same as before;
answer checks.
p
37. - Question An engineer deposits a “X” amount every 6 months for 3 years so that she’ll have $10,000 at the end of this period. The interest rate is 5% per year which is rebalanced every 6-month. The amount deposited is most nearly:
a. $1,565.50 b. $1,585.50 c. $1,595.50 d. $1,600.50
Solution: Determine the components and apply the equation:
n = 6 deposits i = 2.5% per 6-month period
F = $10,000(A/F, 2.5%, 6) = 0.15655 A = $1,565.50 (answer is a) COMPOUND INTEREST – NOMINAL AND EFFECTIVE
38. - Question A credit card advertises a nominal rate of 18%
compounded monthly. If no monthly payments are made, the effective annual interest rate is most nearly:
a. 18.00%
b. 18.25%
c. 18.75%
d. 19.56%
The actual rate is, (18% / 12) = 1.5% per month. The effective annual rate is:
i = (1 + .015)12 – 1 = 0.1956 or 19.56% if you do not pay anything each month. (answer is d)
fast facts
This question illustrates that 18% interest per month is equivalent to 19.56% effective annual rate. The terms are synonymous as they are dependent upon a “point in time”.
p
p
fast facts
Compound interest is the concept of adding accumulated interest back to the principal, so that interest is earned on interest
Compound interest is the concept of adding accumulated interest back to the principal, so that interest is earned on interest