solved homework due...

Spreadsheet Problem

Type of Contract Markup Contract Price

1. Lump sum M = (R+R1)E B = (1 + R + R1)E

2. Unit price M = (R+R2)E B = (1 + R + R2)E

3. Cost plus fixed % M = RA = RE B = (1 + R)E

4. Cost plus fixed fee M = RE B = (1 + R)E

5. Cost plus variable % M = R(2E - A) = RE B = (1 + R)E

6. Target estimate M = RE + N(E - A) = RE B = (1 + R)E

7. Guaranteed maximum price M = (R+R3)E B = (1 + R + R3)E


1. Lump sum =(0.1+0.02)*6000 =(1+0.1+0.02)*6000

2. Unit price =(0.1+0.01)*6000 =(1+0.1+0.01)*6000

3. Cost plus fixed % =0.1*6000 =(1+0.1)*6000

4. Cost plus fixed fee =0.1*6000 =(1+0.1)*6000

5. Cost plus variable % =0.1*(2*6000-6000) =(1+0.1)*6000

6. Target estimate =0.1*6000+0.5*(6000-6000) =(1+0.1)*6000

7. Guaranteed maximum price =(0.1+0.05)*6000 =(1+0.1+0.05)*6000

Solved Homework due 27-Feb-02

Original Estimated Contract Prices (in $ thousands)

Analyze the following construction project, comparing the relative costs of each of the seven

contracts types presented in class. The contractor's original estimate is $6,000,000, and the markup

percentages are: R = 10%, R1 = 2%, R2 = 1%, and R3 = 5%. The shared savings factor is .5. Determine the contractor's markup and the original contract price (B) for each of the contract types.

Original Estimated Contract Prices

University of Delaware

Department of Civil and Environmental Engineering

Construction Methods & Management

CIEG 467/667-012

Solved Homework due 27-Feb-02

Spreadsheet Problem

Analyze the following construction project, comparing the relative costs of each of the seven

contracts types presented in class. The contractor's original estimate is $6,000,000, and the markup percentages are: R = 10%, R1 = 2%, R2 = 1%, and R3 = 5%. The shared savings factor is .5.

Determine the contractor's markup and the original contract price (B) for each of the contract types.

Original Estimated Contract Prices


1. Lump sum M = (R+R1)E B = (1 + R + R1)E

2. Unit price M = (R+R2)E B = (1 + R + R2)E

3. Cost plus fixed % M = RA = RE B = (1 + R)E

4. Cost plus fixed fee M = RE B = (1 + R)E

5. Cost plus variable % M = R(2E - A) = RE B = (1 + R)E

6. Target estimate M = RE + N(E - A) = RE B = (1 + R)E

7. Guaranteed maximum price M = (R+R3)E B = (1 + R + R3)E

Original Estimated Contract Prices (in $ thousands)


1. Lump sum 720 6720

2. Unit price 660 6660

3. Cost plus fixed % 600 6600

4. Cost plus fixed fee 600 6600

5. Cost plus variable % 600 6600

6. Target estimate 600 6600

University of Delaware

Department of Civil and Environmental Engineering

Construction Methods & Management

CIEG 467/667-012

3-1. Estimate the actual bucket load in bank measure for a hydraulic excavator – backhoe whose heaped bucket capacity is 2.10 CY (1.6 CM). The machine is excavating sand and gravel

Bucket Fill Factor (average) = 0.95 (Table 3-2)

Load Factor = 0.89 (Table 2-5)

3-1. Estimate the actual bucket load in bank measure for a hydraulic excavator – backhoe whose heaped bucket capacity is 2.10 CY (1.6 CM). The machine is excavating sand and gravel

Bucket Loadbank = heaped volume x bucket fill factor x

load factor

= 2.10 x 0.95 x 0.89 = 1.78 BCY

= 1.6 x 0.95 x 0.89 = 1.35 BCM

3-3. A 2-CY (1.53 CM) dragline is being used to excavate a canal in common earth. The average swing angle is 70º, the average depth of cut is 8.9 ft. (2.7 m), and the job efficiency is 50 min/hr. Estimate the dragline’s hourly production in loose measure.

Job efficiency = 50/60 = 0.83

Swing-depth factor = 1.06 (Table 3-9)

(70º swing, 90% optimum depth)

2.7

3.0% optimum depth = x 100 = 90%

8.9

9.990%x 100 =% optimum depth =

Ideal output = 230 BCY/hr (176 BCM/hr) (Table 3-7)

Optimum depth = 9.9 ft. (3.0 m) (Table 3-8)

Production = ideal output x swing-depth x job efficiency

= 230 x 1.06 x 0.83 = 202 BCY/hr

= 176 x 1.06 x 0.83 = 155 BCM/hr

3-5. An hydraulic excavator-backhoe is excavating the basement for a building. Heaped bucket capacity is 1.5 CY (1.15 CM). The material is common earth with a bucket fill factor of 0.90. Job efficiency is estimated to be 50min/hr. The machine’s maximum depth is of cut 24 ft (7.3 m) and the average digging depth is 13 ft (4.0 m). Average swing angle is 90º. Estimate the hourly production in bank measure.

Standard cycles/hr = 160 (Table 3-3)

% maximum depth = 13

24x 100 = 54%

% maximum depth = 4.0

7.3x 100 = 54%


(90º swing, 54% maximum depth)

Heaper bucket volume = 1.5 LCY or 1.15 LCM

Bucket Fill Factor = 0.90 (Table 3-2)

Job efficiency = 50/60 = 0.83

Load factor = 0.80

Production (loose) = C x S x V x B x E

= 160 x 1.08 x 1.5 x 0.90 x 0.83 = 194 LCY/hr

194 LCY x 0.80 = 155 BCY/hr

= 160 x 1.08 x 1.15 x 0.90 x 0.83 x 0.80 = 119 BCM/hr

3-6. A small hydraulic excavator will be used to dig a trench in soft clay (bucket fill factor = 0.90). The minimum trench size is 24” (0.61 m) wide by 6’ (1.83 m) deep. The excavator bucket available is 30” (0.76) wide and has a heaped capacity of ¾ CY (0.57 CM). The maximum digging depth of the excavator is 17.5’ (5.3 m). The average swing angle is expected to be 90º. Estimate the hourly trench production in LF (m) if the job efficiency is 50 min/hr.

30 1

12 27x 6 x Actual volume/LF of trench =

= 0.55 BCY

= 1.39 BCM

or

(Table 2-5)

Standard cycles/hr = 200 (Table 3-3)

Load factor = 0.77


17.5x 100 = 34%


5.3x 100 = 34%


(90º swing, 34% maximum depth)

Heaped bucket volume = 0.75 LCY (0.57 LCM)

Job efficiency factor = 0.83

Trench adjustment factor = 0.92 (Table 3-5)

Bucket fill factor = 0.90

= 200 x 1.14 x 0.75 x 0.90 x 0.83 = 128 LCY/hr

128 LCY x 0.77 = 99 BCY/hr

= 200 x 1.14 x 0.57 x 0.90 x 0.83 x 0.77 = 75 BCM/hr

Production (loose) = C x S x V x B x E

99

0.55Trench production = = 180 LF/hr

75

1.39Trench production = = 54 m/hr

4-2. The tractor-scraper whose travel-time curves are shown in Figures 4-4 and 4-5 hauls its rated payload 4,000’ (1220 m) up a 5% grade from the cut to the fill and returns empty over the same route. The rolling resistance factor for the haul road is 120 lb/ton (60kg/t). Estimate the scraper travel time.

Effective grade:

120

20

60

10Haul = 5 + = 11%

11%=5 +Haul =

Effective grade:

120

20

60

10

1%

Return = -5 + = 1%

Return = -5 + =

Haul time = 6.4 min (Figure 4-4)

Return time = 1.6 min (Figure 4-5)

Travel time = 8.0 min

4-6. How many hours should it take an articulated wheel loader equipped with a 4 CY (3.06 CM) bucket to load 3,000 CY (2294 CM) of gravel from a stockpile into rail cars if the average haul distance is 300’ (91.5 m) one way? The area is level with a rolling resistance factor of 120 lb/ton (60 kg/t). Job efficiency is estimated at 50 min/hr.

Bucket fill factor = 0.95 (Table 3-2)

Bucket volume = 4.0 x 0.95 = 3.8 LCY

Bucket volume = 3.06 x 0.95 = 2.91 LCM

Basic cycle time = 0.35 min (Table 4-6)

120

20 Effective grade = = 6%

60

10 Effective grade = = 6%

Travel time = 0.5 min (Figure 4-14)

Cycle time = 0.35 + 0.50 = 0.85 min

50

0.85224 LCY/hr Production = 3.8 x =

50

0.85 Production = 2.91 x = 171 LCM/hr

5-1. The data in the accompanying table resulted from performing Modified Proctor Tests on a soil. Plot the data and determine the soil’s laboratory optimum moisture content. What minimum field density must be achieved to meet job specifications which require compaction to 90% of Modified AASHTO Density?

5-5. Estimate the production in compacted CY (CM) per hour of a self-propelled tamping foot roller under the following conditions: average speed = 5 mph (8.0 km/hr), compacted lift thickness = 6” (15.2 cm), effective roller width = 10’ (3.05 m), job efficiency = 0.75, and number of passes = 8.

16.3 x W x S x L x E

P

10 x W x S x L x E

P

CCY/hr

CCM/hr

Production = =

Production = =

16.3 x 10 x 5 x 6 x 0.75

8

10 x 3.05 x 8.0 x 15.2 x 0.75

P

Production = =

Production = =

458 CCY/hr

348 CCM/hr

solved homework due...

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