Transcript Document

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
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
M = (R+R3 )E
B = (1 + R + R3 )E
7. Guaranteed maximum price
Original Estimated Contract Prices (in $ thousands)
Type of Contract
Markup
Contract Price
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
=(0.1+0.05)*6000
=(1+0.1+0.05)*6000
7. Guaranteed maximum price
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
Type of Contract
1.
2.
3.
4.
5.
6.
7.
Lump sum
Unit price
Cost plus fixed %
Cost plus fixed fee
Cost plus variable %
Target estimate
Guaranteed maximum price
Markup
M = (R+R1)E
M = (R+R2)E
M = RA = RE
M = RE
M = R(2E - A) = RE
M = RE + N(E - A) = RE
M = (R+R3)E
Contract Price
B = (1 + R + R1)E
B = (1 + R + R2)E
B = (1 + R)E
B = (1 + R)E
B = (1 + R)E
B = (1 + R)E
B = (1 + R + R3)E
Original Estimated Contract Prices (in $ thousands)
Type of Contract
1.
2.
3.
4.
5.
6.
Lump sum
Unit price
Cost plus fixed %
Cost plus fixed fee
Cost plus variable %
Target estimate
Markup
Contract Price
720
660
600
600
600
600
6720
6660
6600
6600
6600
6600
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.
Ideal output = 230 BCY/hr (176 BCM/hr)
(Table 3-7)
Optimum depth = 9.9 ft. (3.0 m)
(Table 3-8)
8.9
% optimum depth =
9.9
2.7
% optimum depth =
3.0
x 100 = 90%
x 100 = 90%
Swing-depth factor = 1.06
(Table 3-9)
(70º swing, 90% optimum depth)
Job efficiency = 50/60 = 0.83
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
x 100 = 54%
24
% maximum depth = 4.0
x 100 = 54%
7.3
Swing-depth factor = 1.08
(Table 3-3)
(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.
Actual volume/LF of trench =
30
12
x
6
= 0.55 BCY
or
= 1.39 BCM
x
1
27
Load factor = 0.77
Standard cycles/hr = 200
(Table 2-5)
(Table 3-3)
% maximum depth = 6.0
x 100 = 34%
17.5
% maximum depth = 1.83
x 100 = 34%
5.3
Swing-depth factor = 1.14
(Table 3-3)
(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
Production (loose) = C x S x V x B x E
= 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
Trench production =
Trench production =
99
0.55
75
1.39
= 180 LF/hr
= 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:
Haul = 5 +
Haul = 5 +
120
20
60
10
= 11%
= 11%
Effective grade:
Return =
Return =
-5
-5
+
+
120
20
60
10
=
1%
=
1%
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)
Effective grade =
Effective grade =
120
20
60
10
=
6%
=
6%
Travel time = 0.5 min
(Figure 4-14)
Cycle time =
0.85 min
Production =
0.35 + 0.50 =
3.8
Production = 2.91
x
x
50
0.85
50
0.85
= 224 LCY/hr
= 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.
Production =
Production =
Production =
Production =
16.3 x W x S x L x E
P
10 x W x S x L x E
P
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
= CCY/hr
= CCM/hr
= 458 CCY/hr
= 348 CCM/hr