All Chapters Covered
SOLUTIONS
, Traffic 5 Edition
Engineering,
th
Roess, R.P., Prassas, E.S., and McShane, W.R.
Solutions to Homework No. 2
Problem 5‐1
A volume of 1,200 veh/h is observed at an intersection approach. Find
the peak flow rate within the hour for the following peak-hour
factors: 1.00, 0.90., 0.80, 0.70. Plot and comment on the results.
The peak flow rate of flow is computed as v = V/PHF. The table below
summarizes the results for the information given. A plot follows.
Even with the same hourly
volume, a small difference
in PHF leads to an
enormous difference in peak
flow rates. Traffic engineers
must be able to deal with
this peaking characteristic
on a regular basis.
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,Problem 5‐2
A traffic stream displays average vehicle headways of 2.4s at 55
mph. Compute the density and rate of flow for this traffic stream.
A headway can be converted to a flow rate as follows:
v = 3600 = 3600 = 1,500 veh/hr/ln
h 2.4
Knowing both flow rate and speed (given), the density may now be
computed as:
D = v = 1500 = 27.3 veh/hr/ln
S 55
Problem 5‐3
A freeway detector records occupancy of 0.26 for a 15- minute period.
If the detector is
3.5 ft long, and the average vehicle has a length of 18 ft. what is the
density implied by this measurement?
Density is obtained from occupancy as follows:
Such a high value is indicative of highly congested conditions within a
queue.
,Problem 5‐4
The following traffic count data were taken from a permanent detector
location on a major state highway. From this data, determine (a) the
AADT, (b) the ADT for each month. (c) the AAWT, and (d) the AWT for
each month. From this information, what can be discerned about the
character of the facility and the demand it serves?
The table below illustrates the computation of monthly ADT and AWT
values.
The AADT is computed as the total annual volume divided by 365 days,
or:
AADT = 2,365,000 =
6,479 veh/day
365
The AAWT is computed as the total weekday volume divided by 260
days, or:
AAWT = 2,067,000 =
7,950 veh/day
@@SeSiesimsmici
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,260
@@SeSiesimsmici
icsiosloaltaiotinon
,Because the average weekday volume is higher than the total average
volume, it is likely that this is a commuter route. The difference is even
clearer if the average weekend traffic is computed. The total weekend
volume for the year is 2,365,000 – 2,067,000 = 298,000 vehs. There
are 365‐260 = 105 Saturdays and Sundays in the year. Then, the
average weekend traffic is computed as:
AAWT = 298,000 = 2,838 veh/day
105
This is clearly NOT a recreational route, but one that serves a
substantial proportion of regular commuters.
Problem 5‐5
A lane on the freeway displays the following characteristics: (a) the
average headway between vehicles is 2.8 s, and (b) the average
spacing between vehicles is 235 ft. What is the rate of flow for the
lane? What is the average speed (in mph)?
Headway and Spacing can be converted to the macroscopic measures
of flow rate and density, as follows:
v = 3600 = 3600 = 1,286 veh/h/ln
ha 2.8
D = 5280 = 5280 = 22.5 veh/mi/ln
da 235
Speed is then computed as:
S = v = 1286 =
57.2 mi/h D
22.5
@@SeSiesimsmici
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,Problem 5‐6
The following counts were taken on a major
arterial during the evening peak period.
From this data determine
a) The peak hour
b) The peak hour volume
c) The peak flow rate within the peak hour
d) The peak hour factor
a) The highest hourly volume (within
the study period) occurs between
4:30 and 5:30 PM.
b)The hourly volume is the volume
for this hour, or 1,999 vehs/h.
c) The highest flow rate is the 15‐minute interval within the peak
hour with the highest 15‐minute volume. This is the period
between 5:15 and 5:30 PM. The flow rate within this period is
506 * 4 = 2,024 veh/h.
d)The peak hour factor is 1999/2024 = 0.988.
@@SeSiesimsmici
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,Problem 5‐7
A peak hour volume of 1 200 veh/h is observed on a freeway lane.
What is the peak flow rate within this hour if the PHF is 0.87?
The peak flow rate is found as:
v = V = 1200 = 1379
veh/h PHF 0.87
Problem 5‐8
The flow rate on an arterial lane is 1,300 veh/h. If the average speed in
the same lane is 35 mph, what is the density?
The density is found as:
D = V = 1,300 = 37.1
veh/mi/ln S 35
Problem 5‐9
The AADT for a section of suburban arterial is 50,000 veh/day.
Assuming that this is an urban radial facility, what range of directional
design hour volumes would be expected?
From textbook Table 5‐2, Page 109, for an urban radial facility, K
factors range from 0.07 to 0.12. D factors range from 0.55 to 0.60.
Then:
DDHV = AADT * K * D
DDHVlow = 50,000 * .07 * .55 = 1925 veh/hr
DDHVhigh = 50,000 * .12 * .60 = 3,600 veh/hr
This is a very broad range, and highlights the danger in using such
generalized factors for estimating demand.
@@SeSiesimsmici
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,Problem 5‐10
The following travel times were measured for vehicles traversing a
1,000 ft. segment of an arterial.
Determine the time mean speed (TMS) and space
mean speed (SMS) for these vehicles.
HINT: Convert ft/s to
mi/hr by conversion
factor of 1.47
The TMS is now merely the average of the vehicle speeds, or
266.8/8 = 33.35 mi/h. The SMS is based upon the average travel
speed, or 163.8/8 = 20.475 s/veh. Then:
@@SeSiesimsmici
icsiosloaltaiotinon
, th
Traffic Engineering, 4 Edition
Roess, R.P., Prassas, E.S., and McShane, W.R.
FALL 2012
Solutions to Homework No. 3 – Chapter 8
Problem 8-1
Traffic volumes on a four-lane freeway (two lanes in each direction)
were counted manually from an overhead location, resulting in the data
shown here. The desire was to obtain continuous 15- minute counts for
each lane of the freeway for a two-hour period surrounding the
morning peak hour.
Data for Problem 1
From the data shown, determine the following:
a) Continuous 15 minute volumes for each period and each lane.
b) The peak hour, peak hour volume, and peak hour factor (PHF) for
each direction of flow and for the freeway as a whole.
c) Directional flow rates during each 15-minite period.
@@SeSiesimsmici
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SOLUTIONS
, Traffic 5 Edition
Engineering,
th
Roess, R.P., Prassas, E.S., and McShane, W.R.
Solutions to Homework No. 2
Problem 5‐1
A volume of 1,200 veh/h is observed at an intersection approach. Find
the peak flow rate within the hour for the following peak-hour
factors: 1.00, 0.90., 0.80, 0.70. Plot and comment on the results.
The peak flow rate of flow is computed as v = V/PHF. The table below
summarizes the results for the information given. A plot follows.
Even with the same hourly
volume, a small difference
in PHF leads to an
enormous difference in peak
flow rates. Traffic engineers
must be able to deal with
this peaking characteristic
on a regular basis.
@@SeSiesimsmici
icsiosloaltaiotinon
,Problem 5‐2
A traffic stream displays average vehicle headways of 2.4s at 55
mph. Compute the density and rate of flow for this traffic stream.
A headway can be converted to a flow rate as follows:
v = 3600 = 3600 = 1,500 veh/hr/ln
h 2.4
Knowing both flow rate and speed (given), the density may now be
computed as:
D = v = 1500 = 27.3 veh/hr/ln
S 55
Problem 5‐3
A freeway detector records occupancy of 0.26 for a 15- minute period.
If the detector is
3.5 ft long, and the average vehicle has a length of 18 ft. what is the
density implied by this measurement?
Density is obtained from occupancy as follows:
Such a high value is indicative of highly congested conditions within a
queue.
,Problem 5‐4
The following traffic count data were taken from a permanent detector
location on a major state highway. From this data, determine (a) the
AADT, (b) the ADT for each month. (c) the AAWT, and (d) the AWT for
each month. From this information, what can be discerned about the
character of the facility and the demand it serves?
The table below illustrates the computation of monthly ADT and AWT
values.
The AADT is computed as the total annual volume divided by 365 days,
or:
AADT = 2,365,000 =
6,479 veh/day
365
The AAWT is computed as the total weekday volume divided by 260
days, or:
AAWT = 2,067,000 =
7,950 veh/day
@@SeSiesimsmici
icsiosloaltaiotinon
,260
@@SeSiesimsmici
icsiosloaltaiotinon
,Because the average weekday volume is higher than the total average
volume, it is likely that this is a commuter route. The difference is even
clearer if the average weekend traffic is computed. The total weekend
volume for the year is 2,365,000 – 2,067,000 = 298,000 vehs. There
are 365‐260 = 105 Saturdays and Sundays in the year. Then, the
average weekend traffic is computed as:
AAWT = 298,000 = 2,838 veh/day
105
This is clearly NOT a recreational route, but one that serves a
substantial proportion of regular commuters.
Problem 5‐5
A lane on the freeway displays the following characteristics: (a) the
average headway between vehicles is 2.8 s, and (b) the average
spacing between vehicles is 235 ft. What is the rate of flow for the
lane? What is the average speed (in mph)?
Headway and Spacing can be converted to the macroscopic measures
of flow rate and density, as follows:
v = 3600 = 3600 = 1,286 veh/h/ln
ha 2.8
D = 5280 = 5280 = 22.5 veh/mi/ln
da 235
Speed is then computed as:
S = v = 1286 =
57.2 mi/h D
22.5
@@SeSiesimsmici
icsiosloaltaiotinon
,Problem 5‐6
The following counts were taken on a major
arterial during the evening peak period.
From this data determine
a) The peak hour
b) The peak hour volume
c) The peak flow rate within the peak hour
d) The peak hour factor
a) The highest hourly volume (within
the study period) occurs between
4:30 and 5:30 PM.
b)The hourly volume is the volume
for this hour, or 1,999 vehs/h.
c) The highest flow rate is the 15‐minute interval within the peak
hour with the highest 15‐minute volume. This is the period
between 5:15 and 5:30 PM. The flow rate within this period is
506 * 4 = 2,024 veh/h.
d)The peak hour factor is 1999/2024 = 0.988.
@@SeSiesimsmici
icsiosloaltaiotinon
,Problem 5‐7
A peak hour volume of 1 200 veh/h is observed on a freeway lane.
What is the peak flow rate within this hour if the PHF is 0.87?
The peak flow rate is found as:
v = V = 1200 = 1379
veh/h PHF 0.87
Problem 5‐8
The flow rate on an arterial lane is 1,300 veh/h. If the average speed in
the same lane is 35 mph, what is the density?
The density is found as:
D = V = 1,300 = 37.1
veh/mi/ln S 35
Problem 5‐9
The AADT for a section of suburban arterial is 50,000 veh/day.
Assuming that this is an urban radial facility, what range of directional
design hour volumes would be expected?
From textbook Table 5‐2, Page 109, for an urban radial facility, K
factors range from 0.07 to 0.12. D factors range from 0.55 to 0.60.
Then:
DDHV = AADT * K * D
DDHVlow = 50,000 * .07 * .55 = 1925 veh/hr
DDHVhigh = 50,000 * .12 * .60 = 3,600 veh/hr
This is a very broad range, and highlights the danger in using such
generalized factors for estimating demand.
@@SeSiesimsmici
icsiosloaltaiotinon
,Problem 5‐10
The following travel times were measured for vehicles traversing a
1,000 ft. segment of an arterial.
Determine the time mean speed (TMS) and space
mean speed (SMS) for these vehicles.
HINT: Convert ft/s to
mi/hr by conversion
factor of 1.47
The TMS is now merely the average of the vehicle speeds, or
266.8/8 = 33.35 mi/h. The SMS is based upon the average travel
speed, or 163.8/8 = 20.475 s/veh. Then:
@@SeSiesimsmici
icsiosloaltaiotinon
, th
Traffic Engineering, 4 Edition
Roess, R.P., Prassas, E.S., and McShane, W.R.
FALL 2012
Solutions to Homework No. 3 – Chapter 8
Problem 8-1
Traffic volumes on a four-lane freeway (two lanes in each direction)
were counted manually from an overhead location, resulting in the data
shown here. The desire was to obtain continuous 15- minute counts for
each lane of the freeway for a two-hour period surrounding the
morning peak hour.
Data for Problem 1
From the data shown, determine the following:
a) Continuous 15 minute volumes for each period and each lane.
b) The peak hour, peak hour volume, and peak hour factor (PHF) for
each direction of flow and for the freeway as a whole.
c) Directional flow rates during each 15-minite period.
@@SeSiesimsmici
icsiosloaltaiotinon
