FCE 545: TRANSPORTATION ENGINEERING IIIA
WORKED TUTORIAL QUESTIIONS
1
TRANSPORT PLANNING
Example 1.1
Figure E1 shows a scaled road network for a medium sized town in Western Kenya.
Figure E1
Build a least-cost journey tree on node F using Moore’s algorithm. (Assume all centroid
links are identical in cost, at 5 units)
Solution:
Assume all centroid links are identical in cost, at 5 units, and ignore in tree building
process - therefore starting tree at node 8. Determine tree by moving forward one link
at a time:
Link Cost Link
combination
8-1 5
8-2 5
1 -2 10
8-1-2
1 -4 10
8-1-4
1 - 5 20
8-1-5
1 - 6 15
8-1-6
2-4 10
8-2-4
2 - 3 20
8-2-3
2-9 30
8-2-9
4-3 15
8-1-4-3
4-5 10
8-1-4-5
5 - 3 10
8-1-5-3
5-7 10
8-1-5-7
6-5 20
8-1-6-5
6 - 7 30
8-1-6-7
3-9 10
8-2-3-9
3 - 7 15
8-2-3-7
9 - 3 10
8-2-9-3
Total
cost
Least cost to Node
1
2
15
15
25
20
15
25
35
30
25
35
35
40
50
35
40
45
4
5
6
3
9 (But to D via 3.)
7
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2
JUNCTION DESIGN AND CONTROL
Example 2.1:
Figure E2.1 shows a local intersection. The obstruction is located 14 m from the
centreline of the right lane of a local road and 20 m from the centreline of the right lane
of an intersecting road.
Figure E2.1 Minimum Sight Triangle at a No-Control or Yield-Control Intersection
Cases A and C
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Table 7.7 Suggested lengths and adjustments of Sight-Triangle Leg Case A – No
Traffic
If the maximum speed limit on the intersecting road is 56 km/h, what should the speed
limit on the local road be such that the minimum sight distance is provided to allow the
drivers of approaching vehicles to avoid immine
nt collision by adjusting their speeds? Approach grades are 2%.
Solution:
Step 1:
Determine the distance on the local road at which the driver first sees
traffic on the intersecting road.
Speed limit on intersecting road = 56 km/h
Distance required on intersecting road (da) = 50 m (from Table 7.7)
Calculate the distance available on local road by using the equation
db = a
Step 2:
da
50
= 20 ∗
= 28m
da − b
50 − 14
Determine the maximum speed allowable on the local road.
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•
•
The maximum speed allowable on local road is 32 km/h (from Table
7.7).
No correction is required for the approach grade as it is less than 3%.
Example 2.2
a) Figure E2.2 shows peak-hour volumes for a major intersection on an expressway.
Figure E2.2
Pedestrian volume at the intersection is negligible and a peak-hour factor of 0.95 is
applicable.
The saturation flows for an intended four-phase system are shown below
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The influence of heavy vehicles and turning movements and all other factors that affect
the saturation flow have already been considered. Amber (yellow) time is three seconds
Determine suitable signal timing for the intersection using the Webster method
Solution
Determine equivalent hourly flows by dividing the peak-hour volumes by the PHF
(e.g. for left-turn lane group of phase A. equivalent hourly flow 222/0.95 = 234). See
Figure E2.2.1 for all equivalent hourly flows.
Figure E2.2.1 Equivalent In in rlt flow
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Phase (ϕ)
Critical lane Volume
(veh/h)
488
338
115
371
∑1312
A
B
C
D
Compute the total lost time. Since there is not an all-red phase - hat is. R = 0 - and
there are four phases,
L = ∑l = 4 X 3.5 = 14 sec (assuming lost time per phase is 3.5 sec)
Lane Group
qij
Sj
Qij/Sj
Yi
Phase A
Phase B
Phase C
Phase D
1
2
1
2
1
2
1
2
234
976
676
135
26
194
371
322
1615 3700 3700 1615 1615 3700 1615 3700
0.145 0.264 0.183 0.084 0.016 0.052 0.230 0.087
0.264
0.183
0.052
0.230
Determine Yi and ∑Yi .
∑Yi = (0.264 + 0.183 + 0.052 + 0.230) = 0.729
Determine the optimum cycle length using the equation:
(1.5𝑥14) + 5
1.5𝐿 + 5
𝐶𝑜 =
=
= 95.9 𝑠𝑒𝑐
𝜙
1 − 0.729
1 − ∑ 𝑌𝑖
𝑖=1
Use 100 seconds as cycle lengths are usually multiples of 5 or 10 seconds.
Find the total effective green time:
𝐺𝑡𝑒 = 𝐶 − 𝐿 = 100 − 14 = 86 𝑠𝑒𝑐.
Effective time for phase i is obtained from
𝑌𝑖
𝑌𝑖
𝑌𝑖
𝐺𝑒𝑖 =
𝐺𝑡𝑒 =
𝑥 86 =
𝑥 86
𝑌1 + 𝑌2 + … 𝑌𝑛
0.264 + 0.183 + 0.052 + 0.230
0.729
Yellow time λ = 3.0 sec; the actual green time Gai for each phase is obtained as
follows:
Gai = Gei + li - 3.0
Actual green time for Phase A (GaA) = (0.264/0.729) x 86 + 3 5 - 3 0 = 32 sec
Actual green time for Phase B (GaB) = (0.183/0.729) x 86 + 3,5 - 3.0 = 22 sec
Actual green time for Phase C (GaC) = (0.052/0.729) x 86 + 3.5 - 3.0 = 7 sec
Actual green time for Phase D (GaD) = (0.23/0.729) x 86 + 3.5 - 30 = 27 sec
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Example 2.3
A fixed time 2-phase signal is to be provided at an intersection having a North-South
and an East-West road where only straight-ahead traffic is permitted. The design hour
flows from the various arms and the saturation flows for these arms are given below:
Design hour flow (pcu/hr)
Saturation flow (pcu/hr)
i)
ii)
North
800
2400
South
400
2000
East
750
3000
West
600
3000
Calculate the optimum cycle time and green times for the minimum overall
delay. The minimum inter-green time can be taken as 4 seconds lost time
per phase due to starting delay, 2 seconds and the amber as 2 seconds.
Sketch the timing diagram for each phase.
Solution:
q
s
y =q/s
y (max) values
N
800
2400
0.33
S
400
2000
0.20
0.33
The timing diagram is shown below
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E
W
750
600
3000 3000
0.25
0.20
0.25
3
ECONOMIC ANALYSIS
Example 3.1
An existing highway is 20 Km long connecting two small cities. It is proposed to improve
the alignment by constructing a highway 15 Km long costing $500,000 per kilometre.
Maintenance costs are likely to be $10,000 per kilometre per year. Land acquisition costs
run ($75,000) $60,000 per kilometre. It is proposed to abandon the old road and sell the
land for $10,000 per kilometre. Money can be borrowed at 8% per annum. It has been
estimated that passenger vehicles travel at 35 kph at a cost of 20 cents per kilometre
with a car occupancy of 1.5 persons per car. What should be the traffic demand for this
new road to make this project feasible if the cost of the time of the car's occupants is
assessed at $10 per hour? Give a brief discussion of your result.
Solution:
1. Basic costs:
Construction 15 kilometres at $500,000
=
$7,500,000
Land acquisition 15 kilometres at $75,000 60,000 =
$ 900,000
Sub-total
=
$8,400,000
Deduct sale of land (old alignment)
20 x 10,000 = 200,000
Total
$8,200,000
2. For simplicity, assume the new road will last indefinitely. Therefore, the investment
need not be repaid, only the interest.
Annual cost of initial investment:
8% of $8,200,000 = $656,000
3. Net annual income:
Assume that N vehicles per year use the road. 5 kilometres saved at 20 cents per
kilometre.
5 x N x (20/100) = $N
Time saved per vehicle/trip:
- 5 mi at 35 kph
- Car occupancy
- Value of time
= 5/35 = 0.143 hr
= 1.5 persons/car
= $10/hr
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Therefore,
• Cost saving in time at $10/hr = 0.143 x 1.5 x N x 10 = 2.14N
• Maintenance savings = 5 km x $10,000 = $50,000
• Total savings = $50,000 + $N + 2.14N = 50,000 + 3.14N
4. Benefit-cost ratio:
In order that the project could be justified, the B/C ratio must be equal to or greater
than 1. Because the project is assumed to have an infinite life,
B/C = Annual benefits / Annual Costs
Or
50,000 + 3.14N = 656,000
N = 192,994/yr
or about 528 veh/day
The project, therefore, would be justified if about 528 vehicles used the new road on
a daily basis.
5. Discussion:
The following points should be noted regarding this problem.
i) The social costs of land use have not been considered. For example, the land
on either side of the old road would decrease in value, whereas the land
adjacent to the new route would increase in value. The net result would raise
the value of N.
ii) When one works out the B/C ratio, it is generally advisable to use total benefits
and total costs. In this problem, B/C was assumed to be 1 and is therefore
permissible.
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