1. Page 1 of 5
Intelligent Urban Transport
Management System
Assignment 1
Name Muhammad bin Ramlan
Matrix No. P57600
Subject KA 6423
Session 2012/2013
Lecturer Prof Ir Dr RizaAtiq O.K. Rahmat

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Question
Below is an illustration of two signalised intersections which are to be optimised.
Based on the given data, you are required to determine optimum common cycle
time, green time split for each intersection and offset time.
95 204 102 89 200 92
356 225
521 490
Queue length = 15 cars
80 95
180 190
80 180 90 95 152 95
400m

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Answer
JUNCTION 1
Number Of Saturation Flow Per Lane Saturation Flow Actual Flows Flow / Saturation Flow Green Time
Phase (pcu/Hr) (pcu/Hr) (pcu/Hr)
Lanes Ration Split
1 2 1800 3600 1,005 0.28 0.477
2 2 1800 3600 401 0.11 0.190
3 2 1800 3600 350 0.10 0.166
4 2 1800 3600 350 0.10 0.166
Jumlah, Y 0.59 1.000
Cycle Time
Co =
1.5 L +
1.5 L + 5
5
L=4*4= 16 seconds [lost time per phase 4 seconds] Co =
1-Y
Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 69.88
Take Co = 70 seconds
Effective Green Time = 70 -L
= 54 seconds

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Green Time Split
Phase Green Time Round-Up To
1 25.769 26 Take amber time = 3 seconds
2 10.282 11 Take all red time = 1 seconds
3 8.974 9 Total of all amber and red time = 16 sec
4 8.974 9
total 54 55 Actual CycleTime = 71 sec
JUNCTION 2
Number Of Saturation Flow Per Lane Saturation Flow Actual Flows Flow / Saturation Flow Green Time
Phase (pcu/Hr) (pcu/Hr) (pcu/Hr)
Lanes Ration Split
1 2 1800 3600 857 0.24 0.449
2 2 1800 3600 381 0.11 0.200
3 2 1800 3600 327 0.09 0.171
4 2 1800 3600 342 0.10 0.179
Jumlah, Y 0.53 1.000
Co =
1-Y
Cycle Time 1.5 L + 5
Co =
L=4*4= 16 seconds [lost time per phase 4 seconds] 1-Y
Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 61.67

5. Page 5 of 5
Take Co = 62 seconds
Effective Green Time = 62 -L
= 46 seconds
Green time split
Phase Green Time Round
Round-Up To
1 20.67 21 Teke amber time = 3 seconds
2 9.19 10 Take all red time = 1 seconds
3 7.89 8 Total of all amber and red time = 16 sec
4 8.25 9
total 46 48 Actual CycleTime = 64 sec
OFFSET TIME
L=400 meter
S=10m/s
Q=15 kenderaan
Loss time= 2 second
= 400/10- (15(2)+2)
= 40-32
= 8
# So, Offset Junction 2 is8 Second