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Intelligent Urban Traffic Control System
(KKKA 6424)
Assignment no .1
Traffic Light Setting
Supervisor
Prof. Dr. Riza Atiq Abdullah OK Rehmat
Prepared by: Rasha salah ahmed P64799
Sarah hazim P65407
Fig (1) (The Study area) Location of intersections
First Intersection
Phase Traffic volume Actual
flow
(pcu/hr)
saturation flow per lane (pcu/hr)
saturation
flow
(pcu/hr)
Y= flow/saturation flow
Green
time split
Y / ∑y
(Y / ∑y)
*Ge left straight right
1 111 88 199 1800 3600 0.1 0.33 8 2 95 201 296 1800 3600 0.1 0.33 8 3 85 100 185 1800 3600 0.1 0.33 8 ∑y=0.3 ∑=24
1. L (lost time) =3× (3+2) =15 sec
2. Co=(1.5 𝐿 +5
1−∑𝑌=
1.5 15 +5
1−0.3= 39.28 take Co=40 sec
3. Effective green time (Ge) =Co-L =40-15=25 sec
(Multiply this value in green time split to get the final column).
4. Total green time=24sec
5. Cycle time =green time + lost time
=24+15=39sec
Second intersection
Phase Traffic volume Actual
flow
(pcu/hr)
saturation flow per lane (pcu/hr)
saturation
flow
(pcu/hr)
Y= flow/saturation flow
Green
time split
Y / ∑y
(Y / ∑y)
*Ge left straight right
1 80 328 184 592 1800 3600 0.16 0.31 16 2 84 140 98 322 1800 3600 0.1 0.19 10 3 148 252 160 560 1800 3600 0.16 0.31 16 4 64 92 120 276 1800 3600 0.1 0.19 10 ∑y=0.52 ∑=52
1. L (lost time) =4× (3+2) =20 sec
2. Co=(1.5 𝐿 +5
1−∑𝑌=
1.5 20 +5
1−0.52=72.9 take Co=73 sec
3. Effective green time (Ge) =Co-L =73-20=53 sec
(Multiply this value in green time split to get the final column).
4. Total green time=52 sec
5. Cycle time =green time + lost time
=52+20=72 sec
Third intersection
Phase Traffic volume Actual
flow
(pcu/hr)
saturation
flow per
lane
(pcu/hr)
saturation
flow
(pcu/hr)
Y=
flow/sat
uration
flow
Green
time
split
Y / ∑y
(Y /
∑y)
*Ge left straight right
1 151 215 99 465 1800 3600 0.13 0.217 15 2 90 198 75 363 1800 3600 0.1 0.167 11 3 88 119 90 297 1800 3600 0.1 0.167 11 4 356 501 113 970 1800 3600 0.27 0.45 31 ∑y=0.6 ∑=68
1. L (lost time) =4× (3+2) =20 sec
2. Co=(1.5 𝐿 +5
1−∑𝑌=
1.5 20 +5
1−0.6=87.5 take Co=88 sec
3. Effective green time (Ge) =Co-L =88-20=68 sec
(Multiply this value in green time split to get the final column).
4. Total green time=68 sec
5. Cycle time =green time + lost time
=68+20=88 sec.
To determine (offset time) T ideal from inter1 to 2:
T ideal=𝐿
𝑆− 𝑄 × ℎ + 𝑙𝑜𝑠𝑠 𝑡𝑖𝑚𝑒 Where:
L=450m
S=10m/sec
Q=14 cars
h=2sec
Loss time=2 sec. T ideal=15 sec.
To determine (offset time) T ideal from inter 2 to 3:
L=650m
S=10m/sec
Q=12 cars
h=2 sec
Loss time=2 sec T ideal=39 sec
We choose the maximum value of C₀ = 88 and make recalculation for the green time, and
the new results in the table below:
Co=88 Ge=Co-L=88-20=68 sec
phase Intersection 1 Intersection 2 Intersection 3
Green Time = (Y/∑
Y)*Ge
Green Time = (Y/∑
Y)*Ge
Green Time = (Y/∑
Y)*Ge
Q1 23 21 15
Q2 23 13 11
Q3 22 21 11
Q4 13 31
∑=68 ∑=68 ∑=68
So the new diagram will be like the following:
Pictures from the study area
THANK YOU
