Slids of intellegent assignment
Transcript of Slids of intellegent assignment
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