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