Traffic light setting

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Intelligent Urban Traffic Control System Supervisor Prof. Dr. Riza Atiq Abdullah OK Rehmat prepared by rasha salah p64799 sarah hazim p65407

Transcript of Traffic light setting

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