Slids of intellegent assignment
- 1. 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
- 2. Fig (1) (The Study area) Location of intersections
- 4. First Intersection Phase Traffic volume Actual flow (pcu/hr) saturation flowper lane (pcu/hr) saturation flow (pcu/hr) Y= flow/sat uration flow Green time split Y / ∑y (Y / ∑y) *Geleft 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
- 5. 1. L(losttime)=3×(3+2)=15sec 2. Co= (1.5 𝐿 +5 1−∑𝑌 = 1.5 15 +5 1−0.3 =39.28 takeCo=40sec 3. Effectivegreentime(Ge)=Co-L=40-15=25sec (Multiplythisvalueingreentimesplit to getthefinalcolumn). 4. Totalgreentime=24sec 5. Cycletime=greentime+losttime =24+15=39sec
- 6. Secondintersection Phase Trafficvolume Actual flow (pcu/hr) saturation flowper lane (pcu/hr) saturation flow (pcu/hr) Y= flow/sat uration flow Green timesplit Y/∑y (Y/∑y) *Geleft 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
- 7. 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
- 8. Thirdintersection Phase Trafficvolume Actual flow (pcu/hr) saturation flowper 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
- 9. 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
- 11. We choosethe maximum value ofC₀ = 88 and make recalculation for the greentime, and the new results inthe table below: Co=88 Ge=Co-L=88-20=68 sec phase Intersection 1 Intersection 2 Intersection 3 GreenTime = (Y/∑ Y)*Ge GreenTime = (Y/∑ Y)*Ge GreenTime = (Y/∑ Y)*Ge Q1 23 21 15 Q2 23 13 11 Q3 22 21 11 Q4 13 31 ∑=68 ∑=68 ∑=68 So the new diagramwill be like the following:
- 13. Pictures from the study area
- 17. THANK YOU