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These Answers Are Wrong Do Not Post The Same Answers Can Someone Who Knows Answer These Questions All Answers Are Wron 1 (43.69 KiB) Viewed 48 times
ALL ANSWERS ARE WRONG!!!
These Answers Are Wrong Do Not Post The Same Answers Can Someone Who Knows Answer These Questions All Answers Are Wron 2 (55.35 KiB) Viewed 48 times
Graded Assessment: Traffic Signal Control A Evelensik thit putio Gradod Aescesmert due Hug 9,202212 12 EEST Fancateit Impact of cycle time Q.eron poirts igraced? Consider an isclated arterial junclion with four aperoeches. Due to high denard, long queues are created during the peak period. To address this problem, the fesponsible asthority has decided to increase the cycle nime of the jurction from 90 sec to 120 set. The junction control is based on two phases, ane for approaches 1 and 3 ard arotler for approaches 2 and 4. Both phases have equal green phase durations. Al four approactes have two lames and equal saturation fows of 3600 wehln each. The total lost time is equal to 18 sec. a) What is the additional green time (sec) that wil be allocated to each phase as a sesult of the cycle time increase? ∫secx 30 by How many additioenal vehicles can be served al each cycle in total (foe al four approaches\} Lnder the incregsed cycle lime, assuming that all approaches will conifue being saturated? veh x 1440 c.1) How muchi shor ter lin vehi will the queue of each approach be after one hour of peak persod operation lassuming constant saturated demand? Consider the quede as the total mumber af vehicles waiting to be senved for the wiole road and NOT by lare. ∣ veh x C2) How much shorter is the above queve in meters, assuming that the aver age fength of a queuing vericle equals 6 m ? m x 2160 d) Assume now that the saturation flow of each approach remairs at 3600 vehin for the first 36 sec of the green phase but then drops ta 2160 veh,h. How mamy additional vehicles are served in total ffor al fout approachesi at each increased cycle under these circurnstances? Iveh x
Ans (a) Total Effective Green Time Tg=3600(1−(2∗tl)/C) Since loss time is same for both the cycles Solving the above equation for C1 and C2 We get : Tg1=2160 s per hour Tg2=2520 s per hour Total cycles per hour 1=3600/90=40 Total cycles per hour 2=3600/120=30 Hence green time per cycle =G1=2160/40=54 s green time per cycle =G2=2520/30=84 s Hence increase in green time is 30 s. For b) Effective green time =84 Capacity for each approach =Ci=s∗ effective green time/cycle length =(3600)∗84=2520veh Itherefore Total number of vehicles served for all approach =2520∗4=10080 Similarly calculate number of vehicles served when cycle lenght was 90 s=8640 Itherefore Increase in number of vehicles served =10080−8640=1440 veh For Ans c1) Difference between the capacity of each approach for one hour =(s∗ effective green time/C1 )- (s* effective green time/C2)= 360 veh For Ans (2) Difference in number of vehicles queuing =360 Hence, reduction in length of queue =360∗6=2160 m For Ans d) Total Vehicle served under given circumstances =(3600)∗(36/120)+(2160)∗(48/120)=1944 veh therefore Hence increased number of vehicles served each cycle =1944−2160=−216 veh. (negative
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