PLEASE SHOW ALL WORK CLEARLY. AND IF POSSIBLE EXPLAIN EVERY STEP OF THE WAY THER

Question

PLEASE SHOW ALL WORK CLEARLY. AND IF POSSIBLE EXPLAIN EVERY STEP OF THE WAY THERE AND THE REASONING BEHIND THE PROCEDURE. THANK YOU!!!!!!!

DAta for question 1 can be found below

Problem 2: (50 points) in the intersection presented in problem 1, at 8:00 in the morning, intersection B is completely locked creating a long queue in east-west direction. At 08:10 the length of queue created due to traffic in intersection B goes upto 0.5 miles upstream of intersection A. At 08:10 the intersection B starts clearing up and light for west to east direction turns to green 1 mile Queue Length 0.5 mile End of queue at 08:10 If the cycle length for intersection B and A is 90 seconds and the duration of green west-east direction is 60 seconds, yellow time for all directions is zero At what time the queue behind intersection A, created by the congestion in intersection B will clear up? How many shockwaves you would expect for the time [08:10 to 08:20]? Please explain narratively and quantitatively as well What is the speed of shockwave? Assumptions - Flow rates are the same as the rates in problem 1. ( Just use average flow rates in the computations and don't need to use 90 percent anymore) Assume a triangle fundamental diagram o For both intersections jam density is 140 veh/mile/lane o Critical density is 30 veh/mile/lane o Free flow speed is 60 mph Number of lanes in the east-west direction is 2 - Length of vehicle: 20 ft. - Spacing between vehicles: 5 ft Width if intersection A and B in all directions is 12 foot. But you can ignore this in your computations

Explanation / Answer

Solution 1

At Junction A? The? Own rate is given as (µ) = 1200+500 vph = 1700/ 60 = 29vehicle per minute. Hence, the probability of zero vehicles arriving in one minute p(0) can be computed as follows:
p(0) =µxe?µ x!
=33.e?33 0!
= 44.55

At Junction B?   The?ow rate is given as (µ) = 1200+800 vph = 2000/ 60 = 33vehicle per minute. Hence, the probability of zero vehicles arriving in one minute p(0) can be computed as follows:
p(0) =µxe?µ x!
=27.e?27 0!
= 39

Considering cycle to 90 the vehicle at node A will be 33/1.5 = 22 and at Node B will be 39/1.5 = 26

Likewise cycle to 120, the vehicle moment at node B will be 33/2 = 16 and at Node B will be 39/2 = 19

Probability values of vehicle arrivals computed using Poisson distribution n p(n) p(x ? n) F(n) 0 0.135 0.135 8.120 1 0.271 0.406 16.240 2 0.271 0.677 16.240 3 0.180 0.857 10.827 4 0.090 0.947 5.413 5 0.036 0.983 2.165 6 0.012 0.995 0.722 7 0.003 0.999 0.206 8 0.001 1.000 0.052 9 0.000 1.000 0.011 10 0.000 1.000 0.011

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