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

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

Problem 1: (50 points) Consider a section of the network with 2 consecutive intersections. The arrival of vehicles in these intersections has Poisson distribution and the average arrival rate in each direction is shown in the figure. Saturation flow rate in north-south direction is 2000 (vph) and in the east-west direction is 3600 (vph). 1 mile 1200 voh 1200 voh 03 The cycle length in intersection A and B is equal. If the cycle length can be either 90 or 120 seconds, what is the optimal cycle length such that the total delay for both intersections is minimum. The delay will be computed based on the arrival values at 90% threshold. (hint: 90 percent from the cumulative graphs. You need to first set the cycle length to 90 sec and compute all required information and then set it to 120 seconds and compute the arrivals and the corresponding delay time. The cycle length with the minimum delay time is the best one) In your solution please provide following information: 1 Phasing diagram 2- Green time for each phase 3- Optimal cycle length 4 Total delay 5- Level of service in each intersection You can use any software to solve this problem but please attach all the files.

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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