Module 7 : Traffic Signal Design

Lecture 35 : Signalized Intersection Delay Models

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	HCM 2000 delay models The delay model incorporated into the HCM 2000 includes the uniform delay model, a version of Akcelik's overflow delay model, and a term covering delay from an existing or residual queue at the beginning of the analysis period. The delay is given as, $$d = d_1~PF+d_2+d_3$$ $$d_1 = \frac{c}{2}\frac{(1-\frac{g}{c})^2}{1-[min(1,X)(\frac{g}{c})]}$$ $$d_2 = 900T[(X-1)+\sqrt{(X-1)^2+\frac{8klX}{cT}}]$$ $PF=(\frac{(1-P)}{(1-(g/c)})*f_p$ additional explanation for PF Where, d = control delay, s/veh, d1 = uniform delay component, s/veh, PF = progression adjustment factor, d2 = overflow delay component, s/veh, d3 = delay due to pre-existing queue, s/veh, T = analysis period, h, X = v/c ratio, C = cycle length, s, k = incremental delay factor for actuated controller settings; 0.50 for all pre-timed controllers, l = upstream filtering/metering adjustment factor; 1.0 for all individual intersection analyses, c = capacity, veh/h, P = proportion of vehicles arriving during the green interval and $f_p$ = supplemental adjustment factor for platoon arriving during the green Numerical problems Consider the following situation: An intersection approach has an approach flow rate of 1,400 vph, a saturation flow rate of 2,650 vphg, a cycle length of 102 s, and effective green ratio for the approach 0.55. Assume Progression Adjustment Factor 1.25 and delay due to pre-existing queue, 12 sec/veh. What control delay sec per vehicle is expected under these conditions? Solution: Saturation flow rate =2650 vphg , g/C=0.55, Approach flow rate v=1700 vph, Cycle length C=102 sec, delay due to pre-existing queue =12 sec/veh and Progression Adjustment Factor PF=1.25. The capacity is given as: $$c = s\times \frac{g}{C}$$ $$= 2650\times 0.55$$ $$= 1458~vph v/c$$ Degree of saturation X= v/c= 1700/1458 =1.16. So the uniform delay is given as: $$d_1 = \frac{C}{2}\frac{(1-\frac{g}{C})^2}{[1-min(X,1)(\frac{g}{c})]}$$ $$= \frac{102}{2}\frac{(1-1.16)^2}{[1-min(1.16,1)(.55)]}= 22.95$$ Uniform delay =22.95 and the over flow delay is given as: $$d_2 = \frac{C}{2}*(X-1)+\sqrt{(X-1)^2+\frac{8klX}{cT}}$$ $$= \frac{102}{2}*(1.16-1)+\sqrt{(1.16-1)^2+\frac{(8*5*1*1.16)}{1458}} = 16.81$$ Overflow delay, d2=16.81. Hence, the total delay is" $$d = d_1~PF+d_2+d_3$$ $$= 22.95\times 1.25 + 16.81 + 8 = 53.5~sec/veh.$$ Therefore, control delay per vehicle is 53.5 sec.