Question

The queue length (in number of vehicles) versus time (in seconds) plot for an approach to a signalized intersection with the cycle length of 96 seconds is shown in the figure (not drawn to scale)

At timet=0, the light has just turned red. The effective green time is 36 seconds, during which vehicles discharge at the saturation flow rate, s (in vph). Vehicles arrive at a uniform rate, v (in vph), throughout the cycle. Which one of teh following statements is TRUE

• A. v = 600 vph, and for this cycle, the average stopped delay per vehicle=30 seconds
• B. s=1200 vph, and for this cycle, the average stopped delay per vehicle =28.125 seconds
• C. s= 1800 vph, and for this cycle, the average stopped delay per vehicle=28.125 seconds
• D. v = 600 vph, and for this cycle, the average stopped delay per vehicle=45 seconds

Answer 1

The correct option is C s= 1800 vph, and for this cycle, the average stopped delay per vehicle=28.125 seconds

Vehicle arrived upto 60 sec (Red time)=10
arrival rate=$\frac{10}{60}\times 3600=600v/h$

$V=600v/h$

$\to$ departure at vehicle starts at 60 second and ends at 90 seconds.
So, between 60 second to 90 second total vehicle departed
= Vehicle arrived upto 60 second+ Vehicle arriving between 60 sec to 90 sec
$=10+\frac{600}{3600}\times 30=10+5=15$
So, departure rate = Saturation flow
$S=\frac{15}{30}\times 3600$

$S=1800v/h$

Average delay time is given by
$t_d=\frac{\frac{C}{2}(1-\frac{g}{c}{)}^2}{1-\frac{v}{s}}$
C = 96 seconds
g = 96-60 = 36 seconds
V = 600 Vph
S = 1800 Vph
$t_d=\frac{\frac{96}{2}(1-\frac{36}{96}{)}^2}{1-\frac{600}{1800}}$

$t_d=28.125seconds$

So, (b) option is true
Alternatively:

Total no.of vehicles arriving in 90 sec
$=\frac{10}{60}\times 90=15$
V=Vehicle arrival rate
$=\frac{15}{90}\times 3600=600veh/hr$
S= Vehicle discharge rate
$=\frac{15}{30}\times 3600=1800veh/hr$
Aggregate delay =Area under shaded diagram
$=\frac{1}{2}\times 15\times 60=450veh/sec$
Av. Stop delay per veh$=\frac{450veh.sec}{no.ofveharriving}$
(in one cycle time)
$\frac{450}{\frac{15}{90}\times 96}=28.125sec$
Hence option (b) is correct