Question

Suppose you observed vehicle arrival flows in a single approach for two consecutive cycles at a signalized intersection. The signal has a 60 sec. cycle time with a 30 sec. green interval and a 30 sec. red interval (ignore yellow interval). The arrival rates are assumed to be constant in both cycles. The arrival rate was 720 veh/hour. Assume that the vehicles in the queue formed on red pass through the intersection during the subsequent green interval at the saturation flow rate of 1800 veh/hour immediately after the start of the green interval.

(a) Draw the queuing diagram for the approach during the two cycles

(b) Calculate the total uniform delay per cycle.

(c) Calculate the average delay per vehicle.

(d) Calculate the maximum queue length (i.e. maximum number of vehicles in the queue).

Answer 1

We can set up a table with the information we are given.

Note that vehicles queued up during red pass through in subsequent green interval at 1800 veh/hour.

So this equates to 1800/3600 * 30 = 15 vehicles

So the maximum number of queued vehicles that can be discharged from the previous cycle = 15 vehicles

so in 15 seconds about 7.5 queued vehicles from the previous red can be discharged

Arrival rate in each 15 second interval is 720 veh/hr, so that is 3 vehicles every 15 seconds.

Departure rate is the same as arrival rate during the green time i.e 720 veh/hour. In addition to the normal departures, any queued vehicles from previous red discharge at 1800 veh/hour i.e 7.5 vehicles/15 seconds.

So we can calculate cumulative arrivals and cumulative departures as shown below.

Queue Length = Cumulative Arrivals - Cumulative Departures

a) So our queing diagram can be ploted as below

b) total uniform delay per cycle is calculated using the formula

where d₁ = uniform delay per cycle, and this formula assumes uniform delay, no initial queue

C = cycle length = 60 seconds

g = effective green = 30 seconds

so g/c = 0.5

X = V/C ratio or degree of saturation of lane group, assume that the lane is fully utilized, so this value will be greater than 0.5

So d₁ = 0.5 * 60 * (1-0.5)²/ (1-0.5) = 30 * .5*.5 = 15 seconds/vehicle

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c) Average delay per vehicle

According to Websters Delay Model

average delay per vehicle per cycle is calculated as

where d = average delay

C = cycle length = 60 secondns

g = effective green = 30 seconds

V = 720 veh/hour

S = saturation flow = 1800 veh/hour

So average delay = 60/2 * ( 1-30/60)²/(1-720/1800) = 30 * .5 * .5/(1-0.4) = 12.5 seconds/veh

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d) Maximum Queue Length

This is given by the longest vertical line between the cumulative departures and cumulative arrivals graph. You can also read it from the excel table.

Maximum queue length = 6 vehicles and this happens at the 60 second mark

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