# Fundamentals of Transportation/Traffic Signals/Solution

An approach at a pretimed signalized intersection has an arrival rate of 500 veh/hr and a saturation flow rate of 3000 veh/hr. 30 seconds of effective green are given in a 100-second cycle. Analyze the intersection assuming D/D/1 queueing by describing the proportion of the cycle with a queue, the maximum number of vehicles in the queue, the total and average delay, and the maximum delay.

With the statements in the problem, we know:

• Green Time = 30 seconds
• Red Time = 70 seconds
• Cycle Length = 100 seconds
• Arrival Rate = 500 veh/hr (0.138 veh/sec)
• Departure Rate = 3000 veh/hr (0.833 veh/sec)

Traffic intensity, $\rho$ , is the first value to calculate.

$\rho ={\frac {\lambda }{\mu }}={\frac {500}{3000}}=0.167\,\!$ Time to queue clearance after the start of effective green:

$t_{c}{\rm {}}={\rm {}}{\frac {\rho r}{1-\rho }}={\frac {0.167(70)}{1-0.167}}=14.03\ s\,\!$ Proportion of the cycle with a queue:

$P_{q}{\rm {}}={\rm {}}{\frac {r{\rm {}}+{\rm {}}t_{c}}{C}}{\rm {}}={\rm {}}{\frac {70{\rm {}}+{\rm {}}14.03}{100}}{\rm {}}=0.84\,\!$ Proportion of vehicles stopped:

$P_{s}={\frac {\lambda \left({r+t_{C}}\right)}{\lambda \left({r+g}\right)}}={\frac {0.138\left({70+14.03}\right)}{0.138\left({70+30}\right)}}=0.84\,\!$ Maximum number of vehicles in the queue:

$Q_{\max }{\rm {}}={\rm {}}\lambda r={\rm {}}0.138(70)=9.66\,\!$ Total vehicle delay per cycle:

$D_{t}{\rm {}}={\rm {}}{\frac {\lambda r^{2}}{2\left({1-\rho }\right)}}{\rm {}}={\rm {}}{\frac {0.138(70^{2})}{2\left({1-0.167}\right)}}{\rm {}}=406\ veh-s\,\!$ Average delay per vehicle:

$d_{avg}{\rm {}}={\frac {r^{2}}{2C\left({1-\rho }\right)}}={\frac {(70)^{2}}{2(100)\left({1-0.167}\right)}}=29.41\ s\,\!$ Maximum delay of any vehicle:

$d_{\max }{\rm {}}={\rm {}}r={\rm {}}70\ s\,\!$ Thus, the solution can be determined:

• Proportion of the cycle with a queue = 0.84
• Maximum number of vehicles in the queue = 9.66
• Total Delay = 406 veh-sec
• Average Delay = 29.41 sec
• Maximum Delay = 70 sec