Fundamentals of Transportation/Traffic Signals/Solution

Example

Solution:

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:

$t_c=\frac{\rho r}{1-\rho }=\frac{0.167(70)}{1-0.167}=14.03s$

Proportion of the cycle with a queue:

$P_q=\frac{r+t_c}{C}=\frac{70+14.03}{100}=0.84$

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:

$Q_{\max }=\lambda r=0.138(70)=9.66$

Total vehicle delay per cycle:

$D_t=\frac{\lambda r^2}{2\left(1-\rho \right)}=\frac{0.138(70^2)}{2\left(1-0.167\right)}=406veh-s$

Average delay per vehicle:

$d_{avg}=\frac{r^2}{2C\left(1-\rho \right)}=\frac{(70{)}^2}{2(100)\left(1-0.167\right)}=29.41s$

Maximum delay of any vehicle:

$d_{\max }=r=70s$

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

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