Question

A pre-timed four phase signal has critical lane flow rate for the first three phases as 200, 187 and 210 veh/hr with saturation flow rate of 1800 veh/hr/lane. The lost time is given as 4 second for each phase. If the cycle length is 60 sec, the effective green time (in second) of the fourth phase is_____

• A. 15.755

Answer 1

The correct option is A 15.755
$y_1=\frac{q_1}{s_1}=\frac{200}{1800}$
$y_2=\frac{q_2}{s_2}=\frac{187}{1800}$
$y_3=\frac{q_3}{s_3}=\frac{210}{1800}$
$y=y_1+y_2+y_3+y_4$
$=0.33167+y_4$
$L=4\times 4=16sec$
$C_0=\frac{1.5L+5}{1-y}$
$\Rightarrow 60=\frac{1.5\times 16+5}{1-0.33167-y_4}$
$0.66833-y_4=\frac{1.5\times 16+5}{60}$
$\Rightarrow y_4=0.185$
Eff. green time for 4th Phase
$=\frac{(C_0-L)\times y_4}{y_1+y_2+y_3+y_4}$
$=\frac{(60-16)\times 0.185}{0.33167+0.185}$
$=15.755sec$