A simple four-leg intersection (a single lane per approach) needs a 2-phase fixed-time signal. The critical flows in the N-S and E-W directions are 600 and 400 veh/hr, respectively, with 0.95 peak-hour factor. Saturation flow is 1800 veh/hr and the total lost time per phase is observed to be 5.2 seconds. Amber time and all-red time are 4 and 1 seconds, respectively. Assume that the intersection should operate at 90% of its capacity. Determine the cycle length and distribution of green (green splits). Calculate also effective green times.

Hint: round the cycle length to the next higher 5-sec increment (e.g. 86 -> 90); then round splits to the next integer values.

[tex]=\frac{q_{NS}}{S_{NS}}+\frac{q_{ew}}{S_{ew}}\\\\=\frac{600}{1800}+\frac{400}{1800}\\\\=\frac{6}{18}+\frac{4}{18}\\\\=\frac{3}{9}+\frac{2}{9}\\\\=\frac{1}{3}+\frac{2}{9}\\\\=\frac{3+2}{9}\\\\=\frac{5}{9}\\\\=0.55555555556\\\\=0.57[/tex]

calculating length [tex](C_D)=\frac{1.5 \ L +5}{1-y}\\\\[/tex]

[tex]=\frac{1.5 \times (1.2) +5}{1-0.57}\\\\=\frac{1.8 +5}{0.43}\\\\=\frac{6.8}{0.43}\\\\= 15.8 \ sec[/tex]

critical value:

[tex]\to V_{C_1}= 600 \ \frac{veh}{hr}\\\\\to V_{C_2}= 400 \ \frac{veh}{hr}\\\\[/tex]

Total critical value:

[tex]V_C= V_{C_1}+ V_{C_2}\\\\[/tex]

[tex]= 600 + 400\\\\= 1 000 \ \frac{veh}{hr}\\\\[/tex]

Active green time[tex]= 15.8-1.2=14.6[/tex]

[tex]g_1=\frac{600}{1000}\times 14.6= 8.76 \ sec\\\\g_2=\frac{400}{1000}\times 14.6= 5.84 \ sec[/tex]

Native green time:

[tex]G_1= 8.76 -1.2+4 =11.56 \sec\\\\ G_2= 5.84 -1.2+4 =8.64 \sec[/tex]