An intersection has a three-phase signal with the movements allowed in each phase and corresponding analysis and saturation flow rates shown in the table below.(1) Calculate the sum of the flow ratios for the critical lane groups.(2) calculate the minimum cycle length and the effective green time for each phase (balancing v/c for the critical lane groups). Assume the lost time is 4 seconds per phase and a critical intersection v/c of 0.90 is desired. **Table 7.6: Data for Problem 7.1**This table presents traffic flow data for different phases at an intersection.- **Phase 1:** - **Allowed Movements:** NB L, SB L - **Analysis Flow Rate:** 330, 365 vehicles per hour - **Saturation Flow Rate:** 1700, 1750 vehicles per hour- **Phase 2:** - **Allowed Movements:** NB T/R, SB T/R - **Analysis Flow Rate:** 1125, 1075 vehicles per hour - **Saturation Flow Rate:** 3400, 3300 vehicles per hour- **Phase 3:** - **Allowed Movements:** EB L, WB L; EB T/R, WB T/R - **Analysis Flow Rate:** 110, 80 vehicles per hour; 250, 285 vehicles per hour - **Saturation Flow Rate:** 650, 600 vehicles per hour; 1750, 1800 vehicles per hour**Key:**- **NB**: Northbound- **SB**: Southbound- **EB**: Eastbound- **WB**: Westbound- **L**: Left- **T/R**: Through/Right

The phase comprises the green interval, followed by a change interval and a clearance interval. For this reason, non-conflicting movements belong to each phase during the green interval. In this system, each set of movements is allowed to flow before stopping safely before another set of movements begins.An interaction 3 phase system signal; from the above table 1. flow ratio y y=y1+y2+y3y=normal flow saturation flowy1=3651750=0.208y2=11253400=0.330y3=2851800=0.158 the sum of the flow ratios to the critical lane groups y=y1+y2+y3y=0.208+0.330+0.158y=0.696 the sum of the flow ratios for the critical lane groups = 0.696 2. Minimum cycle length: C0=1.5L+51-yL=3×4=12C0=1.5×12+51-0.696C0=75.657 sec g1=y1yC0-Lg1=0.2080.69675.657-12g1=19.02 secg2=0.3300.69675.657-12g2=30.18 secg3=0.1580.69675.657-12g3=14.45 sec Minimum cycle length = 75.657 seconds Now we calculate the effective green time: