Webster Optimum Cycle Length

C_{0} = \frac{1.5 L + 5}{1 - Y} Y = \sum y_{i} = \sum (q / s)_{i}

L is total lost time per cycle (start-up + clearance, all phases). yi = qi/si is the flow ratio of critical movement in phase i. Y is the sum of critical y. Practical cycle is often 0.75 C0 to 1.5 C0, 30–120 s typical.

$C_{0}$ (s): Optimum cycle
$L$ (s): Lost time per cycle
$Y$: Sum of critical flow ratios

Step-by-step solved example

L=10 s, two phases y=0.40 and 0.30. C0?

Signal cycle splits lost time and green along the intersection timeline.

1. Y
Y=0.70. 1−Y=0.30.
2. C0
C0=(1.5×10+5)/0.30=20/0.30=66.7 s.
$C_{0} = 67 s$

Answer: 67 s

10 practice questions

0/10 correct

1.If Y → 1, C0

2.Lost time L includes

3.yi = q/s uses

4.More phases usually mean

5.Saturation flow s is about

6.Effective green gi is often

7.A cycle much longer than C0

8.Pedestrian minimum green can

9.Units of C0 are

10.Y=0.4+0.25=0.65, L=8 s. C0 nearest