Signal timing: Road 1 14 m, Road 2 7 m

Road 1 and Road 2 each run in one phase, so this is a two-phase signal. The design hour flows of the two approaches of each road are given and the saturation flow is the same for both directions of a road (3675 and 1890 PCU/h).

Step 1: Flow ratios

Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one.

$$Y = 0.231 + 0.132 = 0.364$$

Step 2: Lost time

Lost time per phase = start-up lost time 2 s + amber 3 s = 5 s; no all-red time is given.

$$L = (2.0+3.0) + (2.0+3.0) = 10.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0) + 5}{1 - 0.364} = 31.4\ \text{s}$$

Adopt C = 35 s.

Step 4: Green times

Total effective green $= C - L = 35 - 10.0 = 25.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.231}{0.364}(25.0) = 15.9\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.132}{0.364}(25.0) = 9.1\ \text{s}$$

Displayed green $G = g +$ start-up loss (so that $\sum G + \sum I + R = C$):

Check: 18 + 3 + 11 + 3 = 35 s.

Phase diagram

|      P1 green 18       |   A3  | P2 green 11  |   A3  |