Webster signal: 450 and 350 PCU/hr, all-red 15 s

Saturation flow from headway: $s = 3600/h$. Road 1: 3600/2.5 = 1440 PCU/h; Road 2: 3600/3.75 = 960 PCU/h.

Step 1: Flow ratios

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

$$Y = 0.312 + 0.365 = 0.677$$

Step 2: Lost time

Lost time per phase = start-up loss (2 s for road 1, 3 s for road 2) + amber 2 s. The all-red period $R$ = 15 s for pedestrian crossing occurs once in the cycle and is added.

$$L = (2.0+2.0) + (3.0+2.0) + 15 = 24.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(24.0) + 5}{1 - 0.677} = 127.0\ \text{s}$$

Adopt C = 130 s.

Step 4: Green times

Total effective green $= C - L = 130 - 24.0 = 106.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.312}{0.677}(106.0) = 48.9\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.365}{0.677}(106.0) = 57.1\ \text{s}$$

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

Check: 51 + 2 + 60 + 2 + 15 = 130 s.

Phase diagram

Cycle = 130 s
Road 1:      [ GREEN 51 ][A 2][            RED 77             ]
Road 2:      [     RED 53     ][ GREEN 60 ][A 2][ ALL-RED 15 ]
Pedestrians: [            DON'T WALK 115           ][ WALK 15 ]