Two-phase signal: 400 and 250 PCU/hr, 7 m roads

Both roads are 7 m wide. Pedestrian crossing time (all-red period $R$) = initial walk 6 s + width/speed = $6 + 7/1.0 = 13$ s.

Step 1: Flow ratios

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

$$Y = 0.320 + 0.250 = 0.570$$

Step 2: Lost time

The lost time and amber are not given. As in the usual Webster design, an amber of 2 s per phase is assumed and the lost time per phase is taken as 2 s (equal to the amber), plus the all-red pedestrian period $R$ = 13 s. Because the lost time equals the amber, the displayed green equals the effective green.

$$L = (0.0+2.0) + (0.0+2.0) + 13 = 17.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(17.0) + 5}{1 - 0.570} = 70.9\ \text{s}$$

Adopt C = 75 s.

Step 4: Green times

Total effective green $= C - L = 75 - 17.0 = 58.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.320}{0.570}(58.0) = 32.6\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.250}{0.570}(58.0) = 25.4\ \text{s}$$

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

Check: 33 + 2 + 25 + 2 + 13 = 75 s.

Phase diagram

Cycle = 75 s
Road A:      [ GREEN 33 ][A 2][          RED 40           ]
Road B:      [     RED 35     ][ GREEN 25 ][A 2][ ALL-RED 13 ]
Pedestrians: [          DON'T WALK 62           ][ WALK 13 ]