Fixed-time two-phase signal: optimum cycle and greens

Phase 1 = North-South, Phase 2 = East-West.

Step 1: Flow ratios

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

$$Y = 0.360 + 0.250 = 0.610$$

Step 2: Lost time

Lost time per phase = start-up loss 2 s + amber (3 s for N-S, 4 s for E-W).

$$L = (2.0+3.0) + (2.0+4.0) = 11.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(11.0) + 5}{1 - 0.610} = 55.1\ \text{s}$$

Adopt C = 60 s.

Step 4: Green times

Total effective green $= C - L = 60 - 11.0 = 49.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.360}{0.610}(49.0) = 28.9\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.250}{0.610}(49.0) = 20.1\ \text{s}$$

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

Check: 31 + 3 + 22 + 4 = 60 s.

Phase diagram

|       P1 green 31       | A3 |   P2 green 22   |  A4  |