Two-phase signal cycle length and green times

Two phases: Phase 1 = East-West, Phase 2 = North-South. The question asks for the cycle length and the actual (displayed) green of each phase.

Step 1: Flow ratios

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

$$Y = 0.283 + 0.196 = 0.479$$

Step 2: Lost time

For each phase: start-up lost time 3 s, plus yellow 3 s and all-red 4 s (clearance interval $I$ = 7 s), all treated as lost to traffic movement.

$$L = (3.0+7.0) + (3.0+7.0) = 20.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(20.0) + 5}{1 - 0.479} = 67.2\ \text{s}$$

Adopt C = 70 s.

Step 4: Green times

Total effective green $= C - L = 70 - 20.0 = 50.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.283}{0.479}(50.0) = 29.5\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.196}{0.479}(50.0) = 20.5\ \text{s}$$

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

Check: 33 + 7 + 23 + 7 = 70 s.

Phase diagram

|     P1 green 33     |   A7   | P2 green 23  |   A7   |