The following observations were made for a two phase traffic signal:

Road Average normal flow (veh/hr) Saturation flow (veh/hr)

A 400 1250

B 250 1000

The all red time required for pedestrian crossing is 12 sec and amber time is 2.0 sec for each lane.
The optimum cycle time will be _______ sec. [Use Webster's approach]
67.4

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Solution

The correct option is A 67.4
For road A, $y_{a} = \frac{q_{a}}{S_{a}} = \frac{400}{1250} = 0.32$
For road B, $y_{b} = \frac{q_{b}}{S_{b}} = \frac{250}{1000} = 0.25$

$Y = y_{a} + y_{b}$
= 0.32 + 0.25 = 0.57

$L = 2 n + R = 2 \times 2 + 12 = 16 sec$

$∴ C_{0} = \frac{1.5 L + 5}{1 - Y} = \frac{1.5 \times 16 + 5}{1 - 0.57}$
= 67.4 sec

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Road	Average normal flow (veh/hr)	Saturation flow (veh/hr)
A	400	1250
B	250	1000

The correct option is A 67.4For road A, ya=qaSa=4001250=0.32 For road B, yb=qbSb=2501000=0.25 Y=ya+yb = 0.32 + 0.25 = 0.57 L=2n+R=2×2+12=16sec ∴C0=1.5L+51−Y=1.5×16+51−0.57 = 67.4 sec