If the normal flows on two approach roads at an intersection are respectively 500 pcu per hr and 300 pcu per hr, the saturation flows are 1600 pcu per hr on each road the total lost time per signal cycle is 16s, then the optimum cycle time by Webster's method is

48 S

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72.5 S

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19.3 S

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58 S

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Solution

The correct option is D 58 S
(b)

$C_{0} = \frac{1.5 L + 5}{1 - Y}$

L = Total lost time per cycle
= 16 sec
$Y = y_{1} + y_{2}$

$y_{1} = \frac{500}{1600} a n d y_{2} = \frac{300}{1600}$

$∴ Y = \frac{500 + 300}{1600} = 0.5$

$C_{0} = \frac{1.5 \times 16 + 5}{1 - 0.5} = 58 S$

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If the normal flows on two approach roads at an intersection are respectively 500 pcu per hr and 300 pcu per hr, the saturation flows are 1600 pcu per hr on each road the total lost time per signal cycle is 16s, then the optimum cycle time by Webster's method is

The correct option is D 58 S(b) C0=1.5L+51−Y L = Total lost time per cycle = 16 sec Y=y1+y2 y1=5001600andy2=3001600 ∴Y=500+3001600=0.5 C0=1.5×16+51−0.5=58 S

The correct option is D 58 S
(b)

$C_{0} = \frac{1.5 L + 5}{1 - Y}$

L = Total lost time per cycle
= 16 sec
$Y = y_{1} + y_{2}$

$y_{1} = \frac{500}{1600} a n d y_{2} = \frac{300}{1600}$

$∴ Y = \frac{500 + 300}{1600} = 0.5$

$C_{0} = \frac{1.5 \times 16 + 5}{1 - 0.5} = 58 S$