The uniform arrival and uniform service rates observed on an approach road to a signalized intersection are 20 and 50 vehicles/minutes, respectively. For this signal, the red time is 30 s, the effective green time is 30 s, and the cycle length is 60 s. Assuming that initially there are no vehicles in the queue, the average delay per vehicle using the approach road during a cycle length(in s,round off to 2 decimal places is

12.5

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Solution

The correct option is A 12.5
Average delay per vehicle
$= \frac{C {(1 - \frac{g}{C})}^{2}}{2 (1 - \frac{v}{s})}$
$= \frac{60 {(1 - \frac{30}{60})}^{2}}{2 (1 - \frac{20}{50})} = \frac{50}{4}$
$= 12.5$

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The correct option is A 12.5Average delay per vehicle =C(1−gC)22(1−vs) =60(1−3060)22(1−2050)=504 =12.5

The correct option is A 12.5
Average delay per vehicle
$= \frac{C {(1 - \frac{g}{C})}^{2}}{2 (1 - \frac{v}{s})}$
$= \frac{60 {(1 - \frac{30}{60})}^{2}}{2 (1 - \frac{20}{50})} = \frac{50}{4}$
$= 12.5$