Question

A pre-timed four-phase signal has a critical lane flow rate for the first three phases as 200, 187 and 210 veh/hr with a saturation flow rate of 1800 veh/hr/lane for all phases. The lost time is given as 4 seconds for each phase. If the cycle length is 60 seconds, the effective green time (in seconds) of the fourth phase is ______________ 

1. 15.74
2. 15.84
3. 15.64
4. 15.94

Answer 1

Correct Answer - Option 1 : 15.74

Concept:

Webster’s method for an optimum cycle time of a signal is given by

\({{\rm{C}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5L\ +\ 5}}}}{{{\rm{1\ -\ Y}}}}\)

Effective green time on ith lane

\({{\rm{G}}_{\rm{i}}}{\rm{ = }}\frac{{{{\rm{Y}}_{\rm{i}}}}}{{\rm{Y}}}{\rm{ × }}\left( {{{\rm{C}}_{\rm{o}}}{\rm{ - L}}} \right)\)

L = Lost time per cycle = 2N + R

N = Number of phase = Number of independent movements

R = All red time

Y = Sum of the ratios of normal to saturated flows

Y = Y₁ + Y₂ + Y₃

Where,

\({{\rm{Y}}_1} = \frac{{{{\rm{q}}_1}}}{{{{\rm{s}}_1}}}\), \({{\rm{Y}}_2} = \frac{{{{\rm{q}}_2}}}{{{{\rm{s}}_2}}}\) and \({{\rm{Y}}_3} = \frac{{{{\rm{q}}_3}}}{{{{\rm{s}}_3}}}\)

q₁, q₂ and q₃ is normal flow in phase 1, 2 and 3 respectively
s₁, s₂ and s₃ is saturated flow in phase 1, 2 and 3 respectively

Calculation:

Given,

q₁ = 200 veh/hr, q₂ = 187 veh/hr and q₃ = 210 veh/hr

s = s1 = s2 = s₃ = 1800 veh/hr/lane 

L = Total lost time = 4 × 4 = 16 seconds

C_o = 60 seconds

Y = Y1 + Y2 + Y₃ + Y₄

\({\rm{Y}} = \frac{{{{\rm{q}}_1}}}{{{{\rm{s}}_1}}} + \frac{{{{\rm{q}}_2}}}{{{{\rm{s}}_2}}} + \frac{{{{\rm{q}}_3}}}{{{{\rm{s}}_3}}} + {Y_4}\)

\({\rm{Y}} = \frac{{200}}{{1800}} + \frac{{187}}{{1800}} + \frac{{210}}{{1800}} + {Y_4}\)

Y = 0.332 + Y₄

Now

\({{\rm{C}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5L\ +\ 5}}}}{{{\rm{1\ -\ Y}}}}\)

\({{\rm{60}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5× 16\ +\ 5}}}}{{{\rm{1\ -\ (0.332+Y_4)}}}}\)

Y₄ = 0.185

∴ Y = 0.332 + 0.185 = 0.517

Effective green time for 4^th phase

\({{\rm{G}}_{\rm{4}}}{\rm{ = }}\frac{{{{\rm{Y}}_{\rm{4}}}}}{{\rm{Y}}}{\rm{ × }}\left( {{{\rm{C}}_{\rm{o}}}{\rm{ - L}}} \right)\)

\({{\rm{G}}_{\rm{4}}}{\rm{ = }}\frac{{{{\rm{0.185}}}}}{{\rm{0.517}}}{\rm{ × }}\left( {{{\rm{60}}}{\rm{ - 16}}} \right) = 15.74 \)

Hence the effective green time for 4^th phase = 15.74 seconds