# Traffic management

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### 5.5.2 Example 2 – Displaced Negative Exponential Headways on Major Road

Assume that at the intersection shown in Figure 5.2, all minor road traffic consists of through cars with gap acceptance characteristics as shown for that category in Table 5.2. For this situation, compare the average delays at the stop line and the theoretical absorption capacities that would apply if the major road volumes were as shown and the major road headway distributions were:

• (a) Negative exponential.
• (b) Displaced negative exponential, with minimum headway 1.5 s/veh.

For a negative exponential headway distribution on the major road

The average delay at the stop line over all through cars was calculated in Example 1, using Equation 5.2, with q=1260 veh/h =0.35 veh/s and T=5.0 s/veh , as 8.58 s/veh.

The average delay to only those through cars that are delayed is given by Equation 5.3, with the same values of q and T , as:

 $$d_{av} (d>0) = \frac{1}{0.35 e^{-1.75} } - \frac{5.0}{(1 - e^{-1.75} )} = 10.39$$ s/veh

The theoretical absorption capacity for through cars also was calculated in Example 1, using Equation 5.4, with q=1260 veh/h = 0.35 veh/s, T=5.0 s/veh and T0=2.5 s/veh , as 0.1043 veh/s or 375.5 veh/h.

#### For a displaced negative exponential headway distribution with minimum headway 1.5 s/veh

The average delay at the stop line over all through cars is given by Equation 5.18, with q=1260 veh/h = 0.35 veh/s, T=5.0 s/veh and $$\beta$$ = 1.5 s/veh, as:

 $$d_{av} (d\ge 0) = \frac{1}{0.35 e^{\frac{-0.35(5.0-1.5)}{(1 - 0.35\text{ x }1.5)} } } - \frac{1}{0.35} - (5.0 - 1.5) = 31.31$$ s/veh

The average delay to only those through cars that are delayed is given by Equation 5.19, with the same values of q, T and $$\beta$$, as:

 $$d_{av} (d>0) = \frac{1}{0.35 e^{\frac{-0.35(3.5)}{0.475} } } - \frac{(5.0-1.5)}{(1 - e^{\frac{-0.35(3.5)}{0.475} } )} = 33.88$$  s/veh

The theoretical absorption capacity for through cars is obtained using Equation 5.20, with q=1260 veh/h = 0.35 veh/s, $$\beta$$ =1.5 s/veh, T=5.0 s/veh and T0=2.5 s/veh , as :

 $$C = \frac{0.35 e^{\frac{-0.35(3.5)}{0.475} } }{1 - e^{\frac{-0.35(2.5)}{0.475} } } = 0.03155$$ veh/s = 113.6 veh/h

The significant differences in delays and absorption capacities between the two cases are explained by the fact that with a negative exponential distribution, 17.4% of major road headways are larger than the critical gap, whereas with a displaced negative exponential distribution, this proportion is reduced to only 7.6%.