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# Commentary 10

## C10.1 Derivation of Gap Acceptance Formulae for Displaced Negative Exponential Distribution Of Headways in the Major Traffic Flow

(a) Probabilities of headways of given size

As discussed in Section 3.3.3, the displaced negative exponential headway distribution postulates a minimum possible headway of $$\beta$$ s/veh and is characterised by the probability density function

 $f(t) = \frac{q}{1 - q\beta } e^{\frac{-q(t-\beta )}{1 - q\beta } }$ for $$t\ge \beta$$ C70 and $f(t) = 0$ for $$t<\beta$$ C71

By integration of the function in Equation C70 the probabilities for headway size, $$h$$ are obtained:

 $\Pr (\beta \le t\le h) = \int _{t}^{\infty } f(t).dt = e^{\frac{-q(t-\beta )}{1-q\beta } }$ C72 and $\Pr (\beta \le h\le t) = \int _{\beta }^{t} f(t).dt = 1 - e^{\frac{-q(t-\beta )}{1-q\beta } }$ C73

and, of course, by definition:

 $\Pr (h<\beta ) = 0$ C74

Over a significant period of time, H, there will be qH headways in the major stream flow and it follows from Equations C72 to C74 that the number of headways, N, in each of the size ranges will be:

 $N(\beta \le t\le h) = qH . e^{\frac{-q(t-\beta )}{1-q\beta } }$ C75 and $N(\beta \le h\le t) = qH . (1 - e^{\frac{-q(t-\beta )}{1-q\beta } } )$ C76 $N(h<\beta ) = 0$ C77

(b) Average duration of headways within a given range

Over a period H, the number of headways of duration t to t+dt, where dt is infinitesimally small and $$t\ge \beta$$, will be:

 $N(t\le h\le t+dt) = qH . f(t) . dt = qH . \frac{q}{1-q\beta } . e^{\frac{-q(t-\beta )}{1-q\beta } } . dt$ C78

and the time spent in such headways will be:

 $T(t\le h\le t+dt) = t . N(t\le h\le t+dt) = t . qH . \frac{q}{1-q\beta } . e^{\frac{-q(t-\beta )}{1-q\beta } } . dt$ C79

Therefore, the total time spent in headways ≥ t (where $$t\ge \beta$$ ) is:

 $T(\beta \le t\le h) = qH \int _{t}^{\infty } t . \frac{q}{1-q\beta } . e^{\frac{-q(t-\beta )}{1-q\beta } } . dt$ C80

The integral here is of the form $$\int t . a e^{-at+b} . dt$$, for which the solution is $$\left[-e^{-at+b} . \frac{(at+1)}{a^{2} } \right]$$. In this case, $$a = \frac{q}{1-q\beta }$$ and $$b = \frac{q\beta }{1-q\beta }$$ and Equation C80 becomes:

 $T(\beta \le t\le h) = qH . e^{\frac{-q(t-\beta )}{1-q\beta } } . \left(\frac{1}{q} + t - \beta \right)$ C81

The average duration of headways greater than or equal to t (where t ≥ β) can now be obtained as the total time spent in such headways (Equation C81) divided by the number of such headways (Equation C75), that is:

 $h_{av} (\beta \le t\le h) = \frac{1}{q} + t - \beta$ C82

By a similar analysis, noting that the total time spent in headways ≤ t (where $$t\ge \beta$$ ) is H minus the time given by Equation C81and that the number of such headways is as in Equation C76, Equation C83 is derived:

 $h_{av} (\beta \le h\le t) = \frac{1}{q} - \frac{(t-\beta ) e^{\frac{-q(t-\beta )}{1-q\beta } } }{1 - e^{\frac{-q(t-\beta )}{1-q\beta } } }$ C83

(c) Average delays to minor traffic

As in Commentary 5, the average delay at the gap acceptance point for all minor traffic stream units can be derived by noting that the probability of any unit having to wait for n gaps, each less than T, before being able to proceed is given by the geometric distribution as:

 $$P(n) = (1 - p) p^{n}$$ n = 0, 1, 2, … C84

where (drawing on Equation C73)

 $p = \Pr (gap so that  \[P(n) = e^{\frac{-q(T-\beta )}{1-q\beta } } (1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )^{n}$ C86

The expected number of gaps less than T for which a minor traffic stream unit has to wait before being able to proceed is then given by:

 $$E(n) = 0.P(0) + 1.P(1) + 2.P(2) + 3.P(3) +$$… C87 $= e^{\frac{-q(T-\beta )}{1-q\beta } } (1 - e^{\frac{-q(T-\beta )}{1-q\beta } } ) + 2 e^{\frac{-q(T-\beta )}{1-q\beta } } (1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )^{2} + 3 e^{\frac{-q(T-\beta )}{1-q\beta } } (1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )^{3} +....$ $E(n) = \frac{1 - e^{\frac{-q(T-\beta )}{1-q\beta } } }{e^{\frac{-q(T-\beta )}{1-q\beta } } }$

Now the average duration of headways less than T in the major traffic stream is given by Equation C83 as:

 $h_{av} (\beta \le h\le t) = \frac{1}{q} - \frac{(t-\beta ) e^{\frac{-q(t-\beta )}{1-q\beta } } }{1 - e^{\frac{-q(t-\beta )}{1-q\beta } } }$ C87

Then the average delay experienced by all minor traffic stream units at the gap acceptance point is:

 $d_{av} (d\ge 0) = E(n) . [h_{av} (h which simplifies to:  \[d_{av} (d\ge 0) = \frac{1}{q e^{\frac{-q(T-\beta )}{1-q\beta } } } - \frac{1}{q} - (T-\beta )$ C88

The average delay at the gap acceptance point to only those minor traffic stream units that do experience such delay is obtained by dividing the average delay to all minor stream units (as in Equation C88) by the proportion experiencing non-zero delay (equal to the probability of a headway < T, as given by Equation C73) to give:

 $d_{av} (d>0) = \frac{(1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )}{e^{\frac{-q(T-\beta )}{1-q\beta } } } \frac{\left(1 - e^{\frac{-q(T-\beta )}{1-q\beta } } - q (T-\beta ) e^{\frac{-q(T-\beta )}{1-q\beta } } \right)}{q . (1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )} \frac{1}{(1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )}$

which simplifies to:

 $d_{av} (d>0) = \frac{1}{q e^{\frac{-q(T-\beta )}{1-q\beta } } } - \frac{(T-\beta )}{(1 - e^{\frac{-q(T-\beta )}{1-q\beta } } )}$ C89

(d) Absorption capacity

To derive the theoretical absorption capacity, assume that a queue of minor traffic stream vehicles is waiting to cross or enter a major traffic stream that has volume q and a displaced negative exponential headway distribution, and let ti be the minimum gap required to allow i minor traffic stream vehicles to carry out the manoeuvre within the one gap, where i = 1, 2, 3,....

During a significant period of time, H, the number of major road headways greater than or equal to ti, i = 1, 2, 3,... is:

 No. of headways $$\ge t_{i} = H.qe^{\frac{-q(t_{i} -\beta )}{1-q\beta } }$$ C90

Therefore, the number of major traffic stream headways that allow exactly i minor traffic stream vehicles to cross or enter the major stream, i = 1, 2, 3,... is:

 $n_{i} = H.(q e^{\frac{-q(t_{i} -\beta )}{1-q\beta } } - q e^{\frac{-q(t_{i+1} -\beta )}{1-q\beta } } )$ C91

Hence, the total number of minor stream vehicles able to cross or join the major stream during the period H is:

 $$N = \sum _{i=1}^{\infty }i.n_{i}$$ $$= \sum _{i=1}^{\infty }i.Hq(e^{\frac{-q(t_{i} -\beta )}{1-q\beta } } - e^{\frac{-q(t_{i+1} -\beta )}{1-q\beta } } )$$ $$= H q \sum _{i=1}^{\infty }e^{\frac{-q(t_{i} -\beta )}{1-q\beta } }$$ C92

Taking the critical gap ,T, as the minimum headway that will allow one minor stream vehicle to cross or join the major stream, assume that each additional time interval T0 in the size of the headway is sufficient to allow one additional minor stream vehicle to follow in undertaking the manoeuvre. T0 is known as the follow-up headway and the above assumption implies that:

 $t_{i} = T + (i - 1) T_{0} , i = 1, 2, 3, ...$ C93

Substituting from Equation C93 into Equation C92, gives:

 $$N = H q \sum _{i=1}^{\infty }e^{\frac{-q(T + (i-1)T_{0} -\beta )}{1-q\beta } }$$ $$= H q e^{\frac{-q(T-\beta )}{1-q\beta } } \sum _{k=0}^{\infty }\left(e^{\frac{-qT_{0} }{1-q\beta } } \right)^{k}$$ $$= \frac{H q e^{\frac{-q(T-\beta )}{1-q\beta } } }{1 - e^{\frac{-qT_{0} }{1-q\beta } } }$$ C94

Thus the theoretical maximum rate at which minor stream vehicles can cross or join the major traffic stream, that is, the theoretical absorption capacity, is obtained as:

 $$C = \frac{N}{H} = \frac{q e^{\frac{-q(T-\beta )}{1-q\beta } } }{1 - e^{\frac{-qT_{0} }{1-q\beta } } }$$ C95

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