Delay and queue analysis

In this section the delay faced by vehicles at signalized intersections and the queues developed at signalized intersections are studied. Consider the plot shown in Figure 7. In this figure, the abscissa is time and the ordinate is the cumulative number of arrivals as well as the cumulative number of departures for a given stream (or approach) at an intersection. There are two lines in the figure. One shows a typical graph for cumulative number of arrivals on the given approach at a signalized intersection, the other shows the cumulative number of departures from the given approach at the signalized intersection. On the abscissa, the time is divided into slots named Cycle I, Cycle II, etc. These represent the cycles at the intersection. Each cycle is further subdivided into R and G. The R represents the duration of effective red (i.e., the time during which no vehicle on this particular approach crosses the intersection); similarly the G represents the duration of effective green (i.e., the time during which vehicles on this particular approach crosses the intersection).

図 7: A typical plot of cumulative number of arrivals and departures on an approach to a signalized intersection.

The plot gives a reasonably complete picture of the arrival and departure processes at the intersection. From such a plot one could obtain information about both delay and queues. The horizontal distance at a value of $n$ on the ordinate, for example, will give the delay faced by the $n^{th}$ vehicle to arrive at the intersection. Hence summing all such horizontal distances (or equivalently the area between the two lines) will give the total delay faced by all the vehicles arriving at the intersection. The total delay divided by the total number of arrivals will provide the average delay.

Similarly, the queues on the approach can be easily determined from the figure. For example, the vertical distance between the two lines at time $t$ will give the queue length at the intersection approach at time $t$. This implies that the queue lengths at any given time can be obtained easily from the figure and hence parameters like average queue length, variance of queue lengths, etc. can also be obtained.

However, obtaining such graphs for each and every intersection at all times is not feasible. Hence it is imperative that one analyzes the delay to vehicles and queues with an aim to derive equations which can give these quantities once data on arrival rates, cycle lengths, green times, red times, etc. are known. In the following, such a description of the analyses procedures is provided.

Delay analysis

To begin with assume that the arrival process is deterministic and vehicle arrive at a uniform rate. Further, assume that the system in unsaturated -- that is, the total number of vehicles that arrive in a period is less than the total number of vehicles that can be served by the system. These two assumptions mean that the arrival rate is such that all the vehicles that come in a cycle are cleared within the same cycle (like the situation in Cycle I of Figure 7).

The average delay to vehicles for this case then can be easily determined from the figure shown in Figure 8. The figure shows a typical cumulative arrival / departure graph against time for an unsaturated, uniform arrival rate approach to an intersection. The slope of the cumulative arrival line is $v$, where $v$ is the uniform arrival rate in vehicles per unit time. The slope of the cumulative departure line is sometimes zero (when the light is red) and sometimes $s$ (when the light is green); where $s$ is the saturation flow rate obtained as the reciprocal of the saturation headway explained earlier; $s$ is expressed as vehicles per hour of green per lane or vphgpl.

図 8: A typical plot of cumulative number of arrivals and departures on an unsaturated, uniform arrival rate approach to a signalized intersection.

From the figure (where $C$ is the duration of the cycle length and $g$ is the duration of the effective green period) it can be seen that the total delay (under the assumptions stated above), $TD_{u,us}$, is given by

$$TD_{u,us} = \sum_{i=1}^{i=n} d(i)$$ (6)

Assuming that $n$ is large enough so that the discrete sum of $d(i)$ is equal to the area of the triangle in the figure, the following can be written:

$$TD_{u,us} = 0.5 i^* (C-g)$$ (7)

From the figure, $i^*$ can be easily determined by noting that

$$i^* = vt$$

where,

$$vt = s(t-(C-g))$$

Determining $t$ from the above relation, $i^*$ can be written as

$$i^* = \frac{vs(C-g)}{s-v}$$

Hence,

$$TD_{u,us} = \frac{vs(C-g)^2}{2(s-v)}$$ (8)

From the above and noting that $n$, the total number of vehicles that arrived, is $vC$ the average delay under the above assumptions, $D_{u,us}$, can be obtained as:

$$D_{u,us} =\frac{TD_{u,us}}{vC} = \frac{(C-g)^2s}{2(s-v)C} = \frac{C(1-(g/C))^2}{2(1-(v/s))}$$ (9)

However, an approach to an intersection may not always stay unsaturated. There may be periods of over-saturation during which the arrivals from one cycle spill over to the next and so on. A similar situation can be seen in Cycle II of Figure 7. Under this assumption, and the assumption that arrival rate is still deterministic and uniform, the following plot of cumulative arrivals / departures w.r.t. time can be drawn (see Figure 8). In the figure it is assumed that the arrival rate from 0 to time T is $v$, and that $v$ is large enough to cause over-saturation (i.e., $vC$ greater than, $sg$, the maximum number of vehicles that can be served during a cycle).

図 9: A typical plot of cumulative number of arrivals and departures on an over-saturated, uniform arrival rate approach to a signalized intersection.

From the figure it can be seen that the total delay under these assumptions, $TD_{u,os}$, is given by the sum of the area marked with horizontal stripes (Area I) and the area marked with inclined stripes (Area II). This implies that the average delay, $D_{u,os}$, under these assumptions is the sum of the average delay due to Area I and the average delay due to Area II.

The average delay due to Area II can be easily determined by assuming the dashed line as the cumulative arrival line and using Equation 9. However, first the slope of the dashed line needs to be determined and later substituted for $v$ in Equation 9. If the slope of the dashed line is taken as $\sigma$ (and noting that the slope of the inclined part of the cumulative departure line is $s$, the saturation flow rate) then

$$\sigma C = s g$$

or

$$\sigma = \frac{g}{C}s$$

Substituting this expression of $\sigma$ for $v$ in Equation 9 one can write the average delay due to Area II, $AD_{II}$, as

$$AD_{II} = \frac{C(1-(g/C))^2}{2(1-(gs/Cs))}$$ (10)

$$= \frac{C-g}{2}$$

The average delay due to Area I can be determined by looking at the average time between the cumulative arrival line and the dashed line. If the horizontal distance (i.e., time in this case) between Point P and the dashed line is taken as $Z$, then it can be said that the time between the cumulative arrival line and the dashed line increases linearly from 0 to $Z$ over a time period of 0 to $T$. From this it can be said that the average delay for vehicles arriving between time 0 and $T$ is $Z/2$ (Note that the area of the triangle formed by the cumulative arrival line, a horizontal line from P and the dashed line is given by $Zn/2$, where $n$ is the number of vehicles to arrive till time $T$).

Similarly, the time between the cumulative arrival line and the dashed line decreases linearly from $Z$ to $0$ over a time period of $T$ to $M$. Following the same logic as above, the average delay due to Area I to vehicles arriving between times $T$ and $M$ is $Z/2$. Hence, it can be said that the average delay due to Area I is $Z/2$ irrespective of when a vehicle arrives.

If one assumes the vertical distance of Point P from the dashed line as $y$ then

$$\frac{y}{Z} = \mbox{slope of the dashed line,} \sigma$$

and

$$y=vT - \sigma T$$

hence,

$$Z=\frac{T(v-(gs/C))}{(gs/C)}$$ (11)

or

$$Z=T \left( \frac{v}{c}-1 \right)$$

where, $c=(g/C)s$, represents the maximum number of vehicles that can cross the intersection from a given approach per unit time.

Thus the average delay, $D_{u,os}$ is given as

$$D_{u,os}= \frac{T}{2} \left( \frac{v}{c}-1 \right)+ \frac{C-g}{2}$$ (12)

In reality, however, more often than not the arrival is not deterministic, it is stochastic. Once it is assumed that the arrival is stochastic the above relations cannot be used and can at best function as approximate estimates. Assumption of stochastic arrivals leads us to analysis which is beyond the scope of this book. Hence, in this text, only the relations generally used to determine delay under the above assumptions are described.

One of the relations often used in determining delay is due to Webster [#!web1!#]. Webster assumed that the arrivals are according to a Poisson distribution, departures occur uniformly and at a maximum rate of $s$, the average arrival rate $v$ is such that $vC \leq gs$, and that the queueing process runs under similar arrival and departure conditions long enough for the system to stabilize to a steady state (where, for example, cycle to cycle variations in average delay, average queue length, etc. are minimal). Based on these assumptions and some simulation runs (where the arrival and departure processes at a signalized intersection are simulated for various arrival pattern and signal settings) Webster proposed the following expression for average delay, $D_{s,web}$:

$$D_{s,web} = \frac{C(1-(g/C))^2}{2(1-(v/s))} + \frac{(v/c)^2}{2v(1-(v/c))} - 0.65 (c/v^2)^{1/3}(v/c)^{2+(5g/C)}$$ (13)

The first term in the equation is the same as that given in Equation 9, the second term is the additional term which results from analyzing the process by assuming stochastic arrivals, the third term is a correction factor obtained from simulation studies. It is seen that this term is often between 5% to 15% of the sum of the first two terms. Hence the following simplified form of the above equation is sometimes used:

$$D_{s,web} = 0.9 \left( \frac{C(1-(g/C))^2}{2(1-(v/s))} + \frac{(v/c)^2}{2v(1-(v/c))} \right)$$

In practice, it is found that the $D_{s,web}$ estimates of delay are not good for the entire range of $v/c$ values. In general, when $v/c$ values are close to one (that is, $vC$ is close to $gs$; this implies an increase in the chances of over-saturation when the stochasticity in the arrival rate causes the number of arrivals to be greater than $gs$), $D_{s,web}$ overestimates the average delay. One possible reason for this is that the derivation of the quantity assumes steady state behaviour, which is never achieved at real intersections as over-saturation by design occurs only in short spells. Other researchers around the world have developed other equations which are supposed to predict delays more realistically. All of them predict values which are close to $D_{s,web}$ when $v/c$ is not high (say less than 0.8) while their estimates are much lower when $v/c$ is high (say greater than 0.95). The 1985 Highway capacity manual of USA [#!hcm1!#] proposes the use of one such expression for delay, $D_{s,hcm85}$, reproduced here as Equation 14. The expression was developed based on a large database on intersection delay.

$$D_{s,hcm85}=0.76\frac{C(1-(g/C))^2}{2(1-(v/s))} + 173(v/c)^2 \left[ ((v/c)-1) + \sqrt{((v/c)-1)^2 + 16(v/c^2)} \right]$$ (14)

The 1985 HCM [#!hcm1!#] cautions users against using this expression for $(v/c)>1.2$. In the 1998 HCM [#!hcm98!#] the calculation of delay has been further modified. That expression is not provided here as it uses many site specific empirical constants which are not valid for Indian conditions.

Queue analysis

An involved discussion on analysis of queues at signalized intersections is not possible in this text as it requires substantial knowledge of stochastic queueing processes. However, certain characteristics of the arrival and departure processes from the point of view of queueing analysis are presented here so that the interested reader may pursue this topic further.

The primary purpose of doing a queueing analysis at a signalized intersection is to be able to determine the probability distribution of queues that form. That is, the aim should be to answer questions like, what is the probability that there will be $n$ vehicles in a particular queue at any time. This is important since not only values such as average queue length are important, what is more important is an idea of the queue length distribution so that one can determine lengths of auxiliary lanes ¹ (see next chapter for details) for pre-designated values of probability of overflow (where the queue of vehicles is larger than the space provided by the auxiliary lane) and probability of blockage (where the queue of vehicles on the lane adjacent to the auxiliary lane is so large that it blocks the entrance to the auxiliary lane).

However, the theory of queues available so far are not able to model the queueing process at a signalized intersection in a simple manner. The primary reason for this is that the departure process is not a Poisson process. If one assumes (and correctly so) the time a vehicle waits at the top of the queue to be its service time, then it can be easily seen that the service time duration (to which the departure process is integrally connected) is not negative exponential. In fact the service time is either zero (if one reaches the top of the queue when the light is green) or it is equal to the duration of the red period (if one is the next to the last vehicle to have crossed the intersection); the probability of either of these service times being the service time of a particular vehicle is not easy to calculate. Matters can be further complicated if there is a permitted phase in the signal (as is sometimes the case for turning movements) where a vehicle can cross the intersection (during the non-green time) if a sufficient gap in the opposing stream exists.

Some researchers (see for example Kikuchi et al. [#!kik1!#]), however, have attempted to determine the probability distribution of queues through a Markov chain analysis of the queueing process. The interested reader may refer to Kikuchi et al.'s work cited above for a good understanding of the analysis procedure.

Example

On an approach to a signalized intersection, the effective green time is 30 seconds and the effective red time is 30 seconds. The arrival rate of vehicles on this approach is 360 vph between 0 and 120 seconds; 1800 vph between 120 and 240 seconds ; and 0 vph between 240 and 420 seconds. The saturation flow rate for this approach is 1440 vphgpl. The approach under consideration has one lane. Assume that at time = 0 seconds the light for the approach has just turned red.

1. Plot the arrival rate versus time on a graph paper.
2. Assuming arrival and departure processes to be continuous, plot the cumulative number of arrivals and departures versus time on a graph paper.
3. Determine the average delay to vehicles arriving between 0 seconds and 120 seconds.
4. Determine the average delay to vehicles arriving between 120 seconds and 240 seconds.
5. Determine the average delay to vehicles arriving between 0 seconds and 240 seconds.
6. Determine the delays to the fourth and the sixtieth vehicles that arrive at the intersection.
7. Determine the maximum delay faced by a vehicle on this approach.
8. Determine the maximum queue length on this approach. At what time does the queue length first become equal to the maximum?
9. Determine the percentage of time there exists a queue on this approach.
10. Determine the average queue length from 120 to 420 seconds.

Solution

The following parameter values are provided: $g=30$ s, $C=30+30=60$ s, $s=1440$ vphgpl, $v$ from 0 s to 120 s = 360 vph, $v$ from 120 s to 240 s = 1800 vph, and $v$ from 240 s to 420 s = 0 vph, and $T$, time for which there exists a flow higher than $c$, is 120 s. Note, $c=(g/C)s=720$ vph.

Part 1

Figure 10 gives the required plot.

図 10: Plot of arrival rate versus time for the example problem on delay and queues at signalized intersections.

Part 2

Figure 11 gives the required plot.

図 11: Plot of cumulative number of arrivals and departures versus time for the example problem on delay and queues at signalized intersections.

Part 3

Between 0 second and 120 second the intersection is operating under unsaturated conditions. Further, the arrival is deterministic and uniform. Hence one could use Equation 9 to determine the average delay.

$$D_{u,us} = \frac{C(1-(g/C))^2}{2(1-(v/s))} = \frac{60(1-(30/60))^2}{2(1-(360/1440))} = 10 \;\; \mbox{seconds}$$

One could determine the average delay directly from the graph, by noting that the area of either Triangle I or II in Figure 12 (which is the same as Figure 11 but with few extra annotations) divided by the total number of arrivals during a cycle will give the average delay.

$$\mbox{Average delay} = \frac{\mbox{Area of Triangle I or II}}{\mbox{Number of arrivals in a cycle}}$$

or

$$\mbox{Average delay} = \frac{0.5 \times 30 \times 4}{6} = 10 \;\; \mbox{seconds}$$

図 12: Plot of cumulative number of arrivals and departures versus time for the example problem on delay and queues at signalized intersections with annotations.

Part 4

Between 120 seconds and 240 seconds the intersection is operating under over-saturated conditions. The arrival is deterministic and uniform. Hence one could use Equation 12 to determine the average delay.

$$D_{u,os}= \frac{T}{2} \left( \frac{v}{c}-1 \right)+ \frac{C-g... ...800}{720}-1 \right)+ \frac{60-30}{2} = 105 \;\; \mbox{seconds}$$

One could also determine the average delay directly from the graph (see Figure 12), by noting that,

$$\mbox{Average delay} = \frac{\mbox{Area of Triangle III} + 5 ... ...iangle IV}}{\mbox{Number of arrivals from 120 to 240 seconds}}$$

or

$$\mbox{Average delay} = \frac{0.5 \times 180 \times 60 + 5 \times 0.5 \times 12 \times 30 }{60} = 105 \;\; \mbox{seconds}$$

Part 5

The average delay to all the vehicles between 0 and 240 seconds can be obtained by dividing the total delay (faced by all vehicles) by the total number of vehicles. Hence,

$$\mbox{Average delay} = \frac{n_1d_1 + n_2d_2}{n_1+n_2}$$

where, $n_1$ is the number of vehicles that arrive during 0 to 120 seconds, $d_1$ is the average delay to a vehicle coming during this time, $n_2$ is the number of vehicles that arrive during 120 to 240 seconds, and $d_2$ is the average delay to a vehicle arriving during this time. Hence,

$$\mbox{Average delay} = \frac{12 \times 10 + 60 \times 105}{12+60} = 89.2 \;\; \mbox{seconds}$$

Of course one could find out the average delay here from the graph (the reader should do this).

Part 6

The arrival rate of vehicles from 0 to 120 seconds is 360 vph or 0.1 vps. Assuming that the fourth vehicle arrives before 120 seconds, the time of arrival of the fourth vehicle is $4/0.1=40$ seconds. (Hence the assumption is not violated).

The departure rate of vehicles is $1440/3600=0.4$ vps. The time of departure of the fourth vehicle, assuming that the fourth vehicle gets discharged during the first green, is $30 + 4/0.4 = 40$ seconds. (Since departure time is less than the start of the next red, the assumption is valid).

The delay to the fourth vehicle therefore is

$$\mbox{departure time - arrival time} = 40 - 40 = 0 \;\; \mbox{seconds}$$

The same observation can be made from the graph.

The delay to the sixtieth vehicle can also be read out from the graph (see Figure 12) as 144 seconds.

Part 7

As can be seen from the graph (see Figure 12) the maximum delay is 180 seconds.

Part 8

As can be seen from the graph (see Figure 12) the maximum queue length is 36 vehicles. At time = 240 seconds the queue length first becomes equal to 36 vehicles.

Part 9

As can be seen from the graph (see Figure 12) from 40 to 60 seconds and from 100 to 120 seconds there are no queues. For the rest of the time there is a queue at the intersection. Hence, percentage of time there is no queue at the intersection is $(40/420)100 = 9.52 \%$. Hence percentage of time there exists a queue is $100 - 9.52 = 90.48 \%$.

Part 10

One could determine the average queue length directly from the graph (see Figure 12), by noting that,

$$\mbox{Average queue length} = \frac{\mbox{Area of Triangle II... ...ea of Triangle IV}}{\mbox{Total time from 120 to 420 seconds}}$$

or

$$\mbox{Average queue length} = \frac{0.5 \times 180 \times 60 + 5 \times 0.5 \times 12 \times 30 }{300} = 21 vehicles$$