Traffic signal design-I

Lecture notes in Transportation Systems Engineering

3 August 2009

Overview

The conflicts arising from movements of traffic in different directions is solved by time sharing of the principle. The advantages of traffic signal includes an orderly movement of traffic, an increased capacity of the intersection and requires only simple geometric design. However the disadvantages of the signalized intersection are it affects larger stopped delays, and the design requires complex considerations. Although the overall delay may be lesser than a rotary for a high volume, a user is more concerned about the stopped delay.

Definitions and notations

A number of definitions and notations need to be understood in signal design. They are discussed below:

Cycle: A signal cycle is one complete rotation through all of the indications provided.
Cycle length: Cycle length is the time in seconds that it takes a signal to complete one full cycle of indications. It indicates the time interval between the starting of of green for one approach till the next time the green starts. It is denoted by .
Interval: Thus it indicates the change from one stage to another. There are two types of intervals - change interval and clearance interval. Change interval is also called the yellow time indicates the interval between the green and red signal indications for an approach. Clearance interval is also called all red is included after each yellow interval indicating a period during which all signal faces show red and is used for clearing off the vehicles in the intersection.
Green interval: It is the green indication for a particular movement or set of movements and is denoted by . This is the actual duration the green light of a traffic signal is turned on.
Red interval: It is the red indication for a particular movement or set of movements and is denoted by . This is the actual duration the red light of a traffic signal is turned on.
Phase: A phase is the green interval plus the change and clearance intervals that follow it. Thus, during green interval, non conflicting movements are assigned into each phase. It allows a set of movements to flow and safely halt the flow before the phase of another set of movements start.
Lost time: It indicates the time during which the intersection is not effectively utilized for any movement. For example, when the signal for an approach turns from red to green, the driver of the vehicle which is in the front of the queue, will take some time to perceive the signal (usually called as reaction time) and some time will be lost here before he moves.

Phase design

The signal design procedure involves six major steps. They include the (1) phase design, (2) determination of amber time and clearance time, (3) determination of cycle length, (4)apportioning of green time, (5) pedestrian crossing requirements, and (6) the performance evaluation of the above design. The objective of phase design is to separate the conflicting movements in an intersection into various phases, so that movements in a phase should have no conflicts. If all the movements are to be separated with no conflicts, then a large number of phases are required. In such a situation, the objective is to design phases with minimum conflicts or with less severe conflicts.

There is no precise methodology for the design of phases. This is often guided by the geometry of the intersection, flow pattern especially the turning movements, the relative magnitudes of flow. Therefore, a trial and error procedure is often adopted. However, phase design is very important because it affects the further design steps. Further, it is easier to change the cycle time and green time when flow pattern changes, where as a drastic change in the flow pattern may cause considerable confusion to the drivers. To illustrate various phase plan options, consider a four legged intersection with through traffic and right turns. Left turn is ignored. See figure 1.

**Figure 1:** Four legged intersection
$\begin{figure}\centerline{\epsfig{file=t47-four-legged-intersection.eps,width=8cm}}\end{figure}$

The first issue is to decide how many phases are required. It is possible to have two, three, four or even more number of phases.

Two phase signals

Two phase system is usually adopted if through traffic is significant compared to the turning movements. For example in figure 2, non-conflicting through traffic 3 and 4 are grouped in a single phase and non-conflicting through traffic 1 and 2 are grouped in the second phase.

**Figure 2:** Two phase signal
$\begin{figure}\centerline{\epsfig{file=t48-two-phase-signal.eps,width=8cm}}\end{figure}$

However, in the first phase flow 7 and 8 offer some conflicts and are called permitted right turns. Needless to say that such phasing is possible only if the turning movements are relatively low. On the other hand, if the turning movements are significant ,then a four phase system is usually adopted.

Four phase signals

There are at least three possible phasing options. For example, figure 3 shows the most simple and trivial phase plan.

**Figure 3:** One way of providing four phase signals
$\begin{figure}\centerline{\epsfig{file=t49-illustration-4phase-alternative-1.eps,width=8cm}}\end{figure}$

where, flow from each approach is put into a single phase avoiding all conflicts. This type of phase plan is ideally suited in urban areas where the turning movements are comparable with through movements and when through traffic and turning traffic need to share same lane. This phase plan could be very inefficient when turning movements are relatively low.

Figure 4 shows a second possible phase plan option where opposing through traffic are put into same phase.

**Figure 4:** Second possible way of providing a four phase signal
$\begin{figure}\centerline{\epsfig{file=t50-illustration-4phase-alternative-2.eps,width=8cm}}\end{figure}$

The non-conflicting right turn flows 7 and 8 are grouped into a third phase. Similarly flows 5 and 6 are grouped into fourth phase. This type of phasing is very efficient when the intersection geometry permits to have at least one lane for each movement, and the through traffic volume is significantly high. Figure 5 shows yet another phase plan. However, this is rarely used in practice.

**Figure 5:** Third possible way of providing a four-phase signal
$\begin{figure}\centerline{\epsfig{file=t51-illustration-4phase-alternative-3.eps,width=8cm}}\end{figure}$

There are five phase signals, six phase signals etc. They are normally provided if the intersection control is adaptive, that is, the signal phases and timing adapt to the real time traffic conditions.

Interval design

There are two intervals, namely the change interval and clearance interval, normally provided in a traffic signal. The change interval or yellow time is provided after green time for movement. The purpose is to warn a driver approaching the intersection during the end of a green time about the coming of a red signal. They normally have a value of 3 to 6 seconds.

The design consideration is that a driver approaching the intersection with design speed should be able to stop at the stop line of the intersection before the start of red time. Institute of transportation engineers (ITE) has recommended a methodology for computing the appropriate length of change interval which is as follows:

$\begin{displaymath} y = t+\frac{v_{85}}{2a+19.6g} \end{displaymath}$

(1)

where

is the length of yellow interval in seconds,

is the reaction time of the driver, $v_{85}$ is the 85 $^{th}$ percentile speed of approaching vehicles in m/s,

is the deceleration rate of vehicles in

is the grade of approach expressed as a decimal. Change interval can also be approximately computed as $y = \frac{SSD}{v}$ , where SSD is the stopping sight distance and

is the speed of the vehicle. The clearance interval is provided after yellow interval and as mentioned earlier, it is used to clear off the vehicles in the intersection. Clearance interval is optional in a signal design. It depends on the geometry of the intersection. If the intersection is small, then there is no need of clearance interval whereas for very large intersections, it may be provided.

Cycle time

Cycle time is the time taken by a signal to complete one full cycle of iterations. i.e. one complete rotation through all signal indications. It is denoted by

. The way in which the vehicles depart from an intersection when the green signal is initiated will be discussed now. Figure 6 illustrates a group of N vehicles at a signalized intersection, waiting for the green signal.

**Figure 6:** Group of vehicles at a signalized intersection waiting for green signal
$\begin{figure}\centerline{\epsfig{file=t61-vehicles-wait-for-green.eps,width=8cm}}\end{figure}$

As the signal is initiated, the time interval between two vehicles, referred as headway, crossing the curb line is noted. The first headway is the time interval between the initiation of the green signal and the instant vehicle crossing the curb line. The second headway is the time interval between the first and the second vehicle crossing the curb line. Successive headways are then plotted as in figure 7.

**Figure 7:** Headways departing signal
$\begin{figure}\centerline{\epsfig{file=t62-concept-of-saturation-headway.eps,width=8cm}}\end{figure}$

The first headway will be relatively longer since it includes the reaction time of the driver and the time necessary to accelerate. The second headway will be comparatively lower because the second driver can overlap his/her reaction time with that of the first driver's. After few vehicles, the headway will become constant. This constant headway which characterizes all headways beginning with the fourth or fifth vehicle, is defined as the saturation headway, and is denoted as

. This is the headway that can be achieved by a stable moving platoon of vehicles passing through a green indication. If every vehicles require

seconds of green time, and if the signal were always green, then s vehicles/per hour would pass the intersection. Therefore,

$\begin{displaymath} s=\frac{3600}{h} \end{displaymath}$

(2)

where

is the saturation flow rate in vehicles per hour of green time per lane,

is the saturation headway in seconds. vehicles per hour of green time per lane. As noted earlier, the headway will be more than h particularly for the first few vehicles. The difference between the actual headway and h for the $i^{th}$ vehicle and is denoted as

shown in figure 7. These differences for the first few vehicles can be added to get start up lost time,

which is given by,

$\begin{displaymath} l_1 = \sum_{i=1}^n{e_i} \end{displaymath}$

(3)

The green time required to clear N vehicles can be found out as,

$\begin{displaymath} T = l_1 + h.N \end{displaymath}$

(4)

where

is the time required to clear N vehicles through signal,

is the start-up lost time, and

is the saturation headway in seconds.

Effective green time

Effective green time is the actual time available for the vehicles to cross the intersection. It is the sum of actual green time (

) plus the yellow minus the applicable lost times. This lost time is the sum of start-up lost time (

) and clearance lost time (

) denoted as

. Thus effective green time can be written as,

$\begin{displaymath} g_i = G_i + Y_i - t_L \end{displaymath}$

(5)

Lane capacity

The ratio of effective green time to the cycle length ( $\frac{g_i}{C}$ )is defined as green ratio. We know that saturation flow rate is the number of vehicles that can be moved in one lane in one hour assuming the signal to be green always. Then the capacity of a lane can be computed as,

$\begin{displaymath} c_i = s_i \frac{g_i}{C} \end{displaymath}$

(6)

where

is the capacity of lane in vehicle per hour,

is the saturation flow rate in vehicle per hour per lane,

is the cycle time in seconds.

Problem

Let the cycle time of an intersection is 60 seconds, the green time for a phase is 27 seconds, and the corresponding yellow time is 4 seconds. If the saturation headway is 2.4 seconds/vehicle, the start-up lost time is 2 seconds/phase, and the clearance lost time is 1 second/phase, find the capacity of the movement per lane?

Solution

Total lost time,

= 2+1 = 3 seconds. From equation effective green time,

= 27+4-3 = 28 seconds. From equationsaturation flow rate, $s_i = \frac{3600}{h} = \frac{3600}{2.4}$ = 1500 veh/hr. Capacity of the given phase can be found out from equation, $C_i = 1500\times \frac{28}{60}$ = 700 veh/hr/lane.

Critical lane

During any green signal phase, several lanes on one or more approaches are permitted to move. One of these will have the most intense traffic. Thus it requires more time than any other lane moving at the same time. If sufficient time is allocated for this lane, then all other lanes will also be well accommodated. There will be one and only one critical lane in each signal phase. The volume of this critical lane is called critical lane volume.

Determination of cycle length

The cycle length or cycle time is the time taken for complete indication of signals in a cycle. Fixing the cycle length is one of the crucial steps involved in signal design.

If $t_{Li}$ is the start-up lost time for a phase , then the total start-up lost time per cycle, $L = \sum_{i=1}^N {t_L}$ , where is the number of phases. If start-up lost time is same for all phases, then the total start-up lost time is . If is the cycle length in seconds, then the number of cycles per hour = $\frac{3600}{C}$ The total lost time per hour is the number of cycles per hour times the lost time per cycle and is = $\frac{3600}{C}.L$ Substituting as , total lost time per hour can be written as = $\frac{3600.N.t_l}{C}$ The total effective green time available for the movement in a hour will be one hour minus the total lost time in an hour. Therefore,

$\displaystyle T_g$	$\textstyle =$	$\displaystyle 3600 - \frac{3600.N.t_L}{C}$	(7)
	$\textstyle =$	$\displaystyle 3600\left[1-\frac{N.t_L}{C}\right]$	(8)

Let the total number of critical lane volume that can be accommodated per hour is given by

, then $V_c = \frac{T_g}{h}$ Substituting for

, from equation 9 and

from the maximum sum of critical lane volumes that can be accommodated within the hour is given by,

	$\textstyle =$	$\displaystyle \frac{T_g}{h}$	(9)
$\displaystyle V_c$	$\textstyle =$	$\displaystyle \frac{3600}{h}\left[1-\frac{N.t_L}{C}\right]$	(10)
	$\textstyle =$	$\displaystyle s_i\left[1-\frac{N.t_L}{C}\right]$	(11)
$\displaystyle Therefore C$	$\textstyle =$	$\displaystyle \frac{N.t_L}{1-\frac{V_c}{s}}$	(12)

The expression for

can be obtained by rewriting the above equation. The above equation is based on the assumption that there will be uniform flow of traffic in an hour. To account for the variation of volume in an hour, a factor called peak hour factor, (PHF) which is the ratio of hourly volume to the maximum flow rate, is introduced. Another ratio called v/c ratio indicating the quality of service is also included in the equation. Incorporating these two factors in the equation for cycle length, the final expression will be,

$\begin{displaymath} C = \frac{N.t_L}{1-\frac{V_c}{s_i\times PHF \times \frac{v}{c}}} \end{displaymath}$

(13)

Highway capacity manual (HCM) has given an equation for determining the cycle length which is a slight modification of the above equation. Accordingly, cycle time

is given by,

$\begin{displaymath} C = \frac{N.L.X_C}{X_C-\Sigma (\frac{V}{s})_i} \end{displaymath}$

(14)

where

is the number of phases,

is the lost time per phase, $(\frac{V}{s})_i$ is the ratio of volume to saturation flow for phase

is the quality factor called critical $\frac{V}{C}$ ratio where

is the volume and

is the capacity.

Problem

The traffic flow in an intersection is shown in the figure 8.

**Figure 8:** Traffic flow in the intersection
$\begin{figure}\centerline{\epsfig{file=t52-problem-2phase-flow.eps,width=8cm}}\end{figure}$

Given start-up lost time is 3 seconds, saturation head way is 2.3 seconds, compute the cycle length for that intersection. Assume a two-phase signal.

Solution

If we assign two phases as shown below figure 9, then the critical volume for the first phase which is the maximum of the flows in that phase = 1150 vph.

Figure 9: One way of providing phases
$\begin{figure}\centerline{\epsfig{file=t53-probable-phasing-1.eps,width=8cm}}\end{figure}$

Similarly critical volume for the second phase = 1800 vph. Therefore, total critical volume for the two signal phases = 1150+1800 = 2950 vph.
Saturation flow rate for the intersection can be found out from the equation as $s_i = \frac{3600}{2.3}$ = 1565.2 vph. This means, that the intersection can handle only 1565.2 vph. However, the critical volume is 2950 vph . Hence the critical lane volume should be reduced and one simple option is to split the major traffic into two lanes. So the resulting phase plan is as shown in figure ( 10).

Figure 10: second way of providing phases
$\begin{figure}\centerline{\epsfig{file=t54-probable-phasing-2.eps,width=8cm}}\end{figure}$
Here we are dividing the lanes in East-West direction into two, the critical volume in the first phase is 1150 vph and in the second phase it is 900 vph. The total critical volume for the signal phases is 2050 vph which is again greater than the saturation flow rate and hence we have to again reduce the critical lane volumes.
Assigning three lanes in East-West direction, as shown in figure 11, the critical volume in the first phase is 575 vph and that of the second phase is 600 vph, so that the total critical lane volume = 575+600 = 1175 vph which is lesser than 1565.2 vph.

Figure 11: Third way of providing phases
$\begin{figure}\centerline{\epsfig{file=t55-probable-phasing-3.eps,width=8cm}}\end{figure}$
Now the cycle time for the signal phases can be computed from equation, $C = \frac{2\times3}{1-\frac{1175}{1565.2}}$ = 24 seconds.

**Figure 9:** One way of providing phases
$\begin{figure}\centerline{\epsfig{file=t53-probable-phasing-1.eps,width=8cm}}\end{figure}$

**Figure 10:** second way of providing phases
$\begin{figure}\centerline{\epsfig{file=t54-probable-phasing-2.eps,width=8cm}}\end{figure}$

**Figure 11:** Third way of providing phases
$\begin{figure}\centerline{\epsfig{file=t55-probable-phasing-3.eps,width=8cm}}\end{figure}$

Summary

Traffic signal is an aid to control traffic at intersections where other control measures fail. The signals operate by providing right of way to a certain set of movements in a cyclic order. Depending on the requirements they can be either fixed or vehicle actuated and two or multivalued. The design procedure discussed in this chapter include interval design, determination of cycle time, and computation of saturation flow making use of HCM guidelines.

Problems

Saturation flow rate can be computed as,
1. $\frac{3600}{h}$
2. $\frac{h}{3600}$
3. 3600 $\times$ h
4. none of these
Lane capacity is
1. $c_i=s_i\times\frac{g_i}{C}$
2. $c_i=s_i\times g_i$
3. $c_i=\frac{s_i}{C}$
4. none of these

Solutions

Saturation flow rate can be computed as,
1. $\frac{3600}{h}$ $\surd$
2. $\frac{h}{3600}$
3. 3600 $\times$ h
4. none of these
Lane capacity is
1. $c_i=s_i\times\frac{g_i}{C}$ $\surd$
2. $c_i=s_i\times g_i$
3. $c_i=\frac{s_i}{C}$
4. none of these

Bibliography

1: William R McShane, Roger P Roesss, and Elena S Prassas.
Traffic Engineering.
Prentice-Hall, Inc, Upper Saddle River, New Jesery, 1998.

Prof. Tom V. Mathew 2009-08-03