Module 8 : Specialized Traffic Studies
Lecture 44 : Congestion Studies
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Extent

Extent of congestion is described by estimating the number of people or vehicles affected by congestion and by the geographic distribution of congestion. These measures include:
  1. Number or percentage of trips affected by congestion.
  2. Number or percentage of person or vehicle meters affected by congestion.
  3. Percentage of the system affected by congestion.
Performance measures of extent of congestion can be computed from sum of length of queuing on each segment. Segments in which queue overflows the capacity are also identified. This is useful for ramp metering analysis. To compute queue length, average density of vehicles in a queue need to be known. The default values suggested by HCM 2000 are given in Table 1.
Table 1: Queue density default values
Subsystem Storage density Spacing
  (veh/km/lane) (m)
Free-way 75 13.3
Two lane highway 130 7.5
Urban street 130 7.5

Queue length can be found out using the equation:

$\displaystyle QL_i=\frac{T(v-c)}{N \times ds}$ (1)

where; $ QL_i$ is the queue length (meter), $ v$ is the segment demand (veh/hour), $ c$ is the segment capacity (veh/hour), $ N$ is the number of lanes, $ ds$ is the storage density (veh/meter/lane), and $ T$ is the duration of analysis period (hour). If $ v < c$, $ Q_i$=0 The equation for queue length is similar for both corridor and area-wide analysis.

Numerical example

Consider a road segment of 6 lanes with a capacity of 2400 veh/hr/lane. It is observed that the storage density is 75 veh/meter and the segment demand is found to be 2800 veh/hr/lane. Given that the duration of analysis sub period is 2 hrs calculate the queue length that is formed due to congestion.

Solution

The queue length of a particular road segment is given by,

$\displaystyle QL = \frac{T\times(v-c)}{N \times ds}$ (2)

It is given that Number of lanes, N=6, Duration of analysis sub period, T= 2 hrs, Segment Capacity=c=2400 veh/hr/lane, Segment Demand=v=2800 veh/hr/lane, Storage Density=ds=75 veh/meter. Now,the queue length can be calculated by using the above formula as follows: $ QL= 2*(2800-2400)*6/(6*75)=10.667mts$ Therefore, the extent of congestion in terms of queue length is 10.667mts

Intensity

Intensity of congestion marks the severity of congestion. It is used to differentiate between levels of congestion on transport system and to define total amount of congestion. It is measured in terms of:
  • Delay in person hours or vehicle hours;
  • Average speed of roadway, corridor, or network;
  • Delay per capita or per vehicle travelling in the corridor, or per person or per vehicle affected by congestion;
  • Relative delay rate (relative rate of time lost for vehicles);
Intensity in terms of delay is given by,

$\displaystyle D_{PH}=T_{PH}-T_{PH}^0$ (3)

where, $ D_{PH}$ is the person hours of delay, $ T_{PH}$ is the person hours of travel under actual conditions, and $ T_{PH}^0$ is the person hours of travel under free flow conditions. The $ T_{PH}$ is given by:

$\displaystyle T_{PH}=\frac{O_{AV} \times v \times l}{S}$ (4)

where, $ O_{AV}$ is the average vehicle occupancy, $ v$ is the vehicle demand (veh), $ l$ is the length of link (km), and $ S$ is the mean speed of link (km/hr). The $ T_{PH}$ is given by:

$\displaystyle T_{PH}^0=\frac{O_{AV} \times v \times l}{S_0}$ (5)

where, $ O_{AV}$ is the average vehicle occupancy, $ v$ is the vehicle demand (veh), $ l$ is the length of link (km), and $ S_0$ is the free flow speed on the link (km/hr)

Numerical example

On a 2.8 km long link of road, it was found that the demand is 1000 Vehicles/hour mean speed of the link is 12 km/hr, and the free flow speed is 27 km/hr. Assuming that the average vehicle occupancy is 1.2 person/vehicle, calculate the congestion intensity in terms of total person hours of delay.

Solution:

Given data: Length of the link=l=2.8 km, Vehicle demand=v=1000 veh, Mean Speed of the link=S=12 km/hr, Free flow speed on the link=So=27 km/hr, and Average Vehicle Occupancy=AVO=1.2 person/veh. Person hours of delay is given as
$\displaystyle D_{PH}=T_{PH}-T_{PH}^0$      

Person hours of travel under actual conditions,
$\displaystyle T_{PH}$ $\displaystyle =$ $\displaystyle \frac{O_{AV} \times v \times l}{S}$  
  $\displaystyle =$ $\displaystyle \frac{1.2 \times 1000 \times 2.8}{12}$  
  $\displaystyle =$ $\displaystyle 280 \mathrm{~person~hours}$  

Person hours of travel under free flow conditions,
$\displaystyle T_{PH}^0$ $\displaystyle =$ $\displaystyle \frac{O_{AV} \times v \times l}{S_0}$  
  $\displaystyle =$ $\displaystyle \frac{1.2 \times 1000 \times 2.8}{27}$  
  $\displaystyle =$ $\displaystyle 124.4 \mathrm{~person~hours}$  

Therefore, person hours of delay can be calculated as follows,,
$\displaystyle D_{PH}=$ $\displaystyle =$ $\displaystyle 280-124.4$  
  $\displaystyle =$ $\displaystyle 155.6~\mathrm{~person~hours}$  
  $\displaystyle =$ $\displaystyle 156~\mathrm{person~hours~(approx)}.$  

Hence, the intensity of congestion is determined in terms of person hours of delay as 156 person hours.