Module 4 : Macroscopic And Mesoscopic Traffic Flow Modeling
Lecture 18 : Cell Transmission Models
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Numerical example

Consider a 1.25 km homogeneous road with speed $ v = 50~kmph$, jam density $ k_{j} = 180~veh/km$ and $ q_{max} = 3000~ veh/hr$. Initially traffic is flowing undisturbed at 80% of capacity: $ q = 2400~veh/hr$. Then, a partial lane blockage lasting 2 min occurs on $ 1/3^{rd}$ of the distance from the end of the road. The blockage effectively restricts flow to 20% of the maximum. Clearly, a queue is going to build and dissipate behind the restriction. After 2 minutes, the flow in cell 3 is maximum possible flow. Predict the evolution of the traffic. Take one clock tick as 30 seconds.

Solution

The main purpose of cell transmission model is to simulate the real traffic conditions for a defined stretch of road. The speed and cell length is kept constant and also the cell lengths in cell transmission model. The solution has been divided into 4 steps as follows:

Step 1: Determination of cell length and number of cells Given clock tick, $ t = 30sec = 1/120^{th}$ of an hour. So, cell length = distance travelled by vehicle in one clock tick = $ v \times t = 50 \times (1/120) \approx 5/12~km$. Road stretch given = $ 1.25~km$. Therefore, no of cells = $ 1.25/(5/12)$ = $ 3~cells$

Step 2: Determination of constants (N & Q) N = maximum number of vehicles that can be at time t in cell i, = cell length x jam density, = 180 x (5/12) = 75 vehicles, Q = maximum number of vehicles that can flow into cell I from time t to t+1, = 3000 x (1/120) = 25 vehicles. Now, to simulate the traffic conditions for some time interval, our main aim is to find the occupancies of the 3 cells (as calculated above) with the progression of clock tick. This is easily showed by creating a table. First of all, the initial values in the tables are filled up.

Step 3: Determination of cell capacity in terms of number of vehicles for various traffic flows. For 20% of the maximum = $ 600 \times (1/120)~vehicles$. For 80% of the maximum = $ 2400 \times (1/120)~vehicles$.

Step 4: Initialization of the table The table has been prepared with source cell as a large capacity value and a gate is there which connects and regulates the flow of vehicles from source to cell 1 as per the capacity of the cell for a particular interval. The cell constants ($ Q$ and $ N$) for the 3 cells are shown in the table. Note that the sink can accommodate maximum number of vehicles whichever the cell 3 generates. Q3 is the capacity in terms of number of vehicles of cell 3 . The value from H5 to H7 (i.e 5) corresponds to the 2min time interval with 4 clock ticks when the lane was blocked so the capacity reduced to 20% of the maximum (i.e. 600 $ \times$ (1/120)  vehicles). After the 2 min time interval is passed vehicles flows with full capacity in cell 3. So the value is 25 (i.e 3000 $ \times$ (1/120) vehicles).

Table 1: Entries at the start of the simulation.
  Source(00) Gate(0) Cell1 Cell2 Cell3 Cell4  
Q   20 25 25   25  
N   999 75 75 75 999  
Time             Q3
1 999 20 20 20 20   5
2 999 20         5
3 999 20         5
4 999 20         5
5 999 20         25
6 999 20         25
7 999 20         25
8 999 20         25
9 999 20         25
10 999 20         25
11 999 20         25
12 999 20         25
13 999 20         25
14 999 20         25
15 999 20         25
16 999 20         25
17 999 20         25
18 999 20         25


Step 5: Computation of Occupancies Simulation need not be started in any specific order, it can be started from any cell in the row corresponding to the current clock tick. Now, consider cell circled (cell 2 at time 2) in the final table. Its entry depends on the cells marked with rectangles. By flow conservation law: Occupancy = Storage + Inflow - Outflow. Note that the Storage is the occupancy of the same cell from the preceding clock tick. Also outflow of one cell is equal to the inflow of the just succeeding cell. Here, Storage = 20. For inflow use equation [*] Inflow= min [20,min(25,25),(75-20)]= 20. Outflow= min [20,min(25,5),(75-20)]= 5. Occupancy= 20+20-5=35.

\begin{figure} \centerline{\epsfig{file=qfNumerical1a.eps,width=8 cm}} \end{figure}
Now, For cell 1 at time 2, Inflow= min [20,min(25,25),(75-20)]= 20, Outflow= min [20,min(25,25),(75-20)]= 20, Occupancy= 20+20-20=20.

\begin{figure} \centerline{\epsfig{file=qfNumerical1b.eps,width=8 cm}} \end{figure}
Now, For cell 3 at time 2, Inflow= min [20, min (25,5),(75-20)]= 5. Outflow= 20 (:.sink cell takes all the vehicles in previous cell) Occupancy= 20+5-20=5.

\begin{figure} \centerline{\epsfig{file=qfNumerical1c.eps,width=8 cm}} \end{figure}
Similarly, rest of the entries can be filled and the final result is shown in Table below.

Table 2: Final entries simulating the traffic
  Source(00) Gate(0) Cell 1 Cell 2 Cell 3 Cell 4  
Q   20 25 25   25  
N   999 75 75 75 999  
Time             Q3
1 999 20 20 20 20   5
2 999 20 20 35 5   5
3 999 20 20 50 5   5
4 999 20 20 65 5   5
5 999 20 30 70 5   25
6 999 20 45 50 25   25
7 999 20 40 50 25   25
8 999 20 35 50 25   25
9 999 20 30 50 25   25
10 999 20 25 50 25   25
11 999 20 20 50 25   25
12 999 20 20 45 25   25
13 999 20 20 40 25   25
14 999 20 20 35 25   25
15 999 20 20 30 25   25
16 999 20 20 25 25   25
17 999 20 20 20 25   25
18 999 20 20 20 20   25


From the table it can be seen that the occupancy i.e. the number of vehicles on cell 1 and 2 increases and then decreases simulating the effect of lane blockage in cell 3 on cell 1 and cell 2. The lane blockage lasts 2 minutes in this problem, after that there is no congestion taken into account. So as the time passes by, the occupancy in cell 1 and cell 2 also gets reduced.