Given Traffic Scenario
Figure 1: The intersections of five one-way streets
1) There are five streets making six intersections. All the streets are one-way, that is, cars can move along only one direction on any street. The number of cars between intersections is the taken to be the traffic flows. For optimal flows, it is assumed that the number of cars entering an intersection in a given time must equal the number of cars leaving that intersection in that time. Also the number of cars entering and leaving any street per unit time is as mentioned in Figure 1. It is important to note that the total number of cars entering the entire system must equal the total number of cars leaving the entire system in a given time. In this particular case, the number = 2000. Also, in the forming of a linear model, the number of intersections represents the number of equation that can be formulated.
This kind of linear modeling is very similar to the analysis of circuits. According to Kirchhoff’s current law, the algebraic sum of currents entering and leaving a circuit junction equals zero. The number of junctions is analogous to the number of intersections, and the currents are analogous to the traffic flows. The current branches represent the roads between intersections.
In a linear model, it is assumed that the dependent and independent variables are linearly related. Though this might not be true for many systems, it is the most convenient model, and designers try to approximate other models to the linear model such that error is minimized. In a system of linear equations, if the number of variables equals the number of a equations, then a definite solution set can be obtained.
2) The constraint used for traffic flow is that the number of cars entering an intersection in a given period of time must equal the number of cars leaving the intersection in the same period time. In other word, in flow rate = outflow rate. Applying this constraint to each of the six intersections in the order I1, I2, I3, I4, I5, and I6 , the following equations are obtained respectively:
1) 400 + 450 = 850 = a + f;
2) a + g = b + 350;
3) b + 300 = 450 + c;
4) 350 + 550 = 900 = e + f;
5) d + 350 = e + g;
6) d + 300 = 500 + c;
Where a, b, c, d, e and f are the number of cars between the intersections, as in figure1.
3) Solving the system of linear equations formed in question 2,
a = 850 – f;
b = 500 – f + g;
c = 350 – f + g;
d = 550 – f + g;
e = 900 – f;
4) Acceptable values are those which result is all traffic flows being ≥ 0.
a. First set of values: Let f = 300, and g = 200, then the other traffic flows are,
a = 550; b = 400; c = 250; d = 450; e = 600
Second set of values: Let f = 200, and g = 300, then the other traffic flows become,
a = 650; b = 600; c = 450; d = 650; e = 700
b. The traffic flow on Maple street is given by the variable e = 900 – f. Or, f = 900 – e ≥ 0. Therefore e must be lie between 0 and 900.
c. If g = 100, considering the worst case,
c = 350 – f + 100 = 450 – f ≥ 0. Therefore the maximum value of f can be 450. This will also suit the other cases.
d. If g = 100, then
b = 600 – f ≥ 0; c = 450 – f ≥ 0; d = 650 – f ≥ 0
Therefore b must lie between 0 and 600; c must lie between 0 and 450; and d must lie between 0 and 650. The minimum values of a and e will be when f is maximum = 450, and a = 850 – 450 = 400; and e = 900 – 450 = 450.
e. If the model has five two way streets, then the number of variables representing the traffic flows will double. This means that there will be more number of independent variables. In general, the more the variables for the same number of equations, then the number of independent variable will increase. In this particular case, the number of equations is governed by the number of intersections, which is not going to change if the streets became two-way. Hence the number of independent variables will now be greater than 2. This is explained in the following example. Suppose the following equations represent the one way traffic flow in 3 intersections between four streets represented by a, b, c, and d:
– a + b – c = 50;
a – d = 0;
b – c – d = 50; then, the solution to this system will be a = d; b = c + d + 50. Therefore there are two independent variables c and d. However, even if one of the streets became two-way, then an extra variable added to the equations would then result in the following:
– a + b – c = 50;
a – d + e= 0;
b – c – d + e= 50. Now a = d – e; and b = c + d – e +50. As can be observed, there are now three independent variables: c, d, and e.
Hence if all the streets are made two-way, it can be expected that the number of independent variables will grow according to the number of added variables (in this case 6).
Going back to the comparison of this model with that of an electric circuit: two-way roads represent the presence of more than one current in a given loop. In other words, two currents in opposite directions flow through a branch. Hence the net current needs to be calculated as the algebraic sum of the two currents, in the direction of the greater current. Similarly, in a traffic model problem, the ‘net traffic flow’ between intersections will be the algebraic sum of the up flow and down flow on a given road. It is important to note that while calculating the algebraic sum, a sign convention must be followed based on direction. For instance, if flow is considered positive, then out flow is negative.
