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.
2) Creating linear equations to model continually flowing traffic in the figure provided is a multidimensional aspect. It is pertinent to note that linear system defined by equations can have a single solution, many solutions or no solution at all. However, in this case is to develop a system with one solution enabling rewriting of augmented matrix in the desired format. The traffic model forms a square system. This is where the number of the equations to be formed are qual to the number of the unknowns.
Setting up the system involves anlyizing each node or intersection in the figure at and developing its linear equation.
For intersection (I1) the equation can be generally represented as (450+400= a+f= 850) this is because the number of entering the intersection is (450+400). On the other hand the number of vehicles leaving the intersection is (a+f).
For intersection (I2) the equation can be generally represented as (a + g = b + 350) this is because the number of entering is (a+g). On the other hand the number of vehicles leaving the intersection is (b+350).
For intersection (I3) the equation can be generally represented as (b + 300 = 450 + c) this is because the number of entering is (b+300). On the other hand the number of vehicles leaving the intersection is (450+c).
For intersection (I4) the equation can be generally represented as (350 + 550 = 900 = e + f) this is because the number of entering is (350+550= 900). On the other hand the number of vehicles leaving the intersection is (e+f= 900).
For intersection (I5) the equation can be generally represented as (d + 350 = e + g) this is because the number of entering is (d+350). On the other hand the number of vehicles leaving the intersection is (e+g).
For intersection (I6) the equation can be generally represented as (d + 300 = 500 + c) this is because the number of vehicles entering is (d+300). On the other hand the number of vehicles leaving the intersection is (500+c).
The six equations for the linear system are:
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) Solutions to the systems of equations
The equations can be solved through different approaches which include but not limited to Gaussian elimination and progressive elimination.
The above equations can be written as shown below:
a+ f = 850 ;
b+ f + g = 500 ;
c+ f + g = 350 ;
d+ f + g = 550;
e+f = 900
however, re-writing them as a system of linear equations with the missing variables with a coefficient of 0 they become as shown below:
1a+0b+0c+0d+0e+ 1f+ 0g = 850;
0a +1b+0c+0d+0e+ 1f + 1g = 500;
0a+0b+1c+0d+0e+ 1f + 1g = 350;
0a+0b+0c+1d+0e+ 1f + 1g = 550;
0a+0b+0c+0d+1e+1f +0g= 900;
Creating the augmented matrix yields
100 010 001 001 011 001 011 = 850500350
001 000000001 111 001 110 = 350550900
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). 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. Moreover, at every intersection we will have four figures representing entry and another four representing exit. This will yield a non-square system.
