Traffic Modeling Paper

Concepts of math can be displayed and solved in many different ways. There are equations, graphs, and word problems and several concepts that can be used as examples to help solve mathematical issues. In week four of this College Algebra class Learning Team B was present with an assignment involving a traffic model. The traffic model is the key to helping Learning Team B solve the problems the assignment presents. The traffic model represents a number of cars within the area at any given time represented by letters. Learning team B must solve questions based on this model.
There are several equations that were made in order to prove that the numbers of cars entering and leaving each intersection must be equal to keep traffic moving smoothly. Variables f and g are independent variables. This means that variables a through e are dependent on variables f and g. The number of cars at intersection a depends on how many cars turn from intersection f. The number of cars at intersection b depends on how many cars turn right from intersections f and g. This tells you that there are infinitely many solutions since the number of cars that turn onto Elm from Oak, Pine, and Aspen vary. Similarly the number of cars that turn onto Maple from Oak, Pine, and Aspen also vary.
To list two acceptable traffic flows first let variable f be equal to 365 cars/hr and let variable g be equal to 400 cars/hr. Variable a would be equal to 850-365, which equals 485 cars/hr. Variable b would be equal to 500-f+g, which is 500-365+400 and is equal to 535 cars/hr. Variable c would be equal to 350-f+g, which is 350-365+400 and is equal to 385 cars/hr. Variable d would be equal to 550-f+g, which is 550-365+400 and is equal to 585 cars/hr. Lastly, variable e would be equal to 900-f, which is 900-365 and is equal to 535 cars/hr.
If variable f was equal to 380 cars/hr and variable f was equal to 350 cars/hr, the results would be as the following. Variable a would be 800-f, which is 850-380 and is equal to 470 cars/hr. Variable b would be 500-f+g, which is 500-380+350 and equals 470 cars/hr. Variable c would be 350-f+g, which is 350-380+350 and is equal to 320 cars/hr. Variable d would be 550-f+g, which is 550-380+350 and is equal to 520 cars/hr. Lastly, variable e would be 900-f¬¬, which is 900-380 and equals 520 cars/hr.
Since variable e lies on Maple Street between I5¬ and I6 and the routes exiting I6 is equal to 900 cars/hr, variable e must be less than or equal to 900 cars/hr. Moreover, since 350+d cars/hr enter I5 and g cars/hr exit I5, the traffic flow on Maple Street between I5 and I6 must be the following equation: 900>e>350+d-g.
If g = 100, then a = 850-f, b = 600-f, c = 450-f, d = 650-f, and e = 900-f. The maximum value of variable f would be f=450. Anything larger than 450 and variable c would have a negative number of cars, which is not possible.
Previously it has been proven that if variable g were to be equal to 100 (g=100), then variable f must be less than or equal to 450 (f < 450). Variable d would equal 600-f and variable f would equal 600-d, which gives 600-d < 450; finally giving d > 250 cars/hr. Variable c would equal 450-f and variable f would equal 450-c, which gives 450-c < 450; finally giving c > cars/hr. Variable b would equal 600-f and variable f would equal 600-b, which gives 600-b < 400; finally giving b > 200 cars/hr. Therefore, variable d must be greater than 250 cars/hr, variable c must be great than 0 cars/hr, and variable b must be greater than 200 cars/hr.
Variable a would equal 850-f and variable f would equal 850-a, which gives 850-a < 450; finally giving a > 400. Variable e would equal 900-f and variable f would equal 900-e, which gives 900-e < 450; finally giving e > 450. Therefore, the minimum value for variable a is 400 cars/hr and the minimum value for variable e is 450 cars/hr.
To show what would happen if this model had five two way streets there are two different methods that were used. Method one represents the number of cars moving in a specific direction of travel between the intersections with variable a, b, c, d, e, f, g, h, i, j, k, l, m, and n. Here is a model to prove this method:

There is a system of linear equations representing I¬1 through I6.
I1: a + f = h + m I2: a + i + g = b + h + n I3: b + c = i + j
I4: k + j = d + c I5: d + l + n = e + g + k I6: e + f = m + l

There are infinite many solutions after assigning the five two way streets their own variables in the matrix. The solution set for this method is: {a = r5- r1+ e, b = r7, c = r4+ r3- r2, d = r2, f = r8, g = r6- r3+ r2+ r1- e, h = r5, i = r7+ r3- r2, j = r4, k = r3, l = r1, m = r8- r1+ e, n = r6 }.
In method two represents the number of cars moving in a specific direction of travel between the intersections with variables 2a, 2b, 2c, 2d, 2e,2 f, and 2g. Here is a model to show this method:

The systems of linear equations representing intersections I1 through I6 are followed:
I_1: 400+500=2a+2f I_1: 2a+f2=850 I_1: a=425-f
I_2: 2a+2g=2b+350 I_2: 2a-2b+2g=350 I_2: a-b=175-g
I_3: 2b+300=2c+450 I_3: b2-2c=150 I_3: b-c=75
I_4: c2+500=300+2d I_4: 2c-2d=-200 I_4: c-d=-100
I_5: d2+350=2g+2e I_5: 2d-2e-2g=-350 I_5: d-e=-175+g
I_6: e2+2f=550+350 I_6: 2e+2f=900 I_6: e=450-f With five one way streets, the number of cars per hour at variables a through e was dependent upon the number of cars per hour at variables f and g, respectfully. However, because of the newly added five two way street system, Pine, Oak, Aspen, Elm, and Maple still service the same number of cars per hour, but because of the newly added two way streets, a through e were reduced by half. Still, the numbers of cars per hour at a through e were dependent upon the number of cars per hour at f and g.