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Traffic flow (Linear Algebra)

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data
    the following statement shows the traffic flow(in cars/minute) along 3 one way streets through the intersections A, B and C
    Set up and solve a system of linear equations(using the augmented matrix for the system) to help you find the flows x,y and z. Be sure to generate the RREF of the system
    a link of the drawing:
    http://img18.imageshack.us/img18/2276/trafficflow.jpg [Broken]

    2. Relevant equations
    traffic flow in must equal flow out



    3. The attempt at a solution
    so to start things off, equations i put:
    3 + z = 3
    6 + x = 7
    5 + y = 4
    is this at all right?
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 17, 2009 #2

    Mark44

    Staff: Mentor

    I don't think so. For your first equation, 3 + z = 3 ==> z = 0, which doesn't make any sense. How about giving us the problem exactly as it was worded? Useful information would be what the numbers in your drawing represent.

    I believe that what you have as "relevant equations" means that the traffic flow into and out of an intersection must be equal.
     
  4. Apr 17, 2009 #3
    this is the problem exactly as it was worded, but if u insist...i added "cars/minute" :|
    i also added the label on the intersections in the image A,B and C
    and sorry i thought the numbers were obvious :S, they represent the flow out that that arrow

    i'm guessing i have to go take away some of the flow as well? i'm just not sure how to start the problem
     
  5. Apr 17, 2009 #4

    Mark44

    Staff: Mentor

    If my previous interpretation of what traffic flow in equals traffic flow out is on track, here's what you have:
    Node A (upper left intersection) x + 5 = y + 7
    Node B (upper right int.) z + 6 = x + 3
    Node C (lower int.) y + 3 = z + 4


    Solve this system.
     
  6. Apr 18, 2009 #5
    i got:
    -x + z =-3
    -z +y = 1
    -y + x = z
    to help set up the matrix:
    -x 0 z
    0 y -z
    x -y 0
    augmented matrix:
    -1 0 1 |-3 ---->(1)
    0 1 -1 |1------>(2)
    1 -1 0 |2------>(3)

    after some math:
    (1)/-1
    (3)-(1)
    (3)+(2)

    1 0 -1 | 3
    0 1 -1 | 1
    0 0 0 | 0

    1x -1z = 3
    1y -1z =1
    z=free

    z=1,y=2,x=4
    is this right?
     
  7. Apr 18, 2009 #6

    Mark44

    Staff: Mentor

    The last equation should be -y + x = 2, but otherwise I got the same solution that you showed.
     
  8. Apr 18, 2009 #7
    thx for ur help!
     
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