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Linear Algebra, Water Flow Problem

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    2. Relevant equations
    Flow into a node=flow out of a node.
    Turned relative flows through each node (A,B,C,D,E,F) into system of equations; and entered system of equations into augmented matrix as shown. Reduced matrix to REF.

    3. The attempt at a solution
    [​IMG]
    I am not certain if I can be more exact with part a f_5 and f_6.

    I entered all the pertinent equations into the system possible for each node. (Is there another eautaion that could be entered (to help reduce answer)?)

    Part b, I have no visual explanation, as per the diagram, to why f_1=f_6.
     
    Last edited: Mar 26, 2008
  2. jcsd
  3. Mar 26, 2008 #2
    Okay, this is a lot of work to sift through, so I'll just rework it.. If we end up with the same thing.. It'll just be a lot of work for nothing. Anyway, here goes:

    So, as you said.. flow in = flow out. So, I can set of a system of equations:

    A: [tex]f_{3} + 200 = f_{1} + 100[/tex]
    B: [tex]f_{1} + 150 = f_{2} + f_{4}[/tex]
    C: [tex]f_{2} + f_{5} = 200 + 100[/tex]
    D: [tex]f_{6} + 100 = f_{3} + 200[/tex]
    E: [tex]f_{4} + f_{7} = f_{6} + 100[/tex]
    F: [tex]150 + 100 = f_{5} + f_{7}[/tex]

    Now I can rearrange these to make them easier to be put into a matrix:
    A: [tex]-f_{1} + f_{3} = -100 [/tex]
    B: [tex]f_{1} - f_{2} - f_{4} = -150 [/tex]
    C: [tex]f_{2} + f_{5} = 300 [/tex]
    D: [tex]-f_{3} + f_{6} = 100 [/tex]
    E: [tex]f_{4} - f_{6} + f_{7} = 100 [/tex]
    F: [tex]f_{5} + f_{7} = 250 [/tex]

    Now, writing this as a matrix, we get:
    [tex]\left( \begin{array}{cccccccc}
    -1 & 0 & 1 & 0 & 0 & 0 & 0 & -100 \\
    1 & -1 & 0 & -1 & 0 & 0 & 0 & -150 \\
    0 & 1 & 0 & 0 & 1 & 0 & 0 & 300 \\
    0 & 0 & -1 & 0 & 0 & 1 & 0 & 100 \\
    0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\
    0 & 0 & 0 & 0 & 1 & 0 & 1 & 250 \end{array} \right)[/tex]

    Now, writing this in reduced row echelon form, we get:

    [tex]\left( \begin{array}{cccccccc}
    1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 & -1 & 50 \\
    0 & 0 & 1 & 0 & 0 & -1 & 0 & -100 \\
    0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\
    0 & 0 & 0 & 0 & 1 & 0 & 1 & 250 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)[/tex]

    Now, we can rewrite these as equations to see what we get.. (notice that we have [tex] f_{6} [/tex] and [tex]f_{7}[/tex] being a free variable since the last row is completely zero... therefore [tex]f_{7} = r[/tex] and [tex] f_{6} = s[/tex] where r,s is any real number).

    I am not sure if we can make both arbitrary since we only have 6 equations and there are 7 unknowns, but I think that's what we have to do.... Assuming this is the correct way to do it, the following work is logical..

    Now, we get:
    [tex]f_{1} = s [/tex]
    [tex]f_{2} = r + 50 [/tex]
    [tex]f_{3} = s - 100 [/tex]
    [tex]f_{4} = s - r + 100 [/tex]
    [tex]f_{5} = -r + 250[/tex]
    [tex]f_{6} = s[/tex]
    [tex]f_{7} = r[/tex] where [tex]r,s \geq 0 [/tex] (because there can't be negative flow)

    part b)
    since [tex] f_{1} = s [/tex] and [tex]f_{6} = s[/tex], these two need to be the same, therefore one cannot be 100 and the other be 150.

    part c)
    If [tex] f_{4} = 0[/tex], we get:
    [tex]s - r + 100 = 0[/tex]
    This implies that [tex] s = r - 100 [/tex] and [tex] r = s + 100 [/tex].

    Substituting into the equations, we get:
    [tex] f_{1} = s [/tex]
    [tex] f_{2} = s + 150 [/tex]
    [tex] f_{3} = s - 100 [/tex]
    [tex] f_{4} = 0 [/tex]
    [tex] f_{5} = 150 - s [/tex]
    [tex] f_{6} = s [/tex]
    [tex] f_{7} = s + 100 [/tex].

    Now, knowing that we can't have negative flow, we see that [tex] f_{5} [/tex] tells us that s cannot be greater than 150, and [tex] f_{3} [/tex] tells us that s cannot be less than 100. Therefore, [tex] 100 \leq s \leq 150 [/tex].

    Using this, we can find the range of flow on each:
    [tex] 100 \leq f_{1} \leq 150 [/tex]
    [tex] 250 \leq f_{2} \leq 300 [/tex]
    [tex] 0 \leq f_{3} \leq 50 [/tex]
    [tex] f_{4} = 0 [/tex] (this was given).
    [tex] 0 \leq f_{5} \leq 50 [/tex]
    [tex] 100 \leq f_{6} \leq 150 [/tex]
    [tex] 200 \leq f_{7} \leq 250 [/tex]

    We should get the same ranges if we would have written all the equations in terms of r instead of s.

    I'm not sure if everything I did is correct, but it seems to make sense to me. A second opinion would probably be nice.

    Anyway, that's my input. Good luck.
     
  4. Mar 26, 2008 #3
    Wow jacobpm64, thanks sincerely,,,

    This statement was particularly helpful: "notice that we have f_6 and f_7 being a free variable since the last row is completely zero..."

    Also, knowing that I'm not totally off base gives me confindence to progress forward...

    I originally started with two varaiables; but then I (pointlessly) opted to try to look for correlate them.
     
  5. Mar 26, 2008 #4
    well I mean.. sometimes you'll have extra things that can give you more equations.

    Especially when you're dealing with circuits.

    You have kirschoff's laws.

    There are two different things to look for, so you can easily get more equations.

    I couldn't see anything right off in this problem though to give more equations.

    If you don't have enough information, heck, just make them arbitrary.
     
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