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Math Help - Multiple Unknowns in Simultaneous Equations.

  1. #1
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    Red face Adding matrices assignment.

    \left(\begin{array}{ccc}0&150&300\\240&0&160\\210&  250&0\end{array}\right)

    is the matrix I'm trying to solve with the following unknowns.

    \left(\begin{array}{ccc}0&e+t+v+w&d+e+g+v\\d+e+f+w  &0&f+t+u+w\\g+t+u+v&d+f+g+u&0\end{array}\right)

    so I've put them into 6 simultaneous equations... which I think makes this therefore solvable.

    e+t+v+w=150
    d+e+g+v=300
    d+e+f+w=240
    f+t+u+w=160
    g+t+u+v=210
    d+f+g+u=250

    my calculator can do 6 unknowns... and I've only ever been able to do 2 unknowns algebraically, can anyone help please?
    Last edited by hoppingmad; October 25th 2009 at 06:20 AM. Reason: Wrong topic, wrong question, wrong thread.
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  2. #2
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    Hello, hoppingmad!

    \left(\begin{array}{ccc}0&150&300\\240&0&160\\210&  250&0\end{array}\right)

    is the matrix I'm trying to solve with the following unknowns.

    \left(\begin{array}{ccc}0&e+t+v+w&d+e+g+v\\d+e+f+w  &0&f+t+u+w\\g+t+u+v&d+f+g+u&0\end{array}\right)

    so I've put them into 6 simultaneous equations... which I think makes this therefore solvable.

    e+t+v+w=150
    d+e+g+v=300
    d+e+f+w=240
    f+t+u+w=160
    g+t+u+v=210
    d+f+g+u=250

    my calculator can do 6 unknowns... and I've only ever been able to do 2 unknowns algebraically.
    can anyone help please?

    We have the system:

    . . \begin{array}{ccccccccccccccccc}<br />
d & + & e &&& + & g &&&&& + & v &&& = & 300 \\<br />
d & + & e & + & f & &&&&&&&& + & w & = & 240 \\<br />
d &&&+ & f & + & g &&& + & u &&&&& = & 250 \\<br />
&& e &&&&& + & t &&& + & v & + & w & = & 150 \\<br />
&&&&f &&& + & t & + & u &&& + & w & = & 160 \\<br />
&&&&&& g & + & t & + & u & + & v &&& = & 210\end{array}

    We have 6 equations in 8 variables.
    There will not be a unique solution.

    Could we see the original problem?

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  3. #3
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    Thanks, Soroban, for making it clear that there are 8 unknown values. I was having trouble counting them because I couldn't keep them straight in my mind.

    hoppingmad, as Soroban said, you cannot solve 6 equations for 8 unknown values. But you could solve for 6 of them "in terms of the other two". That is, treat two of them as constants (I don't think it matters which two) and solve the for the other six equal to formulas involving those 2. Exactly how you write the answer depends on exactly what the question is!

    Again, what was the precise wording of the problem?
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  4. #4
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    It's rather complicated... it's really doing my head in. If I give the complete context it will not be a a simultaneous equation, more a matrix/abstract algebra question.

    I'll do my best:
    The following matrix represents the traffic on an intersection during rush hour (SEN - south east and north)

    \left(\begin{array}{cccc}/&S&E&N\\S&0&150&300\\E&240&0&160\\N&210&250&0\end{  array}\right)

    Note:
    sum of elements in any row equal the sum of elements in corresponding column
    Green light times are given as fractions of a minute (1/2 means 30 seconds etc.)
    No U turns allowed (S-S = 0 etc.)
    4 possible traffic flows are shown below (this is the example given to us sort of showing us how to solve.):

    g1 (d)- \left(\begin{array}{cccc}/&S&E&N\\S&0&0&1/2\\E&1/2&0&0\\N&0&1/2&0\end{array}\right)

    g2 (e)- \left(\begin{array}{cccc}/&S&E&N\\S&0&0&0\\E&1/2&0&1/2\\N&0&1/2&0\end{array}\right)

    g3 (f)- \left(\begin{array}{cccc}/&S&E&N\\S&0&0&2/3\\E&0&0&0\\N&2/3&2/3&0\end{array}\right)

    g4 (g)- \left(\begin{array}{cccc}/&S&E&N\\S&0&1/3&1/3\\E&1/3&0&0\\N&0&0&0\end{array}\right)


    Time needed for all light to turn green corresponds to sum of the green light times for the above matrices

    1/2 + 1/2 + 1/3 + 2/3 = 2minutes

    so 4 traffic flows would have green lights 30 times in 1 hour.

    matrix G represents sum

    G = \left(\begin{array}{cccc}/&S&E&N\\S&0&1/3&3/2\\E&4/3&0&1/2\\N&2/3&5/3&0\end{array}\right)

    A matrix G represents a 2 minute time period 30G, matrix T represents 1 hour period

    30G = T = \left(\begin{array}{cccc}/&S&E&N\\S&0&10&45\\E&40&0&15\\N&20&50&0\end{array}  \right)

    10 cars can pass through any green light per minute so 10T = M(max) = maximum number of cars that can pass through this model for green lights in 1 hour

    10T = M(max) \left(\begin{array}{cccc}/&S&E&N\\S&0&100&450\\E&400&0&150\\N&200&500&0\end{  array}\right)
    Earlier stated the real life data rush hour matrix:

    Rush hour = \left(\begin{array}{cccc}/&S&E&N\\S&0&150&300\\E&240&0&160\\N&210&250&0\end{  array}\right)

    As you can see the above model does not support the Rush hour traffic. (example: east to south rush hour traffic needs to support 150 cars while M(max) supports 100)

    refine the green light times using fractions of a minute so M(max) is equal to Rush Hour traffic.
    ---------------------------------------------------------------

    That's the question, now, using the above model (which, when it boils down to it, divide the M(max) by 300 to get matrix G) doesn't quite work, we get a matrix looking like:

    \left(\begin{array}{cccc}/&S&E&N\\S&0&1/2&1\\E&4/5&0&8/15\\N&7/10&5/6&0\end{array}\right)

    To put it algebraically (t,u,v,w):
    \left(\begin{array}{cccc}/&S&E&N\\S&0&w&t+u+w\\E&t+v+w&0&v\\N&u&t+u+v&0\end{  array}\right)

    so we know
    w=
    v = 8/15
    u = 7/10
    so t will equal 4/15 presuming it is still running 2 minute traffic flows (which is why it was divided by 30)

    but M(max) still isn't equal to Rush hour. So this is wrong.

    Which is why I decided to find the 4 other possible traffic flows:

    g5 - \left(\begin{array}{cccc}/&S&E&N\\S&0&t&0\\E&0&0&t\\N&t&0&0\end{array}\right  )

    g6 - \left(\begin{array}{cccc}/&S&E&N\\S&0&u&u\\E&0&0&0\\N&u&0&0\end{array}\right  )

    g7 - \left(\begin{array}{cccc}/&S&E&N\\S&0&v&0\\E&v&0&v\\N&0&0&0\end{array}\right  )

    g8 - \left(\begin{array}{cccc}/&S&E&N\\S&0&0&0\\E&0&0&w\\N&w&w&0\end{array}\right  )

    which is how I got to my first question because I thought I was on the right path (probably not.....)

    \left(\begin{array}{cccc}/&S&E&N\\S&0&150&300\\E&240&0&160\\N&210&250&0\end{  array}\right)

    (g1+g2+...g8)
    \left(\begin{array}{ccc}0&e+t+v+w&d+e+g+v\\d+e+f+w  &0&f+t+u+w\\g+t+u+v&d+f+g+u&0\end{array}\right)

    I suppose I'm on the wrong path. please, if it's blatantly obvious where I'm screwing up, a hint in the right direction, I'm working hard on this but I keep hitting a brick wall. There has to be a better way than trial and error right?
    I need to find the correct traffic flows...
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  5. #5
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    Checking my work for complete stupidity... g6, g7, g8 are all invalid matrices since it causes cars to crash. Not good.

    \left(\begin{array}{ccc}0&e+t&d+e+g\\d+e+f&0&f+t\\  g+t&d+f+g&0\end{array}\right)


    \left(\begin{array}{cccc}/&S&E&N\\S&0&150&300\\E&240&0&160\\N&210&250&0\end{  array}\right)

    Is this solvable =(
    e+t =150
    d+e+g=300
    d+e+f=240
    f+t=160
    g+t=210
    d+f+g=250
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