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Math Help - Factoring response time equation

  1. #1
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    Factoring response time equation

    Remember the calculation of response time in the lecture?

    R = SQ + S(2)

    Based on one Queuing Theorem:

    Q = a * R(3)


    Manipulating equations (2) and (3) using Factoring method, we obtain:

    S
    R = ---------- (4)
    1 – a S


    Show how you manipulate the two equations (2) and (3) in order to get (4).

    The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!
    Last edited by Heather; August 8th 2007 at 03:30 PM. Reason: clarification and additional info
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  2. #2
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    Quote Originally Posted by Heather View Post
    Remember the calculation of response time in the lecture?

    R = SQ + S(2)

    Based on one Queuing Theorem:

    Q = a * R(3)


    Manipulating equations (2) and (3) using Factoring method, we obtain:

    S
    R = ---------- (4)
    1 – a S


    Show how you manipulate the two equations (2) and (3) in order to get (4).

    The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!
    I doubt most of us are taking this class with you!

    Can you give us some background in regard to what you are talking about?

    -Dan
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  3. #3
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    Most of what we are studying right now is factoring and finding GCF. Somehow with this problem you are supposed to be doing factoring, but the only way i see it possible is by starting with substitution.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Heather View Post
    Remember the calculation of response time in the lecture?

    R = SQ + S(2)

    Based on one Queuing Theorem:

    Q = a * R(3)


    Manipulating equations (2) and (3) using Factoring method, we obtain:

    S
    R = ---------- (4)
    1 a S


    Show how you manipulate the two equations (2) and (3) in order to get (4).

    The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!
    Oh okay, from the way you phrased the question I thought there was more to it.

    So we have:
    R = SQ + S

    and

    Q = aR

    So put the bottom equation into the top:
    R = S(aR) + S = aSR + R

    Now we want to solve for R:
    R - aSR =  S

    There is a common R to both terms on the left, so factor it:
    (1 - aS)R =  S

    Now divide:
    R =  \frac{S}{1 - aS}

    -Dan
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  5. #5
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    Thank you soooooo much this has been driving me crazy! But now i feel dumb because this was very simple
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Heather View Post
    Thank you soooooo much this has been driving me crazy! But now i feel dumb because this was very simple
    Don't feel dumb. There have been times I've forgotten how to add. (Seriously!) Everyone has their moments.

    -Dan
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