Results 1 to 10 of 10

Math Help - Increasing at a Decreasing Rate

  1. #1
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Increasing at a Decreasing Rate

    I'm not sure if this is better suited for an Excel forum, but here goes.

    I will know the value of x, which can be anything between 1.00 and 71.32 (inclusive), and can have any number of decimal places to the right, such as 20.123456789.

    An x value of EXACTLY 1.00 must return a value of EXACTLY 0.60
    An x value of EXACTLY 71.32 must return a value of EXACTLY 1.00

    The result values in the continuous range 0.60 to 1.00 must increase at a Decreasing rate. For example,
    (1) An x value of around 2 should return a value greater than 0.60, call it z1
    (2) An x value of around 70.32 should return a value less than 1.00, call it z2
    Because I want a decreasing rate of increase, z1 minus 0.60 should be significantly greater than 1.00 minus z2

    It seems like raising x to a power that is in the range 0 to 1 should provide the increase at a Decreasing rate, but nailing down the correct power that enforces both endpoints is where I'm stuck.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,774
    Thanks
    1140

    Re: Increasing at a Decreasing Rate

    Quote Originally Posted by capheresy View Post
    I'm not sure if this is better suited for an Excel forum, but here goes.

    I will know the value of x, which can be anything between 1.00 and 71.32 (inclusive), and can have any number of decimal places to the right, such as 20.123456789.

    An x value of EXACTLY 1.00 must return a value of EXACTLY 0.60
    An x value of EXACTLY 71.32 must return a value of EXACTLY 1.00

    The result values in the continuous range 0.60 to 1.00 must increase at a Decreasing rate. For example,
    (1) An x value of around 2 should return a value greater than 0.60, call it z1
    (2) An x value of around 70.32 should return a value less than 1.00, call it z2
    Because I want a decreasing rate of increase, z1 minus 0.60 should be significantly greater than 1.00 minus z2

    It seems like raising x to a power that is in the range 0 to 1 should provide the increase at a Decreasing rate, but nailing down the correct power that enforces both endpoints is where I'm stuck.
    $\large a \ln(x/b)$ should get you what you want.

    solving you obtain

    $\large a=\dfrac{2}{5 \log \left(\frac{1783}{25}\right)}$

    $\large b=\dfrac{125}{1783 \sqrt{1783}}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    I haven't used logarithms in many years, so this is a struggle. Anyway, I put the following into Excel and I'm getting a #VALUE! message:

    Cell A1 is a = 2/LOG(1783/25,5)
    Cell A2 is b = 125/(1783*SQRT(1783))
    Cell A3 is x = 20.123456789

    Cell A4 is the result = A1*LN(A3/A2)

    Please tell me where I'm screwing up before I move on to the Excel forum. Thanks much.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    I fixed the value error (a dumb mistake). I tried it with x=1.00, which should return 0.60, but instead I get 4.828
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,774
    Thanks
    1140

    Re: Increasing at a Decreasing Rate

    Quote Originally Posted by capheresy View Post
    I haven't used logarithms in many years, so this is a struggle. Anyway, I put the following into Excel and I'm getting a #VALUE! message:

    Cell A1 is a = 2/LOG(1783/25,5)
    Cell A2 is b = 125/(1783*SQRT(1783))
    Cell A3 is x = 20.123456789

    Cell A4 is the result = A1*LN(A3/A2)

    Please tell me where I'm screwing up before I move on to the Excel forum. Thanks much.
    I have

    a1= =2/(5*LN(1783/25))

    a2 =125/(1783*SQRT(1783))

    a3 = whatever x

    a4 = $a$1*LN(a3/$a$2)

    Increasing at a Decreasing Rate-clipboard01.jpg
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    Fantastic. I was using 5 as the base for Excel's LOG function. Thank you very much. Now, we'll see how well my college football rating system works...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    One more thing, if I may. Each season the 71.32 number will be different. Could you explain where in the formula I would make the change(s)?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,774
    Thanks
    1140

    Re: Increasing at a Decreasing Rate

    Quote Originally Posted by capheresy View Post
    One more thing, if I may. Each season the 71.32 number will be different. Could you explain where in the formula I would make the change(s)?
    Let $var$ be the number that is currently $73.32$

    $a=\dfrac{2}{5 \ln(var)}$

    $b=(var)^{-3/2}$

    so

    a1 = 2/(5*LN(var))

    a2 = (var)^(-3/2)

    maybe what you'd want to do is make

    a1=var

    and shift a1 and a2 down one cell to become a2 and a3, then replace var with a1 in the formulas.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    Thank you again!!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    May 2014
    From
    Cincinnati
    Posts
    7

    Re: Increasing at a Decreasing Rate

    I thought I was done, but no. If I decided that the slope of the rate is too steep for the lower part of the range, how could I play around with different what-if slopes (rates) to find the optimum one (in my mind)? I can see that your 1783 is 25*71.32, and your 125 is 25*5, so I'm guessing that the number 5 and/or the -3/2 power is controlling the steepness. I will appreciate any further help I can get.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Increasing/ Decreasing intervals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 14th 2011, 06:38 PM
  2. increasing or decreasing
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 14th 2009, 08:02 PM
  3. Increasing/Decreasing
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 16th 2009, 07:06 PM
  4. Increasing/Decreasing and Abs Max and Min
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 28th 2009, 06:57 PM
  5. increasing decreasing
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 6th 2007, 02:29 PM

Search Tags


/mathhelpforum @mathhelpforum