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Find Minimum values in cyclical graph

  1. #1
    Kramer1557@hotmail.com
    Guest

    Find Minimum values in cyclical graph

    I hope someone has an idea on how to start this problem.

    I have a lot of cyclical data that diminishes over time. I need to
    extract the minimum values of each cycle. I was thinking something
    along the line of having a start and finish value, and find the minimum
    value between the start and finish rows. This would repeat by setting
    the finish value to the start variable, and obtaining a new finish
    value However, I do not know how to use a variable to select a cell.

    If this sounds strange, I'll write a small example

    A B
    1 5
    2 4
    3 3
    4 4
    5 5
    6 4
    7 3
    8 4
    9 5

    I would initialize start to 1 and finish to 5 (the values in column A).
    The program would then calculate the minimum between the two values in
    clolumn B. Then finish (5) would become start, and 9 would be the new
    finish value.

    Any help is greatly appreciated, thanks in advance.

    -Taylor


  2. #2
    Tom Ogilvy
    Guest

    Re: Find Minimum values in cyclical graph

    Dim i as Long, rng as Range
    for i = 1 to 20 step 5
    set rng = cells(i,1).Resize(5,1)
    msgbox rng.Address & " minimum: " & _
    application.Minimum(rng.offset(0,1))
    Next

    --
    Regards,
    Tom Ogilvy

    <Kramer1557@hotmail.com> wrote in message
    news:1149079246.982551.128630@g10g2000cwb.googlegroups.com...
    > I hope someone has an idea on how to start this problem.
    >
    > I have a lot of cyclical data that diminishes over time. I need to
    > extract the minimum values of each cycle. I was thinking something
    > along the line of having a start and finish value, and find the minimum
    > value between the start and finish rows. This would repeat by setting
    > the finish value to the start variable, and obtaining a new finish
    > value However, I do not know how to use a variable to select a cell.
    >
    > If this sounds strange, I'll write a small example
    >
    > A B
    > 1 5
    > 2 4
    > 3 3
    > 4 4
    > 5 5
    > 6 4
    > 7 3
    > 8 4
    > 9 5
    >
    > I would initialize start to 1 and finish to 5 (the values in column A).
    > The program would then calculate the minimum between the two values in
    > clolumn B. Then finish (5) would become start, and 9 would be the new
    > finish value.
    >
    > Any help is greatly appreciated, thanks in advance.
    >
    > -Taylor
    >




  3. #3
    kounoike
    Guest

    Re: Find Minimum values in cyclical graph

    Is application.Minimum() typo of application.Min() ?

    i'm not sure but i think step 4 instead of step 5, according to Taylor.

    keizi

    "Tom Ogilvy" <twogilvy@msn.com> wrote in message
    news:%234YsQCNhGHA.4864@TK2MSFTNGP05.phx.gbl...
    > Dim i as Long, rng as Range
    > for i = 1 to 20 step 5
    > set rng = cells(i,1).Resize(5,1)
    > msgbox rng.Address & " minimum: " & _
    > application.Minimum(rng.offset(0,1))
    > Next
    >
    > --
    > Regards,
    > Tom Ogilvy
    >
    > <Kramer1557@hotmail.com> wrote in message
    > news:1149079246.982551.128630@g10g2000cwb.googlegroups.com...
    > > I hope someone has an idea on how to start this problem.
    > >
    > > I have a lot of cyclical data that diminishes over time. I need to
    > > extract the minimum values of each cycle. I was thinking something
    > > along the line of having a start and finish value, and find the

    minimum
    > > value between the start and finish rows. This would repeat by

    setting
    > > the finish value to the start variable, and obtaining a new finish
    > > value However, I do not know how to use a variable to select a

    cell.
    > >
    > > If this sounds strange, I'll write a small example
    > >
    > > A B
    > > 1 5
    > > 2 4
    > > 3 3
    > > 4 4
    > > 5 5
    > > 6 4
    > > 7 3
    > > 8 4
    > > 9 5
    > >
    > > I would initialize start to 1 and finish to 5 (the values in column

    A).
    > > The program would then calculate the minimum between the two values

    in
    > > clolumn B. Then finish (5) would become start, and 9 would be the

    new
    > > finish value.
    > >
    > > Any help is greatly appreciated, thanks in advance.
    > >
    > > -Taylor
    > >

    >
    >



  4. #4
    Tom Ogilvy
    Guest

    Re: Find Minimum values in cyclical graph

    Yes it is a typo.

    You know your data better than I do.

    --
    Regards,
    Tom Ogilvy

    "kounoike" <kounoike@nowherembh.nifty.com> wrote in message
    news:ueuBv%23WhGHA.1864@TK2MSFTNGP02.phx.gbl...
    > Is application.Minimum() typo of application.Min() ?
    >
    > i'm not sure but i think step 4 instead of step 5, according to Taylor.
    >
    > keizi
    >
    > "Tom Ogilvy" <twogilvy@msn.com> wrote in message
    > news:%234YsQCNhGHA.4864@TK2MSFTNGP05.phx.gbl...
    > > Dim i as Long, rng as Range
    > > for i = 1 to 20 step 5
    > > set rng = cells(i,1).Resize(5,1)
    > > msgbox rng.Address & " minimum: " & _
    > > application.Minimum(rng.offset(0,1))
    > > Next
    > >
    > > --
    > > Regards,
    > > Tom Ogilvy
    > >
    > > <Kramer1557@hotmail.com> wrote in message
    > > news:1149079246.982551.128630@g10g2000cwb.googlegroups.com...
    > > > I hope someone has an idea on how to start this problem.
    > > >
    > > > I have a lot of cyclical data that diminishes over time. I need to
    > > > extract the minimum values of each cycle. I was thinking something
    > > > along the line of having a start and finish value, and find the

    > minimum
    > > > value between the start and finish rows. This would repeat by

    > setting
    > > > the finish value to the start variable, and obtaining a new finish
    > > > value However, I do not know how to use a variable to select a

    > cell.
    > > >
    > > > If this sounds strange, I'll write a small example
    > > >
    > > > A B
    > > > 1 5
    > > > 2 4
    > > > 3 3
    > > > 4 4
    > > > 5 5
    > > > 6 4
    > > > 7 3
    > > > 8 4
    > > > 9 5
    > > >
    > > > I would initialize start to 1 and finish to 5 (the values in column

    > A).
    > > > The program would then calculate the minimum between the two values

    > in
    > > > clolumn B. Then finish (5) would become start, and 9 would be the

    > new
    > > > finish value.
    > > >
    > > > Any help is greatly appreciated, thanks in advance.
    > > >
    > > > -Taylor
    > > >

    > >
    > >

    >




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