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jbf frylock

  1. #1
    jbf frylock
    Guest

    jbf frylock

    Need some help. This is the formula I have now. It figures a salary range
    and a categorilcally eligible logic test. If a household with x number of
    people (A10) makes n they are free or reduced or paid. The differnce for
    each increment is 4200 for the free category and 6000 for the reduced and
    paid category.

    =IF(L3<=(A10*4200+8203),"Free",IF(L3<=(A10*6000+11674),"Reduced","Paid"))

    What do you do if the difference isn't the same. What if x = 1 and on the
    next cell n = 2 and the jump from income is 1000 to 2004 and then for a
    houselhold of 3, the income is 3007. Then the rest of the household numbers
    have a difference of 1004. How do I work the difference of 1003 into an
    equation where all of the other amounts are 1004? I was told to use: A10 =
    number of people. L4 = income for the people, 353 is the difference and 503
    is the difference. What if the diffference was 354 for all but one (353) and
    503 for all but one (504)? this is the equation I've been playing with.

    =IF(L4<=684+(353*A10)+A10,"Free",IF(L4<=973+(503*A10)+A10,"Reduced","Paid"))

  2. #2
    Gary L Brown
    Guest

    RE: jbf frylock

    This makes NO sense!!!
    --
    Gary Brown
    gary_brown@ge_NOSPAM.com
    If this post was helpful, please click the ''Yes'' button next to ''Was this
    Post Helpfull to you?''.


    "jbf frylock" wrote:

    > Need some help. This is the formula I have now. It figures a salary range
    > and a categorilcally eligible logic test. If a household with x number of
    > people (A10) makes n they are free or reduced or paid. The differnce for
    > each increment is 4200 for the free category and 6000 for the reduced and
    > paid category.
    >
    > =IF(L3<=(A10*4200+8203),"Free",IF(L3<=(A10*6000+11674),"Reduced","Paid"))
    >
    > What do you do if the difference isn't the same. What if x = 1 and on the
    > next cell n = 2 and the jump from income is 1000 to 2004 and then for a
    > houselhold of 3, the income is 3007. Then the rest of the household numbers
    > have a difference of 1004. How do I work the difference of 1003 into an
    > equation where all of the other amounts are 1004? I was told to use: A10 =
    > number of people. L4 = income for the people, 353 is the difference and 503
    > is the difference. What if the diffference was 354 for all but one (353) and
    > 503 for all but one (504)? this is the equation I've been playing with.
    >
    > =IF(L4<=684+(353*A10)+A10,"Free",IF(L4<=973+(503*A10)+A10,"Reduced","Paid"))


  3. #3
    jbf frylock
    Guest

    RE: jbf frylock

    Yes, apparently it isn't easy to explain. Just need a logic test to pop up 3
    different results. 1)Free category 2)Reduced category 3)paid category.
    These will be determined by income. It would be easy if they all had a
    simple relationships, but they don't. The categories are determined by
    income and number of people that income includes. So 1 situation might be a
    household of 3 with a monthly income of 2500, another might be a household of
    5 with a monthly income of 3500. Based on the salary and number of people it
    would place them in a category. I have this figured out for part of my
    spreadsheet. Works great because there is a constant and the increments are
    always the same. I need to figure out how to calculate the same thing when
    the increments aren't the same.

    Here is the formula for the first part of spreadsheet. D10= total income,
    A10 = household size, 4238 = increment, 8203 = constant, 6031 = increment and
    11674 = constant. Need a formula that does this below, but with a variable
    increment.

    =IF(D10<=(A10*4238+8203),"Free",IF(D10<=(A10*6031+11674),"Reduced","Paid"))

    "Gary L Brown" wrote:

    > This makes NO sense!!!
    > --
    > Gary Brown
    > gary_brown@ge_NOSPAM.com
    > If this post was helpful, please click the ''Yes'' button next to ''Was this
    > Post Helpfull to you?''.
    >
    >
    > "jbf frylock" wrote:
    >
    > > Need some help. This is the formula I have now. It figures a salary range
    > > and a categorilcally eligible logic test. If a household with x number of
    > > people (A10) makes n they are free or reduced or paid. The differnce for
    > > each increment is 4200 for the free category and 6000 for the reduced and
    > > paid category.
    > >
    > > =IF(L3<=(A10*4200+8203),"Free",IF(L3<=(A10*6000+11674),"Reduced","Paid"))
    > >
    > > What do you do if the difference isn't the same. What if x = 1 and on the
    > > next cell n = 2 and the jump from income is 1000 to 2004 and then for a
    > > houselhold of 3, the income is 3007. Then the rest of the household numbers
    > > have a difference of 1004. How do I work the difference of 1003 into an
    > > equation where all of the other amounts are 1004? I was told to use: A10 =
    > > number of people. L4 = income for the people, 353 is the difference and 503
    > > is the difference. What if the diffference was 354 for all but one (353) and
    > > 503 for all but one (504)? this is the equation I've been playing with.
    > >
    > > =IF(L4<=684+(353*A10)+A10,"Free",IF(L4<=973+(503*A10)+A10,"Reduced","Paid"))


  4. #4
    Pete
    Guest

    Re: jbf frylock

    Jeff,

    I suspect that whoever has given you the new numbers for the table
    (your boss?) has made a simple mistake in one of them that has
    propagated through.

    When you first posed this question in December, my instinct was to use
    a lookup table, but then I spotted a simple relationship between the
    numbers in the table which made the table redundant. If you definitely
    have to stick with these new numbers, then I would suggest a table this
    time. Assume you have the following table occupying cells N1 to P8
    (I've had to guess the second set of values - correct as necessary).

    1 1037 1476
    2 1390 1979
    3 1744 2483
    4 2097 2986
    5 2450 3489
    6 2803 3992
    7 3156 4495
    8 3509 4998

    The following formula will give you "Free", "Reduced" or "Paid" to suit
    your criteria:

    =IF(L4<VLOOKUP((A10+1),N1:O8,2,0),"Free",
    IF(L4<VLOOKUP((A10+1),N1:P8,3,0),"Reduced","Paid"))

    This is all one formula, and assumes L4 is current income and A10 is
    the number in the household, as in your earlier posting. This will only
    cope with up to 7 in the household - if you have more then you will
    need to extend the table down and adjust the references to O8 and P8 in
    the formula.

    Hope this helps this time - you can always apply it to the previous
    situation.

    Pete


  5. #5
    Pete
    Guest

    Re: jbf frylock

    I'm a bit confused - do these values represent the largest amount that
    can be earned? In other words, can a household of 1 earn up to (and
    including) 1037 and still be regarded as "Free", or up to 1476 and be
    "Reduced"? If so, you need this amended formula:

    =IF(A10<1,"Not valid",IF(L4<=VLOOKUP(A10,N1:O8,2,0),"Free",
    IF(L4<=VLOOKUP(A10,N1:P8,3,0),"Reduced","Paid")))

    This will cover households up to 8 - extend the table and amend the
    ranges as described earlier for more.

    Pete


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