Solving system of inequalities

  1. FysixFox

    FysixFox 19
    Gold Member

    (This isn't homework, so I guess it'd go here...)



    Okay, so I'm trying to solve a system of equations with a bunch of ranges:

    (Some number between 1.2 and 2.0) = (Some number between 5.5y and 9.1y) = (Some number between 5.4x and 10x)

    Where X and Y are ranges made up of two percentages (or ranges of percentages) each. I know one of the percentages for X is a range, and I know that range's approximate value (between 20 and 25%) and I also know one of the percentages for Y (21%). But I want to find the second percentage (or range of percentages) for each one.

    How the heck do I do this? Math with ranges makes my head hurt for some reason. I tried looking it up, but... I just got a bunch of stuff for summing numbers in Excel. :/
     
    Last edited: Jun 20, 2014
  2. jcsd
  3. Mentallic

    Mentallic 3,657
    Homework Helper

    It would probably help if you included your actual problem. Where did this come from?

    At the moment what I think you're trying to say is that there exists some distinct value of n in which

    [tex]1.2\leq n \leq 2.0[/tex]

    and also we want to find the range of x and y such that the above inequality still holds and these inequalities are

    [tex]5.4x \leq n \leq 10x[/tex]

    [tex]5.5y\leq n\leq 9.1y[/tex]

    The range that x and y can take such that n still falls in each range just needs to be chosen at the extreme points.

    5.4x is the lowest value, hence if this value is the highest n value, then 5.4x=2.0, x=37% will force us to choose n=2.0.
    Similarly, if we let 10x=1.2, x=12% then we force n to be the minimum of n=1.2. So the range of values that x can take such that the first two statements still hold true is 12% < x < 37%
    The same can be done for y.

    I'm sure this isn't what you're looking for though, so you'll have to be a little more clear.
     
  4. Mentallic

    Mentallic 3,657
    Homework Helper

    Thinking about it some more, you might instead be looking for a range of x (and similarly, y) values such that the entire range of 1.2<n<2.0 is satisfied.

    For example, if we take 1.2 = 5.4x, x=22% then 10x = 2.2 which is beyond 2.0 so for this x value, all of the range of n is satisfied. But we have "leftovers" between 2.0 and 2.2 so the other extreme that x can take is 2.0 = 10x, x=20%. For this value, 5.4x = 1.08 which of course we expect to be lower than 1.2. So the range of x values you can take is between 20-22%.

    Are we on the right track?
     
  5. FysixFox

    FysixFox 19
    Gold Member

    Yeah, that's part of it. I'm trying to find X and Y, and find out the second component of both X and Y. As in, if the components of Z are 50% and 45%, that would make 0.45 ≤ Z ≤ 0.5 (but I have no idea how to do this if a component is a range)

    The first component of X is 20-25%, and the first component of Y is 21%. So 20-25% and ____% make up X (20-22%), and 21% and ____% make up Y (22%). I've no idea how to get from X and Y to their unknown component.
     
  6. verty

    verty 1,813
    Homework Helper

    Do everything one step at a time and at each step, find the maximum and the minimum possible value.

    So you start with ##1.2 ≤ n ≤ 2.0##, we have a range for ##n##. Now look at x: ##5.4x ≤ n ≤ 10x##, we know the maximum and minimum of ##x## will occur when ##n## is a maximum or minimum, we don't need to use calculus. Find the max and min for those two extreme values of ##n##, then ##x##'s range will be the highest and lowest of those 4 numbers.

    Proceed like this one step further and find the range of ##y##.
     
  7. mfb

    Staff: Mentor

    That does not work, the values are in the wrong order:
    x=20% would fail at 5.4x=1.08 < 1.2, x=22% would fail at 10x=2.2 > 2.

    It would work if your calculated 20% would be larger than the calculated 22%.


    I think it would help a lot if you can give more context to the equations.
     
  8. FysixFox

    FysixFox 19
    Gold Member

    Oh god. Running numbers again from the start, I realize I didn't even need help. Just a break. I ran numbers that weren't even... ugh.

    Running the numbers again and arriving at T, the solution became mind-numbingly simple:

    T = 0.105 = 0.5x = 0.5y

    2T = x = y = 0.21 !!!

    C = 0.5/U = 0.5/I

    C = 2.38

    X and Y ended up equal, instead of slightly different and being inequalities.

    =======

    I had 3 circles overlapping with eachother. Starting with Circle A's Area = Circle B's Area = 1, and knowing the A/B overlap (D), the B/C overlap (E), and the A/C overlap (F), I wanted to find the overlap between all three (T) and the area of circle C. Which ended up actually being ridiculously easy.

    A=B=1
    D=0.21
    E=F=0.5

    T=D*E=D*F=0.21*0.5=0.105

    And then I just set 0.105 equal to E*x and F*y where x and y were the fractions of circle C that were also part of circle A and B, respectively. This made D, y, and x the same number, and I knew then that 0.5/0.21 would get me the area of C.

    I feel really dumb right now. But obviously I'll take a break next time before I make a thread about it and feel dumb later. :/
     
  9. mfb

    Staff: Mentor

    Why?
    That could be true in some special cases, but certainly not in general.

    That does not work at all.

    I don't think there is a general, unique solution to the problem you described. The circle C has 3 degrees of freedom (two for the position of the center and one for the radius), but you just have two constraints on them - E and F.

    Simple extreme example: Let E=F=1. That means circle C encloses both unit circles. How large is it? You have no way to tell.
     
  10. FysixFox

    FysixFox 19
    Gold Member

    E was given as being equal to F. E is where C and A overlap. F is where B and A overlap. Imagine it like three populations. Broguys overlap with Dudemen so that the percent of Dudemen that are Broguys is equal to the percent of Broguys that are Dudemen. This is D, 21% of the two unit circles. Because it's a circle and not an actual population, we can assume that the overlap of D with circle C is D times (0.5E + 0.5F), a perfect average of both circle's overlaps with circle C. Since E and F are the same, this just gives D*E or D*F (0.21*0.5).

    Now what is true for overlap D must also be true for E and F. However, we need one of the two overlaps in terms of C, not in terms of A or B (unit circles). Since this is unknown, this can be represented by variables:

    D(E+F)/2 = E(D+x)/2 = F(D+y)/2

    Which means x=y and comes out to:

    0.105 = 0.105/2 + 0.5x/2
    0.105 = 0.0525 + 0.25x
    0.0525 = 0.25x
    0.21 = x = y

    Remember, since x and y are percents of C, we can take 0.21 and divide E or F by that to get C=0.5/0.21 (multiplied by B if B isn't equal to one)

    Erg, I need a fudging whiteboard to explain this properly. But it definitely SHOULD hold true for any three overlapping circles, if I did this right.
     
  11. mfb

    Staff: Mentor

    :confused:
    I go with the first one.

    This requires the same total population for both. Fine for A and B, but not for C or for intersections.

    You cannot (it would be easier for populations! If you assume the 3 things are independent of each other). Just draw some circles to find counterexamples.


    I think I misread your definitions of x and y.
    x=F/C, y=E/C? Then F=E directly gives x=y.

    Setting F=E:
    DE = E(D+E/C)/2 = E(D+E/C)/2
    D = (D+E/C)/2
    D = E/C
    Why should this be true?
     
  12. FysixFox

    FysixFox 19
    Gold Member

    It doesn't matter what I call the circles. Jimmy, James, and Jack. A, B, C. 1, 2, 3. The only thing that matters is that two are unit circles, one is larger than a unit circle, and how the three overlap.

    When did I contradict this? I specifically defined the two circles as unit circles BECAUSE they had the same "population" or "area" or whichever term you'd like to use.

    I'm defining everything in terms of A and B, as in, if A is really 1,000 and B is 1,000, and the overlap is 500, then it's 50% of either circle. The math is made a hundred times more simple by this fact, and the fact that the overlaps with circle C are the same. If this wasn't the case, it'd become much more complicated.

    x and y are E/C and F/C, yes. In the example with D, though it's not entirely apparent due to how I defined things to simplify the maths, we took E and F (as a percent of A and B) and added them and divided them by two to get the overlap between A and B with C. So for the other overlaps, we must define the numbers just as accordingly.

    Let's take the A-C overlap in example. The overlap of this and B would be A's overlap with B plus C's overlap with B over two, and multiplying this by the A-C overlap would give the A-B-C overlap. Make sense?

    Remember that we ended with C = E/D. This is true and ONLY true because F=E, and A=B. If these were not true... the maths would become more complicated.

    ...

    You know what? I'll attach a diagram I drew this morning. One sec.

    In the diagram, A B and C are the circle areas, while T is the area of all three's overlap and X Y and Z are the areas of the overlaps between A&B, A&C, and A&B.
     

    Attached Files:

  13. mfb

    Staff: Mentor

    Yes it does not matter, but please be consistent. Don't use different notations for different overlaps in different posts.

    That's fine.

    That does not work.

    No.

    Example 1: The overlap AB plus the overlap BC, divided by two, is some non-zero number, but the overlap of (AC overlap) with B is empty.

    [​IMG]


    Example 2: The overlap of (AC) and B is identical to the (AB) overlap, if you multiply this by the (AC) overlap (which is smaller than 1), you don't get the right result.

    [​IMG]

    Which is wrong. The first image is a counterexample, D>E but C>1.

    The equations at the left side are wrong. My first picture is a nice example that has all your given conditions (same areas of A and B, same overlaps AC and BC), but T=0. You can easily modify it to get a very small T, if you know that T cannot be zero. The equations still stay invalid.
     
  14. FysixFox

    FysixFox 19
    Gold Member

    ... wait.

    *does some simple math*

    Never mind, I'm an idiot. Now that I think of it, I've been OVER-thinking it. There's no way to tell what the area of the other circle is. My bad. :/
     
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