# Thread: Solving for Simultaneous Inequalities

1. ## Solving for Simultaneous Inequalities

I noted in my self-introduction post that I forgot much of my maths knowledge way back in the last century, so please forgive me if this is a somewhat basic query. It's based on a real-life problem that I'm trying to resolve.
I have a line with four points on it - A at the beginning, D at the end, and B & C in between. The positions of A, B & C are fixed; D is variable. The relationship between the four points is defined by two formulae:
C - D > x(B - C)
C > 0.5(A - D)
How do I find the minimum value of D, expressed in terms of A, B, C and x? D > . . . . . . . ??

In reality, 'x' is a variable factor. I need to find the minimum value of D using various factors, for example what it was 1.9(B - C), or 2.4(B - C).

Once some kind soul helps me with the expression for determining D then I'll drop it in an Excel spreadsheet and test for the minimum D value using different factors.

Assistance with this will be much appreciated.

2. ## Re: Solving for Simultaneous Inequalities

You talk about a line with points on it, but then treat the points as though they are numbers.

3. ## Re: Solving for Simultaneous Inequalities

They're measurements along the line, at those four points so in effect, yes, the points are numbers. A, B & C are fixed points / fixed values; only D is variable and that's the one I'm trying to optimize

4. ## Re: Solving for Simultaneous Inequalities

My sincere apologies - I've just realized I have repeated the same formula error that I did when I was trying to work this through a couple of days ago. The first formula is correct; the second formula should read:
C > D + 0.5(A - D)
Added to that I made a typo when I referred to finding the minimum value of D. It should of course read: How do I find the minimum value of D, expressed in terms of A, B, C and x? D < . . . . . . . ??

5. ## Re: Solving for Simultaneous Inequalities

$\displaystyle d = (x+1) c-x b$

6. ## Re: Solving for Simultaneous Inequalities

How do I find the minimum value of D, expressed in terms of A, B, C and x? D < . . . . . . . ??
D < ... yields the maximum value of D

the minimum value of D would be an inequality of the form D > ...

7. ## Re: Solving for Simultaneous Inequalities

Thanks Idea, but that's not giving me correct results; no doubt my fault for asking the wrong question. I've since realised that I'm actually looking for the maximum value of D.
The following is an actual data set that satisfies the two criteria above: C > D + 0.5(A - D) and C - D > x(B - C) using a factor of 2 for "x". What I'm trying to achieve is to move D as close as possible towards C and still satisfy these two criteria.

I then used your formula to arrive at a new value for D, substituted those values in the D column of the data set and obtained the following results when I tested against the criteria:

The criteria are not being consistently met.

8. ## Re: Solving for Simultaneous Inequalities

Thanks Skeeter - that's what comes of writing something late in the day after confusing myself by puzzling over it for hours. After a decent night's sleep, I've realised that I'm actually looking for the maximum value of D (see above)

9. ## Re: Solving for Simultaneous Inequalities

the two inequalities ...

C-D > x(B-C)
C > 0.5(A-D)
yields the following interval for the value D ...

A-2C < D < x(C-B)+C

10. ## Re: Solving for Simultaneous Inequalities

Given $\displaystyle C > D + 0.5(A - D)$ and $\displaystyle C - D > x(B - C)$

choose $\displaystyle D$ less than the minimum of $\displaystyle 2C-A$ and $\displaystyle C-x(B-C)$

11. ## Re: Solving for Simultaneous Inequalities

Skeeter / Idea - thank you both so much for that. That gives me something I can plug into Excel to do the number crunching.
It's 40+ years since I last had to deal with inequalities (in a mathematical sense!) and I've long since forgotten even the basics - as I guess was somewhat obvious! I'm in the midst of other issues at the moment so I won't get to trial the formulae until tomorrow morning but the similarity of your responses tells me I'll be able to work it from here.
Much appreciated.