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Fibonacci

Member

Registered: 2009-06-15

Posts: 25

one unknown and three equations

I have three linear equations

(At this point I am not expanding

to avoid needless complication.)

The objective is to find

for which

. For a particular

all lines intersect in one point. Otherwise, I have three intersections, which I can determine by pairing up

etc.:

As I can't equalize all results, I would like to find

for which

are as close as possible.

I'd be grateful for any inspiration how to tackle this problem.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: one unknown and three equations

Hi Fibonacci;

A question, you have 10 unknowns in the first 3 equations. Which of those are constants and which are variables. There is not much hope in solving an underdetermined set like that.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

Sorry. I got so used to my naming conventions I forgot to clarify.

are functions.

is the only variable, all others are constants.

is a constant replacing a more complex expression of

.

I have already solved

. However, there are 3 different possible values depending on which two equations I try to equalize. (That is in the nature of the problem and quite correct.)

Lets call the 3 possible values

. In an ideal world all three lines 

would intersect in one point thus giving me just one

.

In the real world the don't, which is why I have to choose a make-do value for

, which fulfils certain criteria.

Having thought about this a wee bit more, I need to rephrase what I am looking for: I want to find the

for which the lowest 

is the greatest.

I'll post a screenshot to illustrate what I am looking for in the next post.

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

The three purple lines represent

.

They'll always intersect in 3 points forming a triangle. The dotted yellow line is some arbitrarily chosen

. I have marked its intersections with the three

using yellow dots.

On the screenshot, the lowest of the yellow dots has a y-value of 34. If I push the yellow line further right, this yellow dot will eventually coincide with

.

That is as high as the bottommost yellow dot can get for these 3 lines. That's the one I am looking for.

I think, most of the time this will coincide with an 

intersection.

I think what I need is a function that describes the lowest yellow dot and then determine its maximum.

I suspect we're in different timezones. But for me it's goodnight now!

Last edited by Fibonacci (2011-07-18 11:51:12)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: one unknown and three equations

hi Fibonacci,

Now you've posted that diagram, I'm beginning to understand what you are after.  This does look like a linear programming problem.

I think, most of the time this will coincide with an  intersection.

Yes.  You could get a line of possible points if two lines were the same, but, assuming the lines are all distinct, you'll always get your ideal 'x' at an intersection.

Your algorithm for moving the yellow dot and getting its maximum position will always end up at point Ro1.  So you could just solve for those two lines that cross there.  But that doesn't sound right    ??

So what freedom do you have to vary the parameters for the lines.  I could generate an expression using these that will give the single 'x' where all three lines cross in a single point.  If you cannot vary them, then you are stuck with intersections you show.

Bob
997

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

Now you've posted that diagram, I'm beginning to understand what you are after.

Thank you so much for introducing me to GeoGebra! Fantastic for developing and documenting ideas!

assuming the lines are all distinct, you'll always get your ideal 'x' at an intersection

I know it's obvious when looking at the chart. But why?

Your algorithm for moving the yellow dot and getting its maximum position will always end up at point R01.

Or R02, see new screenshot.

I could generate an expression using these that will give the single 'x' where all three lines cross in a single point.

All three lines cross for

.

So what freedom do you have to vary the parameters for the lines?

Within given constraints, I choose the parameters with the objective of a high minimum

. So far I have been doing this from experience.

We know quite a few things about the lines' coefficients:

are positive (and therefore all lines' y-intersects >0)

is negative. I am pretty certain that, through choice, q1 and q2 are always positive because negative slopes give me low r.

Let's zoom in on the q's:

We know:

I choose a scenario where :

Last edited by Fibonacci (2011-07-19 05:09:25)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: one unknown and three equations

hi Fibonacci,

If

and

and

then q1 is negative too.

eg V1 = 0.01;   K2 = K3 = 0.2;     f>0    q1 = pos x (0.01 - 0.2 - 0.2) < 0

Have I misunderstood something here?

Let's see how to get a single point of intersection.

so for a single point for all three lines

and

which leads to

Then substituting for the qs

Oh dear that cannot be right.

In

I'd assumed I could cancel the fs but now I'm thinking that you meant three different fs here.

I'd better stop at this and wait for your response.

Bob

Last edited by Bob (2011-07-20 01:05:21)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

bob bundy, I feel terribly rude for not having replied earlier! But you must have edited your last post rather then creating a new one, which means no email notification is sent to subscribers.

I shall look at your detailed response later when I have some more time.

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

[...]
then q1 is negative too.
eg V1 = 0.01;   K2 = K3 = 0.2;     [...]
Have I misunderstood something here?

Yes:

Let's see how to get a single point of intersection.

The easiest starting point to work the intersection out are the three equations for x in post #1:

The above is true regardless which of the three equations you pair up.

However, finding the intersection is not the objective although it helps me in my real-life scenario to know where it is. Lets put it that way: I have a choice of lines and I choose those three that give me the best possible triangle. And that is one that gives me bottommost yellow dot (as per post #4) with a high y-coordinate.

In

I'd assumed I could cancel the fs

Indeed, the fs are identical.

The question is:
We have three lines that form a triangle where the lines intersect. The y-coordinates of the three intersactions are all different. I want to draw a vertical line that intersects with these three lines.

I want to find a vertical line where the lowest of the three intersects is the highest (as per post # attachment).

Can this be done mathematically? I can do this programmatically by calculating the three triangle points, "drawing" a vertical line through each point, then comparing the y-coordinates of the each vertical line's intersections  with the lines forming the triangle.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: one unknown and three equations

hi Fibonacci

I've re-read all your posts and here's what I think.

I'm assuming that ro always has a negative gradient, that r1 and r2 are always positive gradients with r2 steeper than r1.
Also that the lines all cross in positive co-ordinates.

Firstly, the problem is definitely a linear programming problem, so the optimum point will always be a vertex of the triangle.

Find where r1 and r2 cross.  Call this point A.

There are just two cases; depending on where A is, in relation to the line r0.

Substitute the x co-ordinate of A in the equation for ro, and compare the resulting y with the y co-ord. of A

case 1.  A is above ro.

In this case the point r02 (see your diagram in post #6) will always be the point you want.

case 2.  A is below r0.

In this case the point r01 (see your diagram in post #4) will always be the point you want.

I think it's enough to use the diagrams to say this is obviously 'proved', but you could make the proof rigorous by analysing y values along the line ro.

Incidently I also 'discovered' that for all three intersecting at a point

K1 + V1 = V2 + K2.

Will that suffice?

Bob

Last edited by Bob (2011-08-03 23:01:39)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Fibonacci

Member

Registered: 2009-06-15

Posts: 25

Re: one unknown and three equations

Firstly, the problem is definitely a linear programming problem, so the optimum point will always be a vertex of the triangle.

[...]

Will that suffice?

Yes - thank you.
That's what I implemented.

Incidently I also 'discovered' that for all three intersecting at a point
K1 + V1 = V2 + K2.

That's by design (due to a rule I set when forming the equations).