Linear Algebra, Water Flow Problem

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Homework Help Overview

The discussion revolves around a water flow problem modeled using linear algebra, specifically focusing on the flow into and out of nodes represented by a system of equations. Participants are working with an augmented matrix and exploring the implications of their equations on the flow variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up a system of equations based on flow conservation at nodes. There are attempts to rearrange equations for matrix representation and to reduce the matrix to row echelon form. Questions arise regarding the adequacy of the equations and whether additional equations can be derived. Some participants express uncertainty about the implications of having free variables in the context of the problem.

Discussion Status

There is an ongoing exploration of the relationships between the flow variables, with some participants providing insights that help clarify the situation. The discussion includes considerations of variable constraints and the implications of having fewer equations than unknowns. Participants are sharing their reasoning and seeking validation of their approaches.

Contextual Notes

Participants note the potential for additional equations based on circuit laws, although none are immediately identified in the current problem setup. There is an acknowledgment of the need for assumptions when information is lacking.

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Homework Statement


http://sudokupuzzles.net/IMG_0033g.gif

Homework Equations


Flow into a node=flow out of a node.
Turned relative flows through each node (A,B,C,D,E,F) into system of equations; and entered system of equations into augmented matrix as shown. Reduced matrix to REF.

The Attempt at a Solution


http://sudokupuzzles.net/IMG_0034.jpg
I am not certain if I can be more exact with part a f_5 and f_6.

I entered all the pertinent equations into the system possible for each node. (Is there another eautaion that could be entered (to help reduce answer)?)

Part b, I have no visual explanation, as per the diagram, to why f_1=f_6.
 
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Okay, this is a lot of work to sift through, so I'll just rework it.. If we end up with the same thing.. It'll just be a lot of work for nothing. Anyway, here goes:

So, as you said.. flow in = flow out. So, I can set of a system of equations:

A: [tex]f_{3} + 200 = f_{1} + 100[/tex]
B: [tex]f_{1} + 150 = f_{2} + f_{4}[/tex]
C: [tex]f_{2} + f_{5} = 200 + 100[/tex]
D: [tex]f_{6} + 100 = f_{3} + 200[/tex]
E: [tex]f_{4} + f_{7} = f_{6} + 100[/tex]
F: [tex]150 + 100 = f_{5} + f_{7}[/tex]

Now I can rearrange these to make them easier to be put into a matrix:
A: [tex]-f_{1} + f_{3} = -100[/tex]
B: [tex]f_{1} - f_{2} - f_{4} = -150[/tex]
C: [tex]f_{2} + f_{5} = 300[/tex]
D: [tex]-f_{3} + f_{6} = 100[/tex]
E: [tex]f_{4} - f_{6} + f_{7} = 100[/tex]
F: [tex]f_{5} + f_{7} = 250[/tex]

Now, writing this as a matrix, we get:
[tex]\left( \begin{array}{cccccccc}<br /> -1 & 0 & 1 & 0 & 0 & 0 & 0 & -100 \\<br /> 1 & -1 & 0 & -1 & 0 & 0 & 0 & -150 \\<br /> 0 & 1 & 0 & 0 & 1 & 0 & 0 & 300 \\<br /> 0 & 0 & -1 & 0 & 0 & 1 & 0 & 100 \\<br /> 0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\<br /> 0 & 0 & 0 & 0 & 1 & 0 & 1 & 250 \end{array} \right)[/tex]

Now, writing this in reduced row echelon form, we get:

[tex]\left( \begin{array}{cccccccc}<br /> 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\<br /> 0 & 1 & 0 & 0 & 0 & 0 & -1 & 50 \\<br /> 0 & 0 & 1 & 0 & 0 & -1 & 0 & -100 \\<br /> 0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\<br /> 0 & 0 & 0 & 0 & 1 & 0 & 1 & 250 \\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)[/tex]

Now, we can rewrite these as equations to see what we get.. (notice that we have [tex]f_{6}[/tex] and [tex]f_{7}[/tex] being a free variable since the last row is completely zero... therefore [tex]f_{7} = r[/tex] and [tex]f_{6} = s[/tex] where r,s is any real number).

I am not sure if we can make both arbitrary since we only have 6 equations and there are 7 unknowns, but I think that's what we have to do... Assuming this is the correct way to do it, the following work is logical..

Now, we get:
[tex]f_{1} = s[/tex]
[tex]f_{2} = r + 50[/tex]
[tex]f_{3} = s - 100[/tex]
[tex]f_{4} = s - r + 100[/tex]
[tex]f_{5} = -r + 250[/tex]
[tex]f_{6} = s[/tex]
[tex]f_{7} = r[/tex] where [tex]r,s \geq 0[/tex] (because there can't be negative flow)

part b)
since [tex]f_{1} = s[/tex] and [tex]f_{6} = s[/tex], these two need to be the same, therefore one cannot be 100 and the other be 150.

part c)
If [tex]f_{4} = 0[/tex], we get:
[tex]s - r + 100 = 0[/tex]
This implies that [tex]s = r - 100[/tex] and [tex]r = s + 100[/tex].

Substituting into the equations, we get:
[tex]f_{1} = s[/tex]
[tex]f_{2} = s + 150[/tex]
[tex]f_{3} = s - 100[/tex]
[tex]f_{4} = 0[/tex]
[tex]f_{5} = 150 - s[/tex]
[tex]f_{6} = s[/tex]
[tex]f_{7} = s + 100[/tex].

Now, knowing that we can't have negative flow, we see that [tex]f_{5}[/tex] tells us that s cannot be greater than 150, and [tex]f_{3}[/tex] tells us that s cannot be less than 100. Therefore, [tex]100 \leq s \leq 150[/tex].

Using this, we can find the range of flow on each:
[tex]100 \leq f_{1} \leq 150[/tex]
[tex]250 \leq f_{2} \leq 300[/tex]
[tex]0 \leq f_{3} \leq 50[/tex]
[tex]f_{4} = 0[/tex] (this was given).
[tex]0 \leq f_{5} \leq 50[/tex]
[tex]100 \leq f_{6} \leq 150[/tex]
[tex]200 \leq f_{7} \leq 250[/tex]

We should get the same ranges if we would have written all the equations in terms of r instead of s.

I'm not sure if everything I did is correct, but it seems to make sense to me. A second opinion would probably be nice.

Anyway, that's my input. Good luck.
 
Wow jacobpm64, thanks sincerely,,,

This statement was particularly helpful: "notice that we have f_6 and f_7 being a free variable since the last row is completely zero..."

Also, knowing that I'm not totally off base gives me confindence to progress forward...

I originally started with two varaiables; but then I (pointlessly) opted to try to look for correlate them.
 
well I mean.. sometimes you'll have extra things that can give you more equations.

Especially when you're dealing with circuits.

You have kirschoff's laws.

There are two different things to look for, so you can easily get more equations.

I couldn't see anything right off in this problem though to give more equations.

If you don't have enough information, heck, just make them arbitrary.
 

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