Linear Algebra, Water Flow Problem

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SUMMARY

The discussion centers on solving a water flow problem using linear algebra, specifically through the formulation of a system of equations based on flow conservation principles. Participants utilized an augmented matrix to represent the equations derived from nodes A through F, ultimately reducing it to reduced row echelon form (REF). The analysis revealed that variables f_6 and f_7 are free variables due to the presence of a zero row in the matrix, leading to the conclusion that the flow rates must satisfy specific constraints, particularly 100 ≤ f_1 ≤ 150 and 200 ≤ f_7 ≤ 250.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically systems of equations and matrix operations.
  • Familiarity with reduced row echelon form (REF) and its implications in solving equations.
  • Knowledge of flow conservation principles in network analysis.
  • Basic experience with variable manipulation and constraints in mathematical modeling.
NEXT STEPS
  • Study the application of Kirchhoff's laws in circuit analysis for additional insights into flow problems.
  • Explore advanced techniques in linear algebra, such as singular value decomposition (SVD) for solving underdetermined systems.
  • Learn about optimization methods in linear programming to maximize or minimize flow rates in network problems.
  • Investigate the use of software tools like MATLAB or Python's NumPy library for matrix operations and solving systems of equations.
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are dealing with network flow problems, as well as educators seeking to enhance their understanding of linear algebra applications in real-world scenarios.

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

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

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

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

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

Now, we can rewrite these as equations to see what we get.. (notice that we have f_{6} and f_{7} being a free variable since the last row is completely zero... therefore f_{7} = r and f_{6} = s 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:
f_{1} = s
f_{2} = r + 50
f_{3} = s - 100
f_{4} = s - r + 100
f_{5} = -r + 250
f_{6} = s
f_{7} = r where r,s \geq 0 (because there can't be negative flow)

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

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

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

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

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

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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