Linear Algebra, Water Flow Problem

In summary: That'll give you a range of answers. And usually that's ok.:)In summary, the conversation discusses solving a system of equations using an augmented matrix in order to determine the flow through each node in a circuit. The equations are rearranged and put into a matrix, which is then reduced to find the values for each variable. However, since there are more variables than equations, some values are left as arbitrary or free variables. The conversation also mentions the use of Kirchoff's laws to find additional equations, but in this particular problem, there are not enough clues to do so.
  • #1
RogerDodgr
20
0

Homework Statement


http://sudokupuzzles.net/IMG_0033g.gif [Broken]

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 [Broken]
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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  • #2
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}
-1 & 0 & 1 & 0 & 0 & 0 & 0 & -100 \\
1 & -1 & 0 & -1 & 0 & 0 & 0 & -150 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 300 \\
0 & 0 & -1 & 0 & 0 & 1 & 0 & 100 \\
0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\
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}
1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & -1 & 50 \\
0 & 0 & 1 & 0 & 0 & -1 & 0 & -100 \\
0 & 0 & 0 & 1 & 0 & -1 & 1 & 100 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 250 \\
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.
 
  • #3
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.
 
  • #4
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.
 

1. What is Linear Algebra and how is it used in the context of water flow problems?

Linear Algebra is a branch of mathematics that deals with linear equations and their representations in vector spaces. In the context of water flow problems, Linear Algebra is used to model the flow of water through a network of pipes and nodes, and to solve for the flow rates and pressure at each point in the system.

2. What are the main components of a water flow problem that can be represented using Linear Algebra?

The main components of a water flow problem that can be represented using Linear Algebra are the pipes, nodes, flow rates, and pressure at each point in the system. These components can be represented as matrices and vectors, and can be used to solve for the unknown variables in the system.

3. How does the concept of a matrix relate to water flow problems?

In the context of water flow problems, a matrix is used to represent the relationship between the variables in the system. For example, a matrix can be used to represent the flow rates of water through different pipes, or the change in pressure at different nodes in the system. By manipulating these matrices, we can solve for the unknown variables and analyze the flow of water in the system.

4. What are some common methods used in Linear Algebra to solve water flow problems?

Some common methods used in Linear Algebra to solve water flow problems include Gaussian elimination, LU decomposition, and Gauss-Jordan elimination. These methods involve manipulating matrices and solving systems of linear equations to find the unknown variables in the system.

5. How are water flow problems with multiple sources and sinks handled using Linear Algebra?

Water flow problems with multiple sources and sinks can be handled using Linear Algebra by representing the system as a matrix equation and using methods such as Gaussian elimination to solve for the unknown variables. The sources and sinks can be represented as additional variables in the system, and their values can be adjusted to find the optimal flow rates and pressure in the system.

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