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.