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 & 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)
Now, writing this in reduced row echelon form, we get:
\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)
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.
