Given Information
We have to answer some questions based on the given freeway network.

Step-1: (a) General traffic Pattern
Write system of equations at each Node:

At Node A: \[{x_1} = {x_3} + {x_4} + 40\] At Node B: \[200 = {x_1} + {x_2}\] At Node C: \[{x_2} + {x_3} = {x_5} + 100\] At Node D: \[{x_4} + {x_5} = 60\]

Step-2: The augmented matrix
The system of equations: \[\begin{array}{l} {x_1} + 0.{x_2} - {x_3} - {x_4} + 0.{x_5} = 40\\ {x_1} + {x_2} + 0.{x_3} + 0.{x_4} + 0.{x_5} = 200\\ 0.{x_1} + {x_2} + {x_3} + 0.{x_4} - {x_5} = 100\\ 0{x_1} + 0.{x_2} + 0.{x_3} + {x_4} + {x_5} = 60 \end{array}\] The augmented matrix: \[M = [A|b] = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&{ - 1}&0&{40}\\ 1&1&0&0&0&{200}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&1&1&{60} \end{array}} \right]\]

Step-3: The row-reduced augmented matrix
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&{ - 1}&0&{40}\\ 1&1&0&0&0&{200}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&1&1&{60} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&{ - 1}&0&{40}\\ 0&1&1&1&0&{160}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&1&1&{60} \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} - {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&{ - 1}&0&{40}\\ 0&1&1&1&0&{160}\\ 0&0&0&{ - 1}&{ - 1}&{ - 60}\\ 0&0&0&1&1&{60} \end{array}} \right]::\,\,\left\{ {{R_3} = {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&0&1&{100}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&{ - 1}&{ - 1}&{ - 60}\\ 0&0&0&1&1&{60} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_1} = {R_1} + {R_4}\\ {R_2} = {R_2} + {R_3} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&0&1&{100}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&1&1&{60}\\ 0&0&0&1&1&{60} \end{array}} \right]::\,\,\left\{ {{R_3} = - {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}&0&1&{100}\\ 0&1&1&0&{ - 1}&{100}\\ 0&0&0&1&1&{60}\\ 0&0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - {R_3}} \right\} \end{array}\]

Step-4: The solution
The equations are: \[\begin{array}{l} {x_1} - {x_3} + {x_5} = 100\\ {x_2} + {x_3} - {x_5} = 100\\ {x_4} + {x_5} = 60 \end{array}\] Since there are three equations and 5 variables, we can consider 2 variable as free.

\[\begin{array}{l} {x_1} = 100 + {x_3} - {x_5}\\ {x_2} = 100 - {x_3} + {x_5}\\ {x_4} = 60 - {x_5}\\ {x_3} = {\rm{Free}}\\ {x_5} = {\rm{Free}} \end{array}\]

Step-5: (b): When $x_4$ is closed
Substitute $x_4 = 0$ in above system of equations: \[\begin{array}{l} {x_4} = 60 - {x_5}\\ 0 = 60 - {x_5}\\ {x_5} = 60 \end{array}\] So the solution becomes:

\[\begin{array}{l} {x_1} = 40 + {x_3}\\ {x_2} = 160 - {x_3}\\ {x_4} = 0\\ {x_3} = {\rm{Free}}\\ {x_5} = {\rm{Free}} \end{array}\]

Step-6: (c): Minimum value of $x_1$
From the above system of equations, $x_1$ is given by: \[{x_1} = 40 + {x_3}\] Since, the flow can never be negative, the minimum value of $x_3$ is 0, therefore

The minimum value of $x_1$ is 40