First, write down all the equations at each node. We know that the incoming flow at each node will be equal to the outgoing flow.

For node A:

\({x_1} = {x_2} + 100\)

For node B:

\({x_3} = {x_2} + 50\)

For node C:

\({x_3} = {x_4} + 120\)

For node D:

\({x_4} + 150 = {x_5}\)

Re-arrange all the above equations to get the augmented matrix.

\[\begin{array}{c}{x_1} - {x_2} = 100\\{x_2} - {x_3} = - 50\\{x_3} - {x_4} = 120\\{x_4} - {x_5} = - 150\\{x_5} - {x_6} = 80\\ - {x_1} + {x_6} = - 100\end{array}\]

Write down the equations in an augmented matrix form.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 1}&0&0&0&0&{100}\\0&1&{ - 1}&0&0&0&{ - 50}\\0&0&1&{ - 1}&0&0&{120}\\0&0&0&1&{ - 1}&0&{ - 150}\\0&0&0&0&1&{ - 1}&{80}\\{ - 1}&0&0&0&0&1&{ - 100}\end{array}} \right]\]

Reduce the augmented matrix into an echelon matrix. Hence, the echelon matrix is:

\[\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - 1}&{100}\\0&1&0&0&0&{ - 1}&0\\0&0&1&0&0&{ - 1}&{50}\\0&0&0&1&0&{ - 1}&{ - 70}\\0&0&0&0&1&{ - 1}&{80}\\0&0&0&0&0&0&0\end{array}} \right]\]

Hence, the general solution of the traffic pattern of the network is \[\left\{ \begin{array}{l}{x_1} = 100 + {x_6}\\{x_2} = {x_6}\\{x_3}\, = 50 + {x_6}\\{x_4} = {x_6} - 70\\{x_5} = {x_6} + 80\\{x_6}\,{\rm{is}}\,{\rm{free}}\end{array} \right..\]

Since the value of \({x_4}\)cannot be negative, the value of \({x_6}\) in the equation \({x_4} = {x_6} - 70\) should be equal to or more than 70.

Hence, the smallest possible value of \({x_6}\) is 70.