Question

Short answer

The smallest possible value of \({x_6}\) is 70.

Step-by-step solution

Equation at nodes

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:

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

Reduce the matrix

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.

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

Echelon matrix

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

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

Solution of traffic network

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.