91Ó°ÊÓ

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:

\({x_1} = {x_3} + {x_4} + 40\)

For node B:

\(200 = {x_1} + {x_2}\)

For node C:

\({x_2} + {x_3} = {x_5} + 100\)

For node D:

\({x_4} + {x_5} = 60\)

The total flow into the system:

\(\begin{array}{c}200 = 40 + 60 + 100\\ = 200\end{array}\)

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

\(\begin{array}{c}{x_1} - {x_3} - {x_4} = 40\\{x_1} + {x_2} = 200\\{x_2} + {x_3} - {x_5} = 100\\{x_4} + {x_5} = 60\end{array}\)

Write down the equations in an augmented matrix form.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 1}&0\\1&1&0&0&0\\0&1&1&0&{ - 1}\\0&0&0&1&1\end{array}\,\,\,\begin{array}{*{20}{c}}{40}\\{200}\\{100}\\{60}\end{array}} \right]\)

Reduce the augmented matrix into an echelon matrix.

Apply row operation \({R_2} \to {R_2} - {R_1}\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 1}&0\\0&1&1&1&0\\0&1&1&0&{ - 1}\\0&0&0&1&1\end{array}\,\,\,\begin{array}{*{20}{c}}{40}\\{160}\\{100}\\{60}\end{array}} \right]\)

Apply row operation \({R_3} \to {R_3} - {R_2}\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 1}&0\\0&1&1&1&0\\0&0&0&{ - 1}&{ - 1}\\0&0&0&1&1\end{array}\,\,\,\begin{array}{*{20}{c}}{40}\\{160}\\{ - 60}\\{60}\end{array}} \right]\)

Apply row operation \({R_4} \to {R_4} + {R_3}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 1}&0\\0&1&1&1&0\\0&0&0&{ - 1}&{ - 1}\\0&0&0&0&0\end{array}\,\,\,\begin{array}{*{20}{c}}{40}\\{160}\\{ - 60}\\0\end{array}} \right]\)

a.Hence, the general traffic pattern of the network is \(\left\{ \begin{array}{l}{x_1} = 100 + {x_3} - {x_5}\\{x_2} = 100 - {x_3} + {x_5}\\{x_3}\,{\rm{is}}\,{\rm{free}}\\{x_4} = 60 - {x_5}\\{x_5}\,{\rm{is}}\,{\rm{free}}\end{array} \right.\).

b.When \({x_4} = 0\), the value of \({x_5}\)will be equal to 60, and the pattern of the network will be \(\left\{ \begin{array}{l}{x_1} = 40 + {x_3}\\{x_2} = 160 - {x_3}\\{x_3}\,{\rm{is}}\,{\rm{free}}\\{x_4} = 0\\{x_5} = 60\end{array} \right.\).

c.The value of \({x_3}\)will not be negative. Hence, the minimum value of \({x_1}\) is 40 cars/minute.