Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 13E from Chapter 1.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 13E
Chapter:
Problem:
a. Find the general flow pattern of the network shown in the figure. b. Assuming that the flow must be in the directions indicated, find the minimum flows in the branches denoted by
Step-by-Step Solution
Given Information
We are given with a flow network. We have to find the general flow pattern in the network. We also have to find the minimum flows in the branches $x_2$, $x_3$, $x_4$ and $x_5$
Step-1:
The basic assumption of network flow is that the total flow into the network equal to the total flow out of network and that total flow into a function equal the total flow out of function. Write the flow equation at each node:
Step-2: The system of equations
\[ \begin{aligned} x _ { 1 } - x _ { 2 } & = - 50 \\ x _ { 2 } - x _ { 3 } + x _ { 4 } - x _ { 5 } & = 0 \\ x _ { 5 } - x _ { 6 } & = 60 \\ x _ { 4 } - x _ { 6 } & = 50 \\ x _ { 1 } - x _ { 3 } & = - 40 \end{aligned} \]
Step-3: The augmented Matrix
\[M = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0&0&{ - 50}\\ 0&1&{ - 1}&1&{ - 1}&0&0\\ 0&0&0&0&1&{ - 1}&{60}\\ 0&0&0&1&0&{ - 1}&{50}\\ 1&0&{ - 1}&0&0&0&{ - 40} \end{array}} \right]\] Row Reduce the matrix: \[\left[ {\begin{array}{*{20}{c}} 1&0 &{ - 1}&0 &0 &0 &{ - 40 }\\ 0 &1&{ - 1}&0 &0 &0 &{10 }\\ 0 &0 &0 &1&0 &{ - 1}&{50 }\\ 0 &0 &0 &0 &1&{ - 1}&{60 }\\ 0 &0 &0 &0 &0 &0 &0 \end{array}} \right]\]
Step-4: The General Solution
\[ \begin{array} { l } { x _ { 1 } = x _ { 3 } - 40 } \\ { x _ { 2 } = x _ { 3 } + 10 } \\ { x _ { 3 } = \text { free } } \\ { x _ { 4 } = x _ { 6 } + 50 } \\ { x _ { 5 } = x _ { 6 } + 60 } \\ { x _ { 6 } = \text { free } } \end{array} \]
Step-5: The minimum flow
Since, all the variables are positive, $x_1$ is also positive : \[ \begin{aligned} x _ { 1 } & \geq 0 \\ x _ { 3 } - 40 & \geq 0 \\ x _ { 3 } & \geq 40 \end{aligned} \] So, for $x_2$ \[ \begin{array} { l } { x _ { 2 } = x _ { 3 } + 10 } \\ { x _ { 2 } \geq 50 \quad \left( \text { since } x _ { 3 } \geq 40 \right) } \end{array} \] Using $x_6>0$ \[ \begin{array} { l } { x _ { 4 } - 50 \geq 0 } \\ { x _ { 4 } \geq 50 } \end{array} \] \[ \begin{array} { l } { x _ { 5 } - 60 \geq 0 } \\ { x _ { 5 } \geq 60 } \end{array} \] Therefore, the minimum flows are:
\[ x _ { 2 } = 50 , x _ { 3 } = 40 , x _ { 4 } = 50 , x _ { 5 } = 60 , x _ { 1 } = 0 , x _ { 6 } = 0 \]
We are given with a flow network. We have to find the general flow pattern in the network. We also have to find the minimum flows in the branches $x_2$, $x_3$, $x_4$ and $x_5$
Step-1:
The basic assumption of network flow is that the total flow into the network equal to the total flow out of network and that total flow into a function equal the total flow out of function. Write the flow equation at each node:
| Node | Flow in | Flow out |
| A | $x_2$+30 = | $x_1$+80 |
| B | $x_3$ + $x_5$ = | $x_2$ + $x_4$ |
| C | $x_6$+100 = | $x_5$ + 40 |
| D | $x_4$ +40 = | $x_6$ + 90 |
| E | $x_1$ +60 = | $x_3$ + 20 |
| Total Flow | 230 = | 230 |
Step-2: The system of equations
\[ \begin{aligned} x _ { 1 } - x _ { 2 } & = - 50 \\ x _ { 2 } - x _ { 3 } + x _ { 4 } - x _ { 5 } & = 0 \\ x _ { 5 } - x _ { 6 } & = 60 \\ x _ { 4 } - x _ { 6 } & = 50 \\ x _ { 1 } - x _ { 3 } & = - 40 \end{aligned} \]
Step-3: The augmented Matrix
\[M = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0&0&{ - 50}\\ 0&1&{ - 1}&1&{ - 1}&0&0\\ 0&0&0&0&1&{ - 1}&{60}\\ 0&0&0&1&0&{ - 1}&{50}\\ 1&0&{ - 1}&0&0&0&{ - 40} \end{array}} \right]\] Row Reduce the matrix: \[\left[ {\begin{array}{*{20}{c}} 1&0 &{ - 1}&0 &0 &0 &{ - 40 }\\ 0 &1&{ - 1}&0 &0 &0 &{10 }\\ 0 &0 &0 &1&0 &{ - 1}&{50 }\\ 0 &0 &0 &0 &1&{ - 1}&{60 }\\ 0 &0 &0 &0 &0 &0 &0 \end{array}} \right]\]
Step-4: The General Solution
\[ \begin{array} { l } { x _ { 1 } = x _ { 3 } - 40 } \\ { x _ { 2 } = x _ { 3 } + 10 } \\ { x _ { 3 } = \text { free } } \\ { x _ { 4 } = x _ { 6 } + 50 } \\ { x _ { 5 } = x _ { 6 } + 60 } \\ { x _ { 6 } = \text { free } } \end{array} \]
Step-5: The minimum flow
Since, all the variables are positive, $x_1$ is also positive : \[ \begin{aligned} x _ { 1 } & \geq 0 \\ x _ { 3 } - 40 & \geq 0 \\ x _ { 3 } & \geq 40 \end{aligned} \] So, for $x_2$ \[ \begin{array} { l } { x _ { 2 } = x _ { 3 } + 10 } \\ { x _ { 2 } \geq 50 \quad \left( \text { since } x _ { 3 } \geq 40 \right) } \end{array} \] Using $x_6>0$ \[ \begin{array} { l } { x _ { 4 } - 50 \geq 0 } \\ { x _ { 4 } \geq 50 } \end{array} \] \[ \begin{array} { l } { x _ { 5 } - 60 \geq 0 } \\ { x _ { 5 } \geq 60 } \end{array} \] Therefore, the minimum flows are:
\[ x _ { 2 } = 50 , x _ { 3 } = 40 , x _ { 4 } = 50 , x _ { 5 } = 60 , x _ { 1 } = 0 , x _ { 6 } = 0 \]