Intersections in England are often constructed as one-way "roundabouts," such as the one shown in the figure. Assume that traffic must travel in the directions shown. Find the general solution of the network flow. Find the smallest possible value for X6- x₂ = X3 = X4= 30➤ X5 = X is free 100 x₁ is free x2 = X3 is free X4 is free 120 130 CIND xs X5 = x6 is free X₁ What the general solution of the network flow? Choose the correct answer below and fill in the answer boxes to complete your choice. O A. O B. O C. X, is free E 1x6 F 40 100 x₁ = x2 = x3 = X4 = X5 = xe is free X6 O D. X₁ X₂ is free x3 = X5 is free X is free

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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### Educational Content: Network Flow in Roundabouts

Intersections in England are often constructed as one-way "roundabouts," such as the one shown in the figure. Assume that traffic must travel in the directions shown. Find the general solution of the network flow. Find the smallest possible value for \( x_6 \).

#### Diagram Explanation:

The diagram illustrates a roundabout system with six junctions labeled \( A \) through \( F \). Each junction connects with one another via paths labeled \( x_1 \) through \( x_6 \). The traffic flow values at each entry and exit point are indicated as follows:
- From \( A \) to \( B \), the flow is 100.
- From \( B \) to \( C \), the flow is 30.
- From \( C \) to \( D \), the flow is 120.
- From \( D \) to \( E \), the flow is 130.
- From \( E \) to \( F \), the flow is 40.
- Back to \( F \), the flow is 100.

Each path (\( x_1 \) through \( x_6 \)) represents a section of the roundabout where traffic flow is managed.

#### Problem Statement:

What is the general solution of the network flow? Choose the correct answer below and fill in the answer boxes to complete your choice.

#### Options:

- **Option A:**
  \[
  \begin{align*}
  x_1 & \text{ is free} \\
  x_2 & = \\
  x_3 & = \\
  x_4 & = \\
  x_5 & = \\
  x_6 & \text{ is free} \\
  \end{align*}
  \]

- **Option B:**
  \[
  \begin{align*}
  x_1 & \text{ is free} \\
  x_2 & \text{ is free} \\
  x_3 & = \\
  x_4 & \text{ is free} \\
  x_5 & = \\
  x_6 & \text{ is free} \\
  \end{align*}
  \]

- **Option C:**
  \[
  \begin{align*}
  x_1 & = \\
  x_2 & = \\
  x_3
Transcribed Image Text:### Educational Content: Network Flow in Roundabouts Intersections in England are often constructed as one-way "roundabouts," such as the one shown in the figure. Assume that traffic must travel in the directions shown. Find the general solution of the network flow. Find the smallest possible value for \( x_6 \). #### Diagram Explanation: The diagram illustrates a roundabout system with six junctions labeled \( A \) through \( F \). Each junction connects with one another via paths labeled \( x_1 \) through \( x_6 \). The traffic flow values at each entry and exit point are indicated as follows: - From \( A \) to \( B \), the flow is 100. - From \( B \) to \( C \), the flow is 30. - From \( C \) to \( D \), the flow is 120. - From \( D \) to \( E \), the flow is 130. - From \( E \) to \( F \), the flow is 40. - Back to \( F \), the flow is 100. Each path (\( x_1 \) through \( x_6 \)) represents a section of the roundabout where traffic flow is managed. #### Problem Statement: What is the general solution of the network flow? Choose the correct answer below and fill in the answer boxes to complete your choice. #### Options: - **Option A:** \[ \begin{align*} x_1 & \text{ is free} \\ x_2 & = \\ x_3 & = \\ x_4 & = \\ x_5 & = \\ x_6 & \text{ is free} \\ \end{align*} \] - **Option B:** \[ \begin{align*} x_1 & \text{ is free} \\ x_2 & \text{ is free} \\ x_3 & = \\ x_4 & \text{ is free} \\ x_5 & = \\ x_6 & \text{ is free} \\ \end{align*} \] - **Option C:** \[ \begin{align*} x_1 & = \\ x_2 & = \\ x_3
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