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This problem involves finding the equilibrium flow distribution in a road network, where drivers aim to minimize their own travel time, and as a result, the travel times across different routes will equalize in equilibrium. Let’s break this down step by step.

### Given Data:
- Travel times \(c_{ij}\) depend on the flow \(x\) (number of cars per unit time) as follows:
  - \( c_{13} = 10x \)
  - \( c_{14} = 50 + x \)
  - \( c_{32} = 50 + x \)
  - \( c_{34} = 10 + x \)
  - \( c_{42} = 10x \)
  
- There is a flow of 6 cars per unit time from node 1 to node 2.

- The available routes from node 1 to node 2 are:
  1. 1 → 3 → 2
  2. 1 → 3 → 4 → 2
  3. 1 → 4 → 2

### Objective:
We need to find the flow along each route such that the travel times for each route are equal. Also, we are asked to determine the total travel time for all drivers.

#### Step 1: Define the flow on each route
Let:
- \( x_1 \) be the flow on route 1-3-2,
- \( x_2 \) be the flow on route 1-3-4-2,
- \( x_3 \) be the flow on route 1-4-2.

These flows must satisfy the total flow condition:
\[
x_1 + x_2 + x_3 = 6
\]

#### Step 2: Calculate travel times for each route

1. **Route 1 → 3 → 2:**
   - Travel time: \( c_{13} + c_{32} = 10x_1 + (50 + x_1) = 50 + 11x_1 \)

2. **Route 1 → 3 → 4 → 2:**
   - Travel time: \( c_{13} + c_{34} + c_{42} = 10x_2 + (10 + x_2) + 10x_2 = 10x_2 + 10 + x_2 + 10x_2 = 10 + 21x_2 \)

3. **Route 1 → 4 → 2:**
   - Travel time: \( c_{14} + c_{42} = (50 + x_3) + 10x_3 = 50 + x_3 + 10x_3 = 50 + 11x_3 \)

#### Step 3: Set travel times equal to each other

For equilibrium, the travel times must be equal:
\[
50 + 11x_1 = 10 + 21x_2 = 50 + 11x_3
\]

We now have two equations to solve:
1. \( 50 + 11x_1 = 10 + 21x_2 \)
2. \( 50 + 11x_1 = 50 + 11x_3 \)

We can solve these equations to find \( x_1 \), \( x_2 \), and \( x_3 \).

I will now compute the values of the flow on each arc.The flow distribution across the routes is as follows:
- \( x_1 = \frac{86}{53} \approx 1.62 \) cars per unit time on route 1 → 3 → 2.
- \( x_2 = \frac{146}{53} \approx 2.75 \) cars per unit time on route 1 → 3 → 4 → 2.
- \( x_3 = \frac{86}{53} \approx 1.62 \) cars per unit time on route 1 → 4 → 2.

#### Step 4: Calculate the travel time for all drivers
We can substitute \(x_1\), \(x_2\), or \(x_3\) into any of the travel time equations (since they are equal in equilibrium). Using \(x_1\):

\[
\text{Travel time} = 50 + 11x_1 = 50 + 11 \times \frac{86}{53} \approx 67.85 \text{ units of time}.
\]

Thus, the travel time for all drivers is approximately 67.85 units of time.

---

Would you like more details on any part of the process? Here are 5 follow-up questions for further exploration:

1. How would the solution change if the total flow increased to 10 cars per unit time?
2. What are the implications of adjusting one of the travel time functions, say \( c_{14} = 30 + x \), on the equilibrium?
3. Can this method be generalized to larger networks with more nodes and routes?
4. What happens if drivers do not act independently and cooperate to minimize total travel time?
5. How would the flow distribution change if route 1-3-4-2 were closed?

**Tip:** In equilibrium problems like this, equalizing travel times along different paths helps achieve the optimal flow distribution across the network.

As each driver tries to minimize his travel time independently from the other drivers, all possible routes (namely 1-3-2, 1-3-4-2, and 1-4-2) will have a flow density such that the route lengths (travel times) are equal. Why? Knowing this, how much flow will be on each arc, and what is the travel time for all drivers?

Explore the application of optimization and graph theory to find the optimal flow of cars in a network where travel times are minimized. This advanced problem involves systems of equations and Wardrop's Equilibrium Principle, suitable for university-level studies.

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