Math Problem Statement

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?

Solution

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 cijc_{ij} depend on the flow xx (number of cars per unit time) as follows:

    • c13=10xc_{13} = 10x
    • c14=50+xc_{14} = 50 + x
    • c32=50+xc_{32} = 50 + x
    • c34=10+xc_{34} = 10 + x
    • c42=10xc_{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:

  • x1x_1 be the flow on route 1-3-2,
  • x2x_2 be the flow on route 1-3-4-2,
  • x3x_3 be the flow on route 1-4-2.

These flows must satisfy the total flow condition: x1+x2+x3=6x_1 + x_2 + x_3 = 6

Step 2: Calculate travel times for each route

  1. Route 1 → 3 → 2:

    • Travel time: c13+c32=10x1+(50+x1)=50+11x1c_{13} + c_{32} = 10x_1 + (50 + x_1) = 50 + 11x_1
  2. Route 1 → 3 → 4 → 2:

    • Travel time: c13+c34+c42=10x2+(10+x2)+10x2=10x2+10+x2+10x2=10+21x2c_{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: c14+c42=(50+x3)+10x3=50+x3+10x3=50+11x3c_{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+11x1=10+21x2=50+11x350 + 11x_1 = 10 + 21x_2 = 50 + 11x_3

We now have two equations to solve:

  1. 50+11x1=10+21x250 + 11x_1 = 10 + 21x_2
  2. 50+11x1=50+11x350 + 11x_1 = 50 + 11x_3

We can solve these equations to find x1x_1, x2x_2, and x3x_3.

I will now compute the values of the flow on each arc.The flow distribution across the routes is as follows:

  • x1=86531.62x_1 = \frac{86}{53} \approx 1.62 cars per unit time on route 1 → 3 → 2.
  • x2=146532.75x_2 = \frac{146}{53} \approx 2.75 cars per unit time on route 1 → 3 → 4 → 2.
  • x3=86531.62x_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 x1x_1, x2x_2, or x3x_3 into any of the travel time equations (since they are equal in equilibrium). Using x1x_1:

Travel time=50+11x1=50+11×865367.85 units of time.\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 c14=30+xc_{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.

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Math Problem Analysis

Mathematical Concepts

Optimization
Graph Theory
Systems of Equations

Formulas

c_13 = 10x
c_14 = 50 + x
c_32 = 50 + x
c_34 = 10 + x
c_42 = 10x
x_1 + x_2 + x_3 = 6

Theorems

Wardrop's Equilibrium Principle

Suitable Grade Level

University Level