**Problem:**

Given a flow of six (6) units from origin “o” to destination “r”. Flow on each route ab is designated with Qab in the Time Function. Apply Wardrop's Network Equilibrium Principle (Users Equalize Travel Times on all used routes)

**A.** What is the flow and travel time on each link? (complete the table below) for Network A

Link | Link Performance Function | Flow | Time |
---|---|---|---|

o-p | |||

p-r | |||

o-q | |||

q-r |

**B.** What is the system optimal assignment?

**C.** What is the Price of Anarchy?

**Solution:**

**Part A**

What is the flow and travel time on each link? Complete the table below for Network A:

Link | Link Performance Function | Flow | Time |
---|---|---|---|

o-p | |||

p-r | |||

o-q | |||

q-r |

These four links are really 2 links O-P-R and O-Q-R, because by conservation of flow Qop = Qpr and Qoq = Qqr.

Link | Link Performance Function | Flow | Time |
---|---|---|---|

o-p-r | |||

o-q-r |

By Wardrop's Equilibrium Principle, the travel time (cost) on each used route must be equal. Therefore .

OR

By the conservation of flow principle

By substitution

Check

Check (within rounding error)

Link | Link Performance Function | Flow | Time |
---|---|---|---|

o-p-r | 2.84 | 42.01 | |

o-q-r | 3.15 | 42.01 |

or expanding back to the original table:

Link | Link Performance Function | Flow | Time |
---|---|---|---|

o-p | 2.84 | 14.2 | |

p-r | 2.84 | 27.84 | |

o-q | 3.15 | 26.3 | |

q-r | 3.15 | 15.75 |

User Equilibrium: Total Delay = 42.01 * 6 = 252.06

**Part B**

What is the system optimal assignment?

Conservation of Flow:

Analytic Solution requires minimizing total delay

And we can compute the SO travel times on each path

Note that unlike the UE solution,

Total Delay = 3.04(25+ 6*3.04) + 2.96(20+7*2.96) = 131.45+120.53= 251.98

Note: one could also use software such as a "Solver" algorithm to find this solution.

**Part C**

What is the Price of Anarchy?

User Equilibrium: Total Delay =252.06 System Optimal: Total Delay = 251.98

Price of Anarchy = 252.06/251.98 = 1.0003 < 4/3