1. Traffic assignment

2. Transit person trip table (O&D) Vehicle trip table (O&D)
Trip generation
Trip distribution
Mode split
Transit person trip table (O&D)
Vehicle trip table (O&D)
Trip assignment
Loaded transit network
Loaded highway network

3. Shortest path-min. tree building
what is a tree?
If for each origin node h, at the completion of a pass through the algorithm all other nodes in the network are arrived at by one link only, the set of paths from h to the nodes is called “a tree”. The process of determining the minimum cost path is called “tree building”. The tree together with the minimum path costs from the origin node to all destination nodes is referred to as “skimmed tree” (giving rise to the term “skim table” in TDF)

4. An example tree

5. Dijkstra’s algorithm function Dijkstra(Graph, source):
for each vertex v in Graph: // Initializations
dist[v] := infinity // Unknown distance function from source to v
previous[v] := undefined // Previous node in optimal path from source
dist[source] := 0 // Distance from source to source
Q := the set of all nodes in Graph // All nodes in the graph are unoptimized - thus are in Q
while Q is not empty: // The main loop
u := vertex in Q with smallest dist[]
if dist[u] = infinity:
break // all remaining vertices are inaccessible from source
remove u from Q
for each neighbor v of u: // where v has not yet been removed from Q.
alt := dist[u] + dist_between(u, v)
if alt < dist[v]: // Relax (u,v,a)
dist[v] := alt
previous[v] := u
return dist[]
In the following algorithm, the code u := vertex in Q with smallest dist[], searches for the vertex u in the vertex set Q that has the least dist[u] value. That vertex is removed from the set Q and returned to the user. dist_between(u, v) calculates the length between the two neighbor-nodes u and v. The variable alt on line 13 is the length of the path from the root node to the neighbor node v if it were to go through u. If this path is shorter than the current shortest path recorded for v, that current path is replaced with this alt path. The previous array is populated with a pointer to the "next-hop" node on the source graph to get the shortest route to the source.

6. Dijkstra’s algorithm in plain words
1. to initialize
the origin is the only solved node.
The distance to the origin is 0.
The solved set is empty except the origin node.
2. find all the unsolved nodes that are directly connected by a single arc to any solved node. For each unsolved node, calculate all of the “candidate distances” (there may be several of these for one unsolved node because it may directly connect to more than one solved node):
choose an arc connecting the unsolved node directly to a solved node.
The “candidate distances” is (the distance to solved node) + (length of arc directly connecting the solved and the unsolved node).
3. choose the smallest candidate distance:
add the corresponding unsolved node to the solved set
distance to the newly solved node is the candidate distance
add the corresponding arc to the arc set
4. if the newly solved node is not the destination node, then go to step 2. Else, stop and recover the solution.
length of the shortest route from origin to destination is the distance to the destination node.
Shortest route is found by tracing backwards from the destination node to the origin (or the reverse) using the arcs in the arc set. It is usually easiest to trace backwards because each node is reached from exactly one other node, but may have outwards arcs to several other nodes.

7. An example
Find the shortest paths from node 1 to all other nodes in the network 1 6 2 1 4 2 3 2 1 3 5 6

8. Example solution Step 1: 0 Step 4: 6 1 6 2 1 4 Step 3: 5 2 3 2 1 3 5 6
Solved node list
Dist fr O. to node
Solved arc set 1 3 2
1-3 5
3-2 4 6
2-4 7
4-6 8
4-5

9. Another example Find the shortest paths from node A to all other nodesB C D E F G H 3 5 2 13 6 4

10. Another example Find the shortest paths from node 6 to all other nodes2 1 2 5 3 5 3 4 3 4 6 3 3 3 8 5

11. Inputs to assignment
O-D matrix
Network

12. Terminology (1)
Trip assignment: “loading the network”; volumes are assigned to links
Free flow speed: speed under no congestion
Free flow travel time: travel time under no congestion
Path finding: finding the path with the minimum impedance
Path loading: loading vehicles to links comprising a path
Level of service: a qualitative measure describing the operation conditions
Capacity restraint: the volume loading process is constrained by the capacity of the link

13. Assignment methods Non-equilibrium (heuristic) assignments
All or nothing: link travel times are determined beforehand and trips are assigned at once
Incremental assignment: link travel times are updated through fixed proportions and trips are assigned iteratively
Capacity constraint: link travel times depend on link volumes and final assignment is the average of the last several iterations

14. All or nothing assignment
Simple all-or-nothing method is the fundamental building block in
traffic assignment procedures.
Step 1: find shortest paths between origin and destination zones
Step 2: assign all trips to links comprising the shortest routes

15. Incremental assignment
A simple but inconsistent way to account for capacity and
congestion effects.
Step 1: identify shortest paths between origin and destination zones
Step 2: assign a fixed portion of the trips to links comprising the shortest routes
Step 3: if all assigned, stop, otherwise, continue to step 4
Step 4: update link travel times

16. Example
Consider a simple transportation network that has one origin and one destination and two links/paths that provide access from the origin to the destination. One link is 7.5 miles long and has a capacity of 4000 vehicles per hour and a speed limit of 55 miles per hour. The other link is 5 miles long and has a capacity of 2000 vehicles per hour and a speed limit of 35 miles per hour. Assuming that 5000 drivers wish to make the trip from the origin to the destination, find the loaded network?

17. All or nothing assignment
Link 1
capacity: 4000 vehicles; speed = 55 mph; distance = 7.5 miles
Free-flow travel time = 7.5/55 = 8.18 minutes
Dest.
Link 1
origin
Link 2
Link 2
capacity: 2000 vehicles; speed = 35 mph; distance = 5 miles
Free-flow travel time = 5/35 = 8.57 minutes
All-or-nothing suggests that link 1 is the shortest path from origin to destination and will
be assigned with all 5000 vehicles and link 2 will be assigned 0 vehicles

18. Incremental assignment
Link 1
capacity: 4000 vehicles; speed = 55 mph; distance = 7.5 miles
Free-flow travel time = 7.5/55 = 8.18 minutes
Link 1
Dest.
origin
Link 2
Link 2
capacity: 2000 vehicles; speed = 35 mph; distance = 5 miles
Free-flow travel time = 5/35 = 8.57 minutes
Let us first assign 1000 vehicles to link 1 and then update link travel time, which
will be:
The next 1000 vehicles will still be assigned to link 1, which gives a travel time of The next 1000 vehicles will still be assigned to link 1, which results in 9.48 min.
Thus, link 2 now becomes the shortest path, and therefore, the next 1000 vehicles to link 2, which gives 8.76 min. Thus, the last 1000 vehicles will be loaded on link 2, which will give a travel time of:

19. Example: do all-or-nothing and incremental assignment2
Link 4 4
Link 1 1
Link 3
Link 5
Link 2 3
From\to 1 2 3 4
100 50
Observed O-D flow
Link characteristics
link 1 2 3 4 5
t0l 10 15 Cl
300
500
150
200