AOE Example

The values are computed as follows:

1. Assign the start node an early time of 0.

2. Process the nodes in a topological order to compute the other early times. These are the earliest time each node can be reached.

   For each node, compute for each predecessor the sum of its early time and the weight of the arc joining the nodes. The largest of these sums is the early time of the node under consideration. This follows the notion that you do not reach a node until you have completed all the tasks that must come before it.

   For instance, node 8 sees the sums 32 + 5 = 37, 33 + 6 = 39, and 24 + 10 = 34. The early time of node 8 is 39.

3. Assign the final node a late time equal to its early time.

4. Process the nodes in a reverse topological order to compute the other late times. These are the latest time each node can be reached which will not delay the final completion.

   For each node, compute for each successor the difference of its late time and the weight of the arc joining the nodes. The smallest of these sums is the early time of the node under consideration. This follows the notion that you must reach a node soon enough not to delay any of its successors.

   For instance, node 5 sees 24 - 8 = 16 and 24 - 10 = 14. Its early time is 14.

5. A path from the start to the end in which each node has equal early and late times is a critical path. There may be more than one critical path. If any node on a critical path is delayed, project completion will be delayed. This graph has only one critical path: 4, 5, 10, 7, 8.