Algorithm research: AOE Network and key Path introduction

As we said earlier, topology sequencing is primarily about solving the problem of whether a project can be sequenced, but sometimes we need to solve the problem of the shortest time required to complete the project. If we want to get the shortest time for a flowchart, we have to analyze their topological relationships and find the most critical process, which is the shortest time.

On the basis of the AOV network, we introduce a new concept. In a weighted direction diagram of a representation project, an event is represented by a vertex, an activity is represented by a forward edge, and the duration of the activity is represented by a weighted value on the edge, and the edge of the graph represents the active net, called the AoE net (activity on edge Network). As a result of a project, there is always a beginning, an end, under normal circumstances, AOE net only one source point of a meeting point.

Since the AOE network represents the engineering process, it has obvious engineering properties. Only after an event represented by a vertex occurs does the activity from that vertex begin. Events that are represented by the vertex can occur only if each activity that enters a vertex has ended.

Although AoV and AOE networks are used to model engineering, but they are still very different, mainly reflected in the AOV network is the vertex to represent the activity of the network, it only describes the constraints between activities, and AOE network is used to represent the activity of the network, the edge of the weight of the duration of the activity, As shown in Figure 7-9-3, the comparison of two graphs. Therefore, the AOE network is to establish in the activities between the constraints on the basis of no contradictions, and then to analyze how long it takes to complete the project, or to shorten the time required to complete the project, which activities should be speeded up and so on.

We call the path length after the duration of each activity on the path, and the path from the source point to the meeting point with the maximum length is called the critical path, and the activity completed on the critical path is called the key activity. Obviously on the AoE net of the figure 7-9-3, start-> the engine completes-> the assembly is in place-> Assemble completes is the key path, the path length is 5.5.

If we need to shorten the entire duration to improve the productivity of the wheel, even if the change to 0.1 is not conducive to the entire duration of changes, only to shorten the critical path of critical activity time to reduce the length of the entire duration. For example, if the engine manufacturing is shortened to 2.5 and the whole vehicle assembly is shortened to 1.5, then the critical path is 4.5, which shortens the whole day.

If the earliest start time of an activity is the same as the latest start time, indicating that there is no gap in the middle, this activity is a key activity. To do this, we need to define the following several parameters.

1. The earliest occurrence of the event ETV (earliest time of vertex): The earliest occurrence of the vertex VK.

2, the latest occurrence of the event LTV (latest time of vertex): that is, the shortest occurrence of the vertex VK. That is, the time at which the event needs to start at the latest for each vertex, exceeding this time will delay the entire duration.

3, the earliest activity ete (earliest time of edge): That is, the earliest occurrence of the arc AK time.

4, the activity of the latest start time LTE (Latest times of Edge): That is, the latest time of the arc AK, that is, do not postpone the late start of the construction time.

We first get 1 and 2, and Ete is originally to represent the active <VK, vj> the earliest start time, is for the arc, but only the arc of the ARC vertex VK event occurred, it can start, so ete = Etv[k].

And LTE is the activity <VK, vj> the latest start time, but this activity can not wait until the VJ event occurred before the start, so LTE = Ltv[j]-LEN<VK, vj>.

In the end, let's judge whether Ete and LTE are equal, meaning that the activity is not free, it is a critical path, otherwise it is not.

Now let's talk about how to seek ETV and LTV.

Assuming we have now obtained vertex v0 corresponding etv[0] = 0, vertex v1 corresponding etv[1] = 3, vertex v2 corresponding etv[2] = 4, now we need to find the vertex v3 corresponding ETV [3], in fact, is to seek etv[1] + LEN<V1, v3> With etv[2] + len<v2, v3> 's larger value. Obviously 3+5 < 4+8, get etv[3] = 12, as shown in Figure 7-9-5.