Figure: Finding the critical path

One: Concept:

Before learning the critical path, understand the concept of a AOV and AOE network: as shown in the following figure

If an automobile production plant is to manufacture a car, the approximate event and activity time of the manufacturing process are as follows: AoE
So, obviously for the AOE network above, the so-called critical path:

Start –> engine complete –> parts in place –> assembly completed. The path length is 5.5. If we try to shorten the entire duration to improve the productivity of the wheels, even changing the 0.1 is useless. The length of the entire duration can be reduced only by shortening the critical activity time on the critical path.

For example, if the manufacturing engine is shortened to 2.5 days and the vehicle assembly is reduced to 1.5 days, the critical path is 4.5. The duration has been shortened by a full day.

All right. So what is the point of this critical path?
Assume that the weight of the arc in the AOE network is in hours, and we already know that the dark Dark One is the critical path. Suppose now one o'clock in the morning, in case of a shell completion event, in order not to affect the duration: Shell completion Activities at the earliest, it started at a point, and it must start at two o ' night. The maximum power value of 3 means that all activities must be completed after three hours, while the shell will take only 2 hours to complete. Therefore, the middle of the idle time has one hours, in order not to affect the entire duration, it must start at the latest two points.

It is then possible to ensure completion of the activity at 3 o'clock with the engine and to prepare for subsequent activities. Two: Related terms: AOV Networks (activity on Vertex Network): A direction graph, a vertex representation of activity, an arc representing the sequence of activities AOE Network (activity on Edge): A graph, with vertices representing events, Using arcs to represent activities, using weights to represent activity consumption time (weighted, forward-free graph) activity : Behavior in business logic, representing events with edges: The result of an activity or the trigger condition critical path : The path with the maximum path length (weight), There may be two properties for more than one activity: E (i) The earliest start time, L (i) two properties of the latest start time event: Ve (j) earliest start time, VL (j) Latest Start time

Three: The process of calculating the critical path: steps: The values of VE (the earliest start time) and VL (the latest start time) for each vertex are calculated first These two values allow you to find the E (earliest start time) and L (latest start time) values for each edge. take E (i) =l (i) is the edge of the critical path, connected, you can get the critical path (the key path may be more than one) ①: The value of the VE (j) (the earliest event start time) from the front backward, The VE value of the direct precursor node + the weight of the edge of the current node (possibly multiple, whichever is maximum) the first vertex has ve equals 0

Vertex	V1	V2	V3	V4	V5	V6	V7
VE (j)	0	3	2	6	7	5	10

②: The value of the VL (j) (the event's latest start time)From the back forward (V9), the VL value of the immediate successor node-the weight of the edge of the current node (there may be multiple, take the minimum) the VL of the endpoint is equal to its VE

Vertex	V1	V2	V3	V4	V5	V6	V7
VL (J)	0	3	3	6	7	6	10

③: The value of E (i) (activity earliest start time)E (i): The active AI is represented by arc < VK,VJ >, then the earliest start time of the activity should be equal to the earliest occurrence of the event VK, so there is E (i) =ve (k).
That is, the earliest start time of an edge (activity) equals the earliest occurrence of the vertex (event) it emits

side	A1 (3)	A2 (6)	A3 (2)	A4 (4)	A5 (2)	a6 (1)	A7 (3)	A8 (1)	A9 (3)	A10 (4)
E (i)	0	0	0	3	3	2	2	6	7	5

④: The value of L (i) (the latest start time of the activity)L (i): The active AI is represented by arc < VK,VJ >, then the last occurrence of the AI to ensure that the latest occurrence of VJ is not delayed (VJ at the latest time is 9, the latest AI must be 9-activity time). Therefore, L (i) =VL (i)-len< vk,vj >
That is, the latest occurrence of the activity reaching the vertex minus the weight of the edge

side	A1 (3)	A2 (6)	A3 (2)	A4 (4)	A5 (2)	a6 (1)	A7 (3)	A8 (1)	A9 (3)	A10 (4)
L (i)	0	0	1	3	4	5	3	6	7	6

⑤: Find key edges and critical paths:

When e (i) ==l (i), i.e.: Activity earliest start time = activity Latest start time, the key edges can be obtained as:
A1 A2 A4 A8 A9

Then combine the critical paths according to the key edges:
A1->a4->a9 and A2->A8->A9