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1 Pertemuan 26 Activity Network Matakuliah: T0026/Struktur Data Tahun: 2005 Versi: 1/1

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2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat menjelaskan konsep aplikasi graph dalam activity network

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3 Outline Materi Activity on vertex (AOV) network AOV network(cont.) An AOV network Topology sort Example of topology sort Activity on edge (AOE) network Earliest time Latest time Other factors Critical paths Calculation of earliest times Calculation of Latest times Early and late and critical values

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4 Activity on vertex (AOV) network –An activity on vertex(AOV)network, is a digraph G in which the vertices represent tasks or activities and the edges represent precedence relations between tasks. Predecessor : –Vertex i in an AOV network G is a predecessor of vertex j iff there is a directed path from vertex i to vertex j –Vertex i is an immediate predecessor of vertex j iff is an edge in G Successor : –If i is a predecessor of j, then j is a successor of i. –If i is an immediate predecessor of j, then j is an immediate successor of i

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5 AOV network(cont.) Transitive –A relation " ‧ " is transitive iff for all triples i, j,k, i ‧ j and j ‧ k => i ‧ k Irreflexive –A relation" ‧ "is irreflexive on a set S if i ‧ i is false for all elements, i, in S. Partial order –A partial order is a precedence relation that is both transitive and irreflexive Topological order –A topological order is a linear ordering of the vertices of a graph such that, for any two vertices, i, j, if i is a predecessor of j in the network then i precedes j in the linear ordering.

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6 An AOV network

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7 Topology sort for (i = 0; i

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8 Example of topology sort Topological order generated : V0,V3,V2,V5,V1,V4

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9 Activity on edge (AOE) network AOE network –Edge represents task or activity to be performed on a project –Vertex represents event which signal the completion of certain activities Remarks on AOE –Determine the minimum amount of time required to complete a project –An assessment of the activities whose duration should be shortened to reduce the overall completion time Evaluate performance –activity whose duration time should be shortened –minimum amount of time Critical path –A path that has the longest length –Minimum time required to complete the project

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10 Earliest time The earliest time an event, vi, can occur is the length of the longest path from the start vertex v0 to vertex vi –i.e., the earliest start time for all activities represented by edges leaving that vertex –denote the time by e(i) for activity ai Example : v4 is 7, e(6)= e(7)= 7

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11 Latest time Latest time : –l(i) of activity, ai, is the latest time the activity may start without increasing the project duration Projection duration : –length of the longest path from start to finish Example : e(5)= 5 and l(5)= 8, e(7)= 7 and l(7)= 7

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12 Other factors Critical activity : –An activity for which e(i)= l(i) Remarks on critical : –The difference between l(i) and e(i) is a measure of how critical an activity is –Identify critical activities so that we may concentrate our resources to reduce a project's duration Determine critical paths : –Delete all noncritical activities –Generate all the paths from the start to finish vertex.

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13 Critical paths Critical paths: v0, v1, v4, v7, v8, length= 18 ; v0, v1, v4, v6, v8, length= 18

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14 Calculation of earliest times earliest[j]: the earliest event occurrence time for all event j in the network. latest[j]: the latest event occurrence time for all event j in the network. If activity ai is represented by edge –early(i) = earliest[k] –late(i) = latest[l] - duration of activity ai We compute the times earliest[j] and latest[j] in two stages: a forward stage and a backward stage. During the forwarding stage, we start with earliest[0] = 0 and compute the remaining start times using the formula: –earliest[j] = max{ earliest[i] + duration of }

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15 Calculation of Latest times In the backward stage, we compute the values of latest[i] using a procedure analogous to that used in the forward stage. We start with latest[n-1] = earliest[n-1] and use the equation: –latest[j] = min{ latest[i] - duration of }

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Copyright © 2008 Pearson Education, Inc. Slide 13-1 Unit 13C Scheduling Problems.

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