Chap17-calculation-N - Early start(ES The earliest time an activity can start Assumes all preceding activities start as early as possible For nodes with

# Chap17-calculation-N - Early start(ES The earliest time an...

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1 Early start (ES) The earliest time an activity can start Assumes all preceding activities start as early as possible For nodes with one entering arrow » ES = EF of the entering arrow For activities leaving nodes with multiple entering arrows » ES = the largest of the largest entering EF Early finish (EF) The earliest time an activity can finish EF = ES + t Late Start (LS) The latest time the activity can start and not delay the project The latest starting time for each activity is equal to its latest finishing time minus its expected duration: LS = LF - t Late Finish (LF) The latest time the activity can finish and not delay the project For nodes with one leaving arrow, LF for nodes entering that node equals the LS of the leaving arrow For nodes with multiple leaving arrows, LF for arrows entering node equals the smallest of the leaving arrows 2 1 3 2 5 4 4 9 2 5 6 3 EF 1 =0+9=9 ES 3 =9 ES 4 =14 ES 1 =0 EF 3 =9+5=14 EF 1 =0+4=4 EF 2 =4+6=10 EF 2 =4+2=6 EF 4 =14+3=17 For nodes with one leaving arrow, LF for nodes entering that node equals the LS of the leaving arrow For nodes with multiple leaving arrows, LF for arrows entering node equals the smallest of the leaving arrows LF 3 =14 LF 4 =17 LF 1 =11 LF 2 =17 LF 2 =14 LF 1 =9 LS 2 =17-6=11 LS 2 =14-2=12 LS 3 =14 - 5=9 LS 1 =11-4=7 LS 1 =9-9=0 LS 4 =17-3=14 7 0 8 7 0 0 Critical path The critical path is indicated by the activities with zero slack Critical path = 1 – 3 – 4 - 5 Duration =9+5+3 =17 Problem 2/812 ES 2 = 4 ES 1 =0 ES 2 = 4  #### You've reached the end of your free preview.

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