B 10 WEEKS A D 5 WEEKS 8

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B 10 WEEKS A D 5 WEEKS 8 WEEKS E 6 WEEKS C 12

B 10 WEEKS A D 5 WEEKS 8 WEEKS E 6 WEEKS C 12 WEEKS ES First, have We The figure we calculate a in network the lower early withleft positions five corner activities: for shows all. A, of the B, the different C, activities. D andvalues E. These Thethatshow are activities calculated the earliesthave instart the different points network. (ES) durations and earliest that arefinish shown points in the (EF). figure. For example, activity A has a duration of 5 weeks. EF ACTIVITY FL (t) LS LF Http: //www. prosjektledelse. ntnu. no Prosjektstyring

5 15 B 0 10 WEEKS 5 17 A 25 D 5 WEEKS 5

5 15 B 0 10 WEEKS 5 17 A 25 D 5 WEEKS 5 17 8 WEEKS 25 31 E 6 WEEKS C 12 WEEKS ES First, By The Activity looking duration earliest we E Btake C therefore has at start time finish of the aactivity duration point starting network, forhas activity (ES) that D anof point iswe activity earliest for 10 12 8 Esee weeks, weeks in Aisis activity that week start D 0, can so and the point 31 its start A its’ earliest we because earliest and in earliest can onweek assume is time directly finish thefinish 25. point duration maximum that isfind 25. isfor it in starts the ofof in the earliest activity week the earliest activity 15. 17. timepoint finish A gives finish is 6(EF) weeks. the 0. values by earliest adding for time Btoand itpoint the C. Therefore, duration that bothofactivities activity A: DBstarts and Con 17 + 8 = 25 can start week 17. on. 50 ++12 25 56==531 10 15 17 EF ACTIVITY FL (t) LS LF Http: //www. prosjektledelse. ntnu. no Prosjektstyring

5 15 B 0 10 WEEKS 5 7 17 A 5 WEEKS 0 17

5 15 B 0 10 WEEKS 5 7 17 A 5 WEEKS 0 17 5 12 WEEKS 5 17 31 E 8 WEEKS 17 C 25 D 17 5 25 6 WEEKS 25 25 31 The latest Additionally, Since Activity Now finish we 17 C D B can have isstart date has the calculate subtract by two alatest for ofduration looking the values activity time to project backwards find at of for point Athe 10 12 8 the activity isweeks, network weeks 0 not atlatest and in which given, the A the which start and that network structure activity duration so has the for can gives we latest aallgive latest assume to D we activities. is us find can start 5. can us astart latest start, the that see point latest that start the both is of 7. 5. position latest activity point activities finish. of finish We D 17. for Bat should and point all hasactivities. Cais choose should latest equalfinish This be the to finished the lowest means point earliest in value, by that week finish this we which 25. time. will point. calculate is. Ifvalue Thus, either 5. we the activity set latest the For 517 - -5 activity 10 12 = 0= 57 E, which has a duration of 6 weeks, the latest start is start latest B or points Cfinish are finished and equal latest tolater, 31. finish thepoints. entire project does not finish in week 31. week 25 - 825. = 17 31 - 6 = 25 ES EF ACTIVITY FL (t) LS LF Http: //www. prosjektledelse. ntnu. no Prosjektstyring

5 15 B 0 10 WEEKS 5 7 A 2 17 17 D 0

5 15 B 0 10 WEEKS 5 7 A 2 17 17 D 0 5 WEEKS 0 5 17 C 0 12 WEEKS 5 ES 25 31 E 0 8 WEEKS 17 5 25 0 6 WEEKS 25 25 31 For the floatthe becomes 0. the different activities in the Nowactivity we will. A, E, calculate D, C, B, the becomes float of 0. 2. network. We find the floats by looking at the differences between the 5 - -515 = 0= 20 31 25 17 latest finish points (LF) and the earliest finish points (EF). 17 EF ACTIVITY FL (t) LS LF Http: //www. prosjektledelse. ntnu. no Prosjektstyring

5 15 B 0 10 WEEKS 5 7 A 2 17 17 D 0

5 15 B 0 10 WEEKS 5 7 A 2 17 17 D 0 5 WEEKS 0 5 17 C ES EF ACTIVITY 31 E 0 0 6 WEEKS 25 25 31 0 12 WEEKS 5 25 8 WEEKS 17 5 25 17 This means that all activities in the network are critical, except activity B. Therefore, we find a chain of critical activities A, C , D and E, which form a critical path in the network. At the project start at time 0, the project will have a total duration of 31 weeks. FL (t) LS LF Http: //www. prosjektledelse. ntnu. no Prosjektstyring