(12)

1. D (9) FS0 FF7 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) SF21 FF5 SS3 E SS8 FS7 (18) FF3 G I FF3 (11) (7) 0 12 7 17 7 19 To find the earliest start time for B, we subtract its duration (10), which gives us a value of 7. In activity B, we have a FF5 dependency. This means that in order to find the end time of B,we start with the finish time of A (12), and add the lag time of 5. This gives us an end time of 17 for activity B. A has a duration of 12, so the earliest finish time for A is 12. We then subtract from this the duration of C (12) to get an earliest start of 7. In reality, B and C can start simultaneously, but in the network they are drawn consecutively. Activity C has just one dependency: an SF12 from B. This means that the time from the start of B to the end of C takes 12 time units. The earliest finish time of C is therefore 19. We start with a value of 0 for activity A. We have been given a network as shown, and we will calculate the values for all of its activities. Http://www.prosjektledelse.ntnu.no

2. D (9) FS0 FF7 0 12 7 17 7 19 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) SF21 FF5 SS3 E SS8 FS7 (18) FF3 G I FF3 (11) (7) 19 28 10 28 Activity D has a dependency of FS0 from activity C. The start time of activity D will be equal to the finish time of activity C with an addition of 0, which equals 19. Alternative 1: SF21 from C The start point of C (7) plus 21 gives us 28 as the earliest finish point for activity E. We subtract the duration of E (18) from this value and we get a value of 10 as the earliest start point for activity E. We see that there are two dependencies for activity E, so we should find out which one gives us a higher value. Alternative 1 SF21 from C ES = 10 Alternative 2 SS3 from A ES = 3 10 is the larger value from dependency SF21 from activity C. For activity E, the earliest start is 10 and the earliest finish 28. Alternative 2: SS3 from A The start point of activity A (0) plus 3 gives us a value of 3 for the earliest start date for activity E. The earliest finish time of D equals its start date plus its duration (9), which gives us a value of 28. Http://www.prosjektledelse.ntnu.no

3. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 SF21 FF5 SS3 E SS8 FS7 (18) FF3 G I FF3 (11) (7) 24 35 Alternative 2: FS0 from C The start point is the finish date of activity C, which is linked to the start date of F with an addition of 0. This gives us 19 plus 0, or 19 as the earliest start time. We add to this value the duration of F (11) and we get 30 for the earliest finish time. Alternative 1: FF7 from D The start point is the finish date of activity D, which is linked to the finish date of F with an addition of 7. This gives us 28 plus 7, or 35 as the earliest finish time for F. From this value we subtract 11, which gives us a value of 24 for the earliest start time for F. Alternative 1: FF7 from D ES = 24 Alternative 2: FS0 from C ES = 19 Alternative 3: FF5 from E ES = 22 We should choose the highest value. Therefore, the dependency FF7 from activity D becomes decisive. For activity F, the earliest start point is 24 and earliest finish 35. Alternative 3: FF5 from E The start point is the finish date of activity E, which is linked to the finish date of F with an addition of 5. This gives us 28 plus 5, or 33 as the earliest finish date for F. From this value we subtract 11 to get a value of 22 as the earliest start time for F. Activity F has three dependencies and we should calculate all three alternatives. Http://www.prosjektledelse.ntnu.no

4. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 24 35 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 SF21 FF5 SS3 E SS8 FS7 (18) FF3 G I FF3 (11) (7) 24 35 Alternative 1: FS7 from B The start point is the finish date of B, which is locked to the start date of G with a lag of 7. This gives us 17 plus 7, or a total of 24 as the earliest start date for G. We add this amount to its duration (11) to get an earliest finish time of 35. Alternative 1: FS7 from B ES = 24 Alternative 2: FF3 from E ES = 20 We should choose the highest value, so the dependency FS7 from activity B becomes decisive for G. For activity G, the earliest start time is 24 and the earliest finish 35. Alternative 2: FF3 from E The start point is the finish date of E ,which is locked to the finish date of G with a lag of 3. This gives us 28 plus 3, or a total of 31 as the earliest finish time for G. From this amount we subtract the duration of G (11) which gives us a value of 20 as the earliest start date for G. For activity G there are two dependencies and we should again calculate both alternatives. Http://www.prosjektledelse.ntnu.no

5. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 24 35 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 SF21 FF5 SS3 E SS8 FS7 (18) 24 35 FF3 G I FF3 (11) (7) 32 42 32 39 Alternative 1: FF3 from G The start point is the finish date of G, which is locked to the finish date of I with a lag of 3. This gives us 35 plus 3, or a value of 38 as the earliest finish date for I. We then subtract the duration of I (7) from this amount, which gives us an earliest start date of 31 for I. For activity I there are two dependencies and we should once again calculate both alternatives. Alternative 1: FF3 from G ES = 31 Alternative 2: SS8 from F ES = 32 We should choose the highest value, which comes from dependency SS8 from activity F. For activity I, the earliest start date is 32 and the earliest finish is 39. Alternative 2: SS8 from F The start point is the start date of F which is locked to the start date of I with a lag of 8. This gives us 24 plus 8, or a value of 32 as the earliest start date for G. We then add to this amount its duration (7), which gives us an earliest finish date of 39. Activity H has just one dependency, SS8 from activity F. We take the start date of F (24) and add 8, which gives us an earliest start time of 32 for H. The earliest finish date is 32 plus the duration of the activity (10), which gives us a total of 42. Http://www.prosjektledelse.ntnu.no

6. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 SF21 FF5 SS3 E SS8 FS7 (18) 24 35 32 39 FF3 G I FF3 (11) (7) 24 35 32 42 35 42 For activity F there are two dependencies and we should calculate both alternatives. Alternative 1: SS8 to H The start point is the latest start date of H (32), and we subtract the lag (8) from this value to get a latest start date of 24 for F. We add to this the duration of F (11) to get a latest finish date of 35. For activity I, we repeat this calculation in order to find the latest start date. In this case, we subtract 7 from 42, which gives us a value of 35. Alternative 1: SS8 to H LF = 35 Alternative 2: SS8 to I LF = 38 In this case, we choose the lowest value, which comes from dependency SS8 to activity H. For activity F, the latest start date is 24 and the latest finish is 35. Alternative 2: SS8 to I The start point is the latest start date of I (35), and we subtract from this amount the lag (8) to get a latest start date of 27 for F. We add to this value the activity’s duration (11), which gives us a latest finish date of 38. For activity H we find the latest start date by subtracting its duration (10) from the latest finish date (42), which gives us a value of 32. Now we will calculate backwards and find the latest start and finish dates for all activities. For the finish date, we choose the largest value and set 42 as the latest finish date for both activities H and I. Http://www.prosjektledelse.ntnu.no

7. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 24 35 32 39 FF3 G I FF3 (11) (7) 35 42 19 28 28 39 Activity D also has just one dependency: FF7 to F. FF7 starts in the latest finish date of F, 35. We subtract from this the lag of 7, which gives us a value of 28 as latest finish date for D. From this we subtract the duration of D (9) which gives us a value of 19 as the latest start date for D. From G, there is just one dependency to I. FF3 starts in the latest finish date of I, 42. We subtract from this the lag of 3, which gives us a latest finish date of 39 for G. The latest finish date (39) minus the duration of G (11) gives us a latest start date of 28. Http://www.prosjektledelse.ntnu.no

8. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 24 35 32 39 FF3 G I FF3 (11) (7) 28 39 35 42 12 30 Alternative 1: FF5 to F FF5 starts in the latest finish date of F, 35. We subtract from this the lag of 5, which gives us a latest finish date of 30 for E. The latest finish date minus the duration of E (18) gives us a latest start date of 12. Alternative 1: FF5 to F LF = 30 Alternative 2: FF3 to G LF = 36 In this case, we choose the smallest value, so the dependency FF5 to activity F becomes decisive. For activity E, the latest start date is 12 and the latest finish is 30. Alternative 2: FF3 to G FF3 starts in the latest finish date of E, 39. We subtract from this the lag of 3, which gives us a latest finish date of 36 for E. The latest finish date minus the duration of E (18) gives us a latest start date of 18. For activity E, there are two dependencies and we should calculate both alternatives. Http://www.prosjektledelse.ntnu.no

9. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 G I FF3 (11) (7) 28 39 35 42 7 19 Alternative 2: FS0 to F FS0 starts in the latest start date of F, 24. We subtract from this value the lag of 0, which gives us a latest finish date of 24 for C. We then subtract from this the duration of C (12), which gives us a latest start date of 12. Alternative 1: FS0 to D FS0 starts in the latest start date of D, 19. We subtract from this value the lag of 0, which gives us a latest finish date of 19 for C. We then subtract from this the duration of C (12), which gives us a latest start date of 7. Alternative 3: SF21 to E SF21 starts in the latest finish date of E, 30. We subtract from this value the lag of 21, which gives us a latest start date of 9 for C. We then add to this the duration of C (12), which gives us a latest start date of 9. Alternative 1: FS0 to D LF = 19 Alternative 2: FS0 to F LF = 24 Alternative 3: SF21to E LF = 21 We choose the smallest value, so the dependency FS0 to activity D becomes decisive. For activity C, the latest start date is 7 and the latest finish is 19. There are three dependencies for activity C, and we should calculate all three alternatives. Http://www.prosjektledelse.ntnu.no

10. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 7 19 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 G I FF3 (11) (7) 28 39 35 42 7 17 Alternative 1: SF12 to C SF12 starts in the latest finish date of C, 19. We subtract from this value the lag of 12, which gives us a latest start date of 7 for B. We then add to this number the duration of B (10), which gives us a latest finish date of 17. Alternative 1: SF12 to C LF = 17 Alternative 2: FS7 to G LF = 21 We choose the smallest value, so the dependency SF12 to C becomes decisive. For activity B, the latest start is 7 and the latest finish is 17. Alternative 2: FS7 to G FS7 starts in the latest start date of G, 28. We subtract from this value the lag of 7, which gives us a latest finish date of 21 for B. We then subtract from this number the duration of B (10), which gives us a latest start date of 11. For activity B, there are two dependencies and we should calculated both alternatives. Http://www.prosjektledelse.ntnu.no

11. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 7 17 7 19 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 G I FF3 (11) (7) 28 39 35 42 0 12 Alternative 1: FF5 to B FF5 starts in the latest finish date of B, 17. We subtract from this value the lag of 5, which gives us a latest finish date of 12 for A. We then subtract from this the duration of A (12), which gives us a latest start date of 0. Alternative 1: FF5 to B LF = 12 Alternative 2: SS3 to E LF = 21 We now choose the smallest value, which comes from depency FF5 to B. For activity A, the latest start date is 0 and the latest finish is 12. Alternative 2: SS3 to E The start point is the latest start date of E,1. We subtract from this value the lag of 3, which gives us a latest start date of 9 for A. We add to this the duration of A (12 ), which gives us a latest finish date of 21. There are again two dependencies for activity A, and we should calculate both alternatives. Http://www.prosjektledelse.ntnu.no

12. 19 28 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 A B C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 0 12 7 17 7 19 10 28 24 35 32 42 SF21 FF5 SS3 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 G I FF3 (11) (7) 28 39 35 42 0 0 0 0 0 0 2 4 3 Activity B: 17 - 17 = 0 Now we have calcuated all the values for the entire network. We can now find the float for all activities. We start from end of the network with activity I. In order to calculate the float, we use the formula LF- EF = Float 42 - 39 = 3 Activity H: 42 - 42 = 0 Activity E: 30 - 28 = 2 Activity G: 39 - 35 = 4 Activity C: 19 - 19 = 0 Activity D: 28 - 28 = 0 Activity F: 35 - 35 = 0 Activity A: 12 - 12 = 0 Http://www.prosjektledelse.ntnu.no

13. 19 28 0 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 0 0 0 0 A B 0 C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 0 12 7 17 7 19 10 28 24 35 32 42 SF21 FF5 SS3 2 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 4 3 G I FF3 (11) (7) 28 39 35 42 Now we can mark the critical path in the network: A-B-C-D-F-H Http://www.prosjektledelse.ntnu.no

14. 19 28 0 D (9) FS0 FF7 0 12 7 17 7 19 19 28 24 35 32 42 0 0 0 0 A B 0 C F H FF5 SF12 FS0 SS8 (12) (10) (12) (11) (10) 0 12 7 17 7 19 10 28 24 35 32 42 SF21 FF5 SS3 2 E SS8 FS7 (18) 12 30 24 35 32 39 FF3 4 3 G I FF3 (11) (7) 28 39 35 42 In this case, there is only one dependency between each pair of activities, which makes it relatively easy to find the active dependencies Additionally, we should determine dependencies which are active along the critical path.: Between A and B: FF5 Between B and C: SF12 Between C and D: FS0 Between D and F: FF7 Between F and H: SS8 Http://www.prosjektledelse.ntnu.no

15. FF5 FF7 A B C D F H FS0 (12) (10) (12) (9) (11) (10) SF12 SS8 Additionally, we need to determine normal, neutral and opposite critical activities on the critical path. This is done by looking at what happens if the activities are slightly prolonged. For this purpose, we have re-illustrated the critical path with only the active depencies by which the activities are linked to each other. Http://www.prosjektledelse.ntnu.no

16. FF5 FF7 A B C D F H FS0 (12) (10) (12) (9) (11) (10) SF12 SS8 If we slightly prolong activity C, we will need to start it earlier because its finish point is determined by its dependency to B. There is no dependency linked to the start point of C, so it will not influence the duration of the project. Therefore, activity C is a neutral critical activity. If we slightly prolong activity A, its finish point will move and the duration of the entire project will become longer. Therefore, activity A is a normal critical activity. If we prolong activity B, we will need to start it earlier because its finish point is determined by its dependency to A. If B is started earlier, its dependency with activity C obliges C to start earlier. In this way, the total duration of project is reduced. Therefore, activity B is an opposite critical activity. Http://www.prosjektledelse.ntnu.no

17. FF5 FF7 A B C D F H FS0 (12) (10) (12) (9) (11) (10) SF12 SS8 If we want to slightly prolong activity D, it would be done at the end of the activity, which will increase the duration of whole the project. Therefore activity D is a normal critical activity. If we slightly prolong activity H, the total duration of project will become longer. This happens because the start date of H is determined by its dependency to F. Activity H is therefore a normal critical activity. If we slightly prolong activity F, we will need to start it earlier because its finish point is determined by its dependency to H. If F is started earlier, its dependency to H obliges H to start ealier. In this way, the total duration of the project is reduced. Therefore activity B is an opposite critical activity. Http://www.prosjektledelse.ntnu.no