2. Lesson Objective Understand what critical path analysis is Be able to prioritise events and create a precedence table Begin to use the precedence table to create a network diagram

3. Consider the following problem: You must toast three pieces of bread on both sides. The grill can take two pieces at a time. It takes 30 seconds to toast a side of bread. How quickly can you toast the bread?

4. A solutionA more efficient solution 30 seconds 0 seconds 60 seconds 90 seconds 120 seconds

5. In this problem time can be saved by planning. If you have lots of complicated things to achieve in a set sequence planning to find which things can be run con-currently can save time. The study of this type of problem is called critical path analysis and the critical path is the sequence of events that must run to schedule in order to complete the task as efficiently as possible.

6. A simple Critical path problem Jane, Sue and Meena share a flat. Jane has an interview at 9.00 but she has woken up at 8.10. To get to the interview she must: Shower 3 mins Dry her hair8 mins Fetch the car 7 mins Iron her clothes12 mins Dress and make-up10 mins Drive to the interview20 mins If she does all of these things on her own will she make the interview? What if she uses her friends to help her with some of the tasks?

7. Activity Immediately preceding activities duration AShower-3 BDry hairA8 CFetch car-7 DIron clothes-12 EDress and make-upB,D10 FDrive to interviewC,E20 To solve the problem algorithmically we should first draw up a Precedence table. This is a table that shows how each activity is dependent upon the others:

8. A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 We can then draw a Network diagram to illustrate how each acticty is dependent upon preceding activities:

9. A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 Earliest event time EET Latest event time LET Each event node needs two boxes, to mark in the event times. Event Times:

10. Earliest Event Times: A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 0 3 12 22 42 To find EETs, work forwards through the network from the start node to the finish node. The EET for an event is the earliest time the event can start taking into account the activities that must complete before it can do so.

11. Latest Event Times: A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 3 0 12 22 42 22 12 4 0 To find LETs, work backwards through the network from the finish node to the start node. The LET for an event is the latest time you can leave the event so that you can still complete the remaining activities without over running.

12. Critical Activities: A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 34 00 12 22 42 Critical activities are activities that cannot run late. For critical activities: Latest finish — Earliest start = length of activity In this case D, E and F The green arrows mark the critical activities, which form the critical path. The critical path(s) must form a continuous route from the start node to the finish node.

13. A(3) D(12) B(8) E(10) C(7) F(20) 15 4 3 2 34 00 12 22 42 Non-Critical Activities: Activities that are not critical have some flexibility over when they start/finish. The spare time they have to work in is called the float. In some cases using the float to delay the start time of an activity will not affect the other activities (in the case of C above). The float of C is said to be independent. But in other cases it will (as in the case of A and B). We say that the floats of A and B are interfering.

14. Total float Independent float Interfering float A101 B101 C15 0 Hint: Total float is maximum possible float so take “outside” nos. Independent float is minimum possible float so take “inside” nos. Interfering float is the difference between the two floats. We can put the results of the floats into a table:

15. Have a go at this: Draw an activities network for this situation and find the critical path. TaskDuration (hours)Immediate predecessors A4- B4- C6- D4A E2B F8D,E G3C H4G,F I3B

16. Example from the book wit h an easy diagram and no dummies.

17. Lesson Objective Be able to use the precedence table to create a network diagram Understand the importance and use of dummies when creating a network diagram from a precedence table

18. ActivityDepends on APrepare foundations− BHave foundations passed by inspectorA CObtain bricks− DErect wallsB, C EConstruct roofD FInstall plumbingD GInstall wiringE HPlaster wallsF, G IDecorateH JLandscape gardenE Warm Up – Draw a Network diagram for this precedence table

20. Activity networks TaskDuration (hours)Immediate predecessors A3- B4- C6- D5A E1B F6B G7C, D, E The table below shows the tasks involved in a project, with their durations and immediate predecessors. Draw an activity network for this table.

21. Activity networks Activities A, B and C do not depend on any other activity, so they all begin at node 1. A(3) B(4) 1 C(6) First draw a start node labelled 1.

22. Activity networks Activity D depends on A, so add event node 2 at the end of A. A(3) B(4) 1 C(6) D(5) Now add activity D. 2

23. Activity networks Activities E and F both depend on B, so add event node 3 at the end of B. A(3) B(4) 1 C(6) Now add activities E and F. E(1) D(5) F(6) 3 2

24. C(6) B(4) E(1) F(6) 3 Activity networks Activity G depends on C, D and E, so all these three events need to end at the same node. A(3) This is easiest if you redraw the network so that C is between A and B. D(5) 2 B(4) C(6) E(1) F(6) 3 1

25. Activity networks A(3) D(5) 2 C(6) B(4) E(1) F(6) 3 Now add node 4, with C, D and E leading into it. Now add activity G. 14 D(5) G(7)

26. Activity networks A(3) 2 C(6) B(4) E(1) F(6) 3 A finish node is now needed. Any activities not leading into a node must end at the finish node. 14 D(5) G(7) 5 F(6) G(7)

27. Activity networks A(3) 2 C(6) B(4) E(1) 3 14 D(5) 5 F(6) G(7)

28. Try drawing a network for this table. What problems arise?

30. Using dummy activities Case 1: Activity C and D both have dependency on the same letter, but not exactly the same letters. We need to use a dummy. 3 2 A B D C CB DA,B dummy

31. Using dummy activities Case 2: Activity B and C both share the same start and finish node. This will cause problems when we use our EETs and LETs so we cannot let this happen. We need to use a dummy. A- BA CA DBC

32. Official Practice Paper B

34. Official Practice Paper A

35. D 4

36. Lesson Objective Be able to use a network from a precedence table to find the shortest completion time and the critical paths and events

37. Reminder: The critical path is the path through the Network that takes the minimum amount of time (there may be more than one) but manages to complete all the activities. Critical Activities are those that cannot be delayed (there is no float) if the task is to complete on time. The float of an activity is how long it can be delayed without slowing down the overall completion time. There are three types of float: Total float is maximum possible float so take “outside” nos. Independent float is minimum possible float so take “inside” nos. Interfering/dependent float is the difference between the two floats.

38. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 1 occurs at time zero. 14 D(5) 5 F(6) G(7) 0

39. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 2 cannot occur until A is finished. The earliest time for this is 3. 14 D(5) 5 F(6) G(7) 0 3

40. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 3 cannot occur until B is finished. The earliest time for this is 4. 14 D(5) 5 F(6) G(7) 0 3 4

41. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 4 cannot occur until C, D and E are all finished. 14 D(5) 5 F(6) G(7) 0 3 4 The earliest C can finish is 6. The earliest D can finish is 3 + 5 = 8. The earliest E can finish is 4 + 1 = 5. So the earliest time for event 4 is 8. 8

42. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 5 cannot occur until F and G are both finished. 14 D(5) 5 F(6) G(7) 0 3 4 The earliest F can finish is 4 + 6 = 10. The earliest G can finish is 8 + 7 = 15. So the earliest time for event 5 is 15. 8 15

43. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 The next step is to find the late event times (LETs), working backwards through the network. A LET is the latest time that an event can occur without delaying the project. The LET is found by finding the latest time that each activity leading out of the event can begin – the LET is the earliest of these. 14 D(5) 5 F(6) G(7) 0 3 4 8 15

44. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Event 5 must occur by time 15, or the project will not finish in the minimum possible time. 14 D(5) 5 F(6) G(7) 0 3 4 8 15

45. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 The only activity leading from event 4 is G, which must start by time 8 if the project is not to be delayed. So event 4 must occur by time 8. 14 D(5) 5 F(6) G(7) 0 3 4 8 15 8

46. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 The activities leading from event 3 are E (which must start by time 7) and F (which must start by time 9). So event 3 must occur by time 7. 14 D(5) 5 F(6) G(7) 0 3 4 8 15 8 7

47. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 The only activity leading from event 2 is D, which must start by time 3. So event 2 must occur by time 3. 14 D(5) 5 F(6) G(7) 0 3 4 8 15 8 7 3

48. Activity networks – Example 1 A(3) 2 C(6) B(4) E(1) 3 Finally, event 1 must occur by time zero. 14 D(5) 5 F(6) G(7) 0 3 4 8 15 8 7 3 0

49. Activity networks – Example 1 A(3) C(6) B(4) E(1) 3 D(5) F(6) G(7) 0 3 4 8 15 8 7 3 0 The completed network shows that the project can be completed in 15 hours. The critical activities are the activities (i, j) for which the LET for j – the EET for i is equal to the activity duration. The critical activities are A,Dand G. For analysis of the float in this example, see the Notes and Examples. 1 2 4 5

50. Activity networks – Example 2 TaskDuration (days)Immediate predecessors A2- B3- C5- D6A, B E8C F2C G4D, E The table below shows the tasks involved in a project, with their durations and immediate predecessors. Draw an activity network and use it to find the critical activities and the minimum duration of the project.

51. Activity networks – Example 2 Begin with a start node, labelled 1. Activities A, B and C have no preceding activities, so can all begin at the start node. A(2) B(3) 1 C(5)

52. Activity networks – Example 2 Activity D depends on both A and B. Since A and B must not start and finish at the same node, a dummy activity is needed to ensure unique numbering. The dummy activity has zero duration. Now activity D can be drawn in, following on from both A and B. A(2) B(3) 1 D(6) 2 3 C(5)

53. Activity networks – Example 2 Activities E and F both depend on activity C. A(2) B(3) 1 D(6) 2 3 C(5) 4 E(8) F(2)

54. E(8) Activity networks – Example 2 Since G depends on both D and E, these two activities must both lead into the same node. Activity G can now be drawn in. A(2) B(3) 1 D(6) 2 3 G(4) 5 C(5) 4 E(8) F(2)

55. Activity networks – Example 2 Finally, activities F and G must finish at the end node. A(2) B(3) 1 D(6) 2 3 G(4) 56 E(8) F(2) C(5) 4 F(2)

56. Activity networks – Example 2 The next step is to find the earliest event times (EETs). A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2)

57. Event 1 occurs at time zero. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) The earliest that event 2 can occur is after A has finished, at time 2. 0 2

58. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 Event 3 cannot occur until both A and B have finished, so the earliest time at which event 3 can occur is 3. 3

59. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 Event 4 cannot occur until C has finished, so the earliest time at which event 5 can occur is 5. 5

60. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5The earliest that D can finish is at time 9, and the earliest that E can finish is at time 13, so the earliest that event 5 can occur is time 13. 13

61. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 The earliest that F can finish is at time 7, and the earliest that G can finish is at time 17, so the earliest that event 5 can occur is time 17. 17

62. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 The next step is to find the latest event times (LETs), starting from the finish node and working backwards.

63. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 Event 6 must not occur later than time 17, or the project will be delayed. 17

64. The latest G can start is at time 13, so the latest time for event 5 is 13. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13

65. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13 The latest that E can start is at time 5, and the latest that F can start is at time 15, so the latest possible time for event 4 is 5. 5

66. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13 5The latest that D can start is at time 7, so the latest possible time for event 3 is 7. 7

67. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13 5 7 The latest that the dummy activity (with zero duration) can start is at time 7, so the latest possible time for event 2 is 7. 7

68. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13 5 7 7 The latest that event 1 can start is at time 0. 0

69. Activity networks – Example 2 A(2) B(3) 1 D(6) 2 3 G(4) C(5) E(8) 4 56 F(2) 0 2 3 5 13 17 13 5 7 7 0 The critical activities are activities for which the float is zero: i.e. the latest event time for activity j – the earliest event time for activity i is equal to the activity duration.

70. Activity networks – Example 2 A(2) B(3) D(6) 2 3 G(4) C(5) E(8) F(2) 0 2 3 5 13 17 13 5 7 7 0 The critical activities are C, Eand G. The project can be completed in 17 days. For analysis of the float in this example, see Example 2 in the Notes and Examples. 65 4 1

72. 3