EdExcel Decision Mathematics 1

Transcription

1 EdExcel Decision Mathematics 1 Notes and Examples Critical Path Analysis Section 1: Activity networks These notes contain subsections on Drawing an activity network Using dummy activities Analysing the network Float and slack In these notes several different examples are used to demonstrate the drawing and analysis of an activity network, including the use of dummy activities where required. To help you understand how each network is built up, there are PowerPoint presentations for the shorter examples and video clips for the longer example. Each example will be in two parts: the first part deals with drawing the network (covered in Exercise 5B in the textbook), the second looks at finding the earliest event time, latest event time and critical activities, and also looks at float (covered in Exercise 5C and 5D in the textbook). The same examples will be used in section 2 to look at scheduling activities. The first three examples are given without any context, so that you can concentrate on dealing with the network. The fourth example deals with a practical situation. Drawing an activity network Example 1 shows a very simple network, in which no dummy activities are required. There is a PowerPoint presentation which shows how the network is built up. Note that it is sometimes helpful to redraw a network part way through, if you realise that you can avoid having activities crossing. Example 1 (Part 1) The table below shows the activities required to complete a project, with their durations and immediate predecessors. Activity Duration Immediate (hours) predecessors A - B 4 - C 6 - D 5 A E 1 B F 6 B G 7 C, D, E MEI, 29/06/09 1/8

2 Draw an activity network for the project. A() 2 D(5) C(6) 1 4 E(1) B(4) F(6) G(7) 5 Dummy activities Many networks require dummy activities in order to make them fit the requirements. A dummy activity has zero duration. There are two types of dummy activity. These are shown in the two next examples. In Example 2, a dummy activity is introduced because two activities would otherwise start at the same event and finish at the same event. This means that both activities would be described by the same number pair (i, j). When doing a network by hand, this would not really matter, but when programming a computer it is important that each activity is described by a unique number pair. The dummy activity ensures that this is the case. Even when doing networks by hand you must follow this rule, or you will lose marks in an examination. network is built up. Example 2 (Part 1) The table below shows the activities required to complete a project, with their durations and immediate predecessors. Activity Duration Immediate (days) predecessors A 2 - B - C 5 - D 6 A, B E 8 C F 2 C G 4 D, E Draw an activity network for the project. MEI, 29/06/09 2/8

3 A(2) 2 Activities A and B both start at the same event and need to end at the same event, since D depends on them both. The dummy activity is introduced to ensure unique numbering. 1 B() D(6) G(4) 5 6 C(5) E(8) F(2) 4 Example shows the second type of dummy activity. This is used when it is not otherwise possible to draw a network with the correct dependencies. network is built up. Example The table below shows the activities required to complete a project, with their durations and immediate predecessors. Activity Duration Immediate (days) predecessors A 2 - B 4 - C 5 A, B D B E 6 C F C G 8 D H 2 D, F Draw an activity network for the project. C(5) 2 4 A(2) F() E(6) A dummy activity is needed here as C depends on both A and B but D depends on B only. 1 B(4) D() 5 G(8) H(2) MEI, 29/06/09 /8 6 7 A dummy activity is needed here as G depends on D only but H depends on both F and D.

4 Analysing the network Once you have drawn a precedence network, the next step is to find the earliest event times and latest event times. The earliest event times are found by working forwards through the network and recording the earliest possible times that you can reach each event, and the latest event times are found by working backwards through the network and recording the latest possible time that each event can take place if the project is still to finish on time. The critical path can then be found by identifying those activities (i, j) for which the difference between l j and e i is equal to the duration of the activity. Example 1 (Part 2) Use activity network for the project in Example 1 to find the earliest event times and latest event times. Give the minimum duration of the project and the critical activities. A() 2 D(5) 0 0 C(6) 1 4 B(4) 4 7 E(1) F(6) 8 8 G(7) The critical activities are A, D and G. The minimum time for the project is 15 hours. earliest event times and latest event times are calculated. Float, slack and the critical path Float is the spare time associated with an activity. It is important as it gives us information about the impact that a delay on one activity will have on the whole project. Activities with no float are critical activities and lie on the critical path. Float becomes very important in the next section, when we consider the scheduling of activities. The spare time on non-critical activities can be used to delay the start of some activities. This can enable one resource (such as a MEI, 29/06/09 4/8

5 person or a machine) to carry out more than one activity without affecting the critical path. Slack is the spare time associated with an event. Events with no slack are critical events and lie on the critical path, but an activity that goes from one critical event to another critical event is not necessarily a critical activity. Because of this, the idea of float is often more useful than than the idea of slack. For Example 1 above, the floats for each activity and the slack for each event are as follows: Notice: Activity Total float Event Slack A 0 = = 0 B = 2 = 0 C = = D 8 5 = = 0 E = = 0 F = 5 G = 0 The critical activities, C, E and G have zero float. Any delay in these activities will delay the whole project. Events 1 and 4 are critical events, but activity C, which goes from event 1 to event 4, is not a critical activity. All the float on activity C belongs only to activity C. By looking at the network, you can see that C can take two extra hours without any impact on the whole project or on any other activities. B and E together take up 5 hours, and they need to be complete after 8 hours. So between them they can take an extra hours, but this hours of float is shared between them. If for example B takes an extra hours, E must not take any extra time. F has a total of 5 hours of float. 2 hours of this belongs only to activity F it can take an extra 2 hours without affecting any other activity or the total duration. However, if F takes its maximum possible extra time of 5 hours, then you can see from the activity network that B must not take any extra time. The other hours of belonging to F is therefore shared with B if B takes an extra hours (the maximum allowed by B s float) then F can take only the extra 2 hours that belongs only to F. Example 2 (Part 2) (i) Use the activity network for Example 2 to find the critical activities and the shortest time in which the project can be completed. (ii) (iii) Calculate the total float on each activity. What effect does each of the following delays have on the duration of the whole project and the critical activities? (a) C is delayed by 2 days MEI, 29/06/09 5/8

6 (b) F is delayed by days (c) D is delayed by 5 days (d) A is delayed by 5 days (i) A(2) B() C(5) 7 D(6) E(8) 1 1 G(4) F(2) The critical activities are C, E and G. The total duration of the project is 17 days. (ii) (iii) The float for each activity is as follows: Activity Total float A 5 B 4 C 0 D 4 E 0 F 10 G 0 (a) C is a critical activity, so a delay of 2 days results in the whole project being delayed by 2 days. The critical activities remain the same. (b) F is not a critical activity, and it has float of 10 days, so it can be delayed by days without affecting the overall duration of the project or the critical activities. (c) D has only 4 days of float available, so a delay of 5 days means that the whole project is delayed by 1 day. D and B now become critical activities, and C and E are no longer critical. (d) A has 5 days of float available, so a delay of 5 days makes no difference to the overall duration of the project or the critical activities, but it reduces the float on D to zero. This means that A and D are now critical activities (taking 1 days between them to reach event 5), but C and E are still critical (since they also take 1 days to reach event 5). There are now two critical paths: ADG and CEG). MEI, 29/06/09 6/8

7 Notice: EdExcel D1 CPA Section 1 Notes and Examples A and B share a float of 4, and A has an additional float of 1 which belongs only to itself. So A can take 1 day longer without affecting any other activities or the duration of the whole project, and either D, or both A and B, can take up to 4 days longer (or the 4 days can be split between D and both A and B, e.g. D with an extra days and each of A and B with one extra day). F has float of 10 which belongs only to itself, so it can take up to 10 days longer without affecting any other activities or the duration of the whole project. earliest event times and latest event times are calculated. Example (Part 2) (i) Use the activity network for Example to find the critical activities and the shortest time in which the project can be completed. (ii) (iii) (iv) Calculate the total float on each of the non-critical activities. Activity H is delayed by days. What effect does this have on the duration of the project and the critical activities? Extra resources are available which can be used to shorten the duration of one of activities B, E or F (on the original network) by one day. Which of these activities should be shortened, and why? (i) C(5) 2 4 A(2) F() E(6) B(4) H(2) G(8) D() The critical activities are B, C, D, E and G. The minimum total duration of the project is 15 days. MEI, 29/06/09 7/8

8 (ii) The non-critical activities are A, F and H. A has float of 2 F has float of 1 H has float of 1. (iii) (iv) H has float of 2, so delaying H by days will increase the total duration of the project by 1 day. F and H will now be critical activities, and D, E and G will no longer be critical. Shorten B by one day. F is not a critical activity, so reducing it would not shorten the project. Although E is a critical activity, there are two critical paths (BCE and BDG), so reducing E would not have any effect unless either D or G were reduced as well. Notice that in this example there are two critical paths (BCE and BDG). This is an important consideration in part (iv). earliest event times and latest event times are calculated. MEI, 29/06/09 8/8