1. CPA A group of workers is involved in a building project. The table shows the activities involved. Each worker can perform any of the given activities. Activity Preceding activity Duration (days)Number of workers A-35 BA82 CA73 DB, C84 EC102 FC33 GD, E34 HF61 IG, H23

2. CPA Complete the activity network for the project. 8 D 10 E 3 F 3 G 2 I 8 7 C B 3 A 6 H ActivityPreceding activity Duration (days ) Number of workers A-35 BA82 CA73 DB, C84 EC102 FC33 GD, E34 HF61 IG, H23

3. CPA Find the earliest start time and the latest finish time and insert their values in the appropriate boxes. Duration 3 883 36 2 10 7 C D E F GB IA H Earliest start timeLatest finish time 0 3 3 11 10 20 13 2325 23 20 1710 12 3

4. CPA Find the critical path and state the minimum completion time Duration 3 883 36 2 10 7 C D E F GB IA H Earliest start timeLatest finish time 0 3 3 11 10 20 13 2325 23 20 1710 12 3 Critical Path: A C E G I and the minimum completion time is: 25 days

5. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 ActivityFloat B 1 D 1 F 4 H 4 Activities A, C, E, G and I are “critical” – they must start on time and end on time for the minimum completion time to be 25 days. Some activities do not have to start exactly on time and finish exactly on time without affecting the minimum finish time. They have a "float". Now we can draw a Gantt Chart or Cascade Diagram. This gives us another way of displaying how the activities can be completed. In addition, the diagram can help us to figure out how many workers may be needed to complete the project within the minimum time of 25 days.

6. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 ACEGI B F D H We can draw a Gantt Chart or Cascade Diagram First set down the Critical Path Now set down each activity Including the float

7. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 ACEGI B FDH We can “shuffle” the activities to determine how many workers are needed to complete the project. ActivityNumber of workers A5 B2 C3 D4 E2 F3 G4 H1 I3 DH

8. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 ACEGI B F DH This is for D2, but some UoM students might find it interesting. Add in the number of workers for each activity. ActivityNumber of workers A5 B2 C3 D4 E2 F3 G4 H1 I3 5 42 32 3 4 1 3 Can you see that in order to complete the project on time at least 2 and at most 9 workers are needed.

9. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 B D 42 F H 31 ACEGI 53243 D2: We could draw a “Resource Histogram” to show exactly how many workers are needed on each day

10. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 D2: So this is what the resource histogram looks like. It’s a handy diagram to show haw many worker are needed throughout the duration of the project.

11. CPA 3 883 36 2 10 7 C D E F GB IA H 0 3 3 11 10 20 13 2325 23 20 1710 12 3 For D2 only. Use “resource levelling” to explain why the project will overrun if there are only 7 workers available at any one time. Which activities need to be delayed so that the project will be delayed with the minimum extra time. What is the minimum completion time now. A B C D E F G H I B EE F E Activity F is the problem at day 11. Delay the start of D by 2 days, but because there a 1 day of float, this means that G and I are only delayed by 1 day. The new minimum completion time is 26 days. ActivityNumber of workers A5 B2 C3 D4 E2 F3 G4 H1 I3