How often PERT and probabilistic scheduling is asked

Where it was asked

• May/Jun 201213 marks
• Oct/Nov 201114 marks
• May/Jun 20117 marks

The questions

1. May/Jun 2012, Q5.110 marks

   A PERT network for a project is shown in a diagram with five numbered event nodes (1 to 5) connected by five activities: Activity A runs from node 1 to node 2, Activity B runs from node 1 to node 3, Activity C runs from node 2 to node 4, Activity D runs from node 4 to node 5, and Activity E runs from node 3 to node 5. The target date for completing the whole project is 15 weeks. A table gives three time estimates (in weeks) for each activity: Activity A has an optimistic (a) time of 4, a most likely (m) time of 6 and a pessimistic (b) time of 8; Activity B has optimistic 1, most likely 4, pessimistic 5; Activity C has optimistic 2, most likely 3, pessimistic 5; Activity D has optimistic 2, most likely 5, pessimistic 6; Activity E has optimistic 3, most likely 4, pessimistic 5. The table also has blank columns for the Expected time (te) and Standard Deviation (s) of each activity, which must be completed. Using the PERT network diagram (with nodes 1 to 5 and activities A, B, C, D and E as described, where A runs 1→2, B runs 1→3, C runs 2→4, D runs 4→5 and E runs 3→5) together with the table of optimistic (a), most likely (m) and pessimistic (b) time estimates given for each activity, calculate the Expected time (te) and the Standard Deviation (s) for each of the five activities (A, B, C, D and E), and then show these calculated (te) and (s) values marked onto the network diagram.

2. May/Jun 2012, Q5.22 marks

   A PERT network for a project is shown in a diagram with five numbered event nodes (1 to 5) connected by five activities: Activity A runs from node 1 to node 2, Activity B runs from node 1 to node 3, Activity C runs from node 2 to node 4, Activity D runs from node 4 to node 5, and Activity E runs from node 3 to node 5. The target date for completing the whole project is 15 weeks. A table gives three time estimates (in weeks) for each activity: Activity A has an optimistic (a) time of 4, a most likely (m) time of 6 and a pessimistic (b) time of 8; Activity B has optimistic 1, most likely 4, pessimistic 5; Activity C has optimistic 2, most likely 3, pessimistic 5; Activity D has optimistic 2, most likely 5, pessimistic 6; Activity E has optimistic 3, most likely 4, pessimistic 5. The table also has blank columns for the Expected time (te) and Standard Deviation (s) of each activity, which must be completed. Based on the expected times and standard deviations calculated for the PERT network of activities A, B, C, D and E (with the target completion date of 15 weeks), calculate the Z value for the last event (node 5) of the network.

3. May/Jun 2012, Q5.31 mark

   A PERT network for a project is shown in a diagram with five numbered event nodes (1 to 5) connected by five activities: Activity A runs from node 1 to node 2, Activity B runs from node 1 to node 3, Activity C runs from node 2 to node 4, Activity D runs from node 4 to node 5, and Activity E runs from node 3 to node 5. The target date for completing the whole project is 15 weeks. A table gives three time estimates (in weeks) for each activity: Activity A has an optimistic (a) time of 4, a most likely (m) time of 6 and a pessimistic (b) time of 8; Activity B has optimistic 1, most likely 4, pessimistic 5; Activity C has optimistic 2, most likely 3, pessimistic 5; Activity D has optimistic 2, most likely 5, pessimistic 6; Activity E has optimistic 3, most likely 4, pessimistic 5. The table also has blank columns for the Expected time (te) and Standard Deviation (s) of each activity, which must be completed. Using Figure 5.3, a graph plotting the probability (%) of not meeting the target date against the Z value, and the Z value calculated in Question 5.2, read off and state the probability of not meeting the target completion date of 15 weeks for this project.

4. Oct/Nov 2011, Q5.14 marks

   Question 5 is based on a PERT network shown in Diagram 5.1, where the targeted completion date for the project is stated as nine (11) weeks. The network has four nodes, numbered 1 to 4, each drawn as a box divided into quarters for node number, duration, early/late times etc. Node 1 connects to node 2 by activity A, node 2 connects to node 3 by activity B, node 3 connects to node 4 by activity C, and node 1 connects directly to node 4 by activity D; node 4's box also shows the value 11. A table gives three time estimates (in weeks) for each activity: Activity A has an optimistic time of 3, a most likely time of 4 and a pessimistic time of 5; Activity B has optimistic 1, most likely 2, pessimistic 3; Activity C has optimistic 2, most likely 3, pessimistic 4; Activity D has optimistic 4, most likely 5, pessimistic 6. The table also has blank columns for the Expected time (te) and the Standard Deviation (s) for each activity, which must be completed. A further graph, Figure 5.5, plots the 'Probability of not meeting target date (%)' on the vertical axis (from 0 to 100) against 'Z value' on the horizontal axis (ranging from about -3.25 to 3.25), showing an S-shaped curve. Using the three time estimates given in the table for Activities A, B, C and D, calculate the expected time (te) for each of the four activities.

5. Oct/Nov 2011, Q5.24 marks

   Question 5 is based on a PERT network shown in Diagram 5.1, where the targeted completion date for the project is stated as nine (11) weeks. The network has four nodes, numbered 1 to 4, each drawn as a box divided into quarters for node number, duration, early/late times etc. Node 1 connects to node 2 by activity A, node 2 connects to node 3 by activity B, node 3 connects to node 4 by activity C, and node 1 connects directly to node 4 by activity D; node 4's box also shows the value 11. A table gives three time estimates (in weeks) for each activity: Activity A has an optimistic time of 3, a most likely time of 4 and a pessimistic time of 5; Activity B has optimistic 1, most likely 2, pessimistic 3; Activity C has optimistic 2, most likely 3, pessimistic 4; Activity D has optimistic 4, most likely 5, pessimistic 6. The table also has blank columns for the Expected time (te) and the Standard Deviation (s) for each activity, which must be completed. A further graph, Figure 5.5, plots the 'Probability of not meeting target date (%)' on the vertical axis (from 0 to 100) against 'Z value' on the horizontal axis (ranging from about -3.25 to 3.25), showing an S-shaped curve. Using the same table of optimistic, most likely and pessimistic estimates for Activities A, B, C and D, calculate the standard deviation (s) for each of the four activities.