![44
A = [76+4(86)+120]/6 = 90
A = (120 - 76)/6 = 7.33 A
2 = (7.33)2 = 53.78
Activity
A 90 7.33 53.78
B 15 1.00 1.00
C 5 0.33 0.11
D 20 3.00 9.00
E 21 1.00 1.00
F 25 2.33 5.44
G 14 2.00 4.00
H 28 1.33 1.78
I 30 4.67 21.78
J 45 3.67 13.44
2
KLONE COMPUTERS –
Finding activities’ mean and variance](https://image.slidesharecdn.com/projectscheduling-230705220142-562638b2/85/Project-Scheduling-ppt-44-320.jpg)
Project Scheduling.ppt
- 1. 1 PROJECT SCHEDULING Project Schedule is the tool that communicates what work needs to be performed, which resources of the organization will perform the work and the timeframes in which that work needs to be performed. The project schedule should reflect all of the work associated with delivering the project on time. Without a full and complete schedule, the project manager will be unable to communicate the complete effort, in terms of cost and resources, necessary to deliver the project.
- 2. 2 Work Breakdown Structure (WBS) The building blocks of a schedule start with a Work Breakdown Structure (WBS). The WBS is a hierarchical reflection of all the work in the project in terms of deliverables. In order to produce these deliverables, work must be performed. A typical approach in developing a WBS is to start at the highest level, with the product of the project
- 3. 3 Activity Description A Prototype model design B Purchase of materials Manufacturing C Manufacture of prototype model activities D Revision of design E Initial production run F Staff training Training activities G Staff input on prototype models H Sales training Advertising activities I Pre-production advertising campaign J Post-redesign advertising campaign KLONE COMPUTERS, INC
- 4. 4 • There are three major tasks to perform: – Manufacture the new computer. – Train staff and vendor representatives. – Advertise the new computer. • KLONE needs to develop a precedence relations chart. • The chart gives a concise set of tasks and their immediate predecessors. KLONE COMPUTERS, INC
- 5. 5 From the activity description chart, we can determine immediate predecessors for each activity. Activity A is an immediate predecessor of activity B, because it must be competed just prior to the commencement of B. A B KLONE COMPUTERS, INC
- 6. 6 Immediate Estimated Activity Predecessor Completion Time A None 90 B A 15 C B 5 D G 20 E D 21 F A 25 G C,F 14 H D 28 I A 30 J D,I 45 Precedence Relationships Chart KLONE COMPUTERS, INC
- 7. 7 5.3 The PERT/CPM Approach for Project Scheduling • The PERT/CPM approach to project scheduling uses network presentation of the project to – Reflect activity precedence relations – Activity completion time • PERT/CPM is used for scheduling activities such that the project’s completion time is minimized.
- 8. 8 KLONE COMPUTERS, INC. - Continued • Management at KLONE would like to schedule the activities so that the project is completed in minimal time. • Management wishes to know: – The earliest and latest start times for each activity which will not alter the earliest completion time of the project. – The earliest finish times for each activity which will not alter this date. – Activities with rigid schedule and activities that have slack in their schedules.
- 9. 9 Earliest Start Time / Earliest Finish Time • Make a forward pass through the network as follows: – Evaluate all the activities which have no immediate predecessors. • The earliest start for such an activity is zero ES = 0. • The earliest finish is the activity duration EF = Activity duration. – Evaluate the ES of all the nodes for which EF of all the immediate predecessor has been determined. • ES = Max EF of all its immediate predecessors. • EF = ES + Activity duration. – Repeat this process until all nodes have been evaluated • EF of the finish node is the earliest finish time of the project.
- 10. 10 Earliest Start / Earliest Finish – Forward Pass A 90 B 15 C 5 F 25 I 30 G 14 D 20 E 21 H 28 J 45 90,105 90,115 90,120 105,110 110,124 115,129 129,149 149,170 149,177 120,165 149,194 170 194 A 0,90 B I F C G D E H J 177 194 EARLIEST FINISH
- 11. 11 Latest start time / Latest finish time • Make a backward pass through the network as follows: – Evaluate all the activities that immediately precede the finish node. • The latest finish for such an activity is LF = minimal project completion time. • The latest start for such an activity is LS = LF - activity duration. – Evaluate the LF of all the nodes for which LS of all the immediate successors has been determined. • LF = Min LS of all its immediate successors. • LS = LF - Activity duration. – Repeat this process backward until all nodes have been evaluated.
- 12. 12 B F C A I E D G H H 28 166,194 J J 45 149,194 E 21 173,194 90,105 90,115 90,120 105,110 115,129 129,149 149,170 149,177 149,194 153,173 146,166 194 129,149 0,90 129,149 D 20 129,149 129,149 129,149 129,149 129,149 129,149 129,149 G 14 115,129 I 30 119,149 29,119 C 5 110,115 B 15 95,110 5,95 F 25 90, 115 0,90 A 90 Latest Start / Latest Finish – Backward Pass
- 13. 13 • Activity start time and completion time may be delayed by planned reasons as well as by unforeseen reasons. • Some of these delays may affect the overall completion date. • To learn about the effects of these delays, we calculate the slack time, and form the critical path. Slack Times
- 14. 14 – Slack time is the amount of time an activity can be delayed without delaying the project completion date, assuming no other delays are taking place in the project. Slack Time = LS - ES = LF - EF Slack Times
- 15. 15 Critical activities must be rigidly scheduled Activity LS - ES Slack A 0 -0 0 B 95 - 90 5 C 110 - 105 5 D 119 - 119 0 E 173 - 149 24 F 90 - 90 0 G 115 - 115 0 H 166 - 149 17 I 119 - 90 29 J 149 - 149 0 Slack time in the Klonepalm 2000 Project
- 16. 16 • The critical path is a set of activities that have no slack, connecting the START node with the FINISH node. • The critical activities (activities with 0 slack) form at least one critical path in the network. • A critical path is the longest path in the network. • The sum of the completion times for the activities on the critical path is the minimal completion time of the project. The Critical Path
- 17. 17 B F C A I E D G H H 28 166,194 J J 45 149,194 E 21 173,194 90,105 90,115 90,120 105,110 115,129 129,149 149,170 149,177 149,194 D 20 0,90 129,149 G 14 115,129 I 30 119,149 A 90 C 5 110,115 B 15 95,110 F 25 90, 115 0,90 The Critical Path
- 18. 18 • We observe two different types of delays: – Single delays. – Multiple delays. • Under certain conditions the overall project completion time will be delayed. • The conditions that specify each case are presented next. Possible Delays
- 19. 19 • A delay of a certain amount in a critical activity, causes the entire project to be delayed by the same amount. • A delay of a certain amount in a non-critical activity will delay the project by the amount the delay exceeds the slack time. When the delay is less than the slack, the entire project is not delayed. Single delays
- 20. 20 LS =119 A 90 J 45 H 28 E 21 D 20 I 30 G 14 F 25 C 5 B 15 ES=149 LS=173 DELAYED START=149+15=164 ES=90 DELAYED START=90+15 =105 Activity E and I are each delayed 15 days. THE PROJECT COMPLETION TIME IS NOT DELAYED Multiple delays of non critical activities: Case 1: Activities on different paths
- 21. 21 A 90 90 B 15 Gantt chart demonstration of the (no) effects on the project completion time when delaying activity “I” and “E” by 15 days. Activity I F 25 I 30 105 C 5 115 G 14 129 D 20 149 E 21 H 28 J 45 194 194 Activity E
- 22. 22 A 90 B 15 C 5 F 25 I 30 G 14 D 20 E 21 H 28 J 45 ES=149 LS =173 DELAYED START=149+15 =164 ES=90 DELAYED START =94 LS =95 THE PROJECT COMPLETION TIME IS NOT DELAYED Multiple delays of non critical activities: Case 2: Activities are on the same path, separated by critical activities. Activity B is delayed 4 days, activity E is delayed 15 days
- 23. 23 A 90 B 15 C 5 F 25 I 30 G 14 D 20 E 21 H 28 J 45 DELAYED START= 109 + 4 =113; ES= 90 DELAYED START =94 DELAYED FINISH = 94+15=109 LS =110 3 DAYS DELAY IN THE ENTIRE PROJECT Activity B is delayed 4 days; Activity C is delayed 4 days. THE PROJECT COMPLETION TIME IS DELAYED 3 DAYS Multiple delays of non critical activities: Case 2: Activities are on the same path, no critical activities separating them.
- 24. 24 5.6 Gantt Charts • Gantt charts are used as a tool to monitor and control the project progress. • A Gantt Chart is a graphical presentation that displays activities as follows: – Time is measured on the horizontal axis. A horizontal bar is drawn proportionately to an activity’ s expected completion time. – Each activity is listed on the vertical axis. • In an earliest time Gantt chart each bar begins and ends at the earliest start/finish the activity can take place.
- 25. 25 Immediate Estimated Activity Predecessor CompletionTime A None 90 B A 15 C B 5 D G 20 E D 21 F A 25 G C,F 14 H D 28 I A 30 J D,I 45 Immediate Estimated Activity Predecessor CompletionTime A None 90 B A 15 C B 5 D G 20 E D 21 F A 25 G C,F 14 H D 28 I A 30 J D,I 45 A 90 90 B 15 F 25 I 30 105 C 5 115 G 14 129 D 20 149 E 21 H 28 J 45 194 194
- 26. 26 • Gantt chart can be used as a visual aid for tracking the progress of project activities. • Appropriate percentage of a bar is shaded to document the completed work. • The manager can easily see if the project is progressing on schedule (with respect to the earliest possible completion times). Gantt Charts- Monitoring Project Progress
- 27. 27 A 90 B 15 F 25 I 30 C 5 G 14 D 20 E 21 H 28 J 45 194 194 135 Monitoring Project Progress The shaded bars represent completed work BY DAY 135. Do not conclude that the project is behind schedule. Activity “I” has a slack and therefore can be delayed!!!
- 28. 28 • Advantages. – Easy to construct – Gives earliest completion date. – Provides a schedule of earliest possible start and finish times of activities. • Disadvantages – Gives only one possible schedule (earliest). – Does not show whether the project is behind schedule. – Does not demonstrate the effects of delays in any one activity on the start of another activity, thus on the project completion time. Gantt Charts – Advantages and Disadvantages
- 29. 29 5.7 Resource Leveling and Resource Allocation • It is desired that resources are evenly spread out throughout the life of the project. • Resource leveling methods (usually heuristics) are designed to: – Control resource requirements – Generate relatively similar usage of resources over time.
- 30. 30 • A heuristic approach to “level” expenditures – Assumptions • Once an activity has started it is worked on continuously until it is completed. • Costs can be allocated equally throughout an activity duration. Step 1: Consider the schedule that begins each activity at its ES. Step 2: Determine which activity has slack at periods of peak spending. Step 3: Attempt to reschedule the non-critical activities performed during these peak periods to periods of less spending, but within the time period between their ES and LF. Resource Leveling – A Heuristic
- 31. 31 • Management wishes to schedule the project such that – Completion time is 194 days. – Daily expenditures are kept as constant as possible. • To perform this analysis cost estimates for each activity will be needed. Resource Leveling – KLONE COMPUTERS, Inc. - continued
- 32. 32 Total Total Cost Cost Time per Activity Description (x10000) (days) Day A Prototype model design 2250 90 25 B Purchase of materials 180 15 12 C Manufacture of prototype 90 5 18 D Revision of design 300 20 15 E Initial production run 231 21 11 F Staff training 250 25 10 G Staff input on prototype 70 14 5 H Sales traini ng 392 28 14 I Pre-production advertisement 510 30 17 J Post-production advertisement 1350 45 30 Total cost = 5,623 Resource Leveling – KLONE COMPUTERS, Inc. – cost estimates
- 33. 33 Cumulative Daily Expenditure – Earliest Times vs. Latest Times 55 50 45 40 35 30 25 20 15 10 5 20 40 60 80 100 120 140 160 180 200 Earliest Start-Earliest Finish Budget Latest Start-Latest Finish Budget Feasible Budgets Time
- 34. 34 55 50 45 40 35 30 25 20 15 10 5 20 40 60 80 100 120 140 160 180 200 E H J J Cost Leveling E J E J H J H C F A B F I I H 27 15 44 30 Daily Expenditure of the ES Schedule 55 I I I I I 25 I I 39 45 I ES = 90 I LS = 110 22 32 I I I F G D
- 35. 35 G 55 50 45 40 35 30 25 20 15 10 5 20 40 60 80 100 120 140 160 180 200 E H J J I E J E J H J C F A B F I I H 25 27 22 15 44 30 32 55 H H Cost Leveling I D I F
- 36. 36 5.8 The Probability Approach to Project Scheduling • Activity completion times are seldom known with 100% accuracy. • PERT is a technique that treats activity completion times as random variables. • Completion time estimates are obtained by the Three Time Estimate approach
- 37. 37 • The Three Time Estimate approach provides completion time estimate for each activity. • We use the notation: a = an optimistic time to perform the activity. m = the most likely time to perform the activity. b = a pessimistic time to perform the activity. The Probability Approach – Three Time Estimates
- 38. 38 = the mean completion time = a + 4m + b 6 = the standard deviation = b - a 6 Approximations for the mean and the standard deviation of activity completion time are based on the Beta distribution. The Distribution, Mean, and Standard Deviation of an Activity
- 39. 39 To calculate the mean and standard deviation of the project completion time we make some simplifying assumptions. The Project Completion Time Distribution - Assumptions
- 40. 40 • Assumption 2 – The time to complete one activity is independent of the time to complete any other activity. • Assumption 3 – There are enough activities on the critical path so that the distribution of the overall project completion time can be approximated by the normal distribution. The Project Completion Time Distribution - Assumptions • Assumption 1 – A critical path can be determined by using the mean completion times for the activities. – The project mean completion time is determined solely by the completion time of the activities on the critical path.
- 41. 41 Mean = Sum of mean completion times along the critical path. The three assumptions imply that the overall project completion time is normally distributed, the following parameters: The Project Completion Time Distribution Variance = Sum of completion time variances along the critical path. Standard deviation = Variance
- 42. 42 Activity Optim istic Most Likely Pessim istic A 76 86 120 B 12 15 18 C 4 5 6 D 15 18 33 E 18 21 24 F 16 26 30 G 10 13 22 H 24 18 32 I 22 27 50 J 38 43 60 The Probability Approach – KLONE COMPUTERS
- 43. 43 • Management at KLONE is interested in information regarding the completion time of the project. • The probabilistic nature of the completion time must be considered. The Probability Approach – KLONE COMPUTERS
- 44. 44 A = [76+4(86)+120]/6 = 90 A = (120 - 76)/6 = 7.33 A 2 = (7.33)2 = 53.78 Activity A 90 7.33 53.78 B 15 1.00 1.00 C 5 0.33 0.11 D 20 3.00 9.00 E 21 1.00 1.00 F 25 2.33 5.44 G 14 2.00 4.00 H 28 1.33 1.78 I 30 4.67 21.78 J 45 3.67 13.44 2 KLONE COMPUTERS – Finding activities’ mean and variance
- 45. 45 • The mean times are the same as in the CPM problem, previously solved for KLONE. • Thus, the critical path is A - F- G - D – J. – Expected completion time = A +F +G +D +J=194. – The project variance =A 2 +F 2 +G 2 +D 2 +J 2 = 85.66 – The standard deviation = = 9.255 2 KLONE COMPUTERS – Finding mean and variance for the critical path