Your SlideShare is downloading. ×
0
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Project planning and control
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Project planning and control

412

Published on

Published in: Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
412
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
21
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.
  • 2.  How is it different?  Limited time frame  Narrow focus, specific objectives  Less bureaucratic  Why is it used?  Special needs  Pressures for new or improves products or services
  • 3.  What are the Key Metrics  Time  Cost  Performance objectives  What are the Key Success Factors?  Top-down commitment  Having a capable project manager  Having time to plan  Careful tracking and control  Good communications
  • 4.  What are the Major Administrative Issues?  Executive responsibilities  Project selection  Project manager selection  Organizational structure  Organizational alternatives  Manage within functional unit  Assign a coordinator  Use a matrix organization with a project leader
  • 5.  What are the tools?  Work breakdown structure  Network diagram  Gantt charts  Risk management
  • 6. Project X Level 1 Level 2 Level 3 Level 4
  • 7. MAR APR MAY JUN JUL AUG SEP OCT NOV DEC Locate new facilities Interview staff Hire and train staff Select and order furniture Remodel and install phones Move in/startup Gantt Chart
  • 8. PERT: Program Evaluation and Review Technique CPM: Critical Path Method  Graphically displays project activities  Estimates how long the project will take  Indicates most critical activities  Show where delays will not affect project
  • 9.  Network (precedence) diagram  Diagram of project activities that shows a sequential relationships by use of arrows and nodes  Activity-on-arrow (AOA)  Network diagram convention in which arrows designate activities.  Activity-on-node (AON)  Network diagram convention in which nodes designate activities.
  • 10. 1 2 3 4 5 6 Locate facilities Order furniture Furniture setup Interview Hire and train Remodel Move in AOA
  • 11. 1 2 3 5 6 Locate facilities Order furniture Furniture setup Interview Remodel Move in 4 Hire and train 7S AON
  • 12.  Activities  Project steps that consume resources and/or time.  Events  The starting and finishing of activities, designated by nodes AOA convention.  Path  A sequence of activities that leads from the starting node to the finishing node  Critical path  The longest path; determines expected project duration  Critical activities  Activities on the critical path  Slack  Allowable slippage for path; the difference the length of path and the length of critical path
  • 13.  Deterministic  Time estimates that are fairly certain  Probabilistic  Estimates of times that allow for variation
  • 14. 1 65 4 3 2 1 WEEKS MOVE IN Determine the ff: a. The length of each path c. The expected length of time b. The critical path d. The amount of slack time for each path
  • 15. 1 65 4 3 2 1 WEEKS MOVE IN PATH LENGTH(WEEKS) SLACK 1-2-4-5-6 8+6+3+1=18 20-18=2 1-2-5-6 8+11+1=20 20-20=0 1-3-5-6 4+9+1=14 20-14=6
  • 16.  Network activities  ES: early start  EF: early finish  LS: late start  LF: late finish  Used to determine  Expected project duration  Slack time  Critical path
  • 17.  FOUR PIECES OF INFORMATION ABOUT THE NETWORK ACTIVITIES:  ES-the earliest time activity can start  EF-the earliest time the activity can finish  LS-the latest time the activity can start and not delay the project  LF-the latest time the activity can finish and not delay the project
  • 18. Computation of earliest starting and finishing time is aided by two simple rules: 1. The earliest finish time for any activity is equal to its earliest start time plus its expected duration, t: EF = ES + t 2. ES for activities at nodes with one entering arrow is equal to EF of the entering arrow. ES for activities leaving nodes with multiple entering arrows is equal to the largest EF of the entering arrow.
  • 19. Computation of latest starting and finishing time is aided by the use two rules: 1. The latest starting time for each activity is equal to its latest finishing time minus its expected duration, t: LS = LF - t 2. For nodes with one leaving arrow, LF for arrows entering that node equals the LS of the leaving arrow. For nodes with multiple leaving arrows, LF for arrows entering that node equals the smallest LS of leaving arrows.
  • 20. 1 65 4 3 2 1 WEEKS MOVE IN Determine the ff: a. ES c. EF b. LS d. LF
  • 21. 1 65 4 3 2 1 WEEKS MOVE IN 19 20 19 20
  • 22. Slack = LS – ES Or Slack = LF- EF
  • 23. ACTIVITY LS ES SLACK 1-2 0 0 0 1-3 6 0 6 2-4 10 8 2 2-5 8 8 0 3-5 10 4 6 4-5 16 14 2 5-6 19 19 0
  • 24. Consider the following consulting project: Develop a critical path diagram and determine the duration of the critical path and slack times for all activities. Activity Designation Immed. Pred. Time (Weeks) Assess customer's needs A None 2 Write and submit proposal B A 1 Obtain approval C B 1 Develop service vision and goals D C 2 Train employees E C 5 Quality improvement pilot groups F D, E 5 Write assessment report G F 1
  • 25. A(2) B(1) C(1) D(2) E(5) F(5) G(1) A None 2 B A 1 C B 1 D C 2 E C 5 F D,E 5 G F 1 Act. Imed. Pred. Time
  • 26. ES=9 EF=14 ES=14 EF=15 ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=4 EF=9 ES=4 EF=6 A(2) B(1) C(1) D(2) E(5) F(5) G(1) Hint: Start with ES=0 and go forward in the network from A to G.
  • 27. ES=9 EF=14 ES=14 EF=15 ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=4 EF=9 ES=4 EF=6 A(2) B(1) C(1) D(2) E(5) F(5) G(1) LS=14 LF=15 LS=9 LF=14 LS=4 LF=9 LS=7 LF=9 LS=3 LF=4 LS=2 LF=3 LS=0 LF=2 Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.
  • 28. ES=9 EF=14 ES=14 EF=15 ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=4 EF=9 ES=4 EF=6 A(2) B(1) C(1) D(2) E(5) F(5) G(1) LS=14 LF=15 LS=9 LF=14 LS=4 LF=9 LS=7 LF=9 LS=3 LF=4 LS=2 LF=3 LS=0 LF=2 Duration = 15 weeks Slack=(7-4)=(9-6)= 3 Wks
  • 29.  Optimistic time  Time required under optimal conditions  Pessimistic time  Time required under worst conditions  Most likely time  Most probable length of time that will be required
  • 30. Activity start Optimistic time Most likely time (mode) Pessimistic time to tptm te
  • 31. te = to + 4tm +tp 6 te = expected time to = optimistic time tm = most likely time tp = pessimistic time
  • 32. 2 = (tp – to)2 36 2 = variance to = optimistic time tp = pessimistic time
  • 33. The network diagram for a project is shown in the accompanying figure, with three time estimates for each activity. Activity times are in weeks. Do the following: a. Compute the expected time for each activity and the expected duration for each path b. Identify the critical path c. Compute the variance of each activity and the variance and standard deviation of each path
  • 34. 3-4-5 d 3-5-7 e 5-7-9 f 2-4-6 b 4-6-8 h Optimistic time Most likely time Pessimistic time
  • 35. a.
  • 36. b. The path that has the longest expected duration is the critical path. Because path d-e-f has the largest path total, it is the critical path. c.
  • 37. Z = Specified time – Path mean Path standard deviation Z indicates how many standard deviations of the path distribution the specified tine is beyond the expected path duration.
  • 38. Using information given from the previous example answer the ff: a. What is the probability that the project can be completed within 17 weeks of its start? b. What is the probability that the project will be completed within 15 weeks of its start? c. What is the probability that the project will not be completed within 15 weeks of its start?
  • 39. 17 Weeks Weeks Weeks Weeks 10.0 16.0 13.5 1.00 1.00 a-b-c d-e-f g-h-i
  • 40. a. Path Probability of Completion in 17 weeks a-b-c + 7.22 1.00 d-e-f +1.00 .8413 g-h-I +3.27 1.00 P(Finish by week 17) = 1.00 x .8413 x 1.00 = .8413 Z = 17– Path Duration Path standard deviation
  • 41. b. Path Probability of Completion in 17 weeks a-b-c + 5.15 1.00 d-e-f - 1.00 .1587 g-h-I + 1.40 .9192 P(Finish by week 15) = 1.00 x .1587 x .9192 = .1459 c. 1 - .1459 = .8541 Z = 15– Path Duration Path standard deviation
  • 42. Task Immediate Predecesors Optimistic Most Likely Pessimistic A None 3 6 15 B None 2 4 14 C A 6 12 30 D A 2 5 8 E C 5 11 17 F D 3 6 15 G B 3 9 27 H E,F 1 4 7 I G,H 4 19 28
  • 43. ET(A)= 3+4(6)+15 6 ET(A)=42/6=7 Task Immediate Predecesors Expected Time A None 7 B None 5.333 C A 14 D A 5 E C 11 F D 7 G B 11 H E,F 4 I G,H 18 Task Immediate Predecesors Optimistic Most Likely Pessimistic A None 3 6 15 B None 2 4 14 C A 6 12 30 D A 2 5 8 E C 5 11 17 F D 3 6 15 G B 3 9 27 H E,F 1 4 7 I G,H 4 19 28 Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time 6
  • 44. Example 2. Network A(7) B (5.333) C(14) D(5) E(11) F(7) H(4) G(11) I(18) Duration = 54 Days
  • 45. Example 2. Probability Exercise What is the probability of finishing this project in less than 53 days? p(t < D) TE = 54 Z = D - TE cp 2  t D=53
  • 46. Activity variance, = ( Pessim. - Optim. 6 )2 2  Task Optimistic Most Likely Pessimistic Variance A 3 6 15 4 B 2 4 14 C 6 12 30 16 D 2 5 8 E 5 11 17 4 F 3 6 15 G 3 9 27 H 1 4 7 1 I 4 19 28 16 (Sum the variance along the critical path.) 2  = 41
  • 47. There is a 43.8% probability that this project will be completed in less than 53 weeks. p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156) Z = D - T = 53- 54 41 = -.156E cp 2  TE = 54 p(t < D) t D=53
  • 48.  What is the probability that the project duration will exceed 56 weeks?
  • 49. t TE = 54 p(t < D) D=56 Z = D - T = 56 - 54 41 = .312E cp 2  p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312))

×