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  Network analysis is the general name given to
  certain specific techniques which can be used
  for the planning, management and control of
  projects.
 Use of nodes and arrows:-

Arrows  An arrow leads from tail to head
  directionally
       Indicate ACTIVITY, a time consuming effort that is
        required to perform a part of the work.
•   NODE:-
    A  node is represented by a circle
    - Indicate EVENT, a point in time where one or
      more activities start and/or finish.
•   Activity:-
    –   A task or a certain amount of work required in the
        project
    –   Requires time to complete
    –   Represented by an arrow
•   Dummy Activity:-
    – Indicates only precedence relationships
    – Does not require any time of effort.
   Event:-
       Signals the beginning or ending of an activity
       Designates a point in time
       Represented by a circle (node)
   Network:-
       Shows the sequential relationships among
        activities using nodes and arrows
   Activity-on-node (AON):-
    nodes represent activities, and arrows show precedence
      relationships
   Activity-on-arrow (AOA):-
    arrows represent activities and nodes are events for points
      in time
SITUATIONS IN NETWORK DIAGRAM:-
             B
                         A must finish before either B or C
             C           can start.
    A
                 C       Both A and B must finish before C
    B                    can start.
    A        C            Both A and B must finish before
                         either of C or D can start.
    B        D


                         A must finish before B can start
         B
A                        both A and C must finish before
        Dummy            D can start.

    C                D
 illustration of network analysis of a minor
  redesign of a product and its associated
  packaging.
The key question is: How long will it take to
complete this project ?
   Path
       A connected sequence of activities leading from the
        starting event to the ending event
   Critical Path
       The longest path (time); determines the project
        duration
   Critical Activities
       All of the activities that make up the critical path.
Forward Pass:-
   Earliest Start Time (ES)
     earliest time an activity can start
     ES = maximum EF of immediate predecessors


   Earliest finish time (EF)
     earliest time an activity can finish
     earliest start time plus activity time
                            EF= ES+t
Backward Pass:-
   Latest Start Time (LS)
    Latest time an activity can start without delaying critical
      path time
      LS= LF - t
   Latest finish time (LF)
    latest time an activity can be completed without
       delaying critical path time
    LS = minimum LS of immediate predecessors
   Draw the CPM network
   Analyze the paths through the network
   Determine the float for each activity
     Compute the activity’s float
                float = LS - ES = LF - EF
     Float is the maximum amount of time that this
      activity can be delay in its completion before it
      becomes a critical activity, i.e., delays completion of
      the project
   Find the critical path is that the sequence of activities
    and events where there is no “slack” i.e.. Zero slack
     Longest path through a network
   Find the project duration is minimum project
   CPM Network:-
                            f, 15


                              g, 17              h, 9
       a, 6
                                      i, 6


              b, 8

                             d, 13       j, 12


                     c, 5
                            e, 9
   ES and EF Times:-
                            f, 15

                              g, 17              h, 9
      a, 6
      0 6                             i, 6

             b, 8
           0 8                           j, 12
                            d, 13


       0     5      c, 5
                           e, 9
   ES and EF Times:-
                                         6     21
                                             f, 15


                                                  g, 17                   h, 9
            a, 6
                                     6       23
        0        6                                             i, 6


                     b, 8
                 0        8                   d, 13               j, 12

                                         8        21
                              c, 5
             0        5
                                             e, 9

                                                    5     14
   ES and EF Times:-
                                          6   21
                                    f, 15
                                                                         21 30
       0     6
                                      g, 17                               h, 9
      a, 6
                               6     23
                                                          i, 6
                    0      8                            23 29
             b, 8

                                     d, 13 8       21            j, 12

                                                                         21 33
                    c, 5
       0     5
                                   e, 9
                                                         Project’s EF = 33
                                          5   14
   LS and LF Times:-
                                              6    21
                                              0     0
                                  f, 15                                21 30
          0     6                                                      24 33
          0     0
                                    g, 17                              h, 9
         a, 6
                              6     23
                                                             i, 6 23 29
                              0      0                            27 33
                 0     8
                 0     0
                b, 8                               8    21     j, 12
                                  d, 13
                                                   0     0
                                                                       21 33
                       c, 5                                            21 33
     0     5
     0     0                  e, 9

                                          5       14
                                          0        0
   LS and LF Times:-
                                            6 21
                                           18 24
                                                                    21 30
                                   f, 15
         0      6                                                   24 33
         4     10
                              6 23 g, 17                           h, 9
        a, 6                                          23 29
                             10 27
                                                      27 33 i, 6

                 0     8
                 0     8      8   21
               b, 8           8   21                       j, 12
                                           d, 13
                                                                   21 33
    0    5
                      c, 5                                         21 33
    7   12
                                  e, 9
                                               5 14
                                              12 21
FLOAT:-
                                                 6     21
                                        03
                                                 9     24
                                              f, 15                              21 30
                                                                             3
                                                                                 24 33
                 0 6
        3
                 4 10                       g, 17                                h, 9
                a, 6                     6 23
                                     04
                                        10 27
                                                                    23 29 i, 6
                                                                4
                          0      8                                  27 33
                  0                      8      21
                          0      8   0
                      b, 8               8      21
                                                      d, 13              j, 12

                                                                                     21 33
                                                                                 0
            0     5           c, 5                                                   21 33
    7
            7    12                      e, 9            5 14
                                                 7
                                                        12 21
   Critical Path:-
                              f, 15


                                  g, 17                h, 9
      a, 6
                                                i, 6



             b, 8
                                      d, 13   j, 12


                    c, 5
                           e, 9
   PERT is based on the assumption that an activity’s
    duration follows a probability distribution instead
    of being a single value
   Three time estimates are required to compute the
    parameters of an activity’s duration distribution:
     pessimistic time (tp ) - the time the activity
      would take if things did not go well
     most likely time (tm ) - the consensus best
      estimate of the activity’s duration
     optimistic time (to ) - the time the activity would
      take if things did go well
       te = a+4m+b
                6
   Draw the network.
   Analyze the paths through the network and find
    the critical path.
   The length of the critical path is the mean of the
    project duration probability distribution which is
    assumed to be normal
   The standard deviation of the project duration
    probability distribution is computed by adding
    the variances of the critical activities (all of the
    activities that make up the critical path) and
    taking the square root of that sum
   Probability computations can now be made using
    the normal distribution table.
Determine probability that project is completed
 within specified time
Z=   x-



   where   = tp = project mean time
      = project standard mean time
    x = (proposed ) specified time
Probability




       Z




= tp       x             Time
   Useful at many stages of project management
   Mathematically simple
   Give critical path and slack time
   Provide project documentation
   Useful in monitoring costs
•   How long will the entire project take to be
    completed? What are the risks involved?
•   Which are the critical activities or tasks in the
    project which could delay the entire project if they
    were not completed on time?
•   Is the project on schedule, behind schedule or
    ahead of schedule?
•   If the project has to be finished earlier than
    planned, what is the best way to do this at the least
    cost?
   Parallel paths-identifying a single path is difficult
    when there are parallel paths with similar duration.
   Time consuming-critics note that it takes too much
    time to identify all activities and inter-relate them to
    get multiple projects paths.
   First time projects-CPM is not suitable if projects
    cannot be broken down into discrete activities with
    known completion times.
PRESENTED BY:-
BHUPENDRA SINGH SHEKHAWAT
ANKIT VINOD AGARWAL

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Project Management PERT and CPM

  • 1.
  • 2.  Network analysis is the general name given to certain specific techniques which can be used for the planning, management and control of projects.  Use of nodes and arrows:- Arrows  An arrow leads from tail to head directionally  Indicate ACTIVITY, a time consuming effort that is required to perform a part of the work.
  • 3. NODE:- A node is represented by a circle - Indicate EVENT, a point in time where one or more activities start and/or finish. • Activity:- – A task or a certain amount of work required in the project – Requires time to complete – Represented by an arrow • Dummy Activity:- – Indicates only precedence relationships – Does not require any time of effort.
  • 4. Event:-  Signals the beginning or ending of an activity  Designates a point in time  Represented by a circle (node)  Network:-  Shows the sequential relationships among activities using nodes and arrows  Activity-on-node (AON):- nodes represent activities, and arrows show precedence relationships  Activity-on-arrow (AOA):- arrows represent activities and nodes are events for points in time
  • 5. SITUATIONS IN NETWORK DIAGRAM:- B A must finish before either B or C C can start. A C Both A and B must finish before C B can start. A C Both A and B must finish before either of C or D can start. B D A must finish before B can start B A both A and C must finish before Dummy D can start. C D
  • 6.  illustration of network analysis of a minor redesign of a product and its associated packaging. The key question is: How long will it take to complete this project ?
  • 7.
  • 8.
  • 9. Path  A connected sequence of activities leading from the starting event to the ending event  Critical Path  The longest path (time); determines the project duration  Critical Activities  All of the activities that make up the critical path.
  • 10. Forward Pass:-  Earliest Start Time (ES)  earliest time an activity can start  ES = maximum EF of immediate predecessors  Earliest finish time (EF)  earliest time an activity can finish  earliest start time plus activity time EF= ES+t
  • 11. Backward Pass:-  Latest Start Time (LS) Latest time an activity can start without delaying critical path time LS= LF - t  Latest finish time (LF) latest time an activity can be completed without delaying critical path time LS = minimum LS of immediate predecessors
  • 12. Draw the CPM network  Analyze the paths through the network  Determine the float for each activity  Compute the activity’s float float = LS - ES = LF - EF  Float is the maximum amount of time that this activity can be delay in its completion before it becomes a critical activity, i.e., delays completion of the project  Find the critical path is that the sequence of activities and events where there is no “slack” i.e.. Zero slack  Longest path through a network  Find the project duration is minimum project
  • 13. CPM Network:- f, 15 g, 17 h, 9 a, 6 i, 6 b, 8 d, 13 j, 12 c, 5 e, 9
  • 14. ES and EF Times:- f, 15 g, 17 h, 9 a, 6 0 6 i, 6 b, 8 0 8 j, 12 d, 13 0 5 c, 5 e, 9
  • 15. ES and EF Times:- 6 21 f, 15 g, 17 h, 9 a, 6 6 23 0 6 i, 6 b, 8 0 8 d, 13 j, 12 8 21 c, 5 0 5 e, 9 5 14
  • 16. ES and EF Times:- 6 21 f, 15 21 30 0 6 g, 17 h, 9 a, 6 6 23 i, 6 0 8 23 29 b, 8 d, 13 8 21 j, 12 21 33 c, 5 0 5 e, 9 Project’s EF = 33 5 14
  • 17. LS and LF Times:- 6 21 0 0 f, 15 21 30 0 6 24 33 0 0 g, 17 h, 9 a, 6 6 23 i, 6 23 29 0 0 27 33 0 8 0 0 b, 8 8 21 j, 12 d, 13 0 0 21 33 c, 5 21 33 0 5 0 0 e, 9 5 14 0 0
  • 18. LS and LF Times:- 6 21 18 24 21 30 f, 15 0 6 24 33 4 10 6 23 g, 17 h, 9 a, 6 23 29 10 27 27 33 i, 6 0 8 0 8 8 21 b, 8 8 21 j, 12 d, 13 21 33 0 5 c, 5 21 33 7 12 e, 9 5 14 12 21
  • 19. FLOAT:- 6 21  03 9 24 f, 15 21 30 3 24 33 0 6 3 4 10 g, 17 h, 9 a, 6 6 23 04 10 27 23 29 i, 6 4 0 8 27 33 0 8 21 0 8 0 b, 8 8 21 d, 13 j, 12 21 33 0 0 5 c, 5 21 33 7 7 12 e, 9 5 14 7 12 21
  • 20. Critical Path:- f, 15 g, 17 h, 9 a, 6 i, 6 b, 8 d, 13 j, 12 c, 5 e, 9
  • 21. PERT is based on the assumption that an activity’s duration follows a probability distribution instead of being a single value  Three time estimates are required to compute the parameters of an activity’s duration distribution:  pessimistic time (tp ) - the time the activity would take if things did not go well  most likely time (tm ) - the consensus best estimate of the activity’s duration  optimistic time (to ) - the time the activity would take if things did go well te = a+4m+b 6
  • 22. Draw the network.  Analyze the paths through the network and find the critical path.  The length of the critical path is the mean of the project duration probability distribution which is assumed to be normal  The standard deviation of the project duration probability distribution is computed by adding the variances of the critical activities (all of the activities that make up the critical path) and taking the square root of that sum  Probability computations can now be made using the normal distribution table.
  • 23. Determine probability that project is completed within specified time Z= x-  where = tp = project mean time  = project standard mean time  x = (proposed ) specified time
  • 24. Probability Z = tp x Time
  • 25. Useful at many stages of project management  Mathematically simple  Give critical path and slack time  Provide project documentation  Useful in monitoring costs
  • 26. How long will the entire project take to be completed? What are the risks involved? • Which are the critical activities or tasks in the project which could delay the entire project if they were not completed on time? • Is the project on schedule, behind schedule or ahead of schedule? • If the project has to be finished earlier than planned, what is the best way to do this at the least cost?
  • 27. Parallel paths-identifying a single path is difficult when there are parallel paths with similar duration.  Time consuming-critics note that it takes too much time to identify all activities and inter-relate them to get multiple projects paths.  First time projects-CPM is not suitable if projects cannot be broken down into discrete activities with known completion times.
  • 28. PRESENTED BY:- BHUPENDRA SINGH SHEKHAWAT ANKIT VINOD AGARWAL