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Critical Path Analysis      Presented by
                         Mushahid Khatri
                          Jaysinh Shukla
                           Nimish Doshi
Project


   “A project is a series of activities directed to
    accomplishment of a desired objective.”


    Plan your work first…..then work your plan
Activity Slack

     Each event has two important times associated with it :

   Earliest time , Te , which is a calendar time when a event can occur
    when all the predecessor events completed at the earliest possible times




   Latest time , TL , which is the latest time the event can occur without
    delaying the subsequent events and completion of project.


         Difference between the latest time and the earliest time of an
         event is the slack time for that event
Floats

           Float

Negative   Zero     Positive
Node
Activity
Activity




Starting Event              Completion Event
Activity A




             A
Activity B must follow A




A                     B
ORANGE
BLUE
Example
Activity             Description           Predecessors   Duration (days)
   A       Organize sales office                -               6
   B       Hire salesmen                        A               4
   C       Train salesmen                       B                7
   D       Select advertising agency            A               2
   E       Plan advertising campaign            D               4
   F       Conduct advertising campaign         E               10
   G       Design package                       -               2
   H       Setup packaging facilities           G               10
   I       Package initial stocks              J,H              6
   J       Order stock from manufacturer        -               13
   K       Select distributors                  A               9
   L       Sell to distributors                C,K               3
   M       Ship stocks to distributors         I,L               5
4


                                  L(3)
                           6             9
    G(2)
1              3
                                             10
                           C(7)


                       5

      2
                                  8


                   7
E4=13
                    L4=14
                4
                                                           E9=20
                                                           L9=20
E1=0                                            L(3)
L1=0                    E3=2            6                 9
         G(2)           L3=4                   E6=17
   1                3
                                               L6=17
                                                                   10
                                        C(7)
                                                                   E10=25
                                        E5=10
                                                                   L10=25
                                   5    L5=10

           2
  E2=6                                           8     E8=12
  L2=6
                                                       L8=15

                               7
                                       E7=8
                                       L7=11
E4=13
                    L4=14
                4
                                                           E9=20
                                                           L9=20
E1=0                                            L(3)
L1=0                    E3=2            6                 9
         G(2)           L3=4                   E6=17
   1                3
                                               L6=17
                                                                   10
                                        C(7)
                                                                   E10=25
                                        E5=10
                                                                   L10=25
                                   5    L5=10

           2
  E2=6                                           8     E8=12
  L2=6
                                                       L8=15

                               7
                                       E7=8
                                       L7=11
Benefits of CPM
   Useful at many stages of project management

   Mathematically simple

   Give critical path and slack time

   Provide project documentation

   Useful in monitoring costs
Questions Answered by CPM

   Completion date?


   On Schedule?


   Within Budget?


   Critical Activities?


   How can the project be finished early at the least cost?
Critical path analysis11

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Critical path analysis11

  • 1. Critical Path Analysis Presented by Mushahid Khatri Jaysinh Shukla Nimish Doshi
  • 2. Project  “A project is a series of activities directed to accomplishment of a desired objective.” Plan your work first…..then work your plan
  • 3. Activity Slack Each event has two important times associated with it :  Earliest time , Te , which is a calendar time when a event can occur when all the predecessor events completed at the earliest possible times  Latest time , TL , which is the latest time the event can occur without delaying the subsequent events and completion of project. Difference between the latest time and the earliest time of an event is the slack time for that event
  • 4. Floats Float Negative Zero Positive
  • 7. Activity Starting Event Completion Event
  • 9. Activity B must follow A A B
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 17.
  • 18.
  • 19.
  • 20.
  • 21.
  • 22.
  • 23. BLUE
  • 24.
  • 25.
  • 26.
  • 27.
  • 28.
  • 29. Example Activity Description Predecessors Duration (days) A Organize sales office - 6 B Hire salesmen A 4 C Train salesmen B 7 D Select advertising agency A 2 E Plan advertising campaign D 4 F Conduct advertising campaign E 10 G Design package - 2 H Setup packaging facilities G 10 I Package initial stocks J,H 6 J Order stock from manufacturer - 13 K Select distributors A 9 L Sell to distributors C,K 3 M Ship stocks to distributors I,L 5
  • 30.
  • 31. 4 L(3) 6 9 G(2) 1 3 10 C(7) 5 2 8 7
  • 32. E4=13 L4=14 4 E9=20 L9=20 E1=0 L(3) L1=0 E3=2 6 9 G(2) L3=4 E6=17 1 3 L6=17 10 C(7) E10=25 E5=10 L10=25 5 L5=10 2 E2=6 8 E8=12 L2=6 L8=15 7 E7=8 L7=11
  • 33. E4=13 L4=14 4 E9=20 L9=20 E1=0 L(3) L1=0 E3=2 6 9 G(2) L3=4 E6=17 1 3 L6=17 10 C(7) E10=25 E5=10 L10=25 5 L5=10 2 E2=6 8 E8=12 L2=6 L8=15 7 E7=8 L7=11
  • 34. Benefits of CPM  Useful at many stages of project management  Mathematically simple  Give critical path and slack time  Provide project documentation  Useful in monitoring costs
  • 35. Questions Answered by CPM  Completion date?  On Schedule?  Within Budget?  Critical Activities?  How can the project be finished early at the least cost?