Cost Consideration in PERT & CPM

Program Evaluvation & Review Technique PERT

Critical Path Method Example

CRITICAL PATH METHOD EXAMPLE:

TIME ANALYSIS

TIME ANALYSIS:

NETWORK SCHEDULING BY PERT/CPM

PERT- Program Evaluation Review Technique
CPM-Critical Path Method

NETWORK SCHEDULING
Its a technique used for planning and scheduling large project.Its a method for minimizing troubleshoots such as delays, interruption.
There are two basic planning & Control techniques.:
   PERT
   CPM

NETWORK:
Its the graphical representation of logically and sequentially connected arrows and nodes representing activities and events in the project.Its an arrow diagram.

ACTIVITY:
An activity represents some action and its  a time concuming effort necessary to complete a parrticular part of the overall project.

EVENT:
The beginning and the end points of an activity are called events/nodes.

MERGE EVENT:
Its not necessary for an event to be ending event for only one activity as it can be the ending event of two or more activity such a event is defined as merge event.

BURST EVENT:
If the event happens to be the beginning event of two or more activities then they are called as Burst event.

We have the following types of activities

1. Preceding
2. Succeeding
3.Concurrent
4.Dummy Activity

DUMMY ACTIVITY:
Certain activities which neither consume type or resources but are used simply to represent a connection/link between events are known as Dummy Activity.

CONSTRUCTION OF NETWORK:
Construct the network for the project whose activities and preceding relationships are shown below

ACTIVITIES            A              B               C                D             E             F
IMMEDIATE    
PREDECESSOR     -                 A              A                -               D           B,C,E

DOMINANCE PROPERTY

Q1. Solve

                        Player B
                          B1    B2    B3
Player A   A1   [ 1      7        2
                 A2     6      2        7
                 A3     5      1        6]

Soln:
STEP 1:
Find out the maximin and minimax values

 [ 1      7        2     1
    6      2        7    2     maximin=2
    5      1        6]   1

   6        7        7
minimax=6

maximin != minimax
Therefore no saddle point.

STEP II:
Now compare each row and check whether they are minimum values. If so then delete that row.

Compare A1 and A2 , here few values are minimum and few are maximum.
Now compare  A2 and A3, A3 has minimum values therefore delete A3.

By dominance Property,       B1       B2        B3
                                    A1    1           7         2
                                    A2     6          2         7
STEP III:
Now compare each column and check whether they are maximum values. If so then delete that column.

Compare B1 and B2 , here few values are minimum and few are maximum.
Now compare  B1 and B3, B3 has maximum values therefore delete B3.

          B1         B2
A1     1             7        1
A2     6             2        2         maximin=2
           6             7
minimax=6

maximin!=minimax
Therefore no saddle point

STEP IV:
                a22-a12
p1=--------------------------
       (a22+a11)-(a12+a21)

p1=1/2     p2=1/2

             a22-a21
q1=--------------------------
       (a22+a11)-(a12+a21)

q1=2/5    q2=3/5

                            a11a22-a21a12
value of game=-------------------------- =4
                        (a22+a11)-(a12+a21)

SA= [ A1       A2        A3
          1/2        1/2        0]

SB=[B1       B2         B3
        2/5      3/5          0]

GRAPHICAL METHOD FOR GAME THEORY

Q1.Solve the following 2 x 3 game graphically

                               Player B
Player A    [ 1          3         11
                    8          5          2]



STEP I :
 A's expected pay off against B's pure move is given by:


B's PURE MOVE                     A's EXPECTED PAY OFF E(P)
     B1                                              E1(P)=8-7P1
     B2                                              E2(P)=5-2P1
     B3                                              E3(P)=2P1+2

STEP II:
Now for the graph consider Axis 1 as  A2 and Axis 2 as A1

STEP III:
The reduced Pay off matrix is given by

          B2           B3    --------> the intersection point
A1 [   3              11       3
A2      5               2]      2     maxmin=3

           5               11
minimax=5

maxmin != minmax
Therefore  there exists no saddle point

P1=9/11          p2=2/11

q1=3/11           q2=8/11


Value of game = 49/11

The optimal Strategy is given by
            [  A1          A2
SA  =      9/11       2/11]

           [  B1          B2       B3
SB =      0             3/11     8/11]


NOTE:
The left out matrix value is given as '0'
For 2 x n matrix find lower envelope, find maximin (maximum values)
For m x 2 matrix find upper envelope, find minimax (minimum values)