Pert Network

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    PERT NETWORK

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    CPM : CRITICAL PATH METHOD

    PERT : USED FOR PROJECT PLANNING & MONITORING

    GANTT CHART

    EVENT

    ACTIVITY

    NETWORK

    AOAACTIVITY ON ARROW

    AONACTIVITY ON NODE

    SLACK/FLOAT

    CRITICAL PATH

    t= Duration of activity

    1 2

    EF = ES + t LF = LS + t

    LS = LFt Slack = LSES or LF - EF

    ES LS EF LF

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    EXPECTED COMPLETION TIME =+4+

    6

    STANDARD DEVIATION, = [ (b a)/6]2

    NETWORK CONVENTION :

    1. AOAACTIVITY ON ARROW

    2

    1 43

    2. AON - ACTIVITY ON NODE

    B

    A

    C

    D

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    RULES FOR COMPUTING EF(EARLIEST FINISH TIME) & LF(LATEST FINISH

    TIME) IN NETWORK :

    1.MOVING FORWARD FIND EF TIME (CHOOSING THE MAXIMUM AT ACTIVITY

    INTERSECTION.

    2.AT LAST NODE, MAXIMUM EF = LF

    3.USE RETURN PATH TO FIND LF (CHOOSING MINIMUM AT EVERY

    INTERSECTION.

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    b

    Problem 1. Project with following data are to be completed. Draw the network

    and critical path.

    (a)What is the minimum duration of the project?

    (b)Draw the Gantt Chart for early start schedule

    (c) Determine the peak requirement of money and day on which it occurs in

    above schedule

    ACTIVITY PREDECE-

    SSORS

    DURATION

    (DAYS)

    COST

    (RS/DAY)

    A - 2 50

    B - 4 50

    C A 1 40D B 2 100

    E A,B 3 100

    F E 2 60

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    EF LF

    EF LF

    EF LF C=1

    A=2 EF LF

    t=0 E=3 F=2

    B=4

    D=2

    EF LF

    EF LF

    4 4

    7 7

    1

    3

    4

    5

    2

    0 0

    9 9

    4 4

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    Activity Duration(t)

    EF ES =

    EF - t

    LF LS =

    LF - t

    Event

    Slack =

    LSES

    or

    LF - EF

    On

    critical

    path

    A 2 2 0 4 2 2 No

    B 4 4 0 4 0 0 Yes

    C 1 5 4 9 8 4 No

    D 2 6 4 9 7 3 No

    E 3 7 4 7 4 0 Yes

    F 2 9 7 9 7 0 Yes

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    (c)

    Peak requirement of money on 5thand 6thday for Rs.200/- Ans.

    Day Activity Requirement of money(Rupees)

    1 A + B 50 +50=100

    2 A + B 50 + 50=100

    3 B + C 50 + 40=90

    4 B 50

    5 D + E 100+100=200

    6 D + E 100 + 100=200

    7 E 100

    8 F 60

    9 F 60

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    #2 (a)Draw the network using AON system and find critical path for the

    project.

    Activity Designation Predecessor Time in weeks

    Design A - 21

    Built prototype B A 5

    Evaluate C A 7

    Test prototype D B 2

    Write

    equipment

    report

    E C,D 5

    Write method

    report

    F C,D 8

    Write final

    report

    G E,F 2

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    (b)Using 3 time estimates of activities, find probability of completing project

    in 35 weeks.

    Activity Time estimation

    a=Most

    optimistic time

    b=Pessimistic

    time

    m=Most likely

    time

    A 10 12 28

    B 4 4 10

    C 4 6 14

    D 1 2 3

    E 1 5 9

    F 7 8 9

    G 2 2 2

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    EF LF

    EF LF

    EF LF

    EF LF

    EF LF

    EF LF

    28 2836 36

    38 38

    33 36

    28 28

    26 26

    21 21

    A,21

    C,7F,8

    G,2

    E,5D,2

    B,5

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    Activity Duration

    (t)

    EF ES=

    EF - t

    LF LS= LF -t Slack=

    LS-ES

    On

    critical

    pathA 21 21 O 21 0 0 Yes

    B 5 26 21 26 21 0 Yes

    C 7 28 21 28 21 0 Yes

    D 2 28 26 28 26 0 Yes

    E 5 33 28 36 31 3 No

    F 8 36 28 36 28 0 Yes

    G 2 38 36 38 36 0 Yes

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    CRITICAL PATH :

    THERE ARE TWO CRITICAL PATHS THROUGHOUT THE NETWORK :

    ACFG and ABDFG

    (b) Time Estimate

    Activity a b m Expected

    Project completion time

    Te=(a+4m+b)/6

    A 10 12 28 22.33B 4 4 10 8

    C 4 6 14 11

    D 1 2 3 2.5

    E 1 5 9 7

    F 7 8 9 8.16

    G 2 2 2 2

    Total expected completion time = 60.99 weeks

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    #3 A project has following time schedule :

    (a)Construct a PERT network

    (b) Compute critical path and its duration

    (c) Total Float for each activity

    Activity : 1-2 1-3 1-4 2-5 3-6 3-7 4-6 5-8 6-9 7-8 8-9Time

    (Months)

    2 2 1 4 8 5 3 1 5 4 3

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    EF LF EF LF

    4

    1

    ES LS 2 EF LF EF LF EF LF

    2 5 4

    1 8 3

    EF LF EF LF EF LF

    3 5

    (a) PERT network is shown above

    2 7 6 11

    0 0 2 2 7 8 11 12

    1 710 10 15 15

    1

    2

    3

    4

    7

    5

    8

    96

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    (b & c) Critical path : 136 - 9 with duration of 15 months

    Activity Duration EF ES=

    EF - t

    LF LS =

    LF - t

    Total

    Float =

    LF - EF

    1-2 2 2 0 7 5 5

    1-3 2 2 0 2 0 0

    1-4 1 1 0 7 6 6

    2-5 4 6 2 11 7 5

    3-6 8 10 2 10 2 03-7 5 7 2 8 3 1

    4-6 3 4 1 10 5 6

    5-8 1 7 6 12 11 5

    6-9 5 15 10 15 10 0

    7-8 4 11 7 12 8 1

    8-9 3 14 11 15 12 1