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    ISB ITPM Assignment 7.2 Keithdsouza

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    Keith D'souza
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    ISB ITPM Assignment 7.2 Keithdsouza

    Uploaded by

    Keith D'souza
    4 views3 pages
    ASSSSSS

    Original Title

    ISB_ITPM_Assignment_7.2_keithDsouza

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    ASSSSSS

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    4 views3 pages

    ISB ITPM Assignment 7.2 Keithdsouza

    Uploaded by

    Keith D'souza
    ASSSSSS

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 3

    1.

    Network diagram:

    3 C 9
    9 D 17
    0 A 3 0 0
    0 0 17 E 26
    0 0 3 6 9
    9 8 17 0 0
    0 3 3
    17 9 26

    3 B 10 26 F 31

    16 16 0 0 17 G 23

    19 7 26 26 5 31 8 8

    25 6 31

    26 H 31

    1 0
    31 J 37
    10 I 12
    27 4 31
    0 0
    19 19
    31 6 37
    29 2 31

    37 K 40

    0 0

    37 3 40

    Legend
    ES ID EF

    Slack Slack

    LS Duration LF
    2. Critical Path

    Task A Task C Task D Task E

    Task K Task J Task F

    3. Calculation for (Te) and Variance of all the tasks:

    Weighted (Te) of
    Best Most Worst Activity the Variance of
    Case Likely case Critical Estimate critical the critical
    ID Tasks (To) (Tm) (Tp) Activity (Te) Activities Variance activities
    A Requirements 2 3 4 Yes 3 3 0 0
    B Market assessment 4 7 10 No 7 1
    C Design 5 6 9 Yes 6 6 0 0
    D Development 6 7 16 Yes 8 8 3 3
    E Testing 7 9 10 Yes 9 9 0 0
    F Revising 4 5 6 Yes 5 5 0 0
    G Documentation 3 6 10 No 6 1
    H Quality Assurance 2 4 7 No 4 1
    I Pricing 2 2 2 No 2 0
    J Production 3 4 14 Yes 6 6 4 4
    K Distribution 2 3 4 Yes 3 3 0 0
    Total:7

    std
    TE=40 deviation=2.64

    The estimated completion time is 40 weeks.


    4. The probability that the project could be completed within the deadline of 35 weeks:

    Ts= 35 weeks

    Te=40 weeks

    Std deviation= 2.6

    Area under curve= z= 35-40/2.6 = -1.92

    Probability = NORMSDIST(Z)= 0.02743

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