# Module G Project

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```							                           Module G                      Chapter 8
Project Management – Critical Path Method (CPM)

Activity           Immediate     Activity Time
Predecessor     (weeks)
a                   –             16
b                   –             14
c                   a               8
d                   a               5
e                   b               4
f                   b               6
g                   c             10
h                  d,e            15

Critical path and minimum project completion time can be found manually or by LP.

Earliest Times          Latest Times       Slack
Time
Act.     (i,j)     Pred.           ESij        EFij     LSij        LFij    Sij
a      (1,2)        –               0         16           0         16     0     critical
b      (1,3)        –               0         14           3         17     3
c      (2,4)        a              16         24         18          26     2
d      (2,5)        a              16         21         16          21     0     critical
e      (3,5)        b              14         18         17          21     3
f      (3,6)        b              14         20         30          36    16
g      (4,6)        c              24         34         26          36     2
h      (5,6)       d,e             21         36         21          36     0     critical

Linear Programming Formulation

Let xi  0 be the earliest event time of node i, for i = 1, 2, 3, 4, 5, 6.

Then we have constraints:
x2  x1  16
x3  x1  14

x6  x5  15
and                               xi  0 , for i = 1, 2, 3, 4, 5, 6.

For the objective, we could just minimise x6, but if we want to ensure that all the xi s are
6
earliest event times we just minimise their sum: Z   xi .
i 1
Project Evaluation and Review Technique (PERT)
Problem 18, pages 346-7, Stone River Textile Mill

Time Estimates (weeks)
Activity     Predecessors          a              m               b
a               –                1              2               3
b               –                2              5               8
c               –                1              3               5
d               a                4             10              25
e               a                3              7              12
f               b              10              15              25
g               c                5              9              14
h              d,e               2              3               7
i             d,e,f              1              4               6
j            d,e,f,g             2              5              10
k             h,i,j              2              2               2

Text page 315: The activity times are assumed to have a beta distribution, which has a
minimum value (a), a maximum value (b) and a most likely value (m).
The mean and variance of a beta distribution can be estimated by simple formulae:
a  4m  b                                 ba
2

mean: t             ,                 variance: v     
6                                      6 

Act.          (i,j)      a          m            b        t (mean)      v (variance)
a           (1,2)       1         2             3        2.000           0.111
b           (1,3)       2         5             8        5.000           1.000
c           (1,4)       1         3             5        3.000           0.444
d           (2,5)       4        10            25       11.500         12.250
e           (2,6)       3         7            12        7.167           2.250
f          (3,7)      10        15            25       15.833           6.250
g           (4,8)       5         9            14        9.167           2.250
dummy         (5,6)       0         0             0        0.000           0.000
dummy         (6,7)       0         0             0        0.000           0.000
h           (6,9)       2         3             7        3.500           0.694
dummy         (7,8)       0         0             0        0.000           0.000
i          (7,9)       1         4             6        3.833           0.694
j          (8,9)       2         5            10        5.000           1.778
k          (9,10)       2         2             2        2.000           0.000

Find critical path and expected (mean) duration.

Use normal distribution approximation to calculate probabilities of duration being in
certain ranges.
Project crashing
Problem 27, page 352

Activity Time          Activity Cost
(weeks)                  (\$)
Activity       (i,j)   Predecessor       Normal      Crash     Normal       Crash
a          (1,2)         –              16            8      2000        4400
b          (1,3)         –              14            9      1000        1800
c          (2,4)         a               8            6       500         700
d          (2,5)         a               5            4       600        1300
e          (3,5)         b               4            2      1500        3000
f          (3,6)         b               6            4       800        1600
g          (4,6)         c              10            7      3000        4500
h          (5,6)        d,e             15          10       5000        8000

Activity       (i,j)         Pred.     Max Crash   Crash Cost
(wks)        (\$/wk)
a           (1,2)           –            8           300
b           (1,3)           –            5           160
c           (2,4)           a            2           100
d           (2,5)           a            1           700
e           (3,5)           b            2           750
f           (3,6)           b            2           400
g           (4,6)           c            3           500
h           (5,6)          d,e           5           600

Crash manually or by LP.

```
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