Question 1.
Figure 1. Represents all activities, starting from activity 1 to 12. Event 1 is the beginning and event 12 is the end of my task assigned (thick line represent critical path) 
1.
Describe
PERT/CPM network analysis, applications, advantages and disadvantages.
Project
evaluation and review technique (PERT) is a project – scheduling tool applied to ensure sequence of different
activities eventually lead to a desired completion of a project in
consideration. PERT techniques in network analysis is a method of minimizing
trouble spots, production bottlenecks, delays associated costs, interruptions
and reduce misallocations.
Advantages:
PERT
is a planning and control technique that uses a network for scheduling and budgeting
(time & finance) to accomplish a task.
Disadvantage:
The
disadvantage in PERT network is that, an event cannot be accomplished until and
unless all activities in a network have been completed. So in a PERT network,
we are actually estimating the total project time. The path of the longest
duration is called critical path and activities lying this path are critical
activities. A delay in any of the activities in the critical path will result
in delay, extra costs and resources, redesign, and re – routing of the entire
project.
12.
What
is the difference between an event an activity in a PERT/CPM network?
Activity
is the task allocated to be accomplished towards the total completion oif the
project on time. It is represented by an arrow. The tail of the arrow
represents the beginning of the activity and its head represents its
completion. Whereas,
Event
is designation the beginning of one activity and ending of another activity. So
the starting and ending points of activities are events.
23.
Describe
how you would calculate earliest event (T_{E}) and latest event (T_{L})
times and define what they represent.
The
earliest event time (T_{E}) is the earliest time at which the
activities originating from an event can be started.
The
latest event time (T_{L}) is the latest event time which activities terminating
an event can be competed on scheduled.
So
the earliest event time is calculated by adding earliest event time at the tail
of activity arrow with the duration or earliest time of activity. Note that
when there is more than one activity flowing into an event; choose the maximum
value of T_{Ej }calculated for that event.
The
latest event time is calculated by subtracting the T_{L }of the very
last event by the duration/time for activity ij. Subsequently doing the same for
rest of the events. When there are more than activities lead to an event,
calculate and get the minimum value of T_{L}
_{}
14.
Determine
the critical path of the exploration project schedule.
The
critical path starts from event 1 to event 3 then to event 5 to 6, from
event 6 to event 11 and ends at event 12. At these events the T_{E}
value is equal to T_{L} at each event given above.
15.
What
is a float and slack in PERT/CPM network and state why is having these
advantageous in project scheduling?
Float
in PERT network is time available in the non – critical path that could be
reallocated towards an activity in the critical path. It is advantageous having
float because it represents underutilized time or underutilization of
resources. Not only that but also float represents flexibility of an activity
and disappearance of float signifies a loss of flexibility for non – critical
path activities.
Slack
in PERT network is the difference between the earliest and latest times of any
event. Slack is important as it help us
to see whether it is a critical or non – critical path by analyzing the T_{E}
and T_{L} values.
16.
Calculate
head slack, tail slack, float, free float and independent float for one non –
critical path activity (ij).
Tail
slack Ts = T_{si }=_{ }T_{Li}  T_{Ei} = 15– 11
= 4 weeks (event 7)
Head
slack HS = Tj = T_{Lj}  T_{Ej} = 29 – 18 = 11 weeks (event 7)
Float
F = (TLj  TEi) – teij = (29 – 11) – 7 = 11
weeks
Free
float FF = (TEj – Tei)  teij = (18 –
11) – 7 = 0
Independent
float IF = FF – TS = 0 – 4 =  4 weeks
Note:
a negative independent float is taken as zero for all practical purposes.
17.
Briefly
commend on your understandings of three extremes of expected times : (a)
optimistic, (b) pessimistic and (c) most
likely time (m) as represented by normal probability distribution curves.
·
Optimistic – is the distribution that
satisfy most circumstances at shortest times if execution goes very well.
·
Pessimistic – is the distribution that satisfy most
circumstances at longest times if everything goes bad.
·
Most likely  is the distribution that
satisfies most circumstances at normal times or at middle grounds if execution is
normal. In other words, it is the mean
of the distribution.
18.
Complete
the following table by calculating the variance (Ïƒ^{2}).
Predecessor

Successor

a

M

b

te

Ïƒ^{2}

1

2

3

2.5

5

3

0.11

1*

3

5

4.5

13

6

1.78

2

4

7

2

9

4

0.11

2

6

10

10.75

7

10

0.25

4

6

8

1

12

4

0.44

3*

5

5

6.5

11

7

1.00

5*

6

6

7.5

6

7

0.00

3

7

9

2.75

10

5

0.03

7

8

7

2

9

4

0.11

8

9

4

5

12

6

1.78

6*

11

8

12.5

14

12

1.00

9

11

2

5.75

17

7

6.25

7

10

11

5.75

8

7

0.25

10

12

3

14.25

12

12

2.25

11*

12

6

8

6

9

0.00

*critical
path.
29.
Calculate
Z (number of standard divisions), find corresponding probability value in Z
table a conclusion on the probability of implementing your schedule as planned.
Sum
of critical path Ïƒ^{2} is = Ïƒ^{2} _{(13)} + Ïƒ^{2}_{(35)}
+ Ïƒ^{2}_{(56) }+ Ïƒ^{2}_{(611)} + Ïƒ^{2}_{(1112)}^{
}
1.78 + 1
+ 0 +
1 + 0
Ïƒ^{2} = 3.78
Ïƒ =
(3.78)^{0.5} = 1.9442
Assume
X = 44 weeks, and Âµ = 41 weeks
Z
= X  Âµ = (44 – 41)/1.9442 = 1.5431
Ïƒ
Using
the table (given), probability of success is: Z value of 1.5431 corresponds to probability
of 0.4382 from the table. This means there is 0.9382 (0.4382+0.5) probability
or 94% chance of completing the project in 45 weeks or less.
Basing
on the probability analysis, I am too optimistic that the project will be
completed in 45 weeks or less.
110.
Calculate
the total float for the non – critical path.
Total
float F_{T} = (T_{Lj} – T_{Ei}) – tij = (29 – 11) – 7 =
11 weeks (from event 7 to event 10)
Note
that here, only one total float for only one non – critical path.
_{}