107/8
| Activity
| Optimistic
| Most Likely
| Pesimistic
| Average
| variance
| | A
| 5
| 8
| 11
| 8
| 1.00
| | B
| 4
| 8
| 11
| 7.83
| 1.36
| | C
| 5
| 6
| 7
| 6
| 0.11
| | D
| 2
| 4
| 6
| 4
| 0.44
| | E
| 4
| 7
| 10
| 7
| 1.00
|
Path Total Expected Time
A-C 8+6 = 14.00
A-D-E 8+4+7 = 19.00
B-E 7.83 + 7 = 14.83
The critical path is A-D-E since it is the longest path. Expected project duration is 19.00 (sum of the expected durations of critical activities)
- b) Probablity of completing the project in 21 days: Variance of the project duration is 2.44 (sum of the variances of critical activities)
�� �������������������������������������������������� ������������������������� z = (T - TE)/ standard deviationof project duration = (21 - 19) / (2.440.5) = 1.28
Assuming normal distribution applies, the probability that the projectcan be completed within 21 days is
about 90%.
- c) z = (T - TE)/ standard deviation of project duration = (17 - 19) / (2.440.5) = -1.28 . Probablity of finishing projectwithin 17 days is 10%
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108/10
| Activity
| Start
| A
| B
| C
| D
| E
| F
| Finish
| | Duration
| 0
| 5
| 3
| 2
| 5
| 4
| 7
| 0
| | ES
| 0
| 0
| 0
| 5
| 3
| 8
| 8
| 15
| | EF
| 0
| 5
| 3
| 7
| 8
| 12
| 15
| 15
| | LS
| 0
| 4
| 0
| 9
| 3
| 11
| 8
| 15
| | EF
| 0
| 9
| 3
| 11
| 8
| 15
| 15
| 15
| | Slack
| 0
| 4
| 0
| 4
| 0
| 3
| 0
| 0
|
The critical Path is Start-> B -> D -> F -> Finish
Project Duration is 15
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110/17
Critical path, using normal durations, is A-D-G with a duration of 29. Direct cost of this schedule is $13,050.
|
| start
| A
| B
| C
| D
| E
| F
| G
| H
| I
| Finish
| | Dur
| 0
| 12
| 13
| 18
| 9
| 12
| 8
| 8
| 2
| 4
| 0
| | ES
| 0
| 0
| 0
| 0
| 12
| 12
| 13
| 21
| 24
| 18
| 29
| | EF
| 0
| 12
| 13
| 18
| 21
| 24
| 21
| 29
| 26
| 22
| 29
| | LS
| 0
| 0
| 6
| 7
| 12
| 15
| 19
| 21
| 27
| 25
| 29
| | LF
| 0
| 12
| 19
| 25
| 21
| 27
| 27
| 29
| 29
| 29
| 29
| | S
| 0
| 0
| 6
| 7
| 0
| 3
| 6
| 0
| 3
| 7
| 0
|
Note: Since the project is rather simple we could have calculated the critical path by enumerating all the paths to select the longest one as the critical path.
A-D-G 29
A-E-H 26
B-F-H 23
C-I 22
Clearly, the path A-D-G must be reduced by four time units, A-E-H one to bring all the paths to
within 25 time units.
Cost analysis:
| Activity
| Crash Cost/Day
| Maximum Crash
| | A
| 600
| 1
| | B
| 112.5
| 4
| | C
| 750
| 2
| | D
| 250
| 4
| | E
| 225
| 2
| | F
| 350
| 1
| | G
| 200
| 2
| | H
| 200
| 1
| | I
| 900
| 2
|
| Trial
| Crash
Activity
| Resulting
Critical Path
| Time reduction
| Project Duration
| Crash Cost
| | 0
| -
| A-D-G
| -
| 29
| -
| | 1
| G
| A-D-G
| 2
| 27
| 400
| | 2
| D, H
| A-D-G
A-E-H
| 2, 1
| 25
| 500 +200
|
Note: In trial 2 if we only crash D for 2 days the project durationwill go down to 26, since A-E-H has a normal length of 26. To get the project length down to 25, A-E-H must be shortened to 25 days-- the cheapest activity among A,E,H is H which is crashed for 1 day addinga cost of $200. Thus the project cost is now 13,050 + $700 +400 = $14,150
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What is the probability that the project will be completed by 27 days?
Therefore:•Probability of completing the project within 27days is equal to the area under the normal curve tothe left of z = –0.82. Referring to Table (a), the probability is about 21percent.
How do you calculate project completion probability?
1 Answer. 3.5kviews. written 6.2 years ago by teamques10 ★ 36k.. Step 1 :- For critical activity only. σ=tp−to6. Activity. T0. Tm. ... . Step 2:- Project duration = 31 days. Critical path =1-2-4-6-7-8.. Step 3:- The probability of completing the project in 35 days is. z = scheduled time-project durationϵ S.T.= 35 Days. P.D.= 31 Days.. What is the probability of completing the project in 23 weeks?
(C - 20) / 2.43 = 0.99 C = 2.44 weeks The project will be completed in 22.4 weeks (≈ 23 weeks) if the probability of completing the project is 0.84.
What is the probability of meeting a desired 26 weeks completion time?
Probability analysis
We can also carry out this probability calculation for a number of other completion times, as below for 25 weeks and 26 weeks. Note especially here how the probability of completing within the expected completion time of 26 weeks is only 50%.
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