9. Network Tecniques ................................................................................................... 118-164
9.14 Probability of Completion Time
We assume a normal distribution table (see Appendix) for calculating probability. For this distribution Z – value tables are available. In a normal distribution curve the chances of the project being completed within the mean time is 50%. This is obtained by dividing the area to the left of mean by total areas.
If TE =mean time = 41.7 days and
TS =Actual or Desired Time = 45 days s =1.2 days
Z =
Deviation Standard
Deviation
= 1.2 41.7) –
(45 = 2.75 Standard Deviation = 1.2
This expresses area under
[X mean + 2.75.s (Standard Deviation)]
From Z – value table in Appendix we get 0.497 x 2 = 0.994 Or Area = 99.4
So the probability of competing project in desired time is 99.4%.
Example1: Draw the network diagram for the activities of a maintenance job of a part of refinery
Table 9.1 Refinery date
Activity Description of Activity Predecessor Activity
A Dismantle the pipe line None
B Disassemble other fittings A
C Remove valves and check them B
D Clean the valves and check them C, E
E Clean the pipe lines and others B
F Replace the defective items C, E
G Layout the assembly lines F
H Assemble the valves G
I Do the final connections H, D
J Test the fittings I
Solution
Fig. 9.7 Network
1 3 4
5
6 7
8 9 10 11
A B C
D
E
F G H I J
2
Activity Predecessor Duration (days) Cost (Rs Day)
A - 2 50
B - 4 50
C A 1 40
D B 2 100
E A, B 3 100
F E 2 60
1. What is minimum duration of project 2. Draw a Gantt chart for early start schedule
3. Determine peak requirement money and day on which it occurs above schedule Solution
Fig. 9.10 Network
Critical Pat 1-2-3-4-5
Minimum time = 9 days
Table 9.2 Activity relationship
Activity t ES EF LS LF Event Slack On Critical
(EF – t) (LF – t) (LS –ES) path
(LF –EF)
A 2 0 2 2 4 2 No
B 4 0 4 0 4 0 Yes
C 1 4 5 8 9 4 No
D 2 4 6 7 9 3 No
E 3 4 7 4 7 0 Yes
F 2 7 9 7 9 0 Yes
4 4
3
1
1
2 2
2
4 5
9 9
0 0
A
B
C
D 4
4 4
7 7
F EF LF
Requirement of Money
Day 1 Activity A+B = 50*2
Day 2 Activity A+B = 50*2
Day 3 Activity B+C = 50+40
Day 4 Activity B = 50
Day 5 Activity D+E = 200
Day 6 Activity D+E = 200
Day 7 Activity E = 100
Day 8 Activity F = 60
Day 9 Activity F = 60
Therefore the peak requirement is Rs.200 on day 5 and 6.
Example 5. Draw the network using AON system and find Critical Path for Computer Design Project.
Table 9.3 Activity relationship
Activity Designation Immediate Time in
Predecessors Weeks
Design A _ 21
Built proto type 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
1 2 3 4 5 6 7 8 9 A
B
C
D
E F
0
Solution
Fig. 9.11 Network
There are one two critical path throughout network. The first critical path includes activities A - C - F - G and the second path includes A – B – D – F – G. Only activity E is not on critical path because there is a slack of 3 on it. All other have zero slack.
Table. 9.4 Activity relationship
Activity t ES EF LS LF Slack On Critical
Path
A 21 0 21 0 21 0 Yes
B 5 21 26 21 26 0 Yes
C 7 21 28 21 28 0 Yes
D 2 26 28 26 28 0 Yes
E 5 28 33 31 36 3 No
F 8 28 36 28 36 0 Yes
G 2 36 38 36 38 0 Yes
Example 6: A project has the following time schedule
Activity 1-2 1-3 1-4 2-5 3-6 3-7 4-6 5-8 6-9 7-8 8-9
Time
(months) 2 2 1 4 8 5 3 1 5 4 3
Construct a PERT network and compute
• Critical path and its duration
• Total float for each activity
0 0
2 7 6 11
5 2
2
2
2 2
1
7 7
4 8
8
8
11 12
3
4 3 6 9
15 15
1
1 5
5
1 7 10 10
Also find the minimum number of cranes the project must have for its activities 2-5, 3-7 and 8-9 without delaying the project. Then is there any change required in the PERT network.
Solution Steps
1. Moving forward find EF times (choosing the Maximum at activity intersection) 2. Maximum EF = LF = Critical Path Time.
3. Return path find LF (Choosing the Minimum at activity intersection) 4. Note LF, EF from network (Except activity intersections)
Fig. 9.12 Network Table 9.5 Activity relationship
Activity Duration ES EF LS LF TF
1 –2 2 0 2 5 7 5
1 –3 2 0 2 0 2 0
1 – 4 1 0 1 6 7 6
2 – 5 4 2 6 7 11 5
3 – 6 8 2 10 2 10 0
3 – 7 5 2 7 3 8 1
4 – 6 3 1 4 5 10 6
5 – 8 1 6 7 11 12 5
6 – 9 5 10 15 10 15 0
7 –8 4 7 11 8 12 1
8 – 9 3 11 14 12 15 1
Critical path is 1 – 3 – 6 – 9 with duration 15 Months Minimum number of cranes
• Finish 3 – 7 at 7 with one crane
• Finish 2 –5 at 7 + 4 =11 with same crane
• Finish 5 – 8 at 11 + 1 =12 with same crane
• Finish 8 – 9 at 12 +3 = 15 with same crane
21 21
26 26 28 28 33
38 38
36 G, 2
28 28
C, 7
36 36
F, 8
A, 21
B, 5 D, 2 E, 5
Therefore one crane will be sufficient if start time of following activities are:
• Activities 2 – 5 — 7
• Activities 5 –8 — 11
• Activities 8 – 9 — 12
Example 7. Considering a small maintenance project as given below, plan it with the help of CPM.
Table 9.6 Activity relationship
Activity Duration Predecessor
A 11 Nil
B 3 Nil
C 5 Nil
D 0 A
E 2 A
F 1 B
G 12 B
H 6 C, F
I 7 D, H
J 3 E
Compute the following for each job:
Early start time (ES), Late start time (LS), Early finish time (EF), Late finish time (LF), Total Float (TF), Free float (FF), Minimum total duration of the project. If all the jobs have been scheduled to start as early as possible and that the work has been in schedule up to the end of week 5. There is strike on week 6 causing a delay of 1 week. Draw a CPM diagram for the jobs remaining to be done when work resumes on week 7.
Solution
Network representing the given project:
Fig. 9.13 Network
11 11 13 15
0 0
1 4 5
5 5
6
7 2
3 3 4
18
C 5 H 6
17 F 1
EF LF A
11
J 3 E 2
D 0
11 11
B 3 G 12
18
The ES, EF, LS, LF, TS, FS have been computed as discuss earlier and entered in the below.
Table 9.7 Activity relationship
Activity Duration ES EF LS LF TS Slack FS
at Head
A 11 0 11 0 11 0 0 -
B 3 0 3 1 4 1 1 -
C 5 0 5 0 5 0 0 -
D 0 11 11 11 11 0 0 -
E 2 11 13 13 15 2 2 -
F 1 3 4 4 5 1 0 1
G 12 3 15 6 18 3 0 3
H 6 5 11 5 11 0 0 -
I 7 11 18 11 18 0 0
J 3 13 16 15 18 2 0 2
FS for ending activities will be taken as 0 (Zero) The minimum duration of the project is 18 weeks.
Figure shown is a squared CPM network, which depicts the jobs yet to be done on week 7.
Example 8: For Network given find Total Float (TF), Free Float (FF) and Independent Float (IF)
Fig.9.14 Network D 0
A 6
H 6 G 12
E 2 K 3
J 7
Weak No 7 8 9 10 11 12 13 14 15 16 17 18 19
1
2
3
4
5
8 10
5 2
3
Solution
Fig. 9.15 Network
Table 9.8 Activity relationship
Activity T ES EF LS LF TF Slack at FF Slack at IF
(1+2) (5-1) (4-2) head (6-7) tail (8-9)
(5-3) event evet
1 2 3 4 5 6 7 8 9 10
1 – 2 8 0 8 0 8 0 0 0 0 0
1 – 3 4 0 4 6 10 6 6 0 0 0
2 – 4 2 8 10 13 15 5 5 0 0 0
2 – 5 10 8 18 8 18 0 0 0 0 0
3 – 4 5 4 9 10 15 6 5 1 6 -5 (taken
as 0)
4 – 5 3 10 13 15 18 5 0 5 5 0
Total float = 0 on the Critical Path, which is 1 – 2 – 5 Example 9: Draw PERT Network
• Find expected time and variance for each activity
• Probability of completing project in 32 days.
• Total project duration
Table. 9.9 Activity relationship
Activity Estimated Time
TO TM TP
1 – 2 6 9 18
1 – 3 5 8 17
2 – 4 4 7 22
3 – 4 4 7 16
4 – 5 4 10 22
2 – 5 4 7 10
3 - 5 2 5 8
1 2
2
4 4
4
10 10 15
3
3
8 8
10
5
5 8
18 18
0 0
EF LF
Solution
Table 9.10 Activity relationship
Activity Direction (Expected Time) ES EF LS LF Float Variance 1 – 2 (6 + 4x9+18) / 6 = 10 0 10 0 10 0 ((18 – 6)/ 6) 2 = 4 1 – 3 (5 + 8 x 4 + 17) / 6 = 9 0 9 16 25 16 ((17 – 5)/ 6) 2 = 4 2 – 4 (4 + 4 x 7 + 22) / 6 = 9 10 19 10 19 0 ((22 – 4)/ 6)2 = 9 3 – 4 (4 + 4 x 7 +16) / 6 = 8 9 17 11 19 2 ((16 –4)/ 6)2 = 4 4 – 5 (4 + 4 x 10 + 22) / 6 =11 19 30 19 30 0 ((22- 4)/ 6) 2 = 9 2 – 5 (4 + 4 x 7 + 10) / 6 = 7 10 17 23 30 13 ((10 –4)/ 6)2 = 1 3 - 5 (2 + 4 x 5 + 8) / 6 = 5 9 14 25 30 16 ((8 – 2)/ 6)2 = 1
Critical Path = 1 – 2 – 4 – 5
Variance along critical path = Σσ2 = σ21-2+ σ22-4+ σ24-5
= σ = 221/2 = 4.69
Z = (TS – TE)/σ = 32 – 30)/4.69 = 0.42 From Normal Distribution Table: P = 65.54 %
Example 10: Using the three time estimates of the activities draw the AON network for Computer Design Project and find probability of completing project in 35 weeks.
Solution
Table 9.11 Activity relationship
Activity Designation Immediate Predecessors Time Estimates
a b c
Design A - 10 22 28
Built Prototype B A 4 4 10
Evaluate equipment C A 4 6 14
Test prototype D B 1 2 3
Write equipment report E C, D 1 5 9
Write method report F C, D 7 8 9
Write final report G E, F 2 2 2
Table 9.12 Activity relationship
Designation Time Estimates Expected Time (te) Activity Variances (a+4m+b)/6 (b-a) 2/ 6
a b c
A 10 22 28 21 9
B 4 4 10 5 1
C 4 6 14 7 2 7/9
D 1 2 3 2 1/9
E 1 5 9 5 1 7/9
F 7 8 9 8 1/9
G 2 2 2 2 0
Fig. 9.16 Network
Where
a = Optimistic Path m = Most Likely Time c = Pessimistic Time
te = Expected Time=(a + 4.m + b)/6 Variances= Σσ2 = Σ (b – a)2 /6
The project network was created the same as done previously with the only difference being that the activity times are weighted averages. We determine critical path as before taking these values as if they were single numbers. The difference between single time estimate and three time estimates (optimistic, most likely, pessimistic) is in computing probabilities for completion.
There are two critical paths throughout network. The first critical path includes activities A, C, F, G the second path includes A, B, D, F, G. Only activity E is not on critical path. Using Conservative approach we choose largest total variance which needs maximum attention. So variance associated with activities A, C, F, and G.
For Critical Path Σ σ2 = 9 + 2 7/9 + 1/9 + 0 =11.89 Probability of completing project in 35 Weeks.
Expected Completion Time (TE )= 38 Weeks D = Actual Completion Time = 35 Weeks Σσ2 = Variance =11.89
Z = (D–Te)/ (Σσ2 Critical Path)½ = (35 – 38)/(11.89)½ = – 0.87 P (D<35) = P (Z < –0.87) = P (Z > –0.87) Since symmetric
= 0.5 – P (0 < Z < –0.87)
= 0.5 – 0.31
= 0.19
28 28
28 28
21 21 38 38
33 36
21 26
36 36
G A
B
C F
D E
ET = 7
= 27/9 σ2
ET = 7
2 = 1 ET = 21 σ
σ2 = 9
ET = 2
= 1/9 σ2
ET = 8
= 1/9 σ2
ET = 5
= 17/9
σ2 ET = 2
2 = 0 σ
From the Normal Distribution Tables we find that at value of Z = - 0.87 gives a probability of 0.19. This means project Manager has only 19 Percent Probability of completing the critical path ACEG. Since there is another critical path and other paths that might become critical, the probability of completing the project in 35 Weeks is actually less than 0.19.
Fig.9.17 Normal distribution curve
Example 11: Consider a project for which the time estimates are given in the table below construct the PERT network what in the critical path. Find the probability of completing the project before 23 days.
Table 9.13 Activity relationship Estimated Time (Days)
Most Optimistic T o Most Likely Time Tm Most Pessimistic T p
1-2 2 5 8
1-3 1 4 7
2-3 0 0 0
2-4 2 4 6
2-6 5 7 12
3-4 3 5 10
3-5 3 6 9
4-5 4 6 10
4-6 2 5 8
5-6 2 4 6
Solution
First calculating the estimated average expected time and variance of each activity.
Activity
32% Area
19% Area
–0.87 0 0.87
Fig. 9.19 Normal distribution
DIFFERENCE BETWEEN CPM AND PERT
CPM originated from construction project while PERT evolved from R & D projects. Both CPM and PERT share same approach for constructing the project network and for determining the critical path of the network.
There are some basic differences between PERT and CPM. PERT is associated with uncertainty in the time estimates for activity while in CPM these estimates are treated as fairly deterministic. CPM is also extended to cost-time trade-off decisions. As the project completion time is squeezed the time for the lowest project cost is the optional decision for project planning PERT is considered event oriented while CPM is mainly activity oriented.
Table 9.15
PERT CPM
1. Time estimate are probabilistic Time estimate are deterministic with with uncertainty in time duration. known time durations. Single time
Three time estimates estimate.
2. Event oriented. Activity oriented.
3. Focused on time Focused on time-cost trade off.
4. More suitable for new projects. More suited for repetitive projects.
5. Most costly to maintain Easy to maintain
6. Suitable for complex projects Suitable where problems of resource where uncertain timing like allocation exist like construction projects research programmes
7. Dummy activity required for Use of dummy activity not necessary proper sequencing