PERT Exercise2 Q&A
19
Uncertain activity duration (P14) Consider the IS development project network given below: 1 2 4 3 5 6 A C B F D E G The systems analyst has made estimates of the optimistic, most probable, and pessimistic times (in days) for these activities as follows:
description
PERT
Transcript of PERT Exercise2 Q&A
Uncertain activity duration (P14)
Consider the IS development project network given below:
1
2
4
3
5 6
A
C
B F
D
E
G
The systems analyst has made estimates of the optimistic, most probable, and pessimistic times (in days) for these activities as follows:
Uncertain duration estimates (P14)
Activity Optimistic Most Probable Pessimistic
A 5 6 7 B 5 12 13 C 6 8 10 D 4 10 10 E 5 6 13 F 7 7 10 G 4 7 10
a) Find the critical path.b) How much slack time, if any, is there for activity C?c) Determine the expected project completion time and the
variance.d) Find the probability that the project will be completed in
30 days or less.
Complete this table (P14)
Most ExpectedActivity Optimistic Probable Pessimistic Time
Variance -----------------------------------------------------------------------------------
A 5 6 7 B 5 12 13C 6 8 10D 4 10 10E 5 6 13F 7 7 10G 4 7 10Remember, the expected time formula is t = (a+4m+b)/6 where a is optimistic, m is most probable, and b is pessimistic estimate.The variance formula is: 2 = [(b - a)/6]2
Expected time and variance (P14)Solution
Most Expected
Activity Optimistic Probable Pessimistic TimeVariance
-----------------------------------------------------------------------------------A 5 6 7 6.0 .11B 5 12 13 11.0 1.78C 6 8 10 8.0 .44D 4 10 10 9.0 1.00E 5 6 13 7.0 1.78F 7 7 10 7.5 .25G 4 7 10 7.0 1.00Remember, the expected time formula is t = (a+4m+b)/6 where a is optimistic, m is most probable, and b is pessimistic estimate.The variance formula is: 2 = [(b - a)/6]2
Uncertain time estimates (P14a-d)
1
2
4
3
5 6
A [0,6]
6 [0,6]
C [6,14]
8 [7.5,15.5]
B [0,11]11 [4,15]F [15,22.5]
7.5 [15,22.5]
D [6,15]
9 [6,15]
E [14,21]
7 [15.5,22.5]
a) Critical path: A-D-F-Gb) Slack for activity C = LF - EF = 15.5 - 14.0 = 1.5 daysc) Expected completion time = 29.5 days variance = .11 + 1.00 + .25 + 1.00 = 2.36 ( 2.36 = 1.54)
d) z = (30 - 29.5)/1.54 = (0.5)/1.54 = .32 Probability = .6255
G [22.5,29.5] 7 [22.5,29.5]
Analyst planning a training (P15)
Activity Description Immediate Optimistic Most Pessimistic
Predecessors Prob. A Plan topic - 1.5 2 2.5 B Arrange speakers A 2 2.5 6 C List locations - 1 2 3 D Select location C 1.5 2 2.5 E Speaker travel plan B,D 0.5 1 1.5 F Finalize speakers E 1 2 3 G Prepare brochure B,D 3 3.5 7 H Take reservations G 3 4 5 I Last-minute details F,H 1.5 2 2.5
a. Show the PERT/CPM network for this project.b. What are the critical path activities and expected project
completion time?c. Prepare the activity schedule for this project.d. If the analyst wants a 0.99 probability of completing the project
on time, how far ahead of the scheduled meeting date should she begin working on the project?
Compete this network (P15b)
1
2
3
A
C
Training plan exercise (P15b)
1 74
2 5
3 6
8
A
C
B
D
E F
G H
I
Training plan exercise (P15b)
1 74
2 5
3 6
8
A2
C2
B3
D2
E1
F2
G4 H
4
I2
Training plan exercise (P15b)
1 74
2 5
3 6
8
A[ 0
,2]
2 [ 0
,2]
C[0,2]
2 [1,3]
B[2,5]
3 [2,5]
D[2
,4]
2 [3
,5]
E[5,
6]1[
10,1
1]
F[6,8]
2[11,13]
G[5,9]
4 [5,9] H[9
,13]
4 [9
,13]
I [13,15]2[13,15]
Training plan exercise (P15 c)
Activity ES LS EF LF Slack C-PathA 0 0 2 2 0 YesB 2 2 5 5 0 YesC 0 1 2 3 1D 2 3 4 5 1E 5 10 6 11 5F 6 11 8 13 5G 5 5 9 9 0 YesH 9 9 13 13 0 YesI 13 13 15 15 0 Yes
Training plan exercise (P15b)
Activity Expected Time VarianceA 2 .03B 3 .44C 2 .11D 2 .03E 1 .03F 2 .11G 4 .44H 4 .11I 2 .03
Critical path activities A-B-G-H-IExpected completion time = (2+3+4+4+2) = 15 weeks.Example: tA = [1.5+4(2)+2.5]/6 = 12/6 = 2The variance formula: 2 = [(b - a)/6]2
Training plan exercise (P15d)
Desired Probability
p = .01
…….…….……...
Critical path variance: = 2
= .03 + .44 + .44 + .11 + .03 = 1.02
At .01 z = 2.33 ( see Table)z = (T-15)/1.02
17.38 weeks ahead of the scheduled completion date.15 T
Training plan exercise (P15d)
Desired Probability
p = .01
…….…….……...
Critical path variance: = 2
= .03 + .44 + .44 + .11 + .03 = 1.02
At .01 z = 2.33 ( see Table)z = (T-15)/1.02T-15 = (2.33)1.02T-15 = 2.3766T = 15 + 2.3766T = 17.38
17.38 weeks ahead of the scheduled completion date.15 T
Critical and non-critical paths (p16)
Q1- Why the probability of completing a project on time is based on the analysis of the critical path?
Q2- In what case, if any, would it be desirable to make the probability computation for a non-critical path?
Critical and non-critical paths (p16)
A1- Critical path is the longest path and generally will have the lowest probability of being completed by the desired time. The non-critical paths should have a higher probability of being completed on time.
A2- It may be desirable to consider the probability calculation for a non-critical path if the path activities have little slack, if the path completion time is nearly the same as the critical path completion time, or the path activity times have relatively high variances. When all of these situations occur, the non-critical path may have a probability of completion on time that is less than the critical path.
Project teams• Form teams of 3 members
• Give your team a name or a number
• Try to sit close to your team members to do team assignments
• Brainstorm and finalize project topic
• Write up your project proposal and hand it in by next week.
• Use word processor for all assignments and course works
