answer the survey Quations .short 3 answers
A Comparison of AON and AOA Representations
Activity on Activity Activity on Node (AON) Meaning Arrow (AOA)
A comes before B, which comes before C.
(a) A B C BA C
A and B must both be completed before C can start.
(b)
A
C C
B
A
B
B and C cannot begin until A is completed.
(c)
B
A
C A
B
C
A Comparison of AON and AOA (Contd.)
Activity on Activity Activity on Node (AON) Meaning Arrow (AOA)
C and D cannot begin until both A and B are completed.
(d) A
B
C
D B
A C
D
C cannot begin until both A and B are completed; D cannot begin until B is completed. A dummy activity is introduced in AOA.
(e) CA
B D
Dummy activity A
B
C
D
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Activity Relationship in a Network
u
1 4
2
3
A: REGISTER
FOR CLASS
B: PAY TUITION E: TAKE EXAMS/OBTAIN GRADE
C: ATTEND
CLASS
B & C must be done before
E can begin
D: TERM PROJECT
CPM Problem u The Selby Builders, Inc. will construct a new and unique office
building for a Glassboro based CPA firm. The project will involve several activities; some can be done concurrently and others must be done in sequence. Based on previous knowledge and experience, Selby’s management has identified the activities and the order in which they must be completed. Though the new office building will be unique, Selby’s construction manager has made a best guess of the completion time for each activity.
u (I) The CPA firm wants to know how long would it take to construct the building. (II) Selby’s subcontractors must know when to schedule each activity. (III) In addition, Selby would like to know how long each activity can be put off without delaying the entire project.
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Selby’s Construction Data
ACTIVITY DESCRIPTION IMMED. PRED. TIME ESTIMATES (DAYS)
A ORDER STRUCTURAL ------- 2 MATERIALS
B OBTAIN STRUCTURAL ------- 4.5 LABOR
C EXCAVATE ------- 5 D POUR FOUNDATION C 6.5 E RECEIVE STRUCTURAL A 4
MATERIALS F FRAME STRUCTURE B, D, E 30 G INSTALL PLUMBING F 20 H INSTALL ELECTRICALS F 10 I INTERIOR FINISHING G, H 15 J EXTERIOR FINISHING G 14.5 K CLEAN-UP I, J 5
Selby Builders, Inc.
B = 4.5 F = 30 K = 5
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Selby Builders, Inc.
B = 4.5 F = 30 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5
B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74
C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5
Selby Builders, Inc.
B = 4.5 F = 30 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5
B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74
C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5
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Selby Builders, Inc.; Designation of the Critical Path
B = 4.5 F = 30 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5
B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74
C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5
u Write activity times on the arrows using labels:
[x, y ] Earliest Start Time Earliest Finish Time
On a chronological scale
u Determine the project completion time using the earliest start and earliest finish times for each activity
u Using the project completion time identified above, find the latest start and latest finish times of each activity by making a backward pass through the network
u Use the difference between the latest start and earliest start times (or latest finish and earliest finish times) to identify the slack time (float) available for each activity.
u The Critical path activities are activities with zero slack times.
CPM Procedure
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Selby Builders, Inc.; Designation of the Critical Path
B = 4.5 F = 30 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5
B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74
C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5
(0, 4.5)
(11.5, 41.5) (76.5, 81.5)
Selby Builders, Inc.; Designation of the Critical Path
B = 4.5 F = 30 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5
B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74
C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5
(0, 4.5)
(11.5, 41.5) (76.5, 81.5)
(7, 11.5)
(11.5, 41.5) (76.5, 81.5)
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Project Scheduling by Program Evaluation and Review
Techniques PERT u Program Evaluation Review Technique (PERT)
is used to plan the scheduling of individual activities that make up a project.
u PERT can be used to determine the earliest/latest start and finish times for each activity, the entire project completion time and the slack time for each activity. These calculations help in scheduling activities.
u A critical path for the network is a path consisting of activities with zero slack.
PERT Network
u A PERT network can be constructed to model the precedence of the activities.
u The arcs of the network represent the activities.
u The nodes of the network represent points in time when an activity or a group of activities have been completed.
u The nodes are numbered so that each activity begins at a lower numbered node and ends at a higher numbered node.
u Dummy activities having 0 completion times can be created to help indicate that the proper set of activities has been completed prior to the start of another activity.
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u In the three-time estimate approach, the time to complete an activity is assumed to follow a Beta distribution.
u An activity’s mean completion time is:
u An activity’s completion time variance is:
– a = the optimistic completion time estimate
– b = the pessimistic completion time estimate
– m = the most likely completion time estimate
PERT
mean expected time 4 6
: a m b
t
2 variance :
6
b a v
PERT Activity Times
Probability of 1 in 100 of > b occurring
Probability of 1 in 100 of < a occurring
P ro
b a
b il
it y
Optimistic Time (a)
Most Likely Time (m)
Pessimistic Time (b)
Activity Time
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PERT
u In the three-time estimate approach, the critical path is determined as if the mean times for the activities were fixed times.
u The overall project completion time is assumed to have a normal distribution with mean equal to the sum of the means along the critical path and variance equal to the sum of the variances along the critical path.
PERT makes two more assumptions: Total project completion times follow a
normal probability distribution
Activity times are statistically independent
Expected Project Duration and Variance
u Expected project time (tp) is the sum of the expected times of the critical path activities.
u Project variance (vp) is the sum of the critical path activities’ variances (v1 + v2 + v3 + v4 + … )
u The expected project time is assumed to be normally distributed (based on central limit theorem).
u In the example, expected project time (tp) and variance (vp) interpreted as the mean () and variance (s2) of a normal distribution.
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A Probability Example
uYou’re a project planner for a special aircraft project. The project has an expected completion time () of 40 weeks, with a standard deviation (s) of 5 weeks. What is the probability of finishing the project in 50 weeks or less?
Converting to Standardized Normal Distribution
Refer to Appendix A.1 (Page # 815)
= 40
s = 5
50 X
Normal Distribution
Z X T=
- =
- =
s 50 40
5 2 0.
z= 0
s
Z2.0
Standardized Normal Distribution
Z X T= =
-
s 50 40
5 2 0
Standardized Normal Distribution
- = .
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What is the probability that the project will be completed in 50 weeks
If the standard deviation
Is 5 weeks?
(50 - 40)/(5) = 2
Z-Value = 0.4772
Thus the probability
= 0.5 + 0.4772 = 0.9772
Probability of Completion
Selby’s Construction Data
ACTIVITY IMMED. PRED. TIME ESTIMATES EXPECTED VARIA- a m b TIME NCE
A ------- 1 2 9 3 1.78 B ------- 2.5 4.5 9.5 5 1.36 C ------- 2 5 14 6 4 D C 4 6.5 18 8 5.44 E A 2 4 18 6 7.11 F B, D, E 22 30 50 32 21.78 G F 15 20 37 22 13.44 H F 4.5 10 21.5 11 8.03 I G, H 12 15 24 16 4 J G 14 14.5 48 20 32.11 K I, J 5 5 5 5 0
Selby’s Management has now decided that the table above captures the uncertainty of the project that they will be undertaking. Find (I) the Expected project duration, critical path(s) and the schedule for each activity.
(II) What are the probability of completing the project in 105 days, 78 days?
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u Step 1: Draw the Network Diagram with the mean activity times on each activity arrow.
PERT Analysis Algorithm
B = 5 F = 32 K = 5
Enumeration of Possible Paths
A-E-F-H-I-K = 73 A-E-F-G-DUM-I-K = 84 A-E-F-G-J-K = 88
B-F-H-I-K = 69 B-F-G-DUM-I-K = 80 B-F-G-J-K = 84
C-D-F-H-I-K = 78 C-D-F-G-DUM-I-K = 89 C-D-F-G-J-K = 93
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