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Topics to cover in 2nd part

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Topics to cover in 2nd part (to p2) Chapter 8 - Project Management Chapter Topics Project Management Questions: Why do we need to study Project Management? – PowerPoint PPT presentation

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Title: Topics to cover in 2nd part


1
Topics to cover in 2nd part
(to p2)
2
Chapter 8 - Project ManagementChapter Topics

(to p3)
3
Project Management
  • Questions
  • Why do we need to study Project Management?
  • How does a project management technique work?

(to p4)
(to p5)
4
Objective
  • The main purpose is to govern the operations of a
    project such that all activities involved are
    well administrated and that we can also control
    its completion time

(to p3)
5
Project management technique
  • Steps to solve a project management
  • problem
  • to represent a project problem graphically
  • to determine its completion time
  • to carry out sensitivity analysis, if any

(to p6)
(to p12)
(to p29)
6
1. Represent a project problem graphically
Event Processing Time Precedent constraints
A B C 20 30 10 -- A B
  • Steps
  • Gather all information and organize them in a
    table format that consists of event, processing
    time, and precedent constraints as follows
  • Draw a semantic network to represent them
  • Special case!

(to p7)
(to p9)
(to p4)
7
Semantic network to represent them
  • Here, we use three symbols
  • node to represent stage
  • line/branch to represent event
  • arrow to represent precedent
  • constraint
  • Example

(to p8)
(to p6)
8
Example
Path Event Proc Time Pred Const
1-2 2-3 3-4 A B C 20 30 10 -- A B
C
A
B
1
2
3
4
30
10
20
Rule1 All nodes must starts from one Node and
ends with one node
(to p7)
9
Special case!
  • When two or events taken places in the same time
    interval
  • (known an concurrent events)
  • Consider the following example!
  • How to draw it?

Event Processing Time Precedent constraints
A B C 3 5 7 -- A A
(to p10)
10
Case 1
A
B
2
3
1
5
3
C
7
Wrong!
Rule2 no node can have two outcomes and end
with the same note
Solution
(to p11)
11
Solutions for Rule 2
A dummy activity shows a precedence
relationship Reflects no processing time
  • Three ways to draw it

3
B
Dummy 10
A
1
2
5
C
Solution 1
4
Dummy 2 0
What one is better?
A
B
1
2
4
Solution 2
C
3
Dummy 0
3
Dummy 0
B
A
(to p6)
1
2
4
Solution 3
C
12
2. Determine its completion time
  • Consider the project network as shown in next
    slide
  • Question Is it an easy way to find out the
  • solution?
  • Answer YES, it knows as
  • Critical Path Method (CPM)

(to p13)
(to p15)
13
The Project Network All Possible Paths for
Obtaining a Solution
Figure 8.3 Expanded network for building a house
showing concurrent activities.


Table 8.1 Possible Paths to complete the
House-Building Network
(to p14)
Then the completion time for paths A, B, C and D
can be computed as
14
The Project Network
Completion time for path A 1?2 ? 3 ? 4
? 6 ? 7, 3 2 0 3 1 9 months (Critical
Path)

path B 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7, 3 2 0
1 1 1 8 months

path C 1 ? 2 ? 4 ? 6 ? 7, 3 1 3 1
8 months path D 1 ? 2 ? 4 ? 5 ? 6 ? 7,
3 1 1 1 1 7 months The critical path is
the longest path through the network the minimum
time the network can be completed.
This is the Solution!
Figure 8.5 Alternative paths in the network
(to p12)
15
Critical Path Method (CPM)
  • General concepts
  • For each branch of the project network, we
    firstly determine four values of ES, EF, LS and
    LF
  • For each branch, we compute their slack time,
  • Slack time (LS-ES) or (LF-EF)
  • The critical path is located at branch that has
  • slack time 0
  • (Do you know the reason why?)
  • How it works?

(to p16)
16
How CPM works?
  • Steps
  • Prepare the project network
  • Construct a table as follows
  • Compute ES and EF
  • Compute LS and LF
  • Compute LS-ES or LF-EF

Branch ES EF LS LF

(to p17)
(to p22)
ESij max (EFi) EFij ESi tij with
EF10
(to p26)
Critical path when LS-ES0
(to p4)
17
Compute ES and EF
  • Note
  • When computing these values, the pattern is like
    moving zic-zac format by firstly computer ES12
    and then adding it to EF12 and move to next
    branch by copying the max values of the branch
    1-2 to say, 2-3
  • We compute them from top to bottom!
  • Their relationship
  • Example 1

(to p18)
(to p22)
(to p19)
18
The starting point of ES and EF
  • Consider
  • Then
  • EF1 0
  • ES12 max (EF1) EF12 ES12 t12
  • 0 0 t12

t12
1
2
(to p17)
19
Branches ESij max(EFi) EFijESijtij
1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 ES12 max(EF1) ES23max(EF2) ES24max(EF2) ES34max(EF3) ES45max(EF4) ES46max(EF4) ES56max(EF5) ES67max(EF6) EF12ES12t12 EF23ES23t23 EF24 EF34 EF45 EF46 EF56 EF67
The overall computation is shown in next slide
(to p20)
20
Complete solution
- ES is the earliest time an activity can
start. ESij Maximum (EFi) - EF is the earliest
start time plus the activity time. EFij ESij
tij
add all ti for note 2
Add all t to note 4 and take the longest time Max
(node 3t34, node2t24) max (50,
31) max(5,4)5
Max(node4t46,node5t56 max(53,51)8
(noteyou can compute these values and show in
the network diagram as well)
(to p4)
(to p21)
21
The Project Network Activity Scheduling-
Earliest Times
- ES is the earliest time an activity can
start. ESij Maximum (EFi) - EF is the earliest
start time plus the activity time. EFij ESij
tij
Figure 8.6 Earliest activity start and finish
times
(to p20)
22
Compute LS and LF
  • Note We compute these values from the bottom to
    top, with assigning
  • LSij LFi -tij LFij min LSj
  • with
  • the end of LFij EFij
  • Example computing Figure 8.3

(to p23)
23
Branches LSij LFij-tij LFijmin(LSj)
1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 LS12 Li12-t12 LS23 LF23-t23 LS24 LF24-t24 LS34 LF34-t34 LS45 LF45-t45 LS46 LF46-i46 LS56 LF56-t56 LS67 LF67-t67 LF12min(LS2) LF23min(LS3) LF24min(LS4) LF34min(LS4) LF45min(LS5) LF46min(LS6) LF56min(LS6) LF67min(LS7)
The overall computational is shown in next slide
(to p24)
24
- LS is the latest time an activity can start
without delaying critical path time. LSij LFij
- tij - LF is the latest finish time
LFij Minimum (LSj)
Min(node3-t23,node4-t24) Min(5-2,5-1)Min(3,4)3
Min(node 6-t46,node5-t45) Min(8-3,7-1) Min(5,6)
5
Min(node 7-t67) Min(9-1)8
Same as EF67 from the previous slide
Start with the end node first
Again, you can place these values onto the
branches
(to p25)
(to p22)
25
The Project Network Activity Scheduling - Latest
Times
- LS is the latest time an activity can start
without delaying critical path time. LSij LFij
- tij - LF is the latest finish time
LFij Minimum (LSj)
Figure 8.7 Latest activity start and finish times
(to p24)
26
Compute LS-ES or LF-EF
  • Two ways you can achieve it
  • by compiling slack, Sij
  • by showing branches

(to p27)
(to p28)
(to p16)
27
The Project Network Calculating Activity Slack
Time
- Slack, Sij, computed as follows Sij LSij
- ESij or Sij LFij - EFij

Table 8.2 Activity
Slack   Figure 8.9 Activity
Slack

What does it mean?
(to p26)
28
The Project Network Activity Slack
  • Slack is the amount of time an activity can be
    delayed without delaying the project.
  • Slack time exists for those activities not on
    the critical path for which the earliest and
    latest start times are not equal.
  • Shared slack is slack available for a sequence
    of activities.

Figure 8.8 Earliest activity start and finish
times
(to p26)
29
Sensitivity Analysis
  • Today, we only consider one case
  • Probabilistic Activity Times
  • Refer to activity time estimates usually can not
    be made with certainty
  • PERT is known as the solution method

(to p30)
30
PERT
  • In PERT, three different time estimations are
    applied
  • most likely time (m),
  • the optimistic time (a) , and
  • the pessimistic time (b).
  • How do we make use of these three values?

(to p31)
31
Probabilistic Activity Times
  • We used these values to estimate the mean and
    variance of a beta distribution
  • mean (expected time)
  • variance
  • How to use these values to solve a project
    network problem?

(to p32)
32
PERT
  • We simply apply t values in CPM and determine the
    values of
  • ES
  • EF
  • LS
  • LF
  • S
  • and branches with slack 0 still consider as
    critical paths
  • Example.

(to p33)
33
Procedures for PERT
  • Step 1 based on the values of a, b and m,
    determine the t and v values for each path
  • Step 2 determine the critical path by using t
    values in the CPM
  • Step 3 compute its corresponding means and
    standard deviations according.
  • Example
  • Result implication
  • Applications

(to p34)
(to p38)
(to p39)
34
PERT Example
  • Step 1 computer t and v values
  • Step 2 determine the CPM
  • Step 3 determine v value

(to p35)
(to p36)
(to p37)
(to p33)
35
Step 1 computer t and v values

Figure 8.11 Network with mean activity times and
variances
Table 8.3 Activity Time Estimates for Figure
8.10
(to p34)
36
Step 2 determine the CPM
Figure 8.12 Earliest and latest activity times
Table 8.4 Activity Earliest and Latest Times and
Slack
(to p34)
37
Step 3 determine v value
  • The expected project time is the sum of the
    expected times of the critical path activities.
  • The project variance is the sum of the variances
    of the critical path activities.
  • The expected project time is assumed to be
    normally distributed (based on central limit
    theorum).
  • In example, expected project time (tp) and
    variance (vp) interpreted as the mean (?) and
    variance (?2) of a normal distribution
  • ? 25 weeks
  • ?2 6.9 weeks

(to p34)
38
Probability Analysis of the Project Network
- Using normal distribution, probabilities are
determined by computing number of standard
deviations (Z) a value is from the mean. - Value
is used to find corresponding probability in
Table A.1, App. A.
Figure 8.13 Normal distribution of network
duration
Critical value
(to p33)
39
  • Consider when
  • x 30
  • x 22
  • Tutorial Assignment

(to p40)
(to p41)
(to p42)
40
Probability Analysis of the Project
NetworkExample 1
  • ?2 6.9 ? 2.63
  • Z (x-?)/ ? (30 -25)/2.63 1.90
  • Z value of 1.90 corresponds to probability of
    .4713 in Appendix A of p715. Probability of
    completing project in 30 weeks or less (.5000
    .4713) .9713,
  • or 97.13 (Why so high a probability
    rate?)

Figure 8.14 Probability the network will be
completed in 30 weeks or less
(to p39)
41
Probability Analysis of the Project
NetworkExample 2
Z (22 - 25)/2.63 -1.14 Z value of 1.14
(ignore negative) corresponds to probability of
.3729 in Table A.1, appendix A. Probability that
customer will be retained is .1271 ( 0.5-
0.3729) , or 12.71 (Again, why so low
probability rate?)
Figure 8.15 Probability the network will be
completed in 22 weeks or less
(to p39)
42
Tutorial Assignment
  • Try to use QM to solve CPM/PERT problems (see
    slide 19)
  • Exercises (Chapter 8)
  • Old 8, 10, 17
  • New 4, 6, 11

(to p43)
43
Probability Analysis of the Project
NetworkCPM/PERT Analysis with QM for Windows
Exhibit 8.1
(to p16)
44
The Project Network Activity Slack
  • Slack is the amount of time an activity can be
    delayed without delaying the project.
  • Slack time exists for those activities not on
    the critical path for which the earliest and
    latest start times are not equal.
  • Shared slack is slack available for a sequence
    of activities.

Figure 8.8 Earliest activity start and finish
times
45
The Project Network Calculating Activity Slack
Time
- Slack, Sij, computed as follows Sij LSij
- ESij or Sij LFij - EFij

Table 8.2 Activity
Slack   Figure 8.9 Activity
Slack
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