PERT/CPM AGENDA MGT 606

1
PERT/CPM AGENDAMGT 606

• 1. Motivation
• PERT/CPM
• 2a. Digraphs (Project Diagrams)
• 2b. PERT
• 3a. CPM--Starts, Finishes, Slacks,
• 3b. Resource Allocation
• 4. CPM with Crashing
• 5. PERT Simulation

2
1. Motivation for PERT/CPM Use
2

• Definition of a Project
• Organizational Trajectories and Change
• The Environment
• Adaptation and Agility
• Reactive Strategy
• Proactive Strategy
• Ubiquity of Projects
• Dealing with Project Complexity

3
2. PERT ExampleA. AON Network DiagramsGeneral
Foundry
3

• Diagramming a projects network of activities
• Installation of air-pollution control equipment _at_
  General Foundry, Milwaukee
• 
  Immediate
• Activity Description Predecessor
• A build internal components ___
• B modify roof and floor ___
• C construct collection stack A
• D pour concrete and install frame B
• E build hi-temp burner C
• F install control system C
• G install air-pollution control device D,E
• H inspect and test F,G

4
4
2. PERT ExampleA. AON Network Diagrams
(continued)General Foundry
A
C
edge or arc
Node A represents Activity A
Node C represents Activity C
The edge or arc represents the precedence
relationship between the two activities
5
5
2. PERT ExampleA. AON Network Diagrams
(continued)General Foundry continued
A
A
C
C
B
D
B
D
F
C
A
E
B
D
6
6
2. PERT ExampleA. AON Network Diagrams
(continued) General Foundry concluded
F
A
C
H
E
B
D
G
7
7
2. PERT Example A. AON Network Diagrams
(continued)

• The General Foundry project network began on
  more than one node and ended on a single node.
  Other variants are

Beginning with one node and ending with more than
one node.
B
E
A
D
F
C
Beginning with more than one node and ending
with more than one node.
E
A
C
D
F
B
8
8
2. PERT ExampleA. AON Network Diagrams
(concluded)

• Sometimes the following convention is used.

A
E
Start
Finish
C
D
B
F
The Start and Finish boxes tie the network
off at its ends and give one a sense that the
network has defined points in time at which the
project begins and ends . The use of such
a convention is not necessary and will generally
be avoided.
9
2. PERT ExampleA2. AOA Network Diagrams
(continued)General Foundry
8b
Activity A
1
2
edge or arc
Node 1 represents the beginning of Activity A
Node 2 represents the ending of Activity A
The edge or arc represents Activity A
10
2. PERT ExampleA2. AOA Network Diagrams
(continued)General Foundry continued
8c
A
1
2
C
2
A
1
B
3
4
C
A
2
E
1
B
5
D
3
11
2. PERT ExampleA2. AOA Network Diagrams
(continued) General Foundry concluded
8d
C
2
4
A
F
7
H
1
E
6
B
D
G
3
5
12
2. PERT Example A2. AOA Network Diagrams
(concluded)
8e

• Note that the General Foundry project network on
  slide 6 began on a single node and ended on a
  single node. When constructing Network diagrams
  using the AOA approach, this convention is
  followed--network diagrams begin on a single node
  and end on a single node.
• Note also a second convention in AOA Network
  diagram construction. Two nodes are connected by
  one and only one activity (edge or arc). Thus,
• Act. I.P.
• A __
• B __
• C A,B
• D B
• Finally, a third convention. An arc cannot
  emanate from or terminate at more
• than one node.
• Act. I.P.
• A __
• B __
• C A,B
• D A
• E B

A
C
C
A
d1
B
B
D
D
No, and why not?
Rather, this is preferred and why?
D
D
d1
NO!
A
A
C
C
d2
B
B
E
E
13
2. PERT ExampleB. Project Completion Times and
ProbabilitiesGeneral Foundry of Milwaukee
9

• The duration time for each of a Projects
  activities in a PERT environment are
  estimated on the basis of most likely,
  pessimistic, and optimistic completion times.
  These times can be arrived at in various ways.
  A number of these ways contain substantial
  subjective components because it is often the
  case that little historical information is
  available to guide those estimates.
• The expected (value) duration time of an
  individual activity and its variation follow what
  is called a beta distribution and are calculated
  as follows
• E(ti) (a 4m b) / 6 and var(ti)
  (b - a)2 / 36
• where a is the activitys optimistic
  completion time, m is the activitys
• most likely completion time, and b is the
  activitys most pessimistic completion time.

14
2. PERT ExampleB. Project Completion Times and
ProbabilitiesGeneral Foundry of Milwaukee
(continued)
10

• Consider the following table of activities
  immediate predecessor(s) (I.P.) optimistic, most
  likely, and pessimistic completion times for
  General Foundry and the E(ti) and var(ti) for
  each activity.
• ACT I.P. Optimistic (a) Most Likely (m)
  Pessimistic(b) E(ti) var(ti)
• A __ 1 week 2
  weeks 3 weeks 2 wks.
  4/36
• B __ 2 3
  4
  3 4/36
• C A 1 2
  3
  2 4/36
• D B 2 4
  6
  4 16/36
• E C 1 4
  7
  4 36/36
• F C 1
  2 9
  3 64/36
• G D,E 3 4
  11
  5 64/36
• H F,G 1 2
  3
  2 4/36

15
11
2. PERT ExampleB. Project Completion Times and
ProbabilitiesGeneral Foundry of Milwaukee
(continued)

• From the reduced version of General Foundys PERT
  table., the E(ti) for each activity can be
  entered into the projects network diagram.
• ACT I.P. (a) (m) (b) E(ti)
  var(ti)
• A __ 1 2 3 2
  4/36
• B __ 2 3 4 3
  4/36
• C A 1 2 3 2
  4/36
• D B 2 4 6 4
  16/36
• E C 1 4 7 4
  36/36
• F C 1 2 9 3
  64/36
• G D,E 3 4 11 5
  64/36
• H F,G 1 2 3 2
  4/36
• Inspection of the network discloses three paths
  thru the project
• A-C-F-H A-C-E-G-H and B-D-G-H. Summing the
  E(ti) on each path yield time thru each path of
  9, 15, and 14 weeks, respectively. With an E(t)
  15 for A-C-E-G-H, this path is defined as the
  critical path (CP) being the path that govens
  the completion time of the project . Despite the
  beta distribution of each activity, the
  assumption is made that the number of activities
  on the CP is sufficient for it to be normally
  distributed with a variance equal to the sum of
  the variances of its activities only, var(t)
  112/36 3.11.

F
3
C
A
2
2
H
E
2
4
B
D
3
4
G
5
16
11b
2. PERT ExampleB2. Project Completion Times and
ProbabilitiesGeneral Foundry of Milwaukee
(continued)

• From the reduced version of General Foundys PERT
  table., the E(ti) for each activity can be
  entered into the projects network diagram.
• ACT I.P. (a) (m) (b) E(ti)
  var(ti)
• A __ 1 2 3 2
  4/36
• B __ 2 3 4 3
  4/36
• C A 1 2 3 2
  4/36
• D B 2 4 6 4
  16/36
• E C 1 4 7 4
  36/36
• F C 1 2 9 3
  64/36
• G D,E 3 4 11 5
  64/36
• H F,G 1 2 3 2
  4/36
• Inspection of the network discloses three paths
  thru the project A-C-F-H
• A-C-E-G-H and B-D-G-H. Summing the E(ti) on
  each path yield time thru each path of 9, 15,
  and 14 weeks, respectively. With an E(t) 15
  for A-C-E-G-H, this path is defined as the
  critical path (CP) being the path that governs
  the completion time of the project . Despite the
  beta distribution of each activity, the
  assumption is made that the number of activities
  on the CP is sufficient for it to be normally
  distributed with a variance equal to the sum of
  the variances of its activities only, var(t)
  112/36 3.11.

4
C2
F3
6
2
H2
A2
7
E4
1
5
G5
3
B3
D4
critical path

17
2. PERT ExampleB. Project Completion Times and
ProbabilitesGeneral Foundry (concluded)
12

• Given General Foundrys E(t) 15 weeks and
  var(t) 3.11, what is the probability of the
  project requiring in excess of 16 weeks to
  complete?
• P( X gt 16) 1 - P ( X lt 16) 1 - P ( Z lt (16
  - 15) / 1.76 .57) 1 - 0.716
• where the value 1.76 is the square root of 3.11,
  the standard deviation of the expected completion
  time of General Foundrys project. The
  assumption being made is that summing up a
  sufficient number of activities following a beta
  distribution yield a result which approximates
  or a approaches a variable which is normally
  distributed--N(?,?2)

18
13
3. CPM ExampleA. Starts, Finishes, Slacks

• E(arly)S(tart) -- earliest possible commencement
  time for a project activity.
• ES calculation -- beginning at the initial
  node(s) of a projects network diagram, conduct
  a forward pass where
• ESj ESi ti

D
C
9
7
B
A
3
0
5
2
3
4
0 0 0
7 3 4
9 7 2
3 0 3
B
3
What if?
4
ESB tB 7
D
?
ESD
5
ESC tC 9
C
7
2
MAX
19
14
3. CPM ExampleA. Starts, Finishes, Slacks

• E(arly)F(inish) -- earliest possible time a
  projects activity can be completed.

EF calculation -- beginning at the initial
node(s) of a projects network diagram ,
conduct a forward pass where EFi ESi ti
D
C
14
7
9
9
B
A
3
7
0
3
5
2
3
4
3 0 3
9 7 2
14 9 5
7 3 4
20
15
3. CPM ExampleA. Starts, Finishes, Slacks

• E(arly) S(tarts), E(arly F(inishes) for
  Milwaukee Foundry .

F
EF
4
7
ES
C
A
2
2
4
0
3
2
2
H
E
13
15
8
4
2
D
4
B
3
0
7
3
3
4
G
8
13
5
21
16
3. CPM ExampleA. Starts, Finishes, Slacks

• L(ate) F(inish) -- latest possible time an
  activity can be completed without delaying the
  completion time of the project.
• LF calculation -- beginning at the final node,
  conduct a backward pass through the networks
  paths where
• LFi LFj - tj

C
A
B
D
3
7
0
14
9
3
3
9
2
7
4
14
5
3 7 - 4
9 14 - 5
7 9 - 2
C
2
9
B
LFC -tC 7
A
LFB
LFD -tD 9
3
4
?
D
MIN
5
14
22
3. CPM ExampleA. Starts, Finishes, Slacks
17

• LS calculation -- beginning with the final
  node(s) of the network make a backward pass
  through the network where
• L(ate) S(tart) -- the latest possible time and
  activity can begin without delaying the
  completion time of the project.

LSi LFi - ti
C
A
D
B
9
14
7
3
7
9
0
3
2
3
4
5
0 3 - 3
3 7 -4
9 14 -5
7 9 -2
23
3. CPM ExampleA. Starts, Finishes, Slacks
18

• E(arly) S(tarts)/ F(inishes) and L(ate)
  S(tarts)/F(inishes) for Milwaukee Foundry .

F
7
4
C
A
4
2
0
2
EF
10
3
13
2
4
2
2
0
2
ES
H
15
13
E
8
4
2
13
15
B
D
3
0
7
3
4
4
8
LS
1
3
4
4
8
4
LF
G
13
8
8
5
13
critical path
24
3. CPM ExampleA2. Starts, Finishes, Slacks
18b

• E(arly)S(tart) -- earliest possible commencement
  time for a project activity.
• ES calculation -- beginning at the initial
  node(s) of a projects network diagram, conduct
  a forward pass where
• ESj ESi ti

C2
D5
3
7
9
A3
B4
0
4
1
2
3
ESj
0 0 0
3 0 3
7 3 4
9 7 2
3
node
B4
2
9
D5
What if?
ESB tB 7
4
ESD
C2
7
ESC tC 9
3
MAX
25
3. CPM ExampleA2. Starts, Finishes, Slacks
18c

• E(arly)F(inish) -- earliest possible time a
  projects activity can be completed.

• EF calculation -- beginning at the initial
  node(s) of a projects network diagram ,
  conduct a forward pass where
• EFi ESi ti

D5
C2
7
9
A3
B4
3
7
4
9
0
3
0
1
2
3
ESj
EFi
3 0 3
7 3 4
9 7 2
3
3
4
D4
node
F5
G3
7
6
12
12
7/6
5
What if?
E2
4
4
3
critical path
26
3. CPM ExampleA2. Starts, Finishes, Slacks
18d

• E(arly) S(tarts), E(arly F(inishes) for
  Milwaukee Foundry .

ESj
EFi
C2
2
4
2
4
2
4
A2
F3
node
7
H2
0
13
15
1
7/13
15
E4
6
B3
D4
3
3
G5
8
8/7
3
5
critical path
27
3. CPM ExampleA2. Starts, Finishes, Slacks
18e

• L(ate) F(inish) -- latest possible time an
  activity can be completed without delaying the
  completion time of the project.
• LF calculation -- beginning at the final node,
  conduct a backward pass through the networks
  paths where

LFi LFj - tj
EFi
ESi
node
LFi
B4
C2
A3
D5
14
3
4
0
2
1
5
7
3
9
14
0
7 9 - 2
9 14 - 5
3 7 - 4
0 3 - 3
14 14 - 0
4
What if?
C2
9
LFC -tC 7
LFB
B4
3
2
LFD -tD 9
D5
?
5
MIN
14
28
3. CPM ExampleA2. Starts, Finishes, Slacks
18f

• L(ate) S(tart) -- the latest possible time and
  activity can begin without delaying the
  completion time of the following activity.
• LS calculation -- beginning with the final
  node(s) of the network make a backward pass
  through the network where

LSi LFi - ti
node
EFi
ESi
LFi
LSj
B4
C2
D5
5
2
4
3
14
14
A3
3
9
3
7
7
9
1
9 14 -5
7 9 -2
3 7 - 4
0
0
4
0 3 - 3
C2
9
LFC -tC 7
What if?
LSC,D
3
B4
LFD -tD 9
D5
2
7/9
5
14
29
3. CPM ExampleA2. Starts, Finishes, Slacks
18g

• E(arly) S(tarts)/ F(inishes) and L(ate)
  S(tarts)/F(inishes) for Milwaukee Foundry .

C2
2
4
2
4
2
4
2
A2
F3
2
4
10/4
7
H2
15
0
13
0
15
1
7/13
E4
6
0
0
15
13
15
13
B3
D4
3
3
G5
8
8/7
3
5
4
4
8
8
critical path
30
19
3. CPM ExampleA. Starts, Finishes, Slacks

• T(otal) S(lack) -- the amount of time an
  activitys completion can be delayed without
  delaying the projects completion where,
• TSi LFi - ESi - ti and where i the ith
  activity.
• F(ree) S(lack) -- the amount of time an
  activitys completion can be delayed without
  delaying the commencement of the next activity
  where,
• FSi ESj - ESi - ti and where i the ith
  activity and j the jth activity.

31
3. CPM ExampleA. Starts, Finishes., Slacks
20

• T(otal) S(lacks) and F(ree) S(lacks) for
  Milwaukee Foundry .

F
4
C
A
0
2
3
13
TSH015-13-2
4
FSH015-13-2
2
2
2
ES
TSE8-4-40
FSE8-4-40
H
13
D
E
3
4
2
15
B
0
4
4
8
8
3
4
TSD18-4-3
FSD18-4-3
LF
TSB14-3-0
G
FSB03-3-0
8
5
13
critical path
32
21
3. CPM ExampleA. Starts, Finishes., Slacks

• Shared Slack -- the slack in a project along a
  non-critical segment of a path which all
  activities on that non-critical segment share.
  Consider the lower path for Milwaukee Foundry.

B
G
D
8
0
3
4
8
13
4
3
5
TSB 1
TSD 1
TSG 0
FSB 0
FSD 1
FSG 0
The slack of one time period along this
non-critical path segment is shared between
activities B and D, i.e., 7 time periods of
activities are scheduled over an 8 time period
segment.
8 time periods (TPs)
Other variations are possible if the one TP of
slock is divided up.
A -- 3 TPs
B -- 4 TPs
B -- 4TPs
A -- 3TPs
B -- 4TPs
A -- 3TPs
33
22
3. CPM ExampleA. Starts, Finishes., Slacks

• Nested Slacks -- when one segment of a
  non-critical path is imbedded in another segment
  of a non-critical path, the free slack of the
  terminal activity in the imbedded non-critical
  segment will not necessarily be 0.

Activity
I.P.
D
8
C
0
E
15
__
A
20
8
7
13
25
5
A
B
__
TS5, FS0
TS5, FS0
TS,FS 5
C
A
D
C
B
0
10
E
D
10
10
15
25
TS,FS0
TS,FS0
NOTE Path C-D-E is the non-critical path and is
nested in A-B (the critical path).
Hence, the ES for activity A is 0 and the LF for
activity E is 25, the ES and LF for
activities A and B , respectively.
34
23
3. CPM ExampleA. Starts, Finishes., Slacks

• Nested Slacks -- when one segment of a
  non-critical path is imbedded in another segment
  of a non-critical path, the free slack of the
  terminal activity in the imbedded non-critical
  segment will not necessarily be 0.

Activity
I.P.
D
8
C
0
E
15
__
A
20
8
7
13
25
5
A
B
__
TS5, FS0
TS5, FS2
TS,FS 3
C
D
C
B
F
0
10
E
D,F
20
17
15
25
__
F
TS3, FS0
TS,FS0
A
0
NOTE Path A-B is the one critical path in this
poject while paths F-E and C-D-E
are non-critcal with C-D-E nested
in F-E and shorter in path length by two time
periods. As a consequence, in
figuring the total and free
slacks on C-D-E, refect the two additional time
periods of slack shared by them
beginning with activity D where
TSD FSD or FSD 0.
10
10
TS,FS0
35
3. CPM ExampleA2. Starts, Finishes., Slacks
23b

• T(otal) S(lacks) and F(ree) S(lacks) for
  Milwaukee Foundry .

early start
TSC 4 - 2 - 2 0 FSC 4 - 2 - 2 0
C2
2
4
2
4
2
4
2
A2
F3
2
4
10/8
7
H2
15
0
13
0
15
1
7/13
E4
6
0
0
15
13
13
TSB 4 - 0 - 3 1 FSB 3 - 0 - 3 0
B3
D4
3
3
G5
8
7/8
3
5
4
4
8
8
TSD 8 - 3 - 4 1 FSD 8 - 3 - 4 1
late finish
36
3. CPM ExampleA2. Starts, Finishes., Slacks
23c

• Shared Slack -- the slack in a project along a
  non-critical segment of a path which all
  activities on that non-critical segment share.
  Consider the lower path for Milwaukee Foundry.

8
D4
1
4
5
B3
2
3
8
5
TSD 1
4
TSB 1
FSD 1
FSB 0
C8
TSC 0
0
FSC 0
4
0
The slack of one time period along this
non-critical path segment is shared between
activities B and D, i.e., 7 time periods of
activities are scheduled over an 8 time period
segment.
8 time periods (TPs)
Other variations are possible if the one TP of
slock is divided up.
B -- 3 TPs
D -- 4 TPs
D -- 4TPs
B -- 3TPs
D -- 4TPs
B -- 3TPs
37
23d
3. CPM ExampleA2. Starts, Finishes., Slacks

• Nested Slacks -- when one segment of a
  non-critical path is imbedded in another segment
  of a non-critical path, the free slack of the
  terminal activity in the imbedded non-critical
  segment will not necessarily be 0.

TS5, FS0
Activity
I.P.
15
D7
2
8
4
__
TS,FS 5
A
20
13
TS5, FS0
E5
A
B
C8
__
C
5
25
D
C
0
1
25
E
D
0
B15
A10
10
3
TS,FS0
TS,FS0
10
NOTE Path C-D-E is the non-critical path and is
nested in A-B (the critical path).
Hence, the ES for activity A is 0 and the LF for
activity E is 25, the ES and LF for
activities A and B , respectively.
38
23e
3. CPM ExampleA2. Starts, Finishes., Slacks

• Nested Slacks -- when one segment of a
  non-critical path is imbedded in another segment
  of a non-critical path, the free slack of the
  terminal activity in the imbedded non-critical
  segment will not necessarily be 0.

TS5, FS2
17
D7
2
8
Activity
I.P.
4
TS,FS 3
__
20
13
TS5, FS0
A
E5
C8
A
B
__
F17
C
5
25
0
TS3, FS0
1
25
D
C
0
E
D,F
A10
B15
__
F
10
3
TS,FS0
TS,FS0
20
10
NOTE Path A-B is the one critical path in this
project while paths F-E and C-D-E
are non-critical with C-D-E
nested in F-E and shorter in path length by two
time periods. As a consequence,
in figuring the total and free
slacks on C-D-E, they reflect the two additional
time periods of slack shared by
them beginning with activity D.
39
3. CPM ExampleB. Resource Allocation Scheduling
(ES)
24
Activity
TP2
TP3
TP5 TP6
TP7
TP1
TP4
TP8
TP9 TP10
TP11
TP12
TP13
TP14
TP15
11K
11K
A (2)
B (3)
10K
10K
10K
C (2)
13K
13K
D (4)
12K
12K
12K
12k
E (4)
14K
14K
14K
14K
F (3)
10K
10K
10K
G (5)
16K
16K
16K
16K
16K
8K
8K
H (2)
21
23
25
36
36
36
14
16
16
8
21
16
16
16
8
21
42
65
90
126
162
198
260
276
292
300
308
228
244
212
Activity Cost
A -- 22K (22,000)
D -- 48K
G -- 80K
B -- 30K
E -- 56K
H -- 16K
C -- 26K
F -- 30K
-- critical path
40
3. CPM ExampleB. Resource Allocation Scheduling
(LS)
25
Activity
TP2
TP3
TP5 TP6
TP7
TP1
TP4
TP8
TP9 TP10
TP11
TP12
TP13
TP14
TP15
11K
11K
A (2)
B (3)
10K
10K
10K
C (2)
13K
13K
D (4)
12K
12K
12k
12K
E (4)
14K
14K
14K
14K
F (3)
10K
10K
10K
G (5)
16K
16K
16K
16K
16K
8K
8K
H (2)
21
23
23
26
26
26
26
16
16
8
11
26
26
26
8
11
32
55
78
104
130
156
240
266
292
300
308
198
214
182
Activity Cost
A -- 22K (22,000)
D -- 48K
G -- 80K
B -- 30K
E -- 56K
H -- 16K
C -- 26K
F -- 30K
-- critical path
41
26
3. CPM ExampleB. Resource Allocation Scheduling
Early/Late Start Resource Allocation Schedules
C
u
m.
P
r
o
j.
C
o
s
t
s
Project Time Periods
42
4. CPM ExampleCPM with Crashing --a
27

• It is sometime necessary to accelerate the
  completion time of a project. This usually leads
  to greater cost in completing the project than
  what might have otherwise been the case because
  of opportunity costs incurred as the result of
  diverting resources away from other pursuits. As
  a result of this increase in cost, it is
  incumbent upon project managers to reduce the
  completion time of the project in the most cost
  effective way possible. The following guidelines
  are designed to achieve that end.
• 1) Reduce duration times of critical activities
  only.
• 2) Do not reduce the duration time of critical
  activity such that its path length (in time)
  falls below the lengths (in time) of other paths
  in the network.
• 3) Reduce critical activity duration times on the
  basis of the least costly first and in case of a
  tie, the least costly activity closest to the
  completion node(s) of the project.
• 4) When two or more critical paths exist, reduce
  the length (in time) of all critical paths
  simultaneously.
• 5) Given two or more critical paths and a cost
  tie between a joint activity and a subset of
  disjoint activities on the same critical path,
  generally reduce the duration time of the one
  joint to the most paths.
• 6) Note, if reducing a joint activity means that
  more critical paths emerge that what would
  otherwise be the case, reduce disjoint activities.

43
4. CPM ExampleCPM with Crashing --b
28

• Consider the crashing of the Milwaukee Foundry
  Project
• Act. N.T. C.T. N.C. C.C. U.C.C.
• A 2 1 22K 23K 1K
• B 3 1 30K 34K 2K
• C 2 1 26K 27K 1K
• D 4 3 48K 49K 1K
• E 4 1 56K 59K 1K
• F 3 2 30K 30.5K 0.5K
• G 5 2 80K 86K 2K
• H 2 1 16K 19K 3K
• - Critical Path
• An inspection of Milwaukee Foundrys project
  network identifies the following three paths and
  of duration,
• A - C - F - H 9
• A - C - E - G -H 15
• B - D - G - H 14

44
4. CPM ExampleCPM with Crashing --c
29

• ACT N.T. C.T. N.C. C.C. U.C.C.
• A 2 1 22K 23K 1K
• ?B 3 2 1 1 30K 34K 2K 5
  3K
• C 2 1 1 26K 27K 1K 6
  3K
• ?D 4 3 3 48K 49K 1K 3
  2K
• E 4 3 2 1 1 56K 59K 1K 1
  1K
• F 3 2 30K 30.5K 0.5K
• ?G 5 2 2 80K 86K 2K 2
  6K
• ?H 2 1 1 16K 19K 3K 4
  3K
• 308 K
  18K
• ? - - SECOND CRITICAL PATH
• A - C - F - H 9 8
• A - C - E - G -H 15 14
• B - D - G - H 14 14
• 
  
• A - C - F - H 9 9 9 9
  8 7 7
• A - C - E - G -H 15 14 11 10 9 8
  7
• ?B - D - G - H 14 14 11 10 9 8
  7

Act. E has the lowest U.C.C. and closest to
the final node of the network-- crash one time
period and two CPs emerge. One can now reduce
Acts. E D or G. Reduce G, three time periods.
It is A joint activity. Now E D can be
reduced one time period for the same U.C.C. as
G. Now reduce H by one time unit. Now reduce
C B by one time unit. Finally, reduce E B
by one time unit.
1
2
3
4
1
5
1
2
3
4
6
5
6
45
4. CPM ExampleCPM with Crashing --d ( Summary)
30

• Why would reduce Activity E in crashing step one
  ( 1 ) and by only one time unit?
• Why do you reduce Activity G and not Activities E
  D in 2 and by three time units?
• Why do you now reduce Activities E D in 3
  and by only one time unit?
• Why do you reduce Activity H in 4 ?
• Why do you now reduce E B in 5 but by only
  one time unit?
• Why do you now reduce C B in 6 and how do
  you know now that you have crashed the project
  down to the minimum possible completion time?
• Why were Activities A F never reduced?
• Why should we concern ourselves with crashing a
  project by always reducing the least costly
  activities first?

46
5. PERT Simulation--agenda--
31

• Motivation
• Illustrative Examples
• Performing Simulations with WinQSB

47
5. PERT Simulation
32

• Motivation
• ? non-stochasticity/stochasticity non-critical
  paths
• ? independence of/interdependence between paths
• ? strength of an assumption

48
5. PERT Simulationmotivation --non-stochasticity
/stochasticity non-critical paths--
33

• Reconsider General Foundry of Milwaukee
• The E(ti) for each activity in this reduced
  version of General Foundys PERT table has been
  entered into the projects network diagram.
• ACT I.P. (a) (m) (b) E(ti)
  var(ti)
• A __ 1 2 3 2
  4/36
• B __ 2 3 4 3
  4/36
• C A 1 2 3 2
  4/36
• D B 2 4 6 4
  16/36
• E C 1 4 7 4
  36/36
• F C 1 2 9 3
  64/36
• G D,E 3 4 11 5
  64/36
• H F,G 1 2 3 2
  4/36
• Inspection of the network discloses three paths
  thru the project
• A-C-F-H A-C-E-G-H and B-D-G-H. Summing the
  E(ti) on each path yields
• the time thru each path to be 9, 15, and 14
  weeks, respectively. With an E(t) 15 for
• A-C-E-G-H, this path is defined as the critical
  path (CP) having a path variance of 3.11
• (112.36). The other paths however, also have
  variances of 2.11 for A-C-F-H and 2.44 for
• B-D-G-H. We assume all three paths to be
  normally distributed.

F
3
C
A
2
2
H
E
2
4
B
D
3
4
G
5
49
5. PERT Simulation--motivation --independence
of/interdependence between paths--
34

• Reconsider General Foundry of Milwaukee

F
3
C
A
C
A
H
E
2
2
2
4
2
2
H
E
2
4
B
D
G
B
G
D
F
3
4
5
3
5
3
4
Digraph with interdependent paths Which paths are
interdependent and why? Why might this path
interdependence complicate estimating the
expected completion time of this project? How
strong might the assumption of path independence
be in this case?
Digraph with independent paths
50
5. PERT Simulationillustrative examples
35

• Reconsider General Foundry of Milwaukee

CASE 1 ACT I.P. (a) (m) (b) E(ti)
var(ti) A __ 1 2
3 2 4/36 B __
2 3 4 3 4/36 C
A 1 2 3 2
4/36 D B 2 4 6
4 16/36 E C 1 4
7 4 36/36 F C 1
2 9 3 64/36 G D,E
3 4 11 5 64/36 H
F,G 1 2 3 2
4/36
CASE 2 ACT I.P. (a) (m) (b) E(ti)
var(ti) A __ 1 2
3 2 4/36 B __
.1 3 5.9 3 33.64/36 C
A 1 2 3 2
4/36 D B .1 4 7.9
4 60.84/36 E C 1 4
7 4 36/36 F C 1
2 9 3 64/36 G D,E
3 4 11 5 64/36 H
F,G 1 2 3 2
4/36
The PERT table above in contrast to the one at
its left while replicating the same path
expected duration times path as before B-D-G-H
now has a variance of 4.513. Relaxing the two
assumptions that 1) non-critical paths are
non-stochastic and 2) paths are independent of
each other, how might estimation results of
completion times with respect to Case 1 and Case
2 differ one from the other?
The PERT table above merely replicates the
results of a previous pagepaths A-C-F-H
A-C-E-G-H and B-D-G-H have E(t)s of 9, 15, and
14 weeks with variances of 2.11, 3.11, and 2.44,
respectively.
51
5. PERT Simulationrunning WinQSB
36

• Why simulation? Consider Case 2
• Running WinQSB
• ? Open PERT/CPM gt Select PERT gt enter problem gt
  Solve Analyze gt perform simulation
• ? The simulation input menu will drop defaulting
  to random seed with the
• estimated completion time of the critical path
  based on the standard method presented in 2b.
• ? Enter 10,000 for the of simulated
  observations to be made.
• ? Enter the desired completion time
• ? Click on the simulation button.
• ? View the results

52
5. PERT Simulationrunning WinQSB
37

• Results Foundry Project
• independent paths/non-stochastic non-critical
  paths assumptions in play
• E(t) 15 CP A C E G - H prob(X lt 17)
  87.16
• A-C-F-H 9, s2 2.11 A-C-E-G-H 15, s2
  3.11 and B-D-G-H 14, s2 2.44
• independent paths/non-stochastic non-critical
  paths assumptions NOT in play
• Simulation
• E(t) 15.19 prob(X lt 17) 86.10
• A-C-F-H 9, s2 2.11 A-C-E-G-H 15, s2
  3.11 and B-D-G-H 14, s2 2.44
• (strength of the above assumptions if in play
  with Case 1 ????)
• independent paths/non-stochastic non-critical
  paths assumptions NOT in play
• E(t) 15.39 prob(X lt 17) 82.51
• A-C-F-H 9, s2 2.11 A-C-E-G-H 15, s2
  3.11 and B-D-G-H 14, s2 4.513
• (strength of the above assumptions if in play
  with Case 2 ????)

53
5. PERT Simulationrunning WinQSB
38

• Generalizations???