PERT/CPM Project Management Techniques

PERT/CPM
AGENDA
MGT 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

1. Motivation for PERT/CPM Use
• Definition of a Project
• Organizational Trajectories and Change
The Environment
Adaptation and Agility
Reactive Strategy
Proactive Strategy
• Ubiquity of Projects
• Dealing with Project Complexity
2

2. PERT Example
A. AON Network Diagrams:
General Foundry
Diagramming a project’s network of activities
Installation of air-pollution control equipment @ 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
3

2. PERT Example
A. AON Network Diagrams (continued):
General Foundry
A C
Node A represents
Activity A
Node C represents
Activity C
edge or arc
The edge or arc represents
the precedence relationship
between the two activities
4

2. PERT Example
General Foundry continued
A C
B D
A
C
B
D
F
E
A C
B D
5

2. PERT Example
General Foundry concluded
A
B
C
D
F
E
G
H
6

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:
A
B
C
D
E
F
Beginning with one
node and ending with
more than one node.
A
B
C D
E
F
Beginning with more
than one node and
ending with more
than one node.
7

2. PERT Example
A. AON Network Diagrams (concluded)
Sometimes the following convention is used.
Start
Finish
A
B
C D
E
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.
8

2. PERT Example
A2. AOA Network Diagrams (continued):
General Foundry
1 2
Node 1 represents the
beginning of Activity A
Node 2 represents the
ending of Activity A
edge or arc
The edge or arc represents Activity A
8b
Activity A

2. PERT Example
General Foundry continued
1
C
1
2
3
2
8c
A
B
A
1
2
3
A
B
4
5
C
D
E

2. PERT Example
General Foundry concluded
1
2
3
4
5
6
8d
7
A
B
C
D
E
F
G
H

2. PERT Example
A2. AOA Network Diagrams (concluded)
• 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
8e
A
B
C
D
No, and why not?
B
A
D
C
d1
Rather, this is preferred and why?
A
B
C
D
E
A
B
C
E
d2
d1
D
NO!

2. PERT Example
B. Project Completion Times and Probabilities:
General Foundry of Milwaukee
• The duration time for each of a Project’s 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 activity’s optimistic completion time, “m” is the activity’s
most likely completion time, and “b” is the activity’s most pessimistic
completion time.
9

2. PERT Example
General Foundry of Milwaukee (continued)
• 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
10

2. PERT Example
A
2
B
3
C
2
D
4
F
3
E
4
G
5
H
2
11
•From the reduced version of General Foundy’s PERT table., the E(ti) for each activity
can be entered into the project’s 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.

• From the reduced version of General Foundy’s PERT table., the E(ti) for
each activity can be entered into the project’s network diagram.
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.
2. PERT Example
B2. Project Completion Times and Probabilities:
11b
5
6 7
4
2
1
3
A2
B3
C2
D4
E4
F3
G5
H2
critical path

2. PERT Example
B. Project Completion Times and Probabilites:
General Foundry (concluded)
• Given General Foundry’s 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 > 16) = 1 - P ( X < 16) = 1 - P ( Z < (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 Foundry’s 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)
12

3. CPM Example
A. 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 project’s network
diagram, conduct a forward pass where
ESj = ESi + ti
A B C D
3 4 2 5
0 3 7 9
0 = 0 + 0 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2
B
C
D
4
2
5
3
7
?
What if?
ESD =
MAX
ESB + tB = 7
ESC + tC = 9
13

3. CPM Example
• E(arly)F(inish) -- earliest possible time a project’s activity can be completed.
EF calculation -- beginning at the initial node(s) of a project’s network
diagram , conduct a forward pass where
EFi = ESi + ti A B C D
3 4 2 5
0 3 7 9
3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 14 = 9 + 5
3 7 9 14
14

3. CPM Example
• E(arly) S(tarts), E(arly F(inishes) for Milwaukee Foundry .
A
2
B
3
C
2
D
4
F
3
E
4
G
5
H
2
0 2
0 3
4
8
4
13
2
3
4
7
8
13
15
7 EF
ES
15

A
3
B
4
C
2
D
5
0 3 7 9 14
1
4
9
7
3
9 = 14 - 5
7 = 9 - 2
3 = 7 - 4
C
2
D
5
9
14
B
4
A
3
LFC -tC = 7
LFD -tD =9
MIN
LFB =
?
16
3. CPM Example
•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 network’s paths where
LFi = LFj - tj

3. CPM Example
LSi = LFi - ti
14
9
7
3
3 4 2 5
A B C D
9 = 14 -5
7 = 9 -2
3 = 7 -4
0 = 3 - 3
0 3 7 9
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.

• E(arly) S(tarts)/ F(inishes) and L(ate) S(tarts)/F(inishes) for Milwaukee
Foundry .
A
B
C
D
E
F
G
H
2
3
2
4
3
4
5
2
0
0
2
3
4
4
8
13 15
13
7
8
4
7
2
3
EF
ES
15
13
8
4
13
8
2
4
0
1
LF
LS
13
8
10
4
2
4
critical path
18
3. CPM Example

3. CPM Example
A2. 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 project’s network
diagram, conduct a forward pass where:
ESj = ESi + ti
A3 C2 D5
1 2 3
0 = 0 + 0 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2
B4
C2
D5
3
7
4
What if?
ESD =
MAX
ESB + tB = 7
ESC + tC = 9
18b
B4
0 3 7 4 9
2
3
9
node #
ESj

3. CPM Example
• E(arly)F(inish) -- earliest possible time a project’s activity can be completed.
• EF calculation -- beginning at the initial node(s) of a project’s network
diagram , conduct a forward pass where:
EFi = ESi + ti
3 = 0 + 3 7 = 3 + 4 9 = 7 + 2
18c
A3 C2
1 2 3
B4
0 3 7 4 9 D5
0 3 7 9
E2
4
3
D4
3
4
5
7
3
4
7/6 F5
What if?
6 12
12 G3
critical path
node #
ESj
EF
i

3. CPM Example
• E(arly) S(tarts), E(arly F(inishes) for Milwaukee Foundry .
18d
1
2
3
4
5
6 7
A2
B3
C2
D4
E4
F3
G5
H2
0
2 4
3 8
13 15
critical path
3 8/7
2 4
7/13 15
node #
ESj
EF
i

3. CPM Example
• 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 network’s paths where:
A3
1
B4
2
C2
3 D5
4
3 7
9 = 14 - 5
7 = 9 - 2
3 = 7 - 4
C2 4
D5
9
B4
3
2
LFC -tC = 7
LFD -tD =9
MIN
LFB =
?
18e
0
0
LFi = LFj - tj
5
14
9
14
0 = 3 - 3
14
5
14 = 14 - 0
What if?
EFi
LFi
ESi
node #

3. CPM Example
• 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
0
1
2 3 4
A3
B4 C2 D5
9 = 14 -5
7 = 9 -2
3 = 7 - 4
0 = 3 - 3
0
3 7 9
18f
9
5
14 14
7
3
C2 4
D5
9
B4 3
2
LFC -tC = 7
LFD -tD = 9
LSC,D
=
7/9
14
5
What if?
EFi
LFi
ESi
LSj
node #

3. CPM Example
• E(arly) S(tarts)/ F(inishes) and L(ate) S(tarts)/F(inishes) for Milwaukee
Foundry .
18g
1
2
3
4
5
6 7
A2
B3
C2
D4
E4
F3
G5
H2
0
2 4
3 8
13 15
critical path
3 8/7
2 4
7/13 15
0
0 0 15
13
8
4
2
4
13
10/4
15
2
8
4

• T(otal) S(lack) -- the amount of time an activity’s completion can be
delayed without delaying the project’s completion where,
TSi = LFi - ESi - ti and where i = the ith activity.
• F(ree) S(lack) -- the amount of time an activity’s 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.
3. CPM Example
19

3. CPM Example
A. Starts, Finishes., Slacks
• T(otal) S(lacks) and F(ree) S(lacks) for Milwaukee Foundry .
A
B
C
D E
F
G
H
2
3
2
4
3
4
5
2
0
0
2
3
4
4
8
13
ES
15
13
8
4
13
8
2
4
LF
critical path
TSH=0=15-13-2
FSH=0=15-13-2
TSE=8-4-4=0
FSE=8-4-4=0
TSD=1=8-4-3
FSD=1=8-4-3
TSB=1=4-3-0
FSB=0=3-3-0
20

• 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.
3. CPM Example
B
3
D
4
G
5
0
4
3
8
8
13
TSB = 1
FSB = 0
TSD = 1
FSD = 1
TSG = 0
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)*
A -- 3 TPs B -- 4 TPs
A -- 3TPs B -- 4TPs
B -- 4TPs
A -- 3TPs
*Other variations
are possible if the
one TP of slock is
divided up.
21

• 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.
3. CPM Example
C
A B
D E
8
10
7 5
15
Activity I.P.
A
B
C
D
E
__
A
__
C
D
25
25
10
10
0
0
13 20
8 15
TS,FS = 5
TS=5, FS=0
TS=5, FS=0
TS,FS=0 TS,FS=0
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.
22

3. CPM Example
C
A
B
D E
8
10
7 5
15
Activity I.P.
A
B
C
D
E
__
A
__
C
D,F
25
25
10
10
0
0
13 20
8 15
TS,FS = 3
TS=5, FS=2
TS=5, FS=0
TS,FS=0
TS,FS=0
F
17
0
20
F __
TS=3, FS=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, ref;ect the two additional time
periods of slack shared by them beginning with
activity D where TSD = FSD or FSD = 0.
23

3. CPM Example
A2. Starts, Finishes., Slacks
• T(otal) S(lacks) and F(ree) S(lacks) for Milwaukee Foundry .
23b
1
2
3
4
5
6 7
A2
B3
C2
D4
E4
F3
G5
H2
0
2 4
3 8
13 15
3 7/8
2 4
7/13 15
0
0 0
15
13
8
4
2
4
13
10/8
2
8
4
early start
late finish
TSC = 4 - 2 - 2 = 0
FSC = 4 - 2 - 2 = 0
TSD = 8 - 3 - 4 = 1
FSD = 8 - 3 - 4 = 1
TSB = 4 - 0 - 3 = 1
FSB = 3 - 0 - 3 = 0

• 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.
3. CPM Example
B3
2
D4
3
5
4
1
5
4
8
8
TSB = 1
FSB = 0
TSD = 1
FSD = 1
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)*
B -- 3 TPs D -- 4 TPs
B -- 3TPs D -- 4TPs
D -- 4TPs
B -- 3TPs
*Other variations
are possible if the
one TP of slock is
divided up.
23c
4
0
0
C8
TSC = 0
FSC = 0

3. CPM Example
C8
A10
B15
D7
E5
1
2 4
3
Activity I.P.
A
B
C
D
E
__
A
__
C
D
10
20
0
10
0
13
8 15
TS,FS = 5
TS=5, FS=0
TS=5, FS=0
TS,FS=0
TS,FS=0
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.
23d
5
25
25

3. CPM Example
Activity I.P.
A
B
C
D
E
__
A
__
C
D,F
20
F17
F __
TS=3, FS=0
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 tim
periods of slack shared by them beginning with
activity D.
23e
C8
A10 B15
D7
E5
1
2 4
3
10
20
0
10
0
13
8 17
TS,FS = 3
TS=5, FS=2
TS=5, FS=0
TS,FS=0 TS,FS=0
5
25
25

3. CPM Example
B. Resource Allocation Scheduling (ES)
Activity
*A (2)
B (3)
*C (2)
D (4)
*E (4)
F (3)
*G (5)
*H (2)
TP1TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11TP12 TP13TP14 TP15
Activity Cost
A -- 22K ($22,000)
B -- 30K
C -- 26K
D -- 48K
E -- 56K
F -- 30K
G -- 80K
H -- 16K
11K 11K
13K 13K
14K 14K 14K 14K
16K 16K 16K 16K
* -- critical path
16K
8K 8K
10K 10K 10K
12K 12K 12K 12k
10K 10K 10K
21
21
21
42
23
65
25 36 36 36 14 16 16 16 16 16 8 8
90 126 162 198 212 228 244 260 276 292 300 308
24

3. CPM Example
B. Resource Allocation Scheduling (LS)
Activity
*A (2)
B (3)
*C (2)
D (4)
*E (4)
F (3)
*G (5)
*H (2)
TP1TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11TP12 TP13TP14 TP15
Activity Cost
A -- 22K ($22,000)
B -- 30K
C -- 26K
D -- 48K
E -- 56K
F -- 30K
G -- 80K
H -- 16K
11K 11K
13K 13K
14K 14K 14K 14K
16K 16K 16K 16K
* -- critical path
16K
8K 8K
10K 10K
12K 12K 12k
11
11
21
32
23
55
23 26 26 26 26 16 16 26 26 26 8 8
78 104 130 156 182 198 214 240 266 292 300 308
10K
12K
10K 10K 10K
25

0
50
100
150
200
250
300
350
TP1 TP3 TP5 TP7 TP9 TP11 TP13 TP15
Early Start
Late Start
3. CPM Example
B. Resource Allocation Scheduling
Early/Late Start Resource Allocation Schedules
Project Time Periods
C
u
m.
P
r
o
j.
C
o
s
t
s
26

4. CPM Example
CPM with Crashing --a
• 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.
27

4. CPM Example
CPM with Crashing --b
• 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 Foundry’s 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
28

4. CPM Example
CPM with Crashing --c
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
1
1 2 3 4 6
5
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
5
6
29

4. CPM Example
CPM with Crashing --d ( Summary)
• 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?
30

5. PERT Simulation
--agenda--
• Motivation
• Illustrative Examples
• Performing Simulations with WinQSB
31

5. PERT Simulation
• Motivation
▪ non-stochasticity/stochasticity & non-critical paths
▪ independence of/interdependence between paths
▪ strength of an assumption
32

5. PERT Simulation—motivation
--non-stochasticity/stochasticity & non-critical paths--
Reconsider General Foundry of Milwaukee
• The E(ti) for each activity in this reduced version of General Foundy’s PERT table has
been entered into the project’s 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.
• Implications of assuming non-stochasticity on the (two) non-critical paths--
how serious?
• How strong is the assumption of non-stochasticity in this case?
33
A
2
B
3
C
2
D
4
F
3
E
4
G
5
H
2

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

5. PERT Simulation—illustrative examples
CASE 1
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
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 merely replicates
the results of a previous page–paths
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.
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?
35

5. PERT Simulation—running WinQSB
• Why simulation? Consider Case 2
• Running WinQSB
▪ Open PERT/CPM > Select PERT > enter problem > Solve & Analyze > 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
36

Results: Foundry Project
• independent paths/non-stochastic non-critical paths assumptions in play
E(t) = 15; CP ≡ A – C – E – G - H; prob(X < 17) = 87.16%
A-C-F-H: 9, σ2 = 2.11; A-C-E-G-H: 15, σ2 = 3.11; and B-D-G-H: 14, σ2 = 2.44
• independent paths/non-stochastic non-critical paths assumptions NOT in play
Simulation
E(t) = 15.19; prob(X < 17) = 86.10%
(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 < 17) = 82.51%
(strength of the above assumptions if in play with Case 2 ????)
37

• Generalizations???
38

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