PERT/CPM AGENDA MGT 606 1. Motivation PERT/CPM 2a. Digraphs (Project Diagrams) 2b. PERT 3a. CPM--...
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Transcript of PERT/CPM AGENDA MGT 606 1. Motivation PERT/CPM 2a. Digraphs (Project Diagrams) 2b. PERT 3a. CPM--...
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 ChangeThe Environment
Adaptation and Agility
Reactive Strategy
Proactive Strategy
• Ubiquity of Projects
• Dealing with Project Complexity
2
2. PERT ExampleA. 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 ExampleA. 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 ExampleA. AON Network Diagrams (continued):
General Foundry continued A C
B D
A C
BD
F
E
A C
B D
5
2. PERT ExampleA. AON Network Diagrams (continued):
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:
AB
C
DE
F
Beginning with onenode and ending withmore than one node.
A
B
C D
E
F
Beginning with morethan one node and ending with more than one node.
7
2. PERT ExampleA. AON Network Diagrams (concluded)
Sometimes the following convention is used.
StartFinish
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 intime at which the project begins and ends . The use of such aconvention is not necessary and will generally be avoided.
8
2. PERT ExampleA2. AOA Network Diagrams (continued):
General Foundry
1 2
Node 1 represents thebeginning 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 ExampleA2. AOA Network Diagrams (continued):
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 ExampleA2. AOA Network Diagrams (continued):
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 morethan one node.
Act. I.P. A __ B __ C A,B D A E B
8e
A
B
C
DNo, 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
DNO!
2. PERT ExampleB. 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 ExampleB. Project Completion Times and Probabilities:
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 ExampleB. Project Completion Times and Probabilities:
General Foundry of Milwaukee (continued)
A2
B3
C2
D4
F3
E
4
G5
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.
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.
2. PERT ExampleB2. Project Completion Times and Probabilities:
General Foundry of Milwaukee (continued)
11b
5
6 7
42
13
A2
B3
C2
D4
E4
F3
G5
H2
critical path
2. PERT ExampleB. 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 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 project’s network
diagram, conduct a forward pass where
ESj = ESi + ti A B C D
3 4 2 50 3 7 9
0 = 0 + 0 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2B
C
D4
2
5
3
7
?
What if?
ESD =
MAX
ESB + tB = 7
ESC + tC = 9
13
3. CPM ExampleA. Starts, Finishes, Slacks
• 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 50 3 7 9
3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 14 = 9 + 5
3 7 9 14
14
3. CPM ExampleA. Starts, Finishes, Slacks
• 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 EFES
15
A3
B
4
C
2
D
50 3 7 9 14
14973
9 = 14 - 57 = 9 - 23 = 7 - 4
C
2
D
5
9
14
B
4A3
LFC -tC = 7
LFD -tD =9MIN
LFB =?
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 network’s paths where
LFi = LFj - tj
3. CPM ExampleA. Starts, Finishes, Slacks
LSi = LFi - ti
149733 4 2 5
A B C D
9 = 14 -57 = 9 -23 = 7 -40 = 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
413
8
2
4
0
1
LF
LS
13
8
10
4
2
4
critical path
183. CPM ExampleA. Starts, Finishes, Slacks
3. CPM ExampleA2. 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
74
What if?
ESD =
MAX
ESB + tB = 7
ESC + tC = 9
18b
B40 3 7 4 9
2
3
9
node #
ESj
3. CPM ExampleA2. Starts, Finishes, Slacks
• 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 3B4
0 3 7 4 9 D50 3 7 9
E2
4
3
D43
4
5 7
3
4
7/6 F5
What if?6 1212 G3
critical path
node #
ESjEFi
3. CPM ExampleA2. Starts, Finishes, Slacks
• 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
H20
2 4
3 8
13 15
critical path
3 8/7
2 4
7/13 15
node #
ESjEFi
3. CPM ExampleA2. 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 network’s paths where:
A3
1B4
2C2
3 D543 7
9 = 14 - 57 = 9 - 2
3 = 7 - 4
C2 4
D5
9B4 32
LFC -tC = 7
LFD -tD =9MIN
LFB =
?
18e
0
0
LFi = LFj - tj
5149
14
0 = 3 - 3
145
14 = 14 - 0
What if?
EFi
LFi
ESinode #
3. CPM ExampleA2. Starts, Finishes, Slacks
• 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
01
2 3 4A3
B4 C2 D5
9 = 14 -57 = 9 -23 = 7 - 4
0 = 3 - 3
0
3 7 9
18f
95
14 1473
C2 4
D5
9
B4 32
LFC -tC = 7
LFD -tD = 9LSC,D =
7/9
145
What if?
EFi
LFi
ESi
LSj
node #
3. CPM ExampleA2. Starts, Finishes, Slacks
• 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
H20
2 4
3 8
13 15
critical path
3 8/7
2 4
7/13 150
0 0 1513
8
42
4
13
10/4
15
2
84
• 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 ExampleA. Starts, Finishes, Slacks
19
3. CPM ExampleA. 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
413
8
2
4
LF
critical path
TSH=0=15-13-2FSH=0=15-13-2
TSE=8-4-4=0FSE=8-4-4=0
TSD=1=8-4-3FSD=1=8-4-3
TSB=1=4-3-0FSB=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 ExampleA. Starts, Finishes., Slacks
B
3
D
4
G
5
0
4
3
8
8
13
TSB = 1FSB = 0
TSD = 1FSD = 1
TSG = 0FSG = 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 arescheduled over an 8 time period segment.
8 time periods (TPs)*A -- 3 TPs B -- 4 TPsA -- 3TPs B -- 4TPs
B -- 4TPsA -- 3TPs
*Other variationsare possible if the one TP of slock isdivided 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 ExampleA. Starts, Finishes., Slacks
C
A B
D E8
10
7 5
15
Activity I.P.ABCDE
__A__CD25
25
10100
0
13 20
8 15
TS,FS = 5TS=5, FS=0TS=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
• 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 ExampleA. Starts, Finishes., Slacks
C
A
B
D E8
10
7 5
15
Activity I.P.ABCDE
__A__CD,F25
25
10
10
0
0
13 20
8 15
TS,FS = 3TS=5, FS=2TS=5, FS=0
TS,FS=0
TS,FS=0
F
170
20F __
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 ExampleA2. 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
H20
2 4
3 8
13 15
3 7/8
2 4
7/13 150
0 01513
8
42
4
13
10/82
84
early start
late finish
TSC = 4 - 2 - 2 = 0FSC = 4 - 2 - 2 = 0
TSD = 8 - 3 - 4 = 1FSD = 8 - 3 - 4 = 1
TSB = 4 - 0 - 3 = 1FSB = 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 ExampleA2. Starts, Finishes., Slacks
B32
D4
35
4
1
5
48
8
TSB = 1FSB = 0
TSD = 1FSD = 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 arescheduled over an 8 time period segment.
8 time periods (TPs)*B -- 3 TPs D -- 4 TPsB -- 3TPs D -- 4TPs
D -- 4TPsB -- 3TPs
*Other variationsare possible if the one TP of slock isdivided up.
23c
40
0
C8
TSC = 0FSC = 0
• 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 ExampleA2. Starts, Finishes., Slacks
C8
A10B15
D7
E5
1
2 4
3
Activity I.P.ABCDE
__A__CD
10
20
0
10
0
138 15
TS,FS = 5
TS=5, FS=0
TS=5, FS=0
TS,FS=0TS,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
525
25
• 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 ExampleA2. Starts, Finishes., Slacks
Activity I.P.ABCDE
__A__CD,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 time periods of slack shared by them beginning with activity D.
23e
C8
A10 B15
D7
E5
1
2 4
310
20
0
10
0
138 17
TS,FS = 3
TS=5, FS=2
TS=5, FS=0
TS,FS=0 TS,FS=0
525
25
3. CPM ExampleB. Resource Allocation Scheduling (ES)
Activity
*A (2)B (3)
*C (2)D (4)
*E (4)
F (3)
*G (5)
*H (2)
TP1 TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11TP12 TP13 TP14 TP15
Activity Cost
A -- 22K ($22,000)B -- 30KC -- 26K
D -- 48KE -- 56KF -- 30K
G -- 80KH -- 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 ExampleB. Resource Allocation Scheduling (LS)
Activity
*A (2)B (3)
*C (2)D (4)
*E (4)
F (3)
*G (5)
*H (2)
TP1 TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11TP12 TP13 TP14 TP15
Activity Cost
A -- 22K ($22,000)B -- 30KC -- 26K
D -- 48KE -- 56KF -- 30K
G -- 80KH -- 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 StartLate Start
3. CPM ExampleB. Resource Allocation Scheduling
Early/Late Start Resource Allocation Schedules
Project Time Periods
Cum.
Proj.
Costs
26
4. CPM ExampleCPM 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 ExampleCPM 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 PathAn 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 ExampleCPM 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 7B - D - G - H : 14 14 11 10 9 8 7
1
1 2 3 4 65
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
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4. CPM ExampleCPM 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?
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5. PERT Simulation--agenda--
• Motivation
• Illustrative Examples
• Performing Simulations with WinQSB
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5. PERT Simulation
• Motivation
▪ non-stochasticity/stochasticity & non-critical paths
▪ independence of/interdependence between paths
▪ strength of an assumption
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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?
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A2
B3
C2
D4
F3
E
4
G5
H 2
5. PERT Simulation--motivation --independence of/interdependence between paths--
Reconsider General Foundry of Milwaukee
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A2
B3
C2
D4
F3
E4
G5
H 2
A2
B3
C2
D4
E4
G5
H 2
F3
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 examplesReconsider 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 4/36The 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 haveE(t)s of 9, 15, and 14 weeks with variancesof 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 1and Case 2 differ one from the other?
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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
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5. PERT Simulation—running WinQSBResults: 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
SimulationE(t) = 15.19; prob(X < 17) = 86.10%
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
(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%
A-C-F-H: 9, σ2 = 2.11; A-C-E-G-H: 15, σ2 = 3.11; and B-D-G-H: 14, σ2 = 4.513
(strength of the above assumptions if in play with Case 2 ????)
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5. PERT Simulation—running WinQSB
• Generalizations???
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