Pert/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 EnvironmentAdaptation and Agility

Reactive StrategyProactive 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

Hinspect and test F,G

3

2. PERT ExampleA. AON Network Diagrams (continued):

General Foundry

A CNode 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


General Foundry continued A C

B D

A C

BD

F

E

A C

B D

5


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

CD

E

F

Beginning with onenode and ending withmore than one node.

A

BC 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.

Start Finish

A

BC 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 2Node 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


General Foundry continued 1

C

1

2

3

2

8c

AB

A

1

2

3

A

B

4

5

C

DE


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

BC

D

E

A

BC

Ed2

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


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


General Foundry of Milwaukee (continued)

A2

B3

C2

D4

F3

E4

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 D3 4 2 5

0 3 7 9

0 = 0 + 03 = 0 + 37 = 3 + 49 = 7 + 2B

CD4

25

3

7?

What if?

ESD = MAX

ESB + tB = 7ESC + tC = 9

13


• 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 D3 4 2 5

0 3 7 9

3 = 0 + 37 = 3 + 49 = 7 + 214 = 9 + 5

3 7 9 14

14


• 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

B4

C2

D5

0 3 7 9 1414

9739 = 14 - 57 = 9 - 23 = 7

- 4 C2

D5

9

14

B4

A3

LFC -tC = 7LFD -tD =9MIN

LFB =?

16


•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


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

DE

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


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 + tiA3 C2 D5

1 2 3

0 = 0 + 03 = 0 + 37 = 3 + 49 = 7 + 2

B4

C2

D5 3

74What if?

ESD = MAX

ESB + tB = 7ESC + tC = 9

18b

B40 3 7 4 9

2

3

9node #

ESj


• 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 + 37 = 3 + 49 = 7 + 2

18c

A3 C21 2 3B40 3 7 4 9 D50 3 7 9

E2

4

3

D43

45 7

3

47/6 F5

What if? 6 1212 G3

critical path

node #

ESjEFi


• 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


• 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:

A31

B42 C23 D54

3 79 = 14 - 57 = 9 - 23 = 7 -

4

C2 4

D5

9B4 32

LFC -tC = 7LFD -tD =9MIN

LFB =?

18e

00

LFi = LFj - tj

514914

0 = 3 - 3

14 5

14 = 14 - 0

What if?

EFi

LFi

ESi node #


• 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 4A3B4 C2 D5

9 = 14 -57 = 9 -23 = 7 - 4

0 = 3 - 3

0

3 7 9

18f

9514 1473

C2 4

D5

9B4 3

2

LFC -tC = 7

LFD -tD = 9LSC,D =

7/914 5

What if?

EFi

LFi

ESi

LSj

node #


• 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 1500 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.


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=15-13-2TSE=8-4-4=0FSE=8-4-4=0

TSD=1=8-4-3FSD=1=8-4-3TSB=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.


B

3

D

4

G

5

04

38

813

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 -- 4TPsB -- 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.


C

A B

D E8

10

7 5

15

ActivityI.P.ABCDE

__A__CD25

25

10100

013 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



C

A

B

D E8

10

7 5

15

ActivityI.P.ABCDE

__A__CD,F25

25

10

10

0

013 20

8 15

TS,FS = 3TS=5, FS=2TS=5, FS=0

TS,FS=0

TS,FS=0

F17

020

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 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 1500 0 1513

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.


B32 D4

3 54

15

48

8

TSB = 1FSB = 0TSD = 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 -- 4TPsD -- 4TPsB -- 3TPs

*Other variationsare possible if the one TP of slock isdivided up.

23c

40

0C8

TSC = 0FSC = 0



C8

A10 B15

D7

E5

1

2 4

3

ActivityI.P.ABCDE

__A__CD

10

20

010

0

138 15

TS,FS = 5TS=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

525

25



ActivityI.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

010

0

138 17

TS,FS = 3TS=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)

TP1TP2 TP3 TP4TP5 TP6 TP7 TP8TP9 TP10TP11TP12TP13TP14TP15

Activity CostA -- 22K ($22,000)B -- 30KC -- 26K

D -- 48KE -- 56KF -- 30K

G -- 80KH -- 16K

11K11K

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

2121

2142

2365

25 36 36 36 14 16 16 16 16 16 8 890 126 162 198 212 228 244 260 276 292 300 308

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)

TP1TP2 TP3 TP4TP5 TP6 TP7 TP8TP9 TP10TP11TP12TP13TP14TP15

Activity CostA -- 22K ($22,000)B -- 30KC -- 26K

D -- 48KE -- 56KF -- 30K

G -- 80KH -- 16K

11K11K

13K 13K

14K 14K 14K 14K

16K 16K 16K 16K

* -- critical path

16K 8K 8K

10K 10K

12K 12K 12k

1111

2132

2355

23 26 26 26 26 16 16 26 26 26 8 878 104 130 156 182 198 214 240 266 292 300 308

10K

12K

10K 10K 10K

25

050100150200250300350

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 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 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

29

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?

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

A2

B3

C2

D4

F3

E4

G5

H 2

5. PERT Simulation--motivation

--independence of/interdependence between paths--Reconsider General Foundry of Milwaukee

34

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 pathsWhich 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 1ACT 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 2ACT 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 merely replicatesthe 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 playE(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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Pert/cpm

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