3. Project Analysis Tools

7/30/2019 3. Project Analysis Tools

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Project Scheduling

Faculty of Applied Engineering and Urban Planning

Civil Engineering Department

Week ( 6 + 7 )

Lec. ( 11 + 12 + 13 + 14 )

2nd Semester 2008/2009

UP Copyrights 2008

ConstructionProje

ct

M

anagement

Eng: Eyad Haddad


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CH4: Project Scheduling

Construction Project management functions:

Scheduling = Planning + Time

Scheduling is the determination of the timing of the activities comprising

the project to enable managers to execute the project in a timely manner.

1. Planning

2. Organization

3. Supervision

4. Control

1. Time

2. Cost

3. Quality

4. Performance

Scheduling


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CH4: Project Scheduling

The project scheduling is used for:

1. Knowing the activities timing and the project completion time.

2. Having resources available on site in the correct time.

3. Making correction actions if schedule shows that the plan will result in

late completion.

4. Assessing the value of penalties on project late completion.

5. Determining the project cash flow.

6. Evaluating the effect of change orders on the project completion time.

7. Determining the value of project delay and the responsible parties.


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4.2 The Critical Path Method (CPM)

The critical path can be defined as

the longest possible path through the "network" of project activities.

(CPM) is the most widely technique used for scheduling, it calculates the

minimum completion time for a project along with the possible start and

finish times for the project activities.


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The critical path itself represents the set or sequence of activities

which will take the longest time to complete.

The duration of the critical path is the sum of the activities'

durations along the path.

Duration of the critical path represents the minimum time required

to complete a project.

Any delays along the critical path would delay the project.

More than one critical path may be among all the project activities,

so completion of the entire project could be delayed by delaying

activities along any one of the critical paths.


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For example,

a project consisting of two activities performed in parallel that each

requires three days would have each activity critical for a completion in

three days.

Critical path scheduling assumes that a project has been divided into

activities of fixed duration and well defined predecessor relationships.

A predecessor relationship implies that one activity must come

before another in the schedule


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The CPM is a systematic scheduling method for a project network and involves

four main steps:

1. A forward path to determine activities early-start times;

2. A backward path to determine activities late-finish times;

3. Float calculations ( Free & Total ) float; and

4. Identifying critical activities.


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4.3.1 Activity-on-node networks calculations

The objective of arrow network analysis is to compute each event in the

network its early and late timings. These times are defined as

Early event time (ET) is the earliest time at which an event can occur,

considering the duration of preceding activities.

Late event time (LT) Is the latest time at which an event can occur if the

project is to be completed on schedule.

i jx

ETj LTjETi LTi

dx

. Forward Path:1

ETj = ETi + dx


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. Forward Path:1

1 3

5

7

9 11

0

Project

Start=0

A

d=3

C4

E

5

B

3

D

6 d2

d13

Es+d=EF

6

3+3=60+3=3

9

6+0=6

3+4=7

9+0=9

14

9+5=14

9

3+6=9

1A

Project

Start=0

1A


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. Backward Path2LS = LF d

1 3

5

7

9 11

0 0

3-3=0

A3

C4 E5

B

3

D

6 d2

d13 3

9-4=5

6 9

9-0=9

9-6=3

9 9

14-5=9

LF-d=LS

14 14

9 9

9-0=9

9-3=6


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3. Float Calculations:

First, let's tabulate the information we have as shown in next Table

One important aspect is Total-Float (TF) calculations, which determine the

flexibility of an activity to be delayed.

Total Float (TF) = LF EF

= LS ES

:TF.

Free Float (FF) = ETj ETi d

or FF = smallest ES (of succeeding activities) EF (of current activity)

:FF.


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3. Float Calculations:

Total Float (TF) = LF EF

= LS ES

:TF.



:FF.

i j

ES EF

A

LS LF

AOA

ES EF

ES EF

A

B

LS LF

LS LF

ES EF

A

LS LF

AON

TF

i j

ES EF

A

LS LF

AOA

i j

ES EF

B

LS LF

AOA


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Total Float (TF) = LF EF = LS ES



Critical

Activity

TotalFloat

(TF)

LateFinish

(LF)

EarlyFinish

(EF)

Late Start

(LS)

EarlyStart

(ES)

DurationActivity

Yes033003A

No396633B

No297534C

Yes099336D

Yes01414995E


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4.3.2 Precedence Diagram Method (PDM):

Precedence Diagram Method (PDM) is the CPM scheduling method used for AON

networks and it follows the same four steps of the CPM for AOA method.

Forward PathForward path can proceed from one activity to the other; the process is as

follow .

3 6

B(3)

3 7

C(4)

3 9

D(6)

0 3

A(3)

9 14

E(5)

Earlystart

Name(duration)

Earlyfinish

Latestart

Latefinish

6,7,or 9

Fig. 4.8: Forward Path in PDM Analysis


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Backward Path:

3 6

B(3)

3 7

C(4)

3 9

D(6)

0 3

A(3)

9 14

E(5)

Earlystart

Name(duration)

Earlyfinish

Latestart

Latefinish

6,5, or 3

149

96

95

93

30


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Floats

Activity A7 days Activity B13 days Activity C4 days

Start CompletionProject duration = 24 days

CASE 1: All activities are critical: total float and free floats for all activities = 0

Activity A

7 days

Activity B

13 days

Activity C

4 days

Start CompletionTotal Float = 5Free Float = 5

CASE 2: Activity sequence in which one activity has total and free float

Activity D

8 days

Activity A

7 days

Activity B

13 days

Activity C

4 days

Start Completion

Total Float of D = 5 Total Float of E = 5

Free Float of D = 0 Free Float = 5

CASE 3: Activity sequence illustrating total and free float

Activity D

5 days

Activity E

3 days


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2-Floats

Areas of shared float

Activity duration Total Float

Start Event Finish Event

TLi TEjTEi TLj

Activity duration Free Float

Activity duration Independent Float


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Float Calculations:

Total Float (TF) = LF EF= LS ES


Duration

3

3

46

5

ES

0

3

33

9

LF

3

9

99

14

LS

0

6

53

9

EF

3

6

79

14

TF

0

3

20

0

Activity

A

B

CD

E

Critical Act.

Yes

No

NoYes

Yes

MDPPDM C l l ti


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A

3

B

3

C

5

D

7

E

4

F

6

G

4

I

9

H

6

J

8

Fn

12 20

10 166 10

6 12

8 12

10 19

3 6

3 8

3 10

0 3

12 20

14 208 12

6 12

8 12

11 20

3 6

3 8

4 11

0 3

Act

Dur

ES EF

LS LF

TF/FF

ethod)MiagramDrecedenceP(PDM =PDM Calculations

Example

0/ 0

0/ 0

0/ 0

1/ 0

2/ 0

0/ 0

0/ 0

1/1

4/4

0/0

TF/FF

TFi = LFi - EFi

Free Float (FF) = ETj

ETi

dor FF = smallest ES (of succeeding activities) EF (of current activity)


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Precedence Relationships - Lead & LagFFij Lag time for a finish-to-finish relationship. (The succeeding

activity finishes this amount of time after the completion of thepreceding activity.)

SSij Lead time for a start-to-start relationship. (The preceding

activity starts this much earlier than the start of the

succeeding activity.)

FSij Lag time for a finish-to-finish relationship. (The succeeding

activity starts this amount of time after the completion of the

preceding activity.)

SFij Lead time for a start-to-finish relationship. (The preceding

activity starts this much earlier than the completion of the

succeeding activity.)


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PRECEDENCE LOGIC1. Preceding Activities.

Which activities must be finished before this activity may begin ?

What is the time lag? (finish to start.)

Which activities must be started before this activity may begin?

What is the lead time (start to start.)

Which activities must be finished before this activity may be completed?

What is the lag time? (Finish to finish)

Which activities must be started before this activity is completed?What is the lead time ? (start to finish.)


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Follow: PRECEDENCE LOGIC

2. Succeeding Activities

Which activities can begin after the finish of this activity?

What is the time lag? (finish to start.)

Which activities can begin after the start of this activity?

What is the lead time? ( Start to start )

Which activities can be completed after the finish of this activity?

What is the lag time? (Finish to finish.)

Which activities can finish after the start of this activity? What is

the lead time? (Start to finish.)3. Concurrent Activities.

Which activities can be carried out at the same time?

(Start to start equals zero, that is, SS = 0 in this case.)


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Lead/Lag Relationships

ES DESC.EF ES DESC. EF

FF ij

FS ij

SS ijSF ij

Forward Pass

Backward Pass

i Di jDj

LS DESC.LF LS DESC. LF

FF jk

FS jk

SS jk

SF jk

j Dj kDk


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PDM Activity Diagramming Methods

Activity No.

Duration RESP.StartSide

Finish

Side

METHOD 1

Activity No.

Duration RESP.StartSide

Finish

Side

METHOD 2

DESCRIPTION DESCRIPTION

ES

LS

EF

LF

Activity No.

LS LF

Sta

rtSide

Finish

Side

METHOD 3

Activity No.

Sta

rtSide

Finish

Side

METHOD 4

DESCRIPTION DESCRIPTION

ES

LS

EF

LF

DUR TF

ES EFDuration RESP.


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Logical Relationships of PDM

12

FINISH - TO - START

Layout & Excavate

2 GO

12Install exterior

Conduit & piping

5 EL

20

FINISH - TO - FINISH

Install fuel tanks

2 GO

12Install exterior

Conduit & piping

5 EL

10Contract

Award

2 GO

10Layout &

Excavate

2 GO

START - TO - FINISH

12

Layout &

Excavate

2 GO

18

Install fuel

tanks

2 ME

START - TO - START

12

Layout & Excavate

2 GO

12Install exterior

Conduit & piping

5 EL

Relationship with Lag

1


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PDM Calculation Procedure(Assumes no splitting of activity is allowed)

FORWARD PATH

Step 1

jiji

jiji

iji

iji

j

DSFES

DFFEF

SSES

FSEF

MaxES

1

Step 2

jjj DESEF


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Follow PDM Calculation Procedure

BACKWARD PATH

Step 1

iijj

iijj

ijj

ijj

i

DSFLF

DSSLS

FFLF

FSLS

alTimeTer

MinLF

min

Step 2

iii DLFLS


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Calculation of Total Float and Free Float

Total Float

iii EFLFTF

Free Float

iijj

iijj

iijj

iijj

ESSFEF

EFFFEF

ESSSES

EFFSES


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A

10

B

7

C

10

D

10

E4

F

3

G

10

H

15

I

7

K

5

L

5

J

3

M

6

N

8

SS5

FS2

FF2

SS5

FF1

FS5

FS3

SF7

SS2

SS10, FF2

FS2

SS3

FF2, SS1

FS1

FF0

1 11

6 13

11 21

11 21

7 11

15 18

21 31

16 31

16 23

23 28

28 33

11 14

31 37

38 46

38 46

ESi + SSij

1 +5 = 6

EFi + FFi - Di

11 + 0 - 4

FS0

FS0

EFi + FSij

13 + 2

ESi + SFij - Dj

15 + 7 -5

EFi + FFij - Dj

28 + 2 - 8

ESi + SSij

23 + 1

EFi + FFij-Dj13 + 2 -10

FS0

ESi +SSij

11 + 5

EFi + FFij - Dj

21 + 1 - 7

EFi + FSij

11 +5

ESi + SSij

21 + 2

ESi + SSij

16 + 10

EFi + FFij - Dj

31 + 2 - 5

EFi+FSij

23 + 2

FS0

EFi + FSij

14 + 3

ESi+SSij

28+3EFij+FSij

37+1

FORWARD PASS


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A

10

B

7

C

10

D

10

E4

F

3

G

10

H

15

I7

K

5

L

5

J

3

M

6

N

8

SS5

FS2

FF2

SS5

FF1

FS5

FS3

SF7

SS2

SS10, FF2

FS2

SS3

FF2, SS1

FS1

FF0

38 46

38 46

31 37

31 37

23 28

37 42

28 33

28 33

11 14

25 28

15 18

35 38

21 31

35 45

16 31

16 31

16 23

19 26

6 13

26 33

11 21

11 21

11 21

15 25

7 11

11 14

1 11

1 11

LFj - FFij

46 - 2

LSj - SSj - Di

38 - 1 + 5

LSj - FSij

38 - 1

LSj - SSij + Di

31 - 3 + 5

LSj - FSij

31 - 3

LFj - Sfij + Di

42 - 7 + 3

LSj - SSij + Di

37 - 2 + 10

LFi - FFij

33 - 2

LSj - SSij + Di

28 - 10 + 15

LSj - FSij

28 - 2LSj - FSij

19 - 5

LFj - FFij

26 - 1

LSj - SSij + Di

16 - 5 + 10

FS0

LSj - FSij

35 - 2

LSj - SSij + Di26 - 5 + 10

SF0

SF0

SF0

LFj - FFij

45 - 2

BACKWARD PASS


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A

10

B

7

C

10

D

10

E

4

F

3

aG

10

H

15

I

7

K

5

L

5

J

3

M

6

N

8

SS5

FS2

FF2

SS5

FF1

FS5

FS3

SF7

SS2

SS10, FF2

FS2

SS3

FF2, SS1

FS1

FF0

38 46

38 46

31 37

31 37

28 33

28 33

11 14

25 28

23 28

37 42

15 18

35 38

21 31

35 45

16 31

16 31

1623

19 26

6 13

26 33

11 21

11 21

11 21

15 25

7 11

11 14

1 11

1 11

ESj - SSij - ESi

6-5-1

ESj - FSij - EFi

15 - 2-13

ESj - Sfij - ESi

28 - 7 - 15 ESj - SSij - ESi

38 - 1 - 23

EFj - FFij - EFi

46 - 2 - 28

ESj - FSij - EFi

38 - 1 - 37

ESj - FSij - EFi31 - 3 - 14

ESj - FSij - EFi

11 - 0 - 11

ESj - FFij -EFi11 - 0 - 11

EFj - FFij - EFi

31 - 2 - 13

ESj - SSij - ESi

16 - 5 - 11

ESj - FSij -EFi

21 - 0 - 21

FS0

ESj - FFij - EFi

23 - 1 - 21

ESj - FSij - EFi

16 - 5 - 11

FS0

ESj - SSij - ESi

23 - 2 - 21

ESj - FSij - EFi

28 - 2 - 23

EFj - FFij - EFi

33 - 2 - 31

ESj - SSij - ESi

28 - 10 - 16

ESj - SSij - ESi

31 - 3 28

0/0

20/0

0/0

4/1

3/0

20/0

14/0

0/0

3/3

14/14

0/0

14/14

0/0

0/0

FS0

FS0

ESj - FSij - ESj

11 - 0 - 1

ESj - FSij - Efi

11 - 0 - 11

TF/FF

TFi = LFi - EFi


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4.4 Time-Scaled Diagrams:

Time-scaled diagrams are used extensively in the construction industry.

Such diagrams enable one

to determine immediately which activities are scheduled to

proceed at any point in time .to monitor field progress.

it can be used to determine resources need.

The time scale used in time-scaled diagrams can be either the calendar datesor the working periods (ordinary dates), or using both at the same time.

Its disadvantage is that it needs a great effort to be modified or updated. Also,it can not be used to

represent overlapping activities.

A

3

C

4

B3

D

6

E

5

3

2

Time-scaleddiagram

1 2 3 4 5 6 7 8 9 10 11 12 13 14


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The TF for activity A equals the smallest of the sum of the floats along all pathsfrom the end of activity A to the end of the project. The float on path ABE = 3,path ACE = 2 and path ADE = 0, then the TF of activity A = 0. The calculationsare shown in Table 4.2.

Table 4.2 Time-scaled diagram calculations


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Schedule Presentation:4.5

After the AOA and AON calculations are made, it is important to present theirresults in a format that is clear and understandable to all the parties involved in

, named afterBar chart or Gantt chartthe project. The simplest form is thethe person who first used it. A bar chart is a time versus activity chart in which

activities are plotted using their early or late times.

a) Earlybar chat

b) Latebar chart


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The bar chart representation:

It shows various details. Float times of activities, critical activities can be shownin a different color, or bold borders, as shown in Figure 4.12. The bar chart canalso be used for accumulating total daily resources and / or costs, as shown atthe bottom part of Figure 6.13. In this figure, the numbers on each activity

represent the number of labors needed.

Figure 4.13: Using bar chart to accumulate resources


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4.6 Criticisms to Network Techniques:

1- Assume all required resources are available:

The CPM calculations do not incorporate resources into their formulation. Also,

as they deal with activity durations only, it can result in large resourcefluctuations. Dealing with limited resources and resource leveling, therefore,has to be done separately after the analysis.

2- Ignore project deadline:

The formulations of CPM and PDM methods do not incorporate a

deadline duration to constrain project duration.

3- Ignore project costs:

Since CPM and PDM methods deal mainly with activities durations, they do notdeal with any aspects related to minimize project cost.

4- Use deterministic durations:

The basic assumption in CPM and PDM formulations is that activity durationsare deterministic. In reality, however, activity durations take certain probabilitydistribution that reflect the effect of project conditions on resource productivityand the level of uncertainty involved in the project.


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Solved Examples4.7

3.1Example

For the project data in Table 4.3, answer the following questions:

a) Draw an AOA network of the project?

b) Perform forward path and backward path calculationsc) What is the effect of delaying activity D by 3 days?


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Solution:

a, b

0

1 A 2

3

5 6

0

2

4

2 2

8 8

14 14 16 16

9 11

B

6

E

6

C

3 F

3

G

21D

8,or10

14,or122,or 8

9,or 5

c) Total float of activity D = LF ES d = 11 8 1 = 2.

3 2l


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3.2Example

Perform PDM calculations for the small project below and determine activitytimes. Durations are shown on the activities.

S l ti 7 9


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Solution:

1 5

B(4)

1 5

5 6

D(1)

5 6

6 7

G(1)

6 7

7 14

J(7)

7 14

14 16

L(2)

14 16

1 2

C(1)

6 7

2 4

E(2)

7 9

4 5

H(1)

9 10

2 4

F(2)

8 10

5 9

K(4)

10 14

0 1

A(1)

0 1

7 9

I(2)

12 14

7or85or4

9or9or14

1or6

12or7

3 3E l


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3.3Example

For the activities listed in the table below, draw the time-scaled diagram andmark the critical path. Determine the completion time for the project. Tabulateactivities times and floats.

S l ti


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Solution:

3 4E l


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3.4Example

Perform PDM calculations for the small AoN network shown here. Pay specialattention to the different relationships and the lag times shown on them.

Solution:

0 3

A(3)

0 3

2 5

B(3)

4 7

3 7

C(4)

3 7

3 9

D(6)

4 10

7 12

E(5)

7 12

SS2

FF2

5 or 7 or

2=9-2-5

12-2=10

4 or 3 or5=4-2+3

4Exercise


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(Cont.)4Exercise


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(Cont.)4Exercise


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(Cont.)4Exercise


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(Cont.)4Exercise


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(Cont.)4Exercise


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(Cont.)4Exercise


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3. Project Analysis Tools

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