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LECTURE #6

Precedence Diagramming and Time Scaled Network

January 24, 2020GE 404 (Engineering Management)

1

ENGINEERING MANAGEMENT

(GE 404)

بسم هللا الرحمن الرحيم

Objectives of the Present lecture


2

To explain how to draw Precedence network

diagrams

To discuss how to calculate Early/Late

Start/Finish Times in Precedence Network

diagrams.

To discuss steps involved in drawing the Time-

scaled network diagrams

Precedence Diagramming


3

Precedence diagramming is an extension to the original Activity-On-Nodeconcept.

Precedence diagramming includes precedence relationships among theactivities.

In addition, one may specify a “lag time” associated with any of theprecedence relationships.

A successor "lags" a predecessor, but a predecessor "leads" a successor.

Lag time can be designated on a dependency line with a positive, negative, orzero value.

The prime relationship used in PERT/CPM network is Finish-to-start type ofdependency, with FSij = 0 .

Activity Relationships


4

Finish-to-Start (FS)

In a finish-to-start relationship, the start of an activity depends on the completion of its

preceding activity.

Example: Footing excavation must be completed prior to placing concrete for the footing

Start-to-Start (SS)

In a start-to-start relationship, an activity can not start before its preceding activity starts

Example: Project management activities can not start before the Project work starts

Finish-to-Finish (FF)

In a finish-to-finish relationship, the two activities are related by the fact that they must be

completed at the same time

Example: In a simple task of setting a flagpole, the backfilling task and the positioning the

pole task will be finished at the same time.

Start-to-Finish (SF)

In start-to-finish relationship, the finish of an activity depends on the start of its preceding

activity (A rare relationship)

Example: In a Hospital emergency, the previous shift can not finish before the new shift starts Say you’re building a new gas pipeline. You would first finish construction and implementation of the new pipeline before you would

begin shutting down and breaking down the old pipeline.

Lag


5

The amount of time that exists between the early finish of an

activity and the early start of a specified succeeding activity

Calculate lags for each link by determining the difference

between the ES of each activity following a link line and the EF

of the activity that precedes it

Limitations and Disadvantages of Lag:

Lag would complicate the scheduling process

Lags are not extensively used except where the time effects

are substantial for special project type

Finish-to-Start Activity Precedence Relationships with Lag Values


6

ijij FSEFES

jjj DESEF

iii DLFLS

ijji FSLSLF

Activity i Activity j Activity kFSij FSjk

jkjkijij FSFS lag and lag :Note

Start-to-Start Activity Precedence Relationships with Lag Values


7

ijij SSESES

jjj DESEF

ijji SSLSLS

iijjiii DSSLSDLSLF

Activity i

Activity j

Activity k

SSij

SSjk

jkjkijij SSSS lag and lag :Note

Finish-to-Finish Activity Precedence Relationships with Lag Values


8

jjjijij DEFDFFEFES

ijij FFEFEF

iiiijji DLFDFFLFLS

ijji FFLFLF

Activity i

Activity j

Activity k

FFij

FFjk

jkjkijij FFFF lag and lag :Note

Start-to-Finish Activity Precedence Relationships with Lag Values


9

jjjijij DEFDFESES S

ijij SFESEF ijji SFLFLS

iiiijji DLSDSFLFLF

Activity i

Activity j

Activity k

SFij

SFjk

jkjkijij SFSF lag and lag :Note

Note on SFij


10

Activity j cannot finish till i

starts (rare)

SFij is equal to the minimum

number of time units that

must transpire from the start

of the predecessor i to the

completion of the successor

j.

Example

For the shown diagram

calculate total lag if

SFij'

i

j

SFij''

1064''' ijijij SFSFSF

The finish of j must lag 10

units after the start of i

6 and 4 ''' ijij SFSF

Summary


11

Finish-to-Start Relationships with Lag Values FSij :

ijij FSEFES

Start-to-Start Relationships with Lag Values SSij :

ijij SSESES

Finish-to-Finish Relationships with Lag Values FFij :

jijij DFFEFES

Start-to-Finish Relationships with Lag Values SFij :

jijij DFESES S

ijji FSLSLF

iijji DSSLSLF

ijji FFLFLF

iijji DSFLFLF

Generalized Equations


12

jiji

jiji

iji

iji

ij

DSFES

DFFEF

SSES

FSEF

ES

Time Initial

)Max(all

jjj DESEF

FORWARD PASS COMPUTATIONS BACKWARD PASS COMPUTATIONS

iijj

iijj

ijj

ijj

ji

DSFLF

DSSLS

FFLF

FSLS

LF

Time Terminal

)Min(all

iii DLFLS

Composite Start-to-Start and Finish-to-Finish (ZZij)Activity Precedence Relationships with Lag Values


13

ZZij is a combination of two constraints, i.e., a start-to-start and

finish-to-finish relationship. It is written with the SSij time units

first, followed by the FFij time units.

Example: ZZij =5, 6

The start of activity j must lag 5 units after the start of activity i & The finish of

activity j must lag 6 units after the finish of activity i

Representation


14

ES Duration EF

Activity (i)

LS Total Float LF

ES Duration EF

Activity (j)

LS Total Float LFConstraints with lag/lead Durations

)( ijijij

ij

ij

ij

ij

FFSSZZ

SF

FS

FF

SS

Free and Total Floats in Overlapping Networks


15

Free Float of an activity can be defined as time span in which the completion of the controlling date of an activity may occur and not delay the termination of the project nor delay the early status of the following activity.

Total float (TF) equation is same as it was developed for the single relationship network.

ijij

ijij

ijij

ijij

ji

SFESEF

FFEFEF

SSESES

FSEFES

FF )Min(all

Problem-1


16

For the given precedence diagram, complete the forward and backward pass calculations.

Assume the project starts at T=0, and no splitting on activities is allowed. Also assume that

the project latest allowable completion time (project duration) is scheduled for 30 working

days.

FS 0

SS 3

FF 4

SS 6

SF 12

FS 0A

Develop

system spec.

(D=8)

C

Collect

system data

(D=4)

D

Test & debug

program

(D=6)

E

Run

program

(D=6)

F

Document

program

(D=12)

B

Write comp.

program

(D=12)

EFDES

Activity ID

LFTFLS

SolutionStep-1: Network diagram


17

A

8

B

12

D

6

C

4

END

0

E

6

F

12

EFDES

Activity ID

LFTFLS

Step-2: Forward Pass Computations


18

Activity A

880

0Time Initial

AD

AES

AEF

AES

Activity B

15123

01248

330

0Time Initial

)(

BBB

BABA

ABAB

DESEF

DFFEF

SSESMaxES A

Activity C

1349

963

0Time Initial)(

CD

CES

CEF

BCSS

BES

MaxES BC

jjj

jiji

jiji

iji

iji

ij

DESEF

DSFES

DFFEF

SSES

FSEF

allMaxES

]2[

Time Initial

)(]1[

January 24, 2020GE 404 (Engineering Management) 19

Activity D21615

15015

0

DD

DES

DEF

BDFS

BEF

meInitial TiMax(B)

DES

Activity E27621

13013,

210210Time Initial

),(

ED

EES

EEF

CEFS

CEFOR

DEFS

DEF

DCMaxE

ES

Activity F271215

13013,

151212150Time Initial

)(

FFF

CFC

FDFDF

DESEF

FSEFOR

DSFESMaxES D

jjj

jiji

jiji

iji

iji

ij

DESEF

DSFES

DFFEF

SSES

FSEF

allMaxES

]2[

Time Initial

)(]1[

Contd.


20

A

8

B

12

D

6

C

4

END

0

E

6

F

12

0 8 3 15

9 13

15 21

21 27

15 27

27 27

FS 0

SS 3

FF 4 SS 6

SF 12

FS 0


Step-3 Backward Pass Computations

Activity F 181230

30Time Terminal

FD

FLF

FLS

FLF

Activity E24630

30Time Terminal

EEE

E

DLFLS

LF

Activity D

18624

2461230

or

2402430Time Terminal

)( ,

DDD

DDFF

DEE

FED

DLFLS

DSFLF

FSLSMinLF

iii

iijj

iijj

ijj

ijj

ji

DLFLS

DSFLF

DSSLS

FFLF

FSLS

allMinLF

]4[

Time Terminal

)(]3[

A

8

B

12

D

6

C

4

END

0

E

6

F

12

0 8 3 15

9 13

15 21

21 27

15 27

27 27

3030 3

30

30

18 3

24 3

2418 3

FS 0

SS 3

FF 4 SS 6

SF 12

FS 0


iii

iijj

iijj

ijj

ijj

ji

DLFLS

DSFLF

DSSLS

FFLF

FSLS

allMinLF

]4[

Time Terminal

)(]3[Activity C

14418

18018

or


),(

CCC

CFF

CEE

C

DLFLS

FSLS

FSLSFEMinLF

Activity B

61218

2012614


),

(

BD

BLF

BLS

BD

BCSS

CLS

OR

BDFS

DLS

DCMin

BLF

Activity A

3811

11836

14418

30Time Terminal

)(

AD

ALF

ALS

AD

ABSS

BLS

ABFF

BLF

BMin

ALF

A

8

B

12

D

6

C

4

END

0

E

6

F

12

0 8 3 15

9 13

15 21

21 27

15 27

27 27

3030 3

30

30

18 3

24 3

2418 3

1814 5

186 3113 3

FS 0

SS 3

FF 4 SS 6

SF 12

FS 0

Step 4: Slack time (Float time)


23

Earliest Earliest Latest Latest OnStart Finish Start Finish Slack Critical

Activity ES EF LS LF LS – ES Path

A 0 8 3 11 3 Yes

B 3 15 6 18 3 Yes

C 9 13 14 18 5 No

D 15 21 15 21 3 Yes

E 21 27 24 30 3 Yes

F 15 27 18 30 3 Yes

Critical path is the path with the least total float = The longest path through the

network.

CP1: A-B-D-E, CP2: A-B-D-F ;

Critical Activities: A, B, D, E, and F

Problem-2


24

Given the precedence network for a small engineering project with activity durations in

working days, it is required to compute the activity times (ES, EF, LS, and LF) and total

floats (TF) and then indicate the critical activities.

58 3

FS 3FF 5

SS 8

106105

5FFFS 2

FS 4SF 3,4

7134SS 5

FS 0

L

C

D

E

F

H

I

J

K

ZZ 3,2A

SolutionStep-1: Calculate ES and EF


25

2451916881138

FS 3FF 5

SS 8

3410241961317107550

5FFFS 2

FS 4SF 3,4

167922139945SS 5

FS 0

L

C

D

E

F

H

I

J

K

ZZ 3,2A

Step-2: Calculate LS and LF


26

2451916881138

FS 3FF 5

24 1916 811 8

SS 8

3410241961317107550

5FFFS 2

34 2429 2327 175 0

FS 4SF 3,4

167922139945SS 5

FS 0

20 1326 1313 9

L

C

D

E

F

H

I

J

K

ZZ 3,2A

Step-3: Calculate Total float


27

2451916881138

FS 3FF 5

2401916081108

SS 8

3410241961317107550

5FFFS 2

34024291023271017500

FS 4SF 3,4

167922139945SS 5

FS 0

20413264131349

K

ZZ 3,2A L

C

D

E

F

H

I

J

Step 4: Indicate critical activities


28

2451916881138

FS 3FF 5

2401916081108

SS 8

3410241961317107550

5FFFS 2

34024291023271017500

FS 4SF 3,4

167922139945SS 5

FS 0

20013264131349

K

ZZ 3,2A L

C

D

E

F

H

I

J

Problem-3


29

Given below the precedence network for a small engineering project with activity

durations in working days. Compute the activity times (ES, EF, LS, and LF), total floats

(TF) and free floats. Also indicate the critical activities and find the critical path.

LEGENDS

ES D EF

3 SF,2 2 Act FF

B D LS TF LF

3

SS, 2 F

5

A

FF,1

6 6

C FS,1 E

SolutionStep-1: ES and EF Calculations


300 timeInitial AES

jjj

jiji

jiji

iji

iji

ij

DESEF

DSFES

DFFEF

SSES

FSEF

allMaxES

]2[

Time Initial

)(]1[

2220

0Time Initial)Max(

ABA

BSSES

AES

00615

0Time Initial)Max(

CABA

CDFFEF

AES

7716

0Time Initial)Max(

CEC

EFSEF

CES

22222

0Time Initial)Max(

DBDB

DDSFES

BES

532 BBB DESEF

550 AAA DESEF

660 CCC DESEF

422 DDD DESEF

1367 EEE DESEF

13

13013

4040Time Initial

),Max(

EFE

DFDF

FSEF

FSEFEDES 16313 FFF DESEF

Step-2: LF and LS Calculations


31

13 13013

16Time Terminal)(Min

EFF

EFSLS

FLF 7613 EEE DLFLS

112-13 DDD DLFLS

iii

iijj

iijj

ijj

ijj

ji

DLFLS

DSFLF

DSSLS

FFLF

FSLS

allMinLF

]4[

Time Terminal

)(]3[

6 617


CEE

CFSLS

ELF 066 CCC DLFLS

13 13013


DFF

DFSLS

FLF

14 143213


BBDD

BDSFLF

DLF 113-14 BBB DLFLS

5

516


),(Min

ACC

AABBA

FFLF

DSSLSCBLF05-5 AAA DLFLS

16 timeTerminal FLF 13316 FFF DLFLS

Contd.


32

2 3 5 SF,2 2 2 4

B D

11 9 14 11 9 13

13 3 16

SS, 2 F

13 0 16

0 5 5

A

0 0 5

FF,1

0 6 6 7 6 13

C FS,1 E

0 0 6 7 0 13

Step-3: FF Calculations


33

02-2-4 BDBDB SFESEFFF

0 0156

0202),(Min

ACAC

ABAB

AFFEFEF

SSESESCBFF

ijij

ijij

ijij

ijij

ji

SFESEF

FFEFEF

SSESES

FSEFES

FF )Min(all

90-4-13 DFDFD FSEFESFF

01-6-7 CECEC FSEFESFF

00-13-13 EFEFE FSEFESFF

Final Answer


34

2 3 5 SF,2 2 2 4

B 0 D 9

11 9 14 11 9 13

13 3 16

SS, 2 F 0

13 0 16

0 5 5

A 0

0 0 5

FF,1

0 6 6 7 6 13

C 0 FS,1 E 0

0 0 6 7 0 13

Critical Activities: A; C; E; and FCritical Path: A-C-E-F

Hammock Activity


35

An activity that extends from one activity to another, but which has noestimated duration of its own.

It is time-consuming and requires resources, but its duration iscontrolled, not by its own nature, but by the two activities betweenwhich it spans. Example: activity “A” ends on day 3 and activity “C” begins on day 10 and there is

a set of activities that are summarized as activity “B” (the hammock activity) thatmust be done between activity “A” and activity “C.”

Activity “B” doesn’t have a specific duration—it has the duration of the timebetween day 3 and day 10.

If activity “A” takes longer than expected and ends on day 4, then the duration ofactivity “B” is now the duration of time between day 4 and day 10.

Its ES and LS times are determined by the activity where it begins andits EF and LF times are dictated by the activity at its conclusion.

Time-Scaled Networks


36

Time-scaled networks are merely extension of bar charts

Each activity is shown as a one dimensional line rather than as a

two dimensional box

The horizontal length is equal to its estimated time duration

(beginning at its ES and ending with its EF values)

Vertical solid (or dashed) lines indicate sequential dependence

of one activity on another

Float times are shown as horizontal dashed lines

Free Float or activity Float Total Float or Path Float


Free float is the amount of time

that an activity’s completion

time may be delayed without

affecting the earliest start of

succeeding activity

Free float is “owned” by an

individual activity

Total float is the amount of time

that an activity’s completion

may be delayed without affecting

the earliest start of any activity

on the network critical path

Total float is shared by all

activities along a slack path

Total path float time for activity

(i-j) is the total float associated

with a path

37

Free and Total Floats

Advantages and Disadvantages


38

Advantages

Very suitable device for checking daily project needs of different

resources, and for the advance detection of conflicting demands among

activities for the same resource

Useful for project financial management applications

Disadvantages

Because it is drawn by manual drafting methods, the level of effort

needed to modify and update them is very large

Dependencies among activities are not always so obvious as they are

on the activity on node network.

Problem-1


39

The following activity list represents the job logic and the durations of activities for a

market survey project. Draw a time- scaled network for the project, determine project

time and calculate the activities float times.

Activity Description predecessors Duration, week

A Plan survey — 1

B Hire personnel A 1

C Design questionnaire A 3

D Train personnel B, C 2

E Select samples of customers C 1

F Print questionnaire C 1

G Conduct survey D,E,F 3

H Analyze results G 2

Solution


40

• Project completion time = 11 working days

• Critical Path: A-C-D-G-H.

Activity A B C D E F G H

Total float 0 2 0 0 1 1 0 0

Free float 0 2 0 0 1 1 0 0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Project completion time = 18 working days

Critical Path: A, D, F, I, J.

Activity A B C D E F G H I J

Total float 0 0 0 5 2 0 0

Free float 0 0 0 5 2 0 0

Activity Depends on Duration (day)

E

B

C

G

F

A

H

D

I

J

B, C

A

A

E

C, D

None

E, F

A

F

G, H, I

2

3

4

3

5

2

2

5

4

2

A D

C

B

F I

E G

H

J

Problem-2

Shortening of Project Duration


42

Reducing scope (or quality)

Adding resources

Concurrency (perform tasks in parallel)

Substitution of activities

Further Reading


43

1. Moder J., Phillips, C., and Davis, E. Project Management with CPM, PERT, and Precedence Diagramming, 3rd Edition.

2. Jimmie W. Hinze. “Construction Planning and Management,” Fourth Edition, 2012, Pearson.

Read more about the scheduling network models from:

Thank You


44

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