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PROJECT MANAGEMENT WITH CPM 1

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PROJECT MANAGEMENT WITH CPM 1
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PROJECT MANAGEMENT WITH CPM 1

Let us consider the following project:

In the heating plant of a palace we have to substitute

an old oil boiler with a new gas boiler, equipped with

electrical valves and an automatic regulator.

The project may be subdivided into activities; an

activity is an autonomous piece of project, executed

under the responsibility of one person, one team or

one firm. Activities’ durations are deterministic.

Among activities there may be some finish-to-start

precedences, which impose the preceding activity

(or activities) shall finish before the preceded one(s)

can begin.

We shall treat the problem by the Critical Path

Method (CPM)

PROJECT MANAGEMENT WITH CPM 2

name activity description duration precedent

A old boiler and pipes disconnection 4 -

B electrical plant disassembly 1 -

C new boiler’s order 40 -

D electrical valves’ order 10 -

E automatic regulator order 30 -

F old boiler removal 2 A

G building works (basement, windows,...) 8 B, F

H electrical plant assembly 2 G

I new boiler assembly 3 C,G

L pipes and electrical valves assembly 3 D, I

M automatic regulator assembly and conn. 1 E, L

N gas plant connection 11 G

O final inspection 1 H,M,N

PROJECT MANAGEMENT WITH CPM 3

AMERICAN GRAPHICAL REPRESENTATION

1 – Layering: activities without predecessors

are put in the first layer, then activities

preceded only by activities in the first layer

are put in the second layer, then activities

preceded only by activities in the first or

second layer are put in the third layer, and so

on till all activities have been layered.

In our example:

A B C D E; F; G; H I N; L; M; O.

PROJECT MANAGEMENT WITH CPM 4

AMERICAN GRAPHICAL REPRESENTATION

2 – Activities on arcs: activities are

represented by directed arcs; if activities A1,

A2 precede activities B1, B2, then precedence

is represented as follows:



A1 B1

A2 B2

Circles are called events and separate

predecessors and successors

PROJECT MANAGEMENT WITH CPM 5

AMERICAN GRAPHICAL REPRESENTATION

3 – All activities of the first layer (initial

activities) take origin from the project

beginning event which is generally numbered

0; in our example:

A

B

0

C

D

E

PROJECT MANAGEMENT WITH CPM 6

AMERICAN GRAPHICAL REPRESENTATION

4 – All activities of the last layer (final

activities) take end on the project ending

event; in our example:







O

9

PROJECT MANAGEMENT WITH CPM 7

AMERICAN GRAPHICAL REPRESENTATION

5 – In our example activities H, I, N are

preceded by activity G, but only activity I is

preceded by activity C: in order to correctly

represent precedences, a dummy (fictitious)

activity of null duration is added as follows:

H

G

Y N

C I

PROJECT MANAGEMENT WITH CPM 8

AMERICAN GRAPHICAL REPRESENTATION

6 – In our example activities H, N are both

preceded by activity G, and both precede

activity O: in order to avoid two arcs with the

same extremal events, a dummy (fictitious)

activity of null duration is added as follows:



G H O



N X

PROJECT MANAGEMENT WITH CPM 9

AMERICAN GRAPHICAL REPRESENTATION

7 – All events are numbered in such a way

that every activity has its starting event with a

number smaller than the one of its ending

event: i < j below:





H

i j

PROJECT MANAGEMENT WITH CPM 10

EXAMPLE AMERICAN REPRESENTATION

F G N

1 2 3 7

X

O

A B Y H 8 9



4

C I M

0 D

5

L

E 6

PROJECT MANAGEMENT WITH CPM 11

MINIMUM PROJECT COMPLETION TIME

8 – The problem consists in finding the

minimum project completion time. The model

is a linear program: in our example:

MIN (t9 – t0) s.t.:

t0=0; t1-t0≥4; t2-t0≥1; t2-t1≥2; t3-t2≥8; t4-t0≥40;

t4-t3≥0; t5-t0≥10; t5-t4≥3; t6-t0≥30; t6-t5≥3;

t7-t3≥11; t8-t3≥2; t8-t6≥1; t8-t7≥0; t9-t8≥1.

As we have a maximum path problem a

specific algorithm is better than simplex!

PROJECT MANAGEMENT WITH CPM 12

SPECIFIC ALGORITHM

The specific algorithm consists of two

phases. In the first forward phase it finds, for

every event i, the minimum time ti, i.e., the

time before which the event cannot take place

because of preceding activities. In the second

backward phase it finds, for every event i, the

maximum time Ti, i.e., the time after which the

event causes a delay in project completion

because of following activities.

PROJECT MANAGEMENT WITH CPM 13

SPECIFIC ALGORITHM FORWARD PHASE

S1: i:=0, ti:=0;

S2: i:=i+1, ti:=MAX{th+dur(hi) | h€Prec(i)}

S3: if i is the project ending event ti=project

completion time and STOP, else go to S2

SPECIFIC ALGORITHM BACKWARD PHASE

S1: i:=n (ending event), Ti:= Tf = tn (final time)

S2: i:=i-1, ti:=Min{tj-dur(ij) | j€Succ(i)}

S3: if i is the project beginning event STOP,

else go to S2

PROJECT MANAGEMENT WITH CPM 14

EVENTS’ TOTAL FLOAT

The difference Ti-ti is defined as the total float

fi of event i, and is the time that event i may

delay with respect to minimum time without

delaying the whole project completion.

Events with null float are critical; all paths

from beginning to ending event which cross

only critical events are called critical paths; all

activities on a critical paths are critical

activities, which shall not increase their

duration, otherwise the whole project delays.

PROJECT MANAGEMENT WITH CPM 15

EVENTS’ MIN TIME, MAX TIME AND FLOAT

i ti Ti fi

0 0 0 0

1 4 26 22

2 6 28 22

3 14 36 22

4 40 40 0

5 43 43 0

6 46 46 0

7 25 47 22

8 47 47 0

9 48 48 0

PROJECT MANAGEMENT WITH CPM 16

EUROPEAN GRAPHICAL REPRESENTATION

1 – Layering: activities without predecessors

are put in the first layer, then activities

preceded only by activities in the first layer

are put in the second layer, then activities

preceded only by activities in the first or

second layer are put in the third layer, and so

on till all activities have been layered.

In our example:

A B C D E; F; G; H I N; L; M; O.

PROJECT MANAGEMENT WITH CPM 17

EUROPEAN GRAPHICAL REPRESENTATION

2 – Activities on nodes: activities are

represented by nodes; if activity A precedes

activity B, then precedence is represented by

an arrow, as follows:







A B

PROJECT MANAGEMENT WITH CPM 18

EUROPEAN GRAPHICAL REPRESENTATION

3 – If there is only one activity in the first layer

(initial activity) then the graph takes origin

from it; if there are more activities in the first

layer then a fictitious initial activity of null

duration, preceding all activities of the first

layer, is added. In our example we have the

second situation: A B

C

X

D

E

PROJECT MANAGEMENT WITH CPM 19

EUROPEAN GRAPHICAL REPRESENTATION

4 – If there is only one activity in the last layer

(ending activity) then the graph ends with it; if

there are more activities in the last layer then

a fictitious ending activity of null duration,

successor of all activities of the last layer, is

added. In our example we have the first

situation:



O

PROJECT MANAGEMENT WITH CPM 20

EXAMPLE EUROPEAN REPRESENTATION



A F H



B G N O



C I



x D

L



E M

PROJECT MANAGEMENT WITH CPM 21

EUROPEAN ACTIVITIES TIME VARIABLES

5 –Four variables are defined for each activity:

ES=early start: time before which the activity

cannot start because of preceding activities

EF=early finish: time before which the activity

cannot end because of preceding activities

LS=late start: time after which the activity

shall not start otherwise it delays the project

LF=late finish: time after which the activity

shall not finish otherwise it delays the project

PROJECT MANAGEMENT WITH CPM 22

EUROPEAN SPECIFIC ALGORITHM

The specific algorithm consists of two

phases. In the first forward phase it finds, for

every activity i, the early start ES(i) and the

early finish EF(i). In the second backward

phase it finds, for every activity i, the late

finish LF(i) and the late start LS(i).

PROJECT MANAGEMENT WITH CPM 23

EUROPEAN ALGORITHM FORWARD PHASE

Starting from the first layer and going on layer

by layer till the last one;

For every activity i of the current layer we

have:

If the activity is the initial one: ES = 0,

else: ES(i) = MAX {EF(h) | h € Prec(i)}



EF(i) = ES(i) + dur(i)

PROJECT MANAGEMENT WITH CPM 24

EUROPEAN ALGORITHM BACKWARD PHASE

Starting from the last layer and going on layer

by layer till the first one;

For every activity i of the current layer we

have:

If the activity is the ending one: LF(i) = LS(i)

else: LF(i) = MIN {LS(j) | j € Succ(i)}



LS(i) = LF(i) - dur(i)

PROJECT MANAGEMENT WITH CPM 25

EUROPEAN ACTIVITIES FLOATS

6 –Two floats are defined for each activity:

TF(i)=total float=EF(i)-ES(i)=LF(i)-LS(i):

increase of the activity duration, or delay in

the activity start, which does not cause any

delay in the whole project completion time

FF(i)=free float= MIN {ES(j)| j € Succ(i)} - EF(i):

increase of the activity duration, or delay in

the activity start, which does not cause any

delay in the successor activities’ start

PROJECT MANAGEMENT WITH CPM 26

EUROPEAN CRITICAL ACTIVITIES

7 – Activities with null total float are critical;

paths from the initial activity to the ending

activity which cross only critical activities are

critical paths



EF(ending activity) = project completion time

PROJECT MANAGEMENT WITH CPM 26

EUROPEAN GRAPHICAL REPRESENTATION

8 – The four characteristic times are usually

written near every activity in the following

format:



ES EF



LS LF

PROJECT MANAGEMENT WITH CPM 28

EXAMPLE EUROPEAN RESULTS 1

Activity ES EF LS LF tf ff

A 0 4 22 26 22 0

B 0 1 27 28 27 5

C 0 40 0 40 0 0

D 0 10 33 43 33 33

E 0 30 16 46 16 16

F 4 6 26 28 22 0

G 6 14 28 36 22 0

H 14 16 45 47 31 31

PROJECT MANAGEMENT WITH CPM 29

EXAMPLE EUROPEAN RESULTS 2

Activity ES EF LS LF tf ff

I 40 43 40 43 0 0

L 43 46 43 46 0 0

M 46 47 46 47 0 0

N 14 25 36 47 22 22

O 47 48 47 48 0 0

PROJECT MANAGEMENT WITH CPM 30

EXAMPLE EUROPEAN RESULTS 3

A graphical presentation of results is the

Gantt diagram, a cartesian diagram where

every activity i is represented by a horizontal

bar, where the extreme points have abscissas

respectively equal to ES(i) and EF(i). If useful

bars may be protract with a segment equal

either to tf(i) or to ff(i).

See our example Gantt diagram where no

floats are evidenced.

PROJECT MANAGEMENT WITH CPM 31

EXAMPLE GANTT DIAGRAM

A

B

C

D

E

F

G

H

I

L

M

N

PROJECT MANAGEMENT WITH CPM 32

LIMITED RESOURCE PROJECT PLANNING

Consider a new example, where activities are

characterized by name, duration, precedent

activities and amount of used resource. This

resource is a hired resource, like for instance

operators, trucks, machines, equipments, etc.

We want to plan the project in the case this

resource is limited. It is obvious that the

elaboration shall be different from the one we

employed in the previous example, where we

considered unlimited resources.

PROJECT MANAGEMENT WITH CPM 33

LIMITED RESOURCE EXAMPLE DATA

activity duration predecessors resource

A 7 - 3

B 5 A 2

C 2 B 4

D 3 - 6

E 2 D 5

F 1 E, H 1

G 4 - 3

H 6 G 5

I 3 H 2

PROJECT MANAGEMENT WITH CPM 34

LIMITED RESOURCE EXAMPLE GRAPH



A B C







x D E F x







G H I

PROJECT MANAGEMENT WITH CPM 35

If the amount of resource is infinite, activity

times are the following:

activity ES EF LS LF

A 0 7 0 7

B 7 12 7 12

C 12 14 12 14

D 0 3 8 11

E 3 5 11 13

F 10 11 13 14

G 0 4 1 5

H 4 10 5 11

I 10 13 11 14

PROJECT MANAGEMENT WITH CPM 36

Use of unlimited resource (resource limit=7)



14



12



10 G H



8



6 D

E H I

4 F

I

2 C

A

0 B

PROJECT MANAGEMENT WITH CPM 37

LIMITED RESOURCE PARALLEL TECHNIQUE

Activities are scheduled according to a

decision set (DS), which is a dynamical list

built up for every instant of time.

Activities put in DS(t) are ordered according

to increasing parallel total float, based on

unlimited resource activity times, given by:

TF(k):=LS(k)-t

Activities at the top of the list have higher

urgency to be scheduled.

PROJECT MANAGEMENT WITH CPM 38

PARALLEL TECHNIQUE ALGORITHM

S1: t:=0; put in DS(0) all activities which can

start at time 0;

S2: order all activities in DS(t);

S3: extract from DS(t), in the order, all activi-

ties which can be scheduled at t because

of limited resource, schedule them and

cancel them from DS(t); put the remai-

ning activities into DS(t+1)

S4: t:=t+1; add to DS(t) all activities which can

start at time t because of preceding ones;

if DS(t)=Φ STOP else go to S2;

PROJECT MANAGEMENT WITH CPM 39

PARALLEL TECHNIQUE ALGORITHM

Take the example:

-DS(0):={A,D,G}, ordered {A,G,D}

-schedule A and G at time 0, put D into DS(1)

-t:=1; no activity added to DS(1); DS(1)={D} is

already ordered; D cannot be scheduled

because of missing resource;

-t:=2, 3 without scheduled activities;

-t:=4; H is added to DS(4), ordered {H,D};

-t:=5, 6 without scheduled activities;

-t:=7; B is added to DS(7), ordered {H;B;D}

PROJECT MANAGEMENT WITH CPM 40

PARALLEL TECHNIQUE ALGORITHM

-H and B are scheduled at time 7; DS(8)={D}

-t:=8, 9, 10, 11, 12 no scheduled activities;

-t:=13 C and I are added to DS(13), ordered {D,I,C};

-D is scheduled at time 13; DS(14)={I,C};

-t:=14, 15 no scheduled activities;

-T:=16 E is added to DS(16), ordered {E,I,C};

-E and I are scheduled at time 16; DS(17)={C};

-t:=17 no scheduled activities;

-t:=18 F is added to DS(18), ordered {C,F};

-C and F are scheduled at time 18; DS(19)=Φ; STOP.

PROJECT MANAGEMENT WITH CPM 41

Use of limited resource=7 (parallel technique):

14



12



10



8



6 B F



4 G E

D C

2 H

A I

0

PROJECT MANAGEMENT WITH PERT 41

PROJECT EVALUATION AND REVIEW TECHNIQUE

(PERT)

It cosiders finish-to-start precedences and random

activity durations.

Conventionally the method considers for every

activity k three estimates:

ak minimum duration of activity k

bk maximum duration of activity k

mk most probable duration of activity k

From the three estimates we assume the mean

activity duration: tk = (ak + bk + 4mk)/6

and the standard deviation: бk = (bk - ak)/6

PROJECT MANAGEMENT WITH PERT 43

PERT TECHNIQUE

Here we solve a problem of unlimited resources

CPM, taking for every activity the mean duration.

The resulting completion time Tf is considered as

the expected project duration.

Then we consider the sum of variances бk2 of

activities on the critical path as the variance of

project duration бf2 .

We assume the project duration has a gaussian

distribution of mean Tf and variance бf2 .

It is obvious that results are easy to obtain but

strongly approximated.

PROJECT MANAGEMENT WITH PERT 44

PERT SIMULATION TECHNIQUE

For every activity a duration probability

distribution is given. From every distribution a

random value is extracted, then CPM is solved

and we obtain a realization of project

completion time. This is iterated a sufficient

number of times (50-100 is generally enough),

i.e., a sample is taken, then the sample

distribution of project completion time is

obtained. The result is more precise than the

one of the conventional method.

PROJECT MANAG. OTHER TECHNIQUES 45

CPM-cost: activity duration may be modified with

change of activity cost. The objective is to

minimize total project cost.

MPM (Metra Potential Method): precedence may be

of the finish-to-start, start-to-start, finish-to-

finish and start-to-finish type


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