pert and cpm_new

8/6/2019 Pert and Cpm_new

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PERT & CPM

Project Scheduling

Project Scheduling Objectives

Phases of Project Scheduling

PERT Diagrams & Dummy Activities

CPM Critical Path Method

Acknowledgements: Prof. Maria Petridou

1


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PROJECT SCHEDULING

It is part ofproject management within the PlanningPlanning phase of

the Product Development Life Cycle.

Project Scheduling: Allocation of resources to execute all

activities in the project.

Project: Set of activities or tasks with a clear beginningbeginning and

endingending points with the amount of available resources (time,

personnel and budget) to carry out the activities usually

limited. 2


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PROJECT SCHEDULING

Objectives:Objectives:

y Establish beginning, ending and duration of each activity in the

project.

y

Calculate overall completion time of the project given theamount of usually limited resources.

y Identify the activities that have high potential for causing delays

in completing the project on schedule

y Determine the critical path and its duration.

y Determine the slack time for all non-critical activities and the

whole project.

y Guide the allocation of resources such as time, staff and budget

3


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Program Evaluation and Review Technique

(PERT)

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Program Evaluation and Review Technique

It is a network model that allows for UNCERTAINITY in activity

completion times.

Determines the probabilities of completing various stages of

the project by specified deadlines, including the expected

time to complete the project.

PERT was developed in the late 1950s for the US Navys

Polaris Project.

First used as a management tool for military projects Mostly used in research and development projects

It has the potential to reduce both the time and cost required

to complete a project.


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CPM CRITICAL PATH METHOD

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CPM was developed independently by E.I. du Pont de

Nemours Company

Major difference between the two techniques is that CPM

does not incorporate uncertainties in job times

Basic assumption is that the activity times are proportional to

the level of resources allocated to them

Assigning additional resources (capital, people, materials and

machines) to an activity reduces its duration to a certain

extent

Shortening the duration of an activity is known as crashing in

the CPM terminology. Additional cost is called Crashing cost


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APPLIATIONS OF PERT AND CPM

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1. Construction projects (buildings, highways, bridges)

2. Maintenance planning of oil refineries, ship repairs and other

large projects

3. Development of new weapon systems and new

manufactured products

4. Manufacture and assembly of large items such as

aeroplanes, ships and computers

5. Preparation of bids and proposals for large projects


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PROJECT SCHEDULING Phases:Phases:

y Define activities or tasks according to the project objectives.

y A task is an individual unit of work with a clear beginning and a clear

end.

y Identify precedence relationships or dependenciesy Estimate time required to complete each task.

y Draw an activity-on-arrow diagram inserting dummy activities if

required.

y Calculate earliestand latest starting times, earliestand latest

completion times, slacktimes, critical path etc.y Construct a GANTT chart.

y Reallocate resources and resolve if necessary.

y Continuously monitor/revise the time estimates along the project

duration.7


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NETWORK DIAGRAMS

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Activity-on-arrow


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NETWORK DIAGRAMS

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Some basic rules for Activity on-arrow:

y Tasks are represented as arrows

y Nodes represent the start and finish points of tasks

y There is only one overall start node

y There is only one overall finish node

y Two tasks cannot share the same start and end node.

2 3A D

C

B

Tasks B & C share

the same start and

end node


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DUMMY ACTIVITIES

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Sometimes it is necessary to insert dummy activities (duration

zero) in order to maintain the clarity of the diagram and the

precedence relationships between activities.

In activity-on-arrow diagrams, each activity must be uniquely

identifiable by its start and end nodes.

However, sometimes multiple tasks have the same predecessors

and successors.


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DUMMY ACTIVITIES

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Inserting a dummy activitycan ensure that multiple

tasks have different successors.

1 2 3 5

A dummy task is

inserted to

preserve theimmediate

predecessors of D

A new node is

inserted to give C

a different finishnode to B

4

BA D 6E


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DUMMY ACTIVITIES

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Inserting a dummy activitycan ensure that multiple

tasks have different successors.

1 2 3 5

A dummy task is

inserted to

preserve theimmediate

predecessors of D

A new node is

inserted to give B

a different finishnode to C

4

CA D 6E


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DUMMY ACTIVITIES


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DUMMY ACTIVITIESCorrect Solution!Correct Solution!

The solution is to insert a dummy task so that the precedence of E

is preserved and activity C remains uniquely identifiable.


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Example 1

Consider seven jobs A, B, C, D, E, F and G with the

following job sequence:

Job A precedes B and C

Jobs C and D precede EJob B precedes D

Jobs E and F precede G

1 2

3

4 5A

B

C

D

E 6G

F


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Example 2

Consider five jobs A, B, C, D, and E with the following job

sequence:

Job A precedes C and D

Job B precedes DJobs C and D precede E

1 2

3

4 5

A

B

C

D

E

3 4 5

1 2


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Some new concepts

Earliest start time

Earliest Finish time

Latest Start time Latest Finish time

Slack

Critical job / Non-critical job

Enumerative method / Mathematical

Programming methods

A (10)

B (8) C (8)

D (12)


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Consider a project consisting of eight jobs (A,B,C,D,E,F,G

and H). About each job, we know the following:

Job Predecessors Normal

time (days)

Crash

time

Cost of

Crashing / day

A - 10 7 4

B - 5 4 2

C B 3 2 2

D A,C 4 3 3

E A,C5 3 3

F D 6 3 5

G E 5 2 1

H F,G 5 4 4


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63

4

5

71

2

A

B C

D

E

F

G

H10

5

3

4

5

6

3

5


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Example 4

Job Predecessors Normaltime (days)

A - 15

B - 10

C A, B 10

D A, B 10

E B 5

F D, E 5

G C, F 20

H D, E 10

I G, H 15


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A

B

C

D

E

F

G

H

I


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PERT

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Three time estimates

- Most probable time (m)

- Optimistic time (a)

- Pessimistic time (b)

Beta distribution

Average time = (a + 4m + b) / 6

Standard deviation = (b a )/ 6


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Consider a project consisting of nine jobs (A,B,C,D,E,F,G, H and

i). About each job, we know the following:

Job Predecessors Optimistic

time (days)

Most

Probable

time

Pessimistic

time

A - 2 5 8

B A 6 9 12

C A 6 7 8

D B,C 1 4 7

E A 8 8 8

F D,E 5 14 17

G C 3 12 21

H F, G 3 6 9

I H 5 8 11


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Job Average time Standard

Deviation

Variance

A 5 1 1

B 9 1 1

C 7 1/3 1/9

D 4 1 1

E 8 0 0

F 13 2 4

G 12 3 9

H 6 1 1

I 8 1 1


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The critical jobs are A, B, D, F, H and I

Let T denote the project duration. Expected length ofthe project is

E (T) = Sum of the expected times of critical jobs

= 5+9+4+13+6+8 = 45 days

Variance of the project duration is

V(T) = 1 + 1 + 1 + 4 + 1 + 1 = 9

Standard Deviation of the project duration = 3 days

As per Central Limit Theorem, T is normally

distributed with mean 45 and standard deviation 3


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There is a 68% chance that the project duration will

be between 42 and 48 days (one sigma) Similarly, there is a 99.7% chance that T will lie

within three standard deviations (b/n 36 and 54)

Prob (T 41) ???

Prob (T 50)???


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Job Predecessors Duration

(weeks)

A - 4

B - 7

C - 8

D A 5

E C 4

F B ,E 4

G C 11

H G, F 4

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