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MANAGING PROJECTS USING NETWORK TECHNIQUES. Project Management. Project Management is one of the world’s most in-demand skill sets and is one of the fastest growing professional disciplines in North America.

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Project management
Project Management

  • Project Management is one of the world’s most in-demand skill sets and is one of the fastest growing professional disciplines in North America.

  • Project Management is used by large corporations, governments, and smaller organizations to standardize and reduce the tasks necessary to complete a project in the most effective and efficient manner.

  • Engineers that master in project management skills may lead improvement initiatives that result in measurable growth in return on investment, economic value added, sales growth, customer satisfaction and retention, market share, time to market, employee satisfaction, and employee motivation.

  • PMI provides certification in project management

M. Sundaram Tenn. Tech


Network based techniques outline
Network based Techniques-Outline

  • Project Planning – An Introduction

  • Development of Project Network

  • Identifying Critical paths

  • Probabilistic Analysis in PERT Networks

  • How to use the Normal table

  • Project Cost Control

  • Resource Allocation

  • EVA for monitoring progress of Projects

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Pert and cpm
PERT and CPM

  • PERT (Program Evaluation Review Technique) was developed by a joint team set up by the U.S. Navy Special Projects Office that included representatives from Lockheed Aircraft Corporation (Prime contractor of the POLARIS program) and from the consulting company Booze, Allen, and Hamilton.

  • The objective of this team was to develop an integrated planning and control system for the Polaris missile submarine program which would help avoid the time and cost overruns that had plagued other such development programs.

  • An important feature of the PERT approach is its statistical treatment of the uncertainty in activity time estimate which involved the collection of three separate time estimates and the calculation of probability estimates of meeting specified schedule dates.

  • The three estimates used are: Optimistic time, Pessimistic time, and Most likely time.

  • PERT networks are usually to manage projects that have several uncertain activities

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Pert and cpm contd
PERT and CPM- Contd.

  • CPM (Critical Path Method) evolved from a parallel joint effort initiated originally at DuPont and later expanded to include Remington Rand Univac and Mauchly Associates.

  • The two key differences of this approach from PERT: (1) the use of only one time estimate for each activity (and thus no statistical treatment of uncertainty) and (2) the inclusion, as an integral part of the overall scheme, of a procedure for time/cost tradeoff to minimize the sum of direct and indirect project costs.

  • An important common feature of both PERT and CPM is the use of a network diagram for project representation in which arrows represent activities ("activity-on­arrow").

  • A modification of this approach involves the representation of activities by circles, with arrows indicating precedence ("activity-on-node").

M. Sundaram Tenn. Tech


Applications of pert and cpm
Applications of PERT and CPM

  • Construction projects (e.g.) buildings, highways, houses, and bridges.)

  • Preparation of bids and proposals for large projects.

  • Maintenance planning of oil refineries, ships, chemical plants and other large scale operations.

  • Planning for relocating a facility

  • Manufacture and assembly of large and complex products such as airplanes, ships, and mainframe computers.

  • Simple projects such as home remodeling, moving to a new house, and home cleaning and painting.

  • Design and development of new products

  • Facilities planning and implementation of new layouts in manufacturing.

  • Development of computer Software packages

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What is a design project

Design Project – Unique sequence of activities (work tasks) required to be performed in developing a product.

cost

time

Design project

performance

What is a design project?

Changing the length of any side of the project triangle affects the other sides!

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Project planning

1

2

3

Design

Fabricate

Project Planning

A Simple Project Network

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Managing a design project

?

Managing a design project?

Design problem –FUNCTION

(customer & company requirements)

Activities

(decision making processes)

Develop a project plan

then execute the plan

Solution - FORM

(manufacturing specifications)

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Why should we plan a design project
Why should we plan a design project?

In planning a design project we make decisions which answer the following questions

WHAT ? ……...scope of work tasks

WHEN ? ……...schedule

HOW MUCH?..budget

WHO?………...organization chart,

responsibilities table

Without a clear roadmap, how will you get where you need to go?

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Work breakdown structure
Work breakdown structure

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Scope of work partial

Table 3.4 Example scope of work for section 1.

Scope of work (partial)

  • 1.0 Design Problem Formulation

    • 1.1 Visit Site,

    • Meet with customers, determine desired attributes and parameters

    • 1.2 Complete QFD/HOQ

    • Determine requirements, engineering characteristics

    • 1.3 Satisfaction Curves,

    • Determine the satisfaction curves for each engineering characteristic.

    • 1.4 Create EDS

    • List in-use purposes for the product

    • List product performance requirements

    • 1.5 Conduct Benchmarking

    • Research existing products that are currently available

    • Contact manufacturers and request brochures

    • Analyze the competition for functionality and performance

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1.0 Design Problem Formulation

1.1 Visit Site,

Meet with customers to determine desired attributes and parameters

1.2 Complete QFD/HOQ

Determine requirements, engineering characteristics

1.3 Satisfaction Curves,

Determine the satisfaction curves for each engineering characteristic.

1.4 Create EDS

List in-use purposes for the product

List product performance requirements

1.5 Conduct Benchmarking

Research existing products that are currently available

Contact manufacturers and request brochures

Analyze the competition for functionality and performance

1.6 Contact Customers

Make phone calls to determine pros and cons of current unit

Set an appointment time to witness existing product operation

1.7 Determine parameters

State problem definition parameters

State design variables

State solution evaluation parameters for satisfaction curves

1.8 Determine Schedule

Refine work breakdown structure

Assign a time value to each task

Prepare Gantt chart

1.9 Calculate Budget

Determine total number of engineering hours

Determine total number of expert faculty hours

Sum all hours and material cost

1.10 Outline Work Scope

1.11 Prepare for and conduct design review meeting.


Responsibilities table
Responsibilities table

assist

Who does what

responsible

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Project schedule
Project schedule

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Project budget
Project Budget

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Elements of project management
Elements of Project Management

  • Project team

    • Individuals from different departments within company

  • Matrix organization

    • Team structure with members from different functional areas depending on skills needed

  • Project manager

    • Leader of project team

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Project planning1
Project Planning

  • Statement of work

    • Written description of goals, work & time frame of project

  • Activities require labor, resources & time

  • Precedence relationship shows sequential relationship of project activities

M. Sundaram Tenn. Tech


Project control
Project Control

  • All activities identified and included

  • Completed in proper sequence

  • Resource needs identified

  • Schedule adjusted

  • Maintain schedule and budget

  • Complete on time

M. Sundaram Tenn. Tech


A gantt chart
A Gantt Chart

  • Popular tool for project scheduling

  • Graph with bar for representing the time for each task

  • Provides visual display of project schedule

  • Also shows slack for activities

    • Amount of time activity can be delayed without delaying project

M. Sundaram Tenn. Tech


A gantt chart1

Month

0 2 4 6 8 10

| | | | |

Activity

Design house and obtain financing

Lay foundation

Order and receive materials

Build house

Select paint

Select carpet

Finish work

1 3 5 7 9

Month

A Gantt Chart

M. Sundaram Tenn. Tech

Figure 6.2


Cpm pert a little history
CPM/PERT- A Little History

  • Critical Path Method (CPM)

    • DuPont & Remington-Rand (1956)

    • Deterministic task times

    • Activity-on-node network construction

  • Project Eval. & Review Technique (PERT)

    • US Navy, Booz, Allen & Hamilton

    • Multiple task time estimates

    • Activity-on-arrow network construction

M. Sundaram Tenn. Tech


The project network

Node

1

2

3

Arc

The Project Network

Network consists of arcs & nodes

Figure 6.3

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Network construction
Network Construction

  • In AON, nodes represent activities & arrows show precedence relationships

  • In AOA, arrows represent activities & nodes are events for points in time

  • An event is the completion or beginning of an activity

  • A dummy shows precedence for two activities with same start & end nodes

M. Sundaram Tenn. Tech


Project network for a house

3

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

1

2

4

6

7

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

5

Project Network for a House

Figure 6.4

M. Sundaram Tenn. Tech


Concurrent activities

3

Lay foundation

Lay

foundation

Dummy

2

3

2

0

2

4

1

Order material

Order material

(a) Incorrect precedence relationship

(b) Correct precedence relationship

Concurrent Activities

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Example problem 1
Example Problem -1

Develop an activity-on-arrow (AOA) type network for the precedence relationship shown below.

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Solution
Solution

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Example problem 2
Example Problem -2

Develop an A-O-A type network from the description below.

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Solution1
Solution

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Critical path
Critical Path

  • A path is a sequence of connected activities running from start to end node in network

  • The critical path is the path with the longest duration in the network

  • Project cannot be completed in less than the time of the critical path

M. Sundaram Tenn. Tech


The critical path

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

5

The Critical Path

A: 1-2-3-4-6-73 + 2 + 0 + 3 + 1 = 9 months

B: 1-2-3-4-5-6-73 + 2 + 0 + 1 + 1 + 1 = 8 months

C: 1-2-4-6-73 + 1 + 3 + 1 = 8 months

D: 1-2-4-5-6-73 + 1 + 1 + 1 + 1 = 7 months

M. Sundaram Tenn. Tech


The critical path1

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

5

The Critical Path

A: 1-2-3-4-6-7

3 + 2 + 0 + 3 + 1 = 9 months

B: 1-2-3-4-5-6-7

3 + 2 + 0 + 1 + 1 + 1 = 8 months

C: 1-2-4-6-7

3 + 1 + 3 + 1 = 8 months

D: 1-2-4-5-6-7

3 + 1 + 1 + 1 + 1 = 7 months

The Critical Path

M. Sundaram Tenn. Tech


The critical path2

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

3

Finish at 9 months

Start at 5 months

2

0

1

3

1

1

2

4

6

7

1

2

4

6

7

3

1

1

Start at 8 months

Start at 3 months

5

5

The Critical Path

Activity Start Times

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Early times

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

2003

5

Early Times

  • ES - earliest time activity can start

  • Forward pass starts at beginning of CPM/PERT network to determine ES times

  • EF = ES + activity time

    • ESij=maximum (EFi)

    • EFij=ESij- tij

    • ES12=0

    • EF12 =ES12+ t12=0 + 3 = 3months

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Computing early times

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

5

Computing Early Times

  • ES23 = max EF2 = 3 months

  • ES46 = max EF4 = max 5,4 = 5 months

  • EF46 = ES46 + t46 = 5 + 3 = 8 months

  • EF67 = 9 months, the project duration

M. Sundaram Tenn. Tech


Computing early times1

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

Early Start and Finish Times

3

3

(ES = 3, EF = 5)

(ES = 5, EF = 5)

2

0

(ES = 5, EF = 8)

1

2

4

6

7

1

3

1

1

2

4

6

7

3

(ES = 0, EF = 3)

(ES = 3, EF = 4)

(ES = 8, EF = 9)

1

1

(ES = 5, EF = 6)

(ES = 6, EF = 7)

5

5

Computing Early Times

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Late times
Late Times

  • LS - latest time activity can start & not delay project

  • Backward pass starts at end of CPM/PERT network to determine LS times

  • LF = LS + activity time

    • LSij=LFij-tij

    • LFij= minimum (LSj)

M. Sundaram Tenn. Tech


Computing late times

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

5

Computing Late Times

  • LF67 = 9 months

  • LS67 = LF67 - t67 = 9 - 1 = 8 months

  • LF56 = minimum (LS6) = 8 months

  • LS56 = LF56 - t56 = 8 - 1 = 7 months

  • LF24 = minimum (LS4) = min(5, 6) = 5 months

  • LS24 = LF24 - t24 = 5 - 1 = 4 months

M. Sundaram Tenn. Tech


Computing late times1

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

1

2

4

6

7

5

Computing Late Times

  • LF67 = 9 months

  • LS67 = LF67 - t67 = 9 - 1 = 8 months

  • LF56 = minimum (LS6) = 8 months

  • LS56 = LF56 - t56 = 8 - 1 = 7 months

  • LF24 = minimum (LS4) = min(5, 6) = 5 months

  • LS24 = LF24 - t24 = 5 - 1 = 4 months

M. Sundaram Tenn. Tech


Early and Late Start and Finish Times

3

( )

( )

ES = 3, EF = 5

LS = 3, LF = 5

ES = 5, EF = 5

LS = 5, LF = 5

( )

ES = 5, EF = 8

LS = 5, LF = 8

2

0

1

3

1

1

2

4

6

7

3

( )

( )

( )

ES = 0, EF = 3

LS = 0, LF = 3

ES = 3, EF = 4

LS = 4, LF = 5

ES = 8, EF = 9

LS = 8, LF = 9

1

1

( )

( )

5

ES = 5, EF = 6

LS = 6, LF = 7

ES = 6, EF = 7

LS =7, LF = 8

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Activity slack
Activity Slack

  • Activities on critical path have ES = LS & EF = LF

  • Activities not on critical path have slack

    • Sij = LSij - ESij

    • Sij = LFij - EFij

    • S24 = LS24 - ES24 = 4 - 3 = 1 month

M. Sundaram Tenn. Tech


Activity slack data

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

Activity LS ES LF EF Slack

*1-2 0 0 3 3 0

*2-3 3 3 5 5 0

2-4 4 3 5 4 1

*3-4 5 5 5 5 0

4-5 6 5 7 6 1

*4-6 5 5 8 8 0

5-6 7 6 8 7 1

*6-7 8 8 9 9 0

* Critical path

1

2

4

6

7

5

Activity Slack Data

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Activity slack data1

Lay foundation

Dummy

Build house

Finish work

2

0

3

1

3

1

Design house and obtain financing

Order and receive materials

1

1

Select paint

Select carpet

3

3

Activity Slack

Activity LS ES LF EF Slacks

*1-2 0 0 3 3 0

*2-3 3 3 5 5 0

2-4 4 3 5 4 1

*3-4 5 5 5 5 0

4-5 6 5 7 6 1

*4-6 5 5 8 8 0

5-6 7 6 8 7 1

*6-7 8 8 9 9 0

* Critical path

S = 0

S = 0

1

2

4

6

7

1

2

4

6

7

2

0

S = 0

1

3

1

S = 0

S = 1

S = 0

3

5

5

1

1

S = 1

S = 1

Activity Slack Data

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Other methods of determining the critical path
Other Methods of Determining the Critical Path

  • Simple Method

  • Tabular Method

  • Enumeration of all paths

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Simple method
Simple Method

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Enumeration of all paths
Enumeration of all paths

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Probabilistic time estimates
Probabilistic Time Estimates

  • Reflect uncertainty of activity times

  • Beta distribution is used in PERT

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Probabilistic time estimates1

a + 4m + b

6

Mean (expected time): t =

2

b - a

6

Variance: 2 =

where

a = optimistic estimate

m = most likely time estimate

b= pessimistic time estimate

Probabilistic Time Estimates

  • Reflect uncertainty of activity times

  • Beta distribution is used in PERT

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Example beta distributions

P(time)

P(time)

a

m

t

b

a

t

m

b

Time

Time

P(time)

a

m = t

b

Time

Example Beta Distributions

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Kat tech company

d

2

6

Equipment testing and modification

Equipment installation

Final debugging

a

e

Dummy

l

System development

f

i

m

1

3

5

7

9

b

Manual Testing

System Training

System changeover

Position recruiting

Job training

System Testing

c

g

j

k

Dummy

h

4

8

Orientation

Kat Tech Company

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Activity estimates

2

6

1

3

5

7

9

4

8

TIME ESTIMATES (WKS) MEAN TIME VARIANCE

ACTIVITYa m b tσ2

1 - 2 6 8 10 8 0.44

1 - 3 3 6 9 6 1.00

1 - 4 1 3 5 3 0.44

2 - 5 0 0 0 0 0.00

2 - 6 2 4 12 5 2.78

3 - 5 2 3 4 3 0.11

4 - 5 3 4 5 4 0.11

4 - 8 2 2 2 2 0.00

5 - 7 3 7 11 7 1.78

5 - 8 2 4 6 4 0.44

7 - 8 0 0 0 0 0.00

6 - 9 1 4 7 4 1.00

7 - 9 1 10 13 9 4.00

Activity Estimates

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Early and late times

2

6

1

3

5

7

9

4

8

6 + 4(8) + 10

6

a + 4m + b

6

2

b - a

6

4

9

10 - 6

6

t = = = 8 weeks

 2 = = = week

2

Early and Late Times

For Activity 1-2

a= 6, m = 8, b = 10

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Early and late times1

2

6

1

3

5

7

9

4

8

ACTIVITYtσ2ES EF LS LF S

1 - 2 8 0.44 0 8 1 9 1

1 - 3 6 1.00 0 6 0 6 0

1 - 4 3 0.44 0 3 2 5 2

2 - 5 0 0.00 8 8 9 9 1

2 - 6 5 2.78 8 13 16 21 8

3 - 5 3 0.11 6 9 6 9 0

4 - 5 4 0.11 3 7 5 9 2

4 - 8 2 0.00 3 5 14 16 11

5 - 7 7 1.78 9 16 9 16 0

5 - 8 4 0.44 9 13 12 16 3

7 - 8 0 0.00 13 13 16 16 3

6 - 9 4 1.00 13 17 21 25 8

7 - 9 9 4.00 16 25 16 25 0

Early and Late Times

M. Sundaram Tenn. Tech


Kal tech company

ES = 8, EF = 13

LS = 16, LF = 21

2

6

5

ES = 0, EF = 8

LS = 1, LF = 9

ES = 8, EF = 8

LS = 9, LF = 9

ES = 13, EF = 17

LS = 21, LF = 25

8

0

4

ES = 0, EF = 6

LS = 0, LF = 6

ES = 6, EF = 9

LS = 6, LF = 9

ES = 9, EF = 16

LS = 9, LF = 16

9

1

3

5

7

9

6

3

7

ES = 16, EF = 25

LS = 21, LF = 25

ES = 3, EF = 7

LS = 5, LF = 9

4

0

3

ES = 0, EF = 3

LS = 2, LF = 5

ES = 9, EF = 13

LS = 12, LF = 16

ES = 13, EF = 13

LS = 16, LF = 16

2

4

8

ES = 3, EF = 5

LS = 14, LF = 16

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company1

ES = 8, EF = 13

LS = 16, LF = 21

2

6

5

Total project variance

ES = 0, EF = 8

LS = 1, LF = 9

ES = 8, EF = 8

LS = 9, LF = 9

ES = 13, EF = 17

LS = 21, LF = 25

8

0

4

2 = 2 + 2 + 2 + 2

= 1.00 + 0.11 + 1.78 + 4.00

= 6.89 weeks

13

35

57

79

ES = 0, EF = 6

LS = 0, LF = 6

ES = 6, EF = 9

LS = 6, LF = 9

ES = 9, EF = 16

LS = 9, LF = 16

9

1

3

5

7

9

6

3

7

ES = 16, EF = 25

LS = 21, LF = 25

ES = 3, EF = 7

LS = 5, LF = 9

4

0

3

ES = 0, EF = 3

LS = 2, LF = 5

ES = 9, EF = 13

LS = 12, LF = 16

ES = 13, EF = 13

LS = 16, LF = 16

2

4

8

ES = 3, EF = 5

LS = 14, LF = 16

Kal Tech Company

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Probabilistic network analysis

x - 

Z =

Probabilistic Network Analysis

Determine probability that project is completed within specified time

where

 = tp = project mean time

 = project standard deviation

x = proposed project time

Z = number of standard deviations xis from mean

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Normal distribution of project time

Probability

Z

 = tp

x

Time

Normal Distribution Of Project Time

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Kal tech company2

What is the probability that the project is completed within 30 weeks?

P(x 30 weeks)

 = 25

x = 30

Time (weeks)

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company3

What is the probability that the project is completed within 30 weeks?

Z =

=

= 1.91

P(x 30 weeks)

 2 = 6.89 weeks

 = 6.89

 = 2.62 weeks

x - 

30 - 25

2.62

 = 25

x = 30

Time (weeks)

Example 6.2

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company4

What is the probability that the project is completed within 30 weeks?

Z =

=

= 1.91

P(x 30 weeks)

 2 = 6.89 weeks

 = 6.89

 = 2.62 weeks

x - 

30 - 25

2.62

 = 25

x = 30

Time (weeks)

From Table A.1, a Z score of 1.91 corresponds to a probability of 0.4719.

Thus P(30) = 0.4719 + 0.5000 = 0.9719

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company5

What is the probability that the project is completed within 22 weeks?

P(x 22 weeks)

0.3729

x = 22

 = 25

Time (weeks)

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company6

What is the probability that the project is completed within 22 weeks?

Z =

=

= -1.14

 2 = 6.89 weeks

 = 6.89

 = 2.62 weeks

P(x 22 weeks)

x - 

0.3729

22 - 25

2.62

x = 22

 = 25

Time (weeks)

Kal Tech Company

M. Sundaram Tenn. Tech


Kal tech company7

What is the probability that the project is completed within 22 weeks?

Z =

=

= -1.14

 2 = 6.89

 = 6.89

 = 2.62

P(x 22 weeks)

x - 

0.3729

22 - 25

2.62

x = 22

 = 25

Time (weeks)

From Table A.1, a Z score of -1.14 corresponds to a probability of 0.3729.

Thus P(22) = 0.5000 - 0.3729 = 0.1271

Kal Tech Company

M. Sundaram Tenn. Tech


Another example problem
Another Example problem 22 weeks?

For the PERT network given below, Determine the critical path and the expected length of the critical path.

i. Compute the probability of completing the project in 20 days.

ii. What is the likely project duration that the project manager can be confident with 95% certainty?

M. Sundaram Tenn. Tech


Solution2
Solution 22 weeks?

M. Sundaram Tenn. Tech


Project crashing
Project Crashing 22 weeks?

  • Crashing is reducing project time by expending additional resources

  • Crash time is an amount of time an activity is reduced

  • Crash cost is the cost of reducing the activity time

  • Goal is to reduce project duration at minimum cost

M. Sundaram Tenn. Tech


Housebuilding network

3 22 weeks?

8

0

12

4

12

4

1

2

4

6

7

4

4

5

Housebuilding Network

M. Sundaram Tenn. Tech


Housebuilding network1

$7,000 – 22 weeks?

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

3

8

0

12

4

12

4

1

2

4

6

7

4

4

Normal activity

5

Normal cost

Normal time

| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Housebuilding Network

M. Sundaram Tenn. Tech


Housebuilding network2

$7,000 – 22 weeks?

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

3

Crash cost

8

0

Crashed activity

12

4

12

4

1

2

4

6

7

4

4

Normal activity

5

Normal cost

Crash time

Normal time

| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Housebuilding Network

M. Sundaram Tenn. Tech


Housebuilding network3

$7,000 – 22 weeks?

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

Total crash cost $2,000

Total crash time 5

= = $400 per week

Crash cost

Crashed activity

Slope = crash cost per week

Normal activity

Normal cost

Crash time

Normal time

| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Housebuilding Network

M. Sundaram Tenn. Tech


Normal activity and crash data

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

Normal Activity and Crash Data

M. Sundaram Tenn. Tech


Normal activity and crash data1

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

3

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

$500

8

0

12

4

12

4

1

2

4

6

7

$400

$3,000

$7,000

$7,000

4

4

$200

$200

5

Normal Activity and Crash Data

M. Sundaram Tenn. Tech


Normal activity and crash data2

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

3

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

$500

8

0

12

4

12

4

1

2

4

6

7

$400

$3,000

$7,000

$7,000

4

4

$200

$200

5

Normal Activity and Crash Data

M. Sundaram Tenn. Tech


Normal activity and crash data3

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

3

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

$500

8

0

7

4

12

4

1

2

4

6

7

$3,000

$7,000

$7,000

4

4

$200

$200

5

Normal Activity and Crash Data

Crash cost = $2,000

M. Sundaram Tenn. Tech


Normal activity and crash data4

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

3

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

$500

8

0

7

4

12

4

1

2

4

6

7

$3,000

$7,000

$7,000

4

4

$200

$200

5

Normal Activity and Crash Data

Crash cost = $2,000

M. Sundaram Tenn. Tech


Normal activity and crash data5

3 22 weeks?

1

2

4

6

7

5

TOTAL

NORMAL CRASH ALLOWABLE CRASH

TIME TIME NORMAL CRASH CRASH TIME COST PER

ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK

3

1-2 12 7 $3,000 $5,000 5 $400

2-3 8 5 2,000 3,500 3 500

2-4 4 3 4,000 7,000 1 3,000

3-4 0 0 0 0 0 0

4-5 4 1 500 1,100 3 200

4-6 12 9 50,000 71,000 3 7,000

5-6 4 1 500 1,100 3 200

6-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

$500

7

0

7

4

12

4

1

2

4

6

7

$3,000

$7,000

$7,000

4

4

$200

$200

5

Normal Activity and Crash Data

Crash cost = $2,000 + $500 = $2,500

M. Sundaram Tenn. Tech


Time cost relationship
Time-Cost Relationship 22 weeks?

  • Crashing costs increase as project duration decreases

  • Indirect costs increase as project duration increases

  • Reduce project length as long as crashing costs are less than indirect costs

M. Sundaram Tenn. Tech


Time cost tradeoff

M 22 weeks?inimum cost = optimal project time

Total project cost

Indirect cost

Cost ($)

Direct cost

Crashing Time

Project duration

Time-Cost Tradeoff

M. Sundaram Tenn. Tech






Earned value analysis eva

Earned -Value Analysis (EVA) 22 weeks?

EVA is a widely used project control method that compares actual Vs budgeted expenses on a period-by-period basis.

Project control is comparing project progress to the plan so that corrective action can be taken when deviation from planned performance occurs.

The first step in performing an earned -value analysis is to calculate the budget cost of work scheduled(BCWS) for each time period.

- For example if weekly periods are chosen, the total budgeted amount for a task will be distributed evenly during the scheduled period.

TheTable next page shows a project with three tasks. Task “A” is budgeted for $3,000. If it is scheduled for three weeks, the budget cost/week is $1,000.

M. Sundaram Tenn. Tech


Earned value analysis eva contd
Earned -Value Analysis (EVA)- Contd. 22 weeks?

  • The next step is to determine the actual cost of work performed (ACWP).

  • This may be entered weekly into the weekly column labeled ACWP.

  • The next step is to calculate the budgeted cost of workperformed (BCWP) by estimating the percent completion for each task and multiplying the total budget amount.

  • Next the schedule variance and the costvariance for eachweek will have to be calculated as below.

    - Schedule variance = BCWP – BCWS

    - Cost Variance = BCWP – ACWP

M. Sundaram Tenn. Tech





Example problem
Example Problem 22 weeks?

Project team Alpha has accumulated $5,000 of expenses as of the end of week#10. The budgeted cost of work scheduled for Week #10 is $6,500. The project has one task, which is about 45% complete at the end of week #10. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?

M. Sundaram Tenn. Tech


Solution3
Solution 22 weeks?

  • ACWP=$5000

  • BCWS=$6500

  • BCWP (for end of wk #10) = 45%(6500) = $2925

  • Schedule Variance = BCWP-BCWS

    = 2925-6500

    = $-3575

  • Cost Variance = BCWP-ACWP

    = 2925-5000

    = $-2075

  • The project is behind by $3575 of work. Also, for the work performed, it is over budget by $2075.

M. Sundaram Tenn. Tech


Another problem
Another Problem 22 weeks?

Team Delta is working on a project that has two work tasks, each worth $5,000, which are 45% and 35% complete as of Week #5. The actual expenses are $2,500 as of the end of Week #5. The budgeted cost of work scheduled for Week #5 is $6,500. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?

M. Sundaram Tenn. Tech


Solution4
Solution 22 weeks?

  • ACWP=$2500

  • BCWS=$6500

  • BCWP (end of wk #5)=45%(5000)+35%(5000) =2250+1750

    = $4000

  • Schedule Variance = BCWP-BCWS

    = 4000-6500

    = $-2500

  • Cost Variance = BCWP-ACWP

    = 4000-2500

    = $ 1500

M. Sundaram Tenn. Tech


Resources
Resources 22 weeks?

  • Manpower

  • Money

  • Machines

  • Materials

  • Method

M. Sundaram Tenn. Tech


Resource allocation
Resource Allocation 22 weeks?

  • What to do if resources are limited?

    Allocate the limited resources so that the project may be completed on time using the resources optimally.

    Two or more activities may require the same resource simultaneously resulting in increased project length.

  • What to do if resources are unlimited?

    Very rarely a project may have the luxury of having unlimited resources.

    If this pleasant situation were to arise, the responsibility of the project team lies in leveling the resources so that the same amount of the resources are used every period.

M. Sundaram Tenn. Tech


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