Chapter 3 part 2

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Chapter 3 part 2. Project Management. CPM With 3 Time Estimates (PERT). In past, time estimate is firm Now, uncertain as to task duration Natural variance Use 3 time estimates to deal with uncertainty a = optimistic estimate m = most likely b = pessimistic. Critical Path.

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Chapter 3 part 2

Project Management

CPM With 3 Time Estimates (PERT)
• In past, time estimate is firm
• Now, uncertain as to task duration
• Natural variance
• Use 3 time estimates to deal with uncertainty
• a = optimistic estimate
• m = most likely
• b = pessimistic
Critical Path
• Path length is sum of expected times of activities on path (not sum of most likely times)
• Longest expected path length is critical
• ET = expected time of project
• Path length is uncertain, and so is project duration
• Path has variance equal to sum of variances of individual activities on path
Formulas
• ET (expected time) = (a + 4m + b)/6
• σ2 (variance) = [(b - a)/6]2
• σp (st dev of project) =

sqrt(S (variances on CP))

• Z = (D - ET)/σ
PERT Probability Example

You’re a project planner for Apple. A new Ipod project has an expected completion time of 40 weeks, with a standard deviation of 5 weeks. What is the probability of finishing the sub in 50 weeks or less?

Converting to Standardized Variable

-

-

X

ET

50

40

=

=

=

Z

2

.

0

s

5

Normal Distribution

Standardized Normal Distribution

s

= 1

s

= 5

Z

m

= 40

50

X

T

Z

= 0

2.0

z

Obtaining the Probability

Standardized Normal Probability Table (Portion)

Z

.00

.01

.02

s

= 1

.50000

.50399

.50798

0.0

Z

:

:

:

:

.97725

.97725

.97784

.97831

2.0

m

Z

= 0

2.0

.98214

.98257

.98300

2.1

z

Probabilities in body

Example 1. Expected Time Calculations

ET(A)= 3+4(6)+15

6

ET(A)=42/6=7

Ex. 1. Expected Time Calculations

ET(B)= 2+4(4)+14

6

ET(B)=32/6=5.333

Ex 1. Expected Time Calculations

ET(C)= 6+4(12)+30

6

ET(C)=84/6=14

Duration = 54 Days

C(14)

E(11)

H(4)

A(7)

D(5)

F(7)

I(18)

B

(5.333)

G(11)

Example 1. Network

D=53

Example 1. Probability Exercise

What is the probability of finishing this project in

less than 53 days?

p(t < D)

t

TE = 54

p(t < D)

t

TE = 54

D=53

p(Z < -.156) = .438, or 43.8 %

There is a 43.8% probability that this project will be completed in less than 53 weeks. (from the z table)

• What is the probability that the project duration will exceed 56 weeks?

p(t < D)

t

TE = 54

D=56

p(Z > .312) = .378, or 37.8 %

Activity List for Example Problem

Required Activity

Immediate

Activity

Description

Predecessors

Time (weeks)

A

3

Select office site

-

B

-

5

Create organization and financial plan

C

B

3

Determine personnel requirements

D

A,C

4

Design facility

E

D

8

Construct the interior

F

2

Select personnel to move

C

G

F

4

Hire new employees

H

F

2

Move records, key personnel, etc.

I

B

5

Make financial arrangements

J

3

H,E,G

Train new personnel

Example 2

• ET = (a + 4m + b)/6
• σ2 = (b - a)2/36 = [(b - a)/6]2

Example 2

• T = (a + 4m + b)/6
• σ2 = (b - a)2/36 = [(b - a)/6]2

Example 2

• T = (a + 4m + b)/6
• σ2 = (b - a)2/36 = [(b - a)/6]2
Calculations
• ET = 23 weeks
• Σσ2CP = σ 2B + σ 2C + σ 2D + σ 2E + σ 2J = 5.8056
• Z = (D - ET)/σ CP = (22-23)/2.4095 = -0.42
• Take .42 to the Z table => 1 – Z[.42] = .337
• When negative – subtract from 1
• Beware - never add standard deviations!
ET = 23

22

34% chance < 22

50% chance

> 23

16% chance between 22 and 23

(from table)

22 is 0.42 SD below Mean

0.4

0.3

0.2

0.1

0

Variability of Completion Time for Noncritical Paths
• Variability of times for activities on noncritical paths must be considered when finding the probability of finishing in a specified time.
• Variation in noncritical activity may cause change in critical path.