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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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Presentation Transcript
slide1

Chapter 3 part 2

Project Management

cpm with 3 time estimates pert
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
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
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
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
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
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
Example 1. Expected Time Calculations

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

6

ET(A)=42/6=7

ex 1 expected time calculations
Ex. 1. Expected Time Calculations

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

6

ET(B)=32/6=5.333

ex 1 expected time calculations1
Ex 1. Expected Time Calculations

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

6

ET(C)=84/6=14

slide12

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

slide13

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

slide15

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)

ex 1 additional probability exercise
Ex 1. Additional Probability Exercise
  • What is the probability that the project duration will exceed 56 weeks?
example 1 additional exercise solution

p(t < D)

t

TE = 54

D=56

Example 1. Additional Exercise Solution

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

activity list for example problem
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

slide19

Example 2

  • ET = (a + 4m + b)/6
  • σ2 = (b - a)2/36 = [(b - a)/6]2
slide20

Example 2

  • T = (a + 4m + b)/6
  • σ2 = (b - a)2/36 = [(b - a)/6]2
slide21

Example 2

  • T = (a + 4m + b)/6
  • σ2 = (b - a)2/36 = [(b - a)/6]2
calculations
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!
22 is 0 42 sd below mean
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 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.