Chapter 10. Activity-on-Node Network Fundamentals X Y Z Y and Z are preceded by X Y and Z can begin...

Chapter 10

Activity-on-Node Network Fundamentals

X

Y

Z

Y and Z are preceded by X

Y and Z can begin at thesame time, if you wish

(B)

A B C

A is preceded by nothingB is preceded by AC is preceded by B

(A)

J

K

L

M

J, K, & L can all begin atthe same time, if you wish(they need not occursimultaneously)

All (J, K, L) must becompleted before M canbegin

but

X Z

AAY

(C)

(D)

Z is preceded by X and Y

AA is preceded by X and Y

Network-Planning Models• A project is made up of a sequence of activities

that form a network representing a project.

• The path taking longest time through this network of activities is called the “critical path.”

• The critical path provides a wide range of scheduling information useful in managing a project.

• Critical Path Method (CPM) helps to identify the critical path(s) in the project networks.

Prerequisites for Critical Path Methodology

A project must have:

well-defined jobs or tasks whose completion marks the end of the project;

independent jobs or tasks;

and tasks that follow a given sequence.

Types of Critical Path Methods

• CPM with a Single Time Estimate– Used when activity times are known with certainty.– Used to determine timing estimates for the project, each

activity in the project, and slack time for activities.

• CPM with Three Activity Time Estimates– Used when activity times are uncertain. – Used to obtain the same information as the Single Time

Estimate model and probability information.

• Time-Cost Models– Used when cost trade-off information is a major consideration

in planning.– Used to determine the least cost in reducing total project time.

Steps in the CPM with Single Time Estimate

• 1. Activity Identification.

• 2. Activity Sequencing and Network Construction.

• 3. Determine the critical path.– From the critical path all of the project and

activity timing information can be obtained.

Example 1. CPM with Single Time Estimate

Consider the following consulting project:

Develop a critical path diagram and determine the duration of the critical path and slack times for all activities

Activity Designation Immed. Pred. Time (Weeks)Assess customer's needs A None 2Write and submit proposal B A 1Obtain approval C B 1Develop service vision and goals D C 2Train employees E C 5Quality improvement pilot groups F D, E 5Write assessment report G F 1

Example 1: First draw the network

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

A None 2

B A 1

C B 1

D C 2

E C 5

F D,E 5

G F 1

Act. Imed. Pred. Time

Example 1: Determine early starts and early finish times

ES = ? EF = ?

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

Example 1: Determine early starts and early finish times

ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

WHAT IS EF OF THE PROJECT?

Example 1: Determine late starts and late finish

times

ES=14EF=15

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

Duration = 15 weeks

Example 1: Determine late starts and late finish

times

ES=14EF=15

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

Duration = 15 weeks

Example 1: Determine late starts and late finish

times

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS = ?LF = ?

Example 1: Determine late starts and late finish

times

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

Example 1: DON’T WRITE DOWN, JUST TO SHOW ALL NUMBERS

ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

NOW:::

• FOR CRITICAL PATH

Example 1: Critical Path & Slack

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

ALL THAT IS NEEDED ES & LS or EF & LF

I PREFER ES & LS

Example 1: Critical Path & Slack

ES=9 ES=14ES=0 ES=2 ES=3

ES=4

ES=4

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LS=9

LS=4

LS=7

LS=3LS=2LS=0

Duration = 15 weeks

Example 1: Critical Path & Slack

ES=9 ES=14ES=0 ES=2 ES=3

ES=4

ES=4

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LS=9

LS=4

LS=7

LS=3LS=2LS=0

Slack=(7-4 = 3 Wks

A CHECKTASK LS - ES CPA 0 - 0 YESB 2 - 2 YESC 3 - 3 YESD 7 - 4 NOE 4 - 4 YESF 9 - 9 YESG 14 - 14 YES

THEREFORE CP = A-B-C-E-F-G

Example 2. CPM with Three Activity Time Estimates

TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

6

4

6

Time Pess. + Time)Likely 4(Most + Time Opt. = Time Expected

bmaET

OR

a m b

Example 2. Expected Time Calculations

TaskImmediate

PredecesorsExpected

TimeA None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

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

6

ET(A)=42/6=7

6

4 bmaET

Example 2. Network

A(7)

B(5.333)

C(14)

D(5)

E(11)

F(7)

H(4)

G(11)

I(18)

Duration = 54 Days

Example 2. Network

A(7)

B(5.333)

C(14)

D(5)

E(11)

F(7)

H(4)

G(11)

I(18)

Duration = 54 Days

ES=0

ES=7

ES=7

ES=21

ES=32

ES=12

ES=36

ES=0 ES=5.333

LS=36

LS=32

LS=21LS=7

LS=0

LS=25LS=20

LS=25LS=19.667

THEREFORE:

• CRITICAL PATH IS:

A-C-E-H-I

Example 2. Probability ExerciseWhat is the probability of finishing this project in less than 53 days?

p(t < D)

TE = 54

Z = D - TE

cp2

tD=53

22

6

a-b = ariance,Activity v

Task Optimistic Most Likely Pessimistic VarianceA 3 6 15 4B 2 4 14C 6 12 30 16D 2 5 8E 5 11 17 4F 3 6 15G 3 9 27H 1 4 7 1I 4 19 28 16

(Sum the variance along the critical path.)

41=2 cp

There is a 43.6% probability that this project will be completed in less than 53 weeks.

p(Z < -.16) = .5 - .0636 = .436, or 43.6 % (Appendix D)

Z = D - T

=53- 54

41= -.156E

cp2

TE = 54

p(t < D)

t

D=53

or - .16

Std Normal Dist.

Example 2. Additional Probability Exercise

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

Example 2. Additional Exercise Solution

tTE = 54

p(t > D)

D=56

Z = D - T

=56 - 54

41= .312E

cp2

p(Z >.31) = .5 - .1217 = .378, or 37.8 % (Appendix D)

or .31

Std Normal Dist.

Time-Cost Models• Basic Assumption: Relationship between

activity completion time and project cost.

• Time Cost Models: Determine the optimum point in time-cost tradeoffs.– Activity direct costs.– Project indirect costs.– Activity completion times.

CPM Assumptions/Limitations • Project activities can be identified as entities.

(There is a clear beginning and ending point for each activity.)

• Project activity sequence relationships can be specified and networked.

• Project control should focus on the critical path.• The activity times follow the beta distribution,

with the variance of the project assumed to equal the sum of the variances along the critical path. Project control should focus on the critical path.