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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.
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