problems on CPM and PERT

1. UNIT 2-NETWORK
OPTIMISATION

2. PROJECT MANAGEMENT
APPLICATIONS
• What is a project?
• Any unique endeavor with specific objectives
• With multiple activities
• With defined precedent relationships
• With a specific time period for completion
• Phases:Planning,scheduling and controlling
• Examples?
• A major event like a wedding
• Any construction project
• Designing a political campaign

3. PROJECT LIFE CYCLE
• Conception: identify the need
• Feasibility analysis or study: costs benefits, and risks
• Planning: who, how long, what to do?, objective and assumptions,
listing the tasks, cost/time estimation
• Scheduling: laying the activity according to precedence
Start and end time for each activity, critical path, floats and slacks
• Controlling: periodical progress reports, review
• Execution: doing the project
• Termination: ending the project

4. NETWORK PLANNING
TECHNIQUES
• A network is a graphic representation of a project’s operations and a
composed of activities and events that must be completed to reach the end
objective of a project, showing the planning sequence of time accomplishment,
their dependence and inter-relationship
• Program Evaluation & Review Technique (PERT):
• Developed to manage the Polaris missile project
• Many tasks pushed the boundaries of science & engineering (tasks’
duration = probabilistic)
• Critical Path Method (CPM):
• Developed to coordinate maintenance projects in the chemical industry
• A complex undertaking, but individual tasks are routine (tasks’ duration
= deterministic)

5. • The basic components of a network are
• Activity- An activity is a task, or item of work to be done, that
consume time, effort, money or other resources. An activity is
represented by an arrow with its head indicating the sequence
in which the events are to occur.
• Event- An event represents the start (beginning) or
completion (end) of some activity and as such it consume no
time. It has no time duration and does not consume any
resources. It is also known as a node. An event is generally
represented on the network by a circle.

6. • The activity can be further classified into the following three categories
• 1. Predecessor activity- An activity which must be completed before one or more other
activities start is known as predecessor activity
• 2. Successor activity- An activity which started immediately after one or more of other
activities are completed is known as successor activity.
• 3. Dummy activity- An activity which does not consume either any resource or time is
known as dummy activity. A dummy activity is depicted by dotted line in the network
diagram
• The dummy activity is inserted in the network to clarify the activity pattern in the
following two situations
To make activities with common starting and finishing points distinguishable
To identify and maintain the proper precedence relationship between activities that is
not connected by events.
For example, consider a situation where A and B are concurrent activities. C is dependent
on A and D is dependent on A and B both. Such a situation can be handled by using a
dummy activity as shown in the figure.
1
2
3
4

7. • The events are classified in to three categories
• 1. Merge event – When more than one activity comes and joins
an event such an event is known as merge event.
• 2. Burst event – When more than one activity leaves an event
such an event is known as burst event.
• 3. Merge and Burst event – An activity may be merge and
burst event at the same time as with respect to some activities
it can be a merge event and with respect to some other
activities it may be a burst event.

8. • Logical sequences in Network diagram:
• All the projects consist of certain activities that can begin only
after certain others are completed.
• In logical sequencing, following two types of errors are most
common while drawing a network diagram.
• 1. Looping: No Activity should start and end on same event.
• 2. Dangling- No activity should end without being joined to
the end event. If it is not so, a dummy activity is introduced in
order to maintain the continuity of the system. Such end event
other than the end of the project as a whole are called
dangling events.

9. • Rules for Network Representation
• Three rules are available for constructing the network
• 1. Each activity is represented by one, and only one arrow (arc)
• 2. Each activity must be identified by two distinct end nodes & no two or more activities can have the same
tail.
• 3. To maintain the correct precedence relationships, the following questions must be answered as each is
added to network:
• (a) What activities must immediately precede the current activity?
• (b) What activities must follow the current activity? (
• c) What activities must occur concurrently with the current activity?
• The answer of these questions may require the use of dummy activities to ensure correct precedences
among the activities.
• Numbering the events
• (a) Event numbers should be unique
• (b) Event numbering should be carried out on a sequential basis from left to right
• (c) The initial event which has all outgoing arrows with no incoming arrow is numbered 0 or 1
• (d) The head of an arrow should always bear a number higher than the one assigned at the tail of the arrow

10. • Example 1:Costruct the network diagram for the project with
the following activities
Activities A B C D E F G
Predecessor - - - A B C D

11. BOTH PERT AND CPM
• Graphically display the precedence
relationships & sequence of activities
• Estimate the project’s duration
• Identify critical activities that cannot be
delayed without delaying the project
• Estimate the amount of slack associated with
non-critical activities

12. • The major differences between PERT and CPM are summarized as given below
• PERT
• Event oriented
• Probabilistic in nature
• Concerned with time
• Used for new projects
• CPM
• Activity oriented
• Deterministic in nature
• Concerned with time and cost
• Used for repetitive projects

13. CRITICAL PATH METHOD IN
NETWORK ANALYSIS
• The notations used are
• (i, j) = Activity with tail event i and head event j
• Ei = Earliest occurrence time of event I
• Lj = Latest allowable occurrence time of event j
• Dij = Duration of activity (i, j)
• (Es)ij = Earliest starting time of activity (i, j)
• (Ef)ij = Earliest finishing time of activity (i, j)
• (Ls)ij = Latest starting time of activity (i, j)
• (Lf)ij = Latest finishing time of activity (i, j)

14. • Forward Pass computation
• Step 1 :The computation begins from the start node and move
towards the end node. Set E1 = 0 to indicate that the project
starts at time 0
• Step 2
• i. Earliest starting time of activity (i, j) is the earliest event
time of the tail end event i.e. (Es)ij = Ei
• ii. Earliest finish time of activity (i, j) is the earliest starting
time + the activity time i.e. (Ef)ij = (Es)ij + Dij or (Ef)ij = Ei +
Dij
• iii. Earliest event time for event j is the maximum of the
earliest finish times of all activities ending in to that event i.e.
Ej = max [(Ef)ij for all immediate predecessor of (i, j)] or Ej
=max [Ei + Dij]

15. • Backward Pass computation (for latest allowable time)
• Step 1 :For ending event assume E = L. (Remember that all E’s
have been computed by forward pass computations)to indicate that
the earliest and latest occurrences of the last node of the project
are the same.
• Step 2 :Latest finish time for activity (i, j) is equal to the latest
event time of event j i.e. (Lf)ij = Lj
• Step 3 :Latest starting time of activity (i, j) = the latest completion
time of (i, j) – the activity time or (Ls)ij =(Lf)ij - Dij or (Ls)ij = Lj –
Dij
• Step 4 :Latest event time for event ‘i’ is the minimum of the latest
start time of all activities originating from that event i.e. Li = min
[(Ls)ij for all immediate successor of (i, j)] = min [(Lf)ij - Dij] = min
[Lj - Dij]

16. DETERMINATION OF FLOATS
AND SLACK TIMES
• In case of non-critical activities, certain amount of spare time
is available and this spare time is called “float”. There are
three type of float
• Total float – The amount of time by which the completion of an
activity could be delayed beyond the earliest expected
completion time without affecting the overall project duration
time. Mathematically (Tf)ij = (Latest start – Earliest start) for
activity ( i – j)
(Tf)ij = (LFj - (ES)i-Dij or (Tf)ij = (Lj - Dij) - Ei

17. • Free float – The time by which the completion of an activity can be delayed
beyond the earliest finish time without affecting the earliest start of a
subsequent activity. Mathematically (Ff)ij = (Earliest time for event j –
Earliest time for event i) – Activity time for ( i, j) (Ff)ij = (Ej - Ei) - Dij
• Independent float – The amount of time by which the start of an activity can
be delayed without effecting the earliest start time of any immediately
following activities, assuming that the preceding activity has finished at its
latest finish time. Mathematically (If)ij = (Ej - Li) - Dij The negative
independent float is always taken as zero.
• Event slack - It is defined as the difference between the latest event and
earliest event times. Mathematically
Head event slack = Lj – Ej , Tail event slack = Li – Ei
Critical event – The events with zero slack times are called critical events. In
other words the event i is said to be critical if Ei = Li
Critical activity – The activities with zero total float are known as critical
activities.
Critical path – The sequence of critical activities in a network is called critical
path.

18. SOME NETWORK DEFINITIONS
• All activities on the critical path have zero slack
• Slack defines how long non-critical activities can be
delayed without delaying the project
• Slack = the activity’s late finish minus its early finish (or
its late start minus its early start)
• Earliest Start (ES) = the earliest finish of the
immediately preceding activity
• Earliest Finish (EF) = is the ES plus the activity time
• Latest Start (LS) and Latest Finish (LF) = the latest an
activity can start (LS) or finish (LF) without delaying
the project completion

19. • Ex1.A small maintenance project consist of the following jobs
whose precedence relationship is given below
Job 1-2 1-3 2-3 2-5 3-4 3-6 4-5 4-6 5-6 6-7
Duration
(days)
15 15 3 5 8 12 1 14 3 14
a)Draw an arrow diagram representing the project
b)Find the total float for each activity
c)Find the critical path and the total project duration

20. Early start(ES) for forward pass(incoming
arrows)
ESj=Max(ESi+Dij) value
ES1=0 0
ES2=ES1+D12 0+15 15
ES3=max(ES1+D13,ES2+D23) Max(0+15,15+3) 18
ES4=ES3+D34 18+8 26
ES5=max(ES2+D25,ES4+D45) Max(15+5,26+1) 27
Es6=max(ES3+D36,ES4+D46,ES5+D56) Max(18+12,26+14,27+3) 40
ES7=ES6+D67 40+14 54
1
2
3
4
5
6
7
A(15)
B(15
))
C(3)
D(5)
E(8)
F(12
)
G(1)
H(14
)
I(3)
J(14)
ES1=0
LC1=0
E
ES3=18
LC3=18
ES4=26
LC4=26
ES5=27
LC5=37
ES6=40
LC6=40
ES7=54
LC7=54

21. Latest Complete(LC) for Backward pass(Outgoing arrows) LCi=min(LCj-Dij) value
LC7=ES7 54
LC6=LC7-D67 54-14 40
LC5=LC6-D56 40-3 37
LC4=min(LC5-D45,LC6-D46)
=min(37-1,40-14)
40-14 26
LC3=min(LC4-D34,LC6-D36)
=min(26-8,40-12)
26-8 18
LC2=min(LC3-D23,LC5-D25)
=min(18-3,37-5)
18-3 15
LC1=min(LC2-D12,LC3-D13)
=min(15-15,18-15)
15-15 0
c)Critical Path conditions
1.ESi=LCi 2.ESj=LCj 3.ESj-ESi=Dij
Critical path:1-2-3-4-6-7
Project duration=15+3+8+14+14=54days

22. Activity Duration Total float=LCj-Dij-Esi Free Float=ESj-Dij-ESi Independent float=ESj-Dij-
LCi
1-2 15 LC2-D12-ES1=15-15-0=0 ES2-D12-Es1=15-15-0=0 ES2-D12-LC1=15-15-0=0
1-3 15 LC3-D13-ES1=18-15-0=3 ES3-D13-ES1=18-15-0=3 ES3-D13-LC1=18-15-0=3
2-3 3 LC3-D23-ES2=18-3-15=0 ES3-D23-ES2=18-3-15=0 ES3-D23-LC2=18-3-15=0
2-5 5 LC5-D25-ES2=37-5-15=17 ES5-D25-ES2=27-5-15=7 ES5-D25-LC2=27-5-15=7
3-4 8 LC4-D34-ES3=26-8-18=0 ES4-D34-ES3=26-8-18=0 ES4-D34-LC3=26-8-18=0
3-6 12 LC6-D36-ES3=40-12-18=10 ES6-D36-ES3=40-12-18=10 ES6-D36-LC3=40-12-18=10
4-5 1 LC5-D45-ES4=37-1-26=10 ES5-D45-ES4=27-1-26=0 Es5-D45-LC4=27-1-26=0
4-6 14 LC6-D46-ES4=40-14-26=0 ES6-D46-ES4=40-14-26=0 ES6-D46-LC4=40-14-26=0
5-6 3 LC6-D56-ES5=40-3-27=10 ES6-D56-ES5=40-3-27=10 ES6-D56-LC5=40-3-37=0
6-7 14 LC7-D67-ES6=54-14-40=0 ES7-D67-ES6=54-14-40=0 Es7-D67-LC6=54-14-40=0
Floats for critical activities are zero

23. • Ex2.A project consist of a series of tasks labelled A to I with
the following constraints.
• A<D, E ; B,D<F ; C<G;B,D<H;F,G<I;
• W<X,Y; means X and Y can’t start until W is completed. You
are required to construct a network diagram using this
notation. Also find the time of completion of the project ,given
the time of completion of each task.
Task A B C D E F G H I
Time
(days)
23 8 20 16 24 18 19 4 10

24. 1
2
1
3
4
5
A(23)
B(8)
C(20
)
D(16)
E(24)
F(18
)
G(19
)
6
H(4)
I(10)
ES1=0
LC1=0
ES2=23
LC2=23
ES3=39
LC3=39
ES4=20
LC4=38
ES5=57
LC5=57
ES3=39
LC3=39
ES6=67
LC6=67
Task A B C D E F G H I
Predecessors - - - A A B,D C B,D F,G
Time 23 8 20 16 24 18 19 4 10

25. ESj=max(ESi+Dij), forward
pass(incoming arrow)
Values for
early start
Latest Completion LCi=min(LCj-
Dij) Backward pass(outgoing
arrow)
Values for latest
completion
ES1=0 0 LC6=ES6=67 67
ES2=Es1+D12=0+23=23 23 LC5=LC6-D56=67-10=57 57
ES3=max(ES1+D13,ES2+D23)
Max(0+8,23+16)=39
39 LC4=LC5-D45=57-19=38 38
ES4=ES1+D14=0+20=20 20 LC3=min(LC5-D35,LC6-D36)
Min(57-18,67-4)=39
39
ES5=max(Es3+D35,ES4+D45)
Max(39+18,20+19)=57
57 LC2=min(LC3-D23,LC6-D26)
Min(39-16,67-24)=23
23
ES6=max(ES2+D26,ES3+D36,ES5+D56)
Max(23+24,39+4,57+10)=67
67 LC1=min(LC2-D12,LC3-
D13,LC4-D14)
Min(23-23,39-8,38-20)=0
0
Critical path conditions:
ESi=LCi
ESj=LCj
ESj-ESi=LCj-LCi=Dij
Critical activities 1-2-3-5-6
Duration 23+16+18+10=67
critical

26. • Ex3.Task A to I constitute a project. The notation X<Y means
that the task X must be completed before Y is started.
• A<D;A<E;B<F;D<F;C<G;C<H;F<I;G<I. Draw a graph to
represent the sequence of tasks and find the time of
completion of the project, when the time(in days) of completion
of each tasks is as follows.
Task A B C D E F G H I
Time
(days)
8 10 8 10 16 17 18 14 9

27. 6
5
4
3
2
1
A(8)
C(8)
B(10)
D(10
)
E(16)
F(17)
H(14
)
G(18)
I(9)
ES1=0
LC1=0
ES2=8
LC2=8
ES3=18
LC3=18
ES4=8
LC4=17
ES5=35
LC5=35
ES6=44
LC6=44
Tasks A B C D E F G H I
predecessors - - - A A B,D C C F,G
Time(days) 8 10 8 10 16 17 18 14 9

28. ESj=max(ESi+Dij), forward
pass(incoming arrow)
Values for
early start
Latest Completion LCi=min(LCj-
Dij) Backward pass(outgoing
arrow)
Values for latest
completion
ES1=0 0 LC6=ES6=44 44
ES2=Es1+D12=0+8=8 8 LC5=LC6-D56=44-9=35 35
ES3=max(ES1+D13,ES2+D23)
Max(0+10,8+10)=18
18 LC4=min(LC5-D45,LC6-D46)
=min(35-18,44-14)=17
17
ES4=ES1+D14=0+8=8 8 LC3=LC5-D35
35-17=18
18
ES5=max(ES3+D35,ES4+D45)
Max(18+17,8+18)=35
35 LC2=min(LC3-D23,LC6-D26)
Min(44-16,18-10)=8
8
ES6=max(ES2+D26,ES4+D46,ES5+D56)
Max(8+16,35+9,8+14)=44
44 LC1=min(LC2-D12,LC3-
D13,LC4-D14)
Min(8-8,18-10,17-8)=0
0
Critical path conditions:
Esi=LCi
ESj=LCj
ESj-ESi=LCj-LCi=Dij
Critical activities 1-2-3-5-6
Duration 8+10+17+9=44

29. • Example 4 .Determine the early start and late start in respect
of all node points and identify critical path for the following
network.

30. • Solution :Calculation of E and L for each node is shown in the
network
The critical nodes are (1, 2, 5,7,8,10)
There are two possible critical paths
i. 1 → 2 → 5 → 8 → 10
ii. 1 → 2 → 5 → 7 → 10

32. PERT FOR DEALING WITH
UNCERTAINTY

33. ESTIMATING ACTIVITY TIMES
• Optimistic time ( t0 ) : is that time estimate of an activity when everything is
assumed to go as per plan. In other words it is the estimate of minimum
possible time which an activity takes in completion under ideal conditions.
• Most likely time ( tm ) : the time which the activity will take most frequently if
repeated number of times.
• Pessimistic time ( tp) : the unlikely but possible performance time if
whatever could go wrong , goes wrong in series. In other words it is the
longest time the can take.
• The times are combined statically to develop the expected time te .
te = to + 4tm + tp
6
Standard deviation of the time of the time required to complete the project
= tp - to
6

34. STEPS INVOLVED IN PERT
• Develop list of activities.
• A rough network for PERT is drawn.
• Events are numbered from left to right.
• Time estimates for each activity are obtained.
• Expected time for each activity is calculated : to+4tm+tp / 6
• Using these expected times calculate earliest & latest finish &
start times of activities.
• Estimate the critical path.
• Using this estimate compute the probability of meeting a
specified completion date by using the standard normal
equation
Z = Due date – expected date of completion =Ts-Te
standard deviation of critical path δ

35. • Example 1:A small project is composed of 7 activities. whose
time estimates are listed in the table.
Activity 1-2 1-3 2-4 2-5 3-5 4-6 5-6
Estimated
duration(we
eks)
to 1 1 2 1 2 2 3
tm 1 4 2 1 5 5 6
tp 7 7 8 1 14 8 15
a)Draw the project network
b) Find the expected duration and variance of each activity
c)Calculate the early and late occurences for each event and the expected
project length.
d) Calculate the variance and standard deviation of project length
e)What is the probability that the project will be completed
1.4 weeks earlier than expected.
2.Not more than 4 weeks later than expected.
3.If the project due date is 19 weeks, what is the probability of meeting
the date.

36. Activity To tm tp Te=(to+4tm+
tp)/6
Variance 2 =[(tp-to)/6]2
1-2 1 1 7 12/6=2 6/6=(1)2=1
1-3 1 4 7 24/6=4 6/6=1
2-4 2 2 8 18/6=3 6/6=1
2-5 1 1 1 6/6=1 0
3-5 2 5 14 36/6=6 12/6= (2)2 =4
4-6 2 5 8 30/6=5 6/6=1
5-6 3 6 15 42/6=7 12/6=(2)2=4
6
4
5
3
2
1
A(2)
B(4)
C(3)
D(1)
E(6)
F(5)
G(7)
ES1=0
LC1=0
ES2=2
LC2=9
ES3=4
LC3=4
ES4=5
LC4=12
ES5=10
LC5=10
ES6=17
LC6=17
ESj=max(ESi+Dij)
LCi=min(LCj-Dij)

37. • c)critical path 1-3-5-6,Expected project duration:4+6+7=17weeks
Critical
activity
te Variance2
1-3 4 1
3-5 6 4
5-6 7 4
17 9
Standard deviation(SD)=sqrt(variance)
=sqrt(9)=3
d)
1)4 weeks earlier than expected
te=17 weeks ts=(17-4)=13weeks
p(Z<=D),Where D=(ts-te)/SD
D=(13-17)/3=-4/3=-1.33(SNDT)
P(Z<D)= .09176=9.17%(out of 100 times 9.1 times, project will be completed 4 weeks earlier than expected)
2)not more than 4 weeks
te=17 ts=17+4 SD=3
D=(ts-te)/SD= (21-17)/3=1.33
P(Z<=D)=0.90824=90.82%
3)te=17 ts=19 SD=3
D=19-17/3=0.67
P(Z<=D)=0.7485=74.8%

38. • Ex2: Find the probability of meeting the schedule date as
given for the network
Activity(i-j) 1-2 1-3 2-4 3-4 4-5 3-5
Optimistic(days) 2 9 5 2 6 8
Most likely(days) 5 12 14 5 6 17
pessimistic(days) 14 15 17 8 12 20
Find the probability that Scheduled project completion date is 30days.
Also find the date on which the project manager can complete the project
with a probability of 0.90.

39. Activity to tm tp Te=(to+4tm+tp)/6 Variance 2=[(tp-to)/6]2
1-2 2 5 14 36/6=6 12/6=22=4
1-3 9 12 15 72/6=12 6/6=12=1
2-4 5 14 17 78/6=13 12/6=22=4
3-4 2 5 8 30/6=5 6/6=12=1
3-5 8 17 20 96/6=16 12/6=22=4
4-5 6 6 12 42/6=7 6/6=12=1
5
4
3
2
1
A(6)
B(12
)
C(13
)
D(5)
E(16
)
B(7)
ES1=0
LC1=0
ES2=6
LC2=8
ES3=12
LC3=12
ES4=19
LC4=21
ES5=28
LC5=28
Critical path=1-3-5
Expected Project duration=12+16=28
Esj=max(ESi+Dij) and Lci=min(LCj-Dij)

40. activities te Variance2
1-3 12 1
3-5 16 4
28 5
SD=sqrt(5)=2.23
a)Probability of completing in 30 days, te=28 ts=30 SD=2.23
D=(ts-te)/SD=(30-28)/2.23=0.89
P(Z<=D)= .81327=81.32 %
b)Given probability=0.90, the number of days required to complete the project.
0.90 from standard normal distribution table =1.29
D=ts-te/SD=1.29
Ts-28/2.23=1.29
Ts=(1.29*2.23)+28
=30.87days

41. • Example 3:Consider the following project, Find the path and
standard deviation. Also find the probability of completing the
project by 18 weeks.
Activity A B C D E F G
Predecessor - - A B B C,D E
Time estimate(weeks) to 3 2 2 2 1 4 1
tm 6 5 4 3 3 6 5
tp 9 8 6 10 11 8 15

42. 6
5
4
3
2
1
A(6)
B(5)
C(4)
D(4)
E(4)
F(6)
G(6)
ES1=0
LC1=0
ES2=6
LC2=6
ES3=5
LC3=6
ES4=10
LC4=10
ES5=9
LC5=10
ES6=16
LC6=16
Activities To(week
s)
Tm(weeks
)
Tp(weeks) Te=(to+4tm+tp)/6 Variance2=[(tp-
to)/4]2
A 3 6 9 36/6=6 1
B 2 5 8 30/6=5 1
C 2 4 6 24/6=4 0.44
D 2 3 10 24/6=4 1.77
E 1 3 11 24/6=4 2.77
F 4 6 8 36/6=6 0.44
G 1 5 15 36/6=6 5.44
Esj=max(Esi+Dij) and Lci=min(LCj-Dij)
Critical path 1-2-4-6, Expected project duration=16weeks

43. Critical Activities Te for critical activities Variance2 for critical activities
A 6 1
C 4 0.44
F 6 0.44
16 1.88
SD=sqrt(Variance2)=sqrt(1.88)=1.374
Probability of completing project by 18 weeks P(Z<=D)
D=(ts-te)/SD=18-16/1.374=1.455(check value from normal distribution table)
P(Z<=D)=0.92647

44. RESOURCE LEVELLING AND
SMOOTHING
• Resource optimization is employed when there is a resource conflict (i.e.
when the schedule requires more than the available resources) or when
there is a need to keep the resource usage at a constant level.
• Two examples of resource optimization techniques are:
1. Resource levelling and
2. Resource smoothing
• Resource smoothing is used when the time constraint takes priority. The
objective is to complete the work by the required date while avoiding peaks
and troughs of resource demand.
• Resource levelling is used when limits on the availability of resources are
dominant.

46. TIME-COST TRADE-OFF
• The objective of the time-cost trade-off analysis is to reduce the original
project duration, determined form the critical path analysis, to meet a specific
deadline, with the least cost. In addition to that it might be necessary to
finish the project in a specific time to:
• - Finish the project in a predefined deadline date.
• - Recover early delays.
• - Avoid liquidated damages.
• - Free key resources early for other projects.
• Avoid adverse weather conditions that might affect productivity.
• - Receive an early completion-bonus.
• - Improve project cash flow

47. TIME-COST TRADE-OFF CONTD..
• The activity duration can be reduced by one of the following actions:
• - Applying multiple-shifts work.
• - Working extended hours (over time).
• - Offering incentive payments to increase the productivity.
• - Working on week ends and holidays.
• - Using additional resources.
• - Using materials with faster installation methods.
• - Using alternate construction methods or sequence.
• selecting some activities on the critical path to shorten their duration.

48. HIGHLIGHTS
• A project is a unique, one time event of some duration that
consumes resources and is designed to achieve an objective in
a given time period.
• Each project goes through a five-phase life cycle: concept,
feasibility study, planning, execution, and termination.
• Two network planning techniques are PERT and CPM. Pert
uses probabilistic time estimates. CPM uses deterministic
time estimates.
• Pert and CPM determine the critical path of the project and
the estimated completion time. On large projects, software
programs are available to identify the critical path.

49. HIGHLIGHTS CON’T
• Pert uses probabilistic time estimates to determine the
probability that a project will be done by a specific time.
• To reduce the length of the project (crashing), we need to
know the critical path of the project and the cost of reducing
individual activity times. Crashing activities that are not on
the critical path typically do not reduce project completion
time.
• The critical chain approach removes excess safety time from
individual activities and creates a project buffer at the end
of the critical path.