11/7/2017

time pLanning
Lecture-2 As we discussed in the previous chapter,
chapter two - time pLanning Time planning may have the following
project work Breakdown process.
activity duration estimation  activity definition (using WBS)
scheduLing  duration estimation
project network anaLysis (modeLing) (cpm,
pert & pna)  schedule development (sequencing, Net working
etc.) and
 controlling.
11/7/2017 Temesgen G. 1 11/7/2017 Temesgen G. 2

Project Breakdowns Project work Breakdown structure

Project work-breakdown methodology enables splitting of
the project work into hierarchical work-breakdown levels of;
Project
sub-projects
tasks
Work packages
Activities
Each activity represents an identifiable lower-level job which
consumes time, and possibly resource.
 A Work Breakdown Structure does not show the sequence in which work is
performed! Such sequencing is determined when a schedule is developed.

11/7/2017 Temesgen G. 3 11/7/2017 Temesgen G. 4

 11/7/2017

Project work Breakdown … Real Estate Development

Mega project
(Programme)

The project work-breakdown process involves breaking Residential Buildings Service Buildings Recreation Centers
down of the project work into manageable parts arranged Sub projects

in a hierarchical order till the desired level is reached. Educational Buildings Health Centre Shopping centre

A WBS is a deliverable-oriented grouping of project

Tasks
components that organizes and defines the total scope of Sub structure Super Structure Roofing Finishing

the project. Any work not in the WBS is outside the scope Base construction Footing Con Plinth wall Con Ground floor Con Work package
of the project. (Eg…)
It is used to develop or confirm a common understanding Earth work Base preparation Blinding

of project scope. Activities

Layout Excavation Leveling/compacting

11/7/2017 Temesgen G. 5 11/7/2017 Temesgen G. 6

Activity Duration
Project work Breakdown …
Duration of an activity is defined as the
Example, Operation involved in concreting are; expected economical transaction time.
Cleaning and preparing inner side of the raft for The estimation of time is based upon the
concreting current practices carried out in an
Pumping concrete organized manner under the normal
Spreading and vibrating concrete prevailing conditions, and its assessment
is done preferably, by the person
Finishing of top concrete surface responsible for its performance.

11/7/2017 Temesgen G. 7 11/7/2017 Temesgen G. 8

 11/7/2017

Duration Estimation methods

Application in Construction Projects :
1. one time estimate
The estimation of duration is based on one of the 1. Planning of the projects especially, at the feasibility stage
following; 2. The Skelton networks enclosed with the tender documents.
i. Planning data 3. The contracted works, where time is the main consideration
ii. Past experience or execution of a similar project for management
iii. Average time assessed by a group of executives.
4. The complex structures, where the exact duration estimate is
2. Three time estimate
Te=[To+4Tm+Tp]/6 To, everything goes difficult to assess.
extremely well with no delays, Tp everything 3. Trapezoidal distribution estimate
goes wrong
 When the exact duration of an activity, like In practice, the profile of most activities takes the shape of trapezoidal
research and development, is not certain, the distribution. The build-up and rundown phase can be expressed in terms
three-times estimate is used to compute its of total activity duration.
11/7/2017
expected duration. Temesgen G. 9 11/7/2017 Temesgen G. 10

Duration Estimation procedure

Estimating the quantity of work
Scheduling
Deciding the labor and material constants Scheduling my have the following processes:
Assessing the effective activity-wise Define Activities – identifying the specific actions to be performed to
11

employment of resource produce the project deliverables

Estimate the activity completion period

 Sequencing Activities – identifying and documenting relationships
Completion period=Quantity of work/[output among the project activities

per unit of resource x resource earmarked] Estimate Activity Resources – estimating the type and quantities of
material, people, equipment, or supplies required to perform each
t=Q/(nXp) activity
11/7/2017 Temesgen G. 12
 11/7/2017

Scheduling Cont… Scheduling Cont…

 Estimate Activity Duration – approximating the number of work periods Schedule documents can be submitted mainly in two types
needed to complete individual activities with estimated resources.
completion of total project is determined by knowing the quantity of work i. Master schedule: is used as the contract baseline, usually
involved and the rate of performing each activity under the form of a milestone chart
ii. Project detailed schedule: is used to the actual progress of of
 Develop Schedule – analyzing activity sequences, durations, resource
requirements, and schedule constraints to create the project schedule the project in detail.

 Control Schedule – monitoring the status of the project to update project

progress and managing changes to the schedule baseline

11/7/2017 Temesgen G. 13 11/7/2017 Temesgen G. 14

Scheduling Cont…
Scheduling Techniques
Scheduling may also Classified as:
Construction schedule Line-of-Balance
Material schedule bar chart/Gantt charts
Labor schedule Net work diagram
Equipment schedule Critical Path Method (CPM)
Expenditure schedule Program Evaluation and Review Technique
(PERT)
Eontrol schedule
Precedence diagramming method (PDM)/PNA
11/7/2017 Temesgen G. 15 11/7/2017 Temesgen G. 16
 11/7/2017

The line-of-balance (LOB) LOB

The LOB technique was originated by the Goodyear In LOB, the time is usually plotted on horizontal axis, and
Company in the early 1940's and was developed by the units or stages of an activity on the vertical axis.
U.S. Navy in the early 1950's for the programming and The production rate of an activity is the slope of the
control of both repetitive and non-repetitive projects production line and is expressed in terms of units per
(Turban 1968, Johnston 1981, Lutz and Halpin 1992). time.
It was first applied to industrial manufacturing and The method uses man-hour estimate and the optimum
production control. The basic concepts of LOB have been crew or gang size. The LOB diagram can be plotted once
applied in the construction industry as a planning and the man hour estimates and optimum size of crew are
scheduling method (Lumsden 1968, Khisty 1970). determined by the planner based on his or her experience
The LOB technique is based on the underlying or in consultation with the implementing agencies such as
assumption that the rate of production for an activity is vendor or subcontractor.
uniform. Two oblique and parallel lines, whose slope is equal to
In other words, the rate of production of an activity is the actual rate of output will denote the start and finish
linear. 17
times respectively of each activity in all the units from the18
first to last.

LOB LOB

This technique can be applied to construction LOB

projects for repetitive unit projects such as mass

12

10
housing, high rise buildings, tunnels, etc. 8

No. of Houses
A typical LOB diagram is shown in the next slide 6

for housing construction. 4

0
0 50 100 150 200 250
TIME

19
20
 11/7/2017

Bar Chart Bar Chart

 Until around 1958, the only tool for scheduling projects was the bar chart.
No Work
description  Bar charts are easy to plot, comprehend and communicate, and are most appropriate for
Residential presentation of schedules. However, as planning technique, the bar chart is not suitable for
Bldg.
complex projects due to the following reasons:
1
2 (a) It does not reflect the relationship between various activities which are a common feature of all
3
4
complex projects.
5
6
(b) It cannot identify and highlight the emerging critical tasks needing special attention for
7 preventing schedule slippages, time overruns, and other bottlenecks.
8
9 (c) In complex projects, time durations are often educated guesses. Any change in schedule or time
10
11 duration would require a redrawing of the multi-task bar chart schedule.
12
13  thus, it can be said that the bar chart format is most useful for presentation of
schedules, but as a planning technique, it is not suitable for scheduling complex
projects.
Months  To overcome this problem, two methods of scheduling (CPM&PERT) were
` Working days developed in the late 1950s and early 1960s, both of which use arrow diagrams
in month
to capture the sequential and parallel relationships among project activities.
Working days
cumulative 21 22

Time planning techniques

SI.N Nature of the project Planning Scheduling Monitoring Displaying
O

1 Simple project/sub
project Bar Chart Bar Chart Bar Chart Bar Chart
(a)Non-repetitive work
(b)Repetitive work LOB LOB LOB LOB/Bar chart Lecture-3
2 Complex sub-projects CPM Time scale Bar Chart
(a)Deterministic Network
(b)Probabilistic PERT CPM >>
>>
3 Complex Projects PNA Bar Chart PNA Bar Chart
(a)Non-repetitive W. PNA LOB LOB Bar Chart/LOB
(b)Repetitive work PERT Time Scale network PERT Bar Chart
(c)Probabilistic
CPM-Critical Path
PERT-Programme Evaluation and Review Techniques BC-Bar Chart
PNA- Precedence Network Analysis TSN-Time Scale Network/Logic Bar Chart
LOB-Line of Balance 23 11/7/2017 Temesgen G. 24
 11/7/2017

Network analysis CPM Network analysis Fundamentals

Network elements
CPM-> is best suited for activities with deterministic
single-time duration Event or Milestone
 A point in time when certain conditions have been fulfilled, such
PERT(program evaluation and review technique) - as the start or completion of one or more activities
>useful for project feasibility reports or tasks  Unlike an activity, does not consume time or resources
involving uncertainties.  Hence, expresses a state of being
PNA>commonly used technique for time planning of  Activities take place between events
Activity
construction projects.
 An item of work that consumes time and resources to produce
some result
11/7/2017 Temesgen G. 25 11/7/2017 Temesgen G. 26

Dummy Activity Illustration of event, activity, and dummy activity

 This activity does not involve consumption of
resources, and therefore does not need any time to
be ‘completed’. A B

 It is used to define interdependence between 10 30 50

activities and included in a network for logical and E

mathematical reasons as will be shown later. C D

20 40 60

11/7/2017 Temesgen G. 27 11/7/2017 Temesgen G. 28

 11/7/2017

Float or Slack Time

Critical Path
The series of activities all of which must  The additional time available to complete a non-
finish on time for the whole project to finish critical activity
on time Leads and Lags
Sometimes described as the longest path  An imposed modification of the logical
through a network, hence the shortest
relationship between activities
project time
A critical path has zero float  To accelerate or delay the apparent natural order
A critical path assumes that the network
logic is sound
11/7/2017 Temesgen G. 29 11/7/2017 Temesgen G. 30

Illustration for TF, IF, and FF calculation

Total Float
 Total Float in an activity (i,j) [TF(i,j)]
TLi T.F TLj
 Total float is the amount of time by which the start of an
Duration
activity may be delayed without causing a delay in the
Duration
completion of the project. This is calculated as (TF(i,j) =
I.F
TEj
[LST(i,j)]– [EST(i,j)]) or ([LFT(i,j)]– [EFT(i,j)])
TEi
Duration F.F

11/7/2017 Temesgen G. 31 11/7/2017 Temesgen G. 32

 11/7/2017

Free Float Network Preparation

 Free float is the amount of time by which the start of an tabulate the network logic drawing the arrow diagrams step-by-step.
activity may be delayed without delaying the start of a Logic activity can be tabulated as
following activity. -> which activity/ies preceded & Succeed this activity?
->Are there any logical constraints imposed on this activity?
 Free Float = (Earliest start time of the following activity – ->Is it the final activity?

Duration of the activity – Earliest start time of the activity) Example

that is Free Float = TEj- TEi -D Activity Preceding Succeeding Remarks
A - B,H
Independent Float B A G,J
C - H
It is defined as the difference in Total Float and Free D
E
-
D
E,G
F
Float. In other words: Interference Float= Total Float – F
G
E
B,D
-
-
Last activity
Last activity
Free Float.
11/7/2017 Temesgen G. 33 11/7/2017
H
J
A,C
B,H
J
- Temesgen G. Last activity 34

Logic diagram of activities

Numbering Events
B B
G G
A A
C H J C H J

D E F D E F

Rearrange to avoid crossing of arrows, inserting events to mark the start

and completion of activities and writing the duration of each activity.
E E
2
2 F 1 2 F
1 1
D G D G
5 5 5 8
3 3
A B J A B J
0 3 4
4 3 4 3
C 2 C 2
H H 7
2 2 6
3 3

11/7/2017 Temesgen G. 35 11/7/2017 Temesgen G. 36

 11/7/2017

Event Timings, Activity Timings and • Latest Finish Time of an activity (i,j) [LFT(i,j)]
Associated Terms –the latest time that an activity needs to be
Start and finish times
• Earliest Start Time of an activity (i,j) [EST(i,j)]
completed in order that there is no delay in the
• This is the earliest that the activity (i,j) can be started, i.e., all the
project completion.
necessary preconditions are met. • Latest Start Time of an activity (i,j) [LST(i,j)]
• Earliest Finish Time of an activity (i,j) [EFT(i,j)]
–the latest time when an activity must be started, in
• This is the earliest that an activity can be completed. Mathematically,
the relationship can be expressed as order that there is no delay in the project
• EFT (i,j)= EST(i,j) + D(i,j) completion.
LST(i,j) = LFT(i,j) – D(i,j)
11/7/2017 Temesgen G. 37 11/7/2017 Temesgen G. 38

Path and critical path Forward and Backward Pass

The forward pass moves from the ‘start’ node towards the
Any series of activities connecting the starting point to ‘finish’ node, and basically calculates the earliest occurrence
the finishing point can be said to define a ‘path’ and times of all events.
indeed in a project having several activities, several Considering that the project starts at time zero, the earliest
occurrence time at each node is found by going from node
such ‘paths’ can be identified. to node in the order of increasing node numbers keeping in
Among these paths, the ‘critical path’ is defined as one mind the logical relationships between the nodes as shown
that gives the longest time of completion (of the by the connecting arrows.
project), which also defines the shortest possible The earliest occurrence time for any node can be estimated
from the (maximum) time taken to reach that node from the
project time. different incoming arrows.
11/7/2017 Temesgen G. 39 11/7/2017 Temesgen G. 40
 11/7/2017

EXAMPLE
Task ID Duration Dependency
Network of the example
A 7 C
B 3
A
2 6 3 G
C 6 A 7
3
H
D 3 B A
B D 6 E 7 8
1 4
3
4 2
E 3 D,F 3 3
F
F 2 B 2

G 3 C 5
H 2 E,G

11/7/2017 Temesgen G. 41 11/7/2017 Temesgen G. 42

Network of the example Computations

Act. Duration EST EFT LST LFT TF
C
2 6
3 A 7 0 7 0 7 0
G
A 3
7
H B 3 0 3 7 10 7
A
B D 6 E 7 8 C 6 7 13 7 13 0
1
4
3
4 2
3 3
F D 3 3 6 10 13 7
2
E 3 6 9 13 16 7
5
F 2 3 5 11 13 8
G 3 13 16 13 16 0
11/7/2017
H 2 16 18
Temesgen G.
16 18 0 44
11/7/2017 Temesgen G. 43
 11/7/2017

Problem 1 a. Draw the network diagram.

Task ID Duration Dependency b. Determine the project completion time and
A 4
isolate the critical path.
B 6 A
C 9 A c. Identify critical and non critical activities.
D 2 A d. Calculate EST,LST,EFT & LFT of each activity.
E 3 B
F 8 B
G 10 C
H 4 G
11/7/2017
I 2 Temesgen G. D,E 45 11/7/2017 Temesgen G. 46

PERT Beta distribution for the activity ‘design foundation’

Example of Three time estimate Expected Time te

For an activity “Design foundation” te=19

 the optimistic time = 14 days

 the most likely time = 18 days and
 the pessimistic time estimates = 28 days
The PERT technique assumes that the three time
estimates of an activity are random variables and
the frequency distribution of duration of an activity to=14 tm=18 tp=28

takes the shape of Beta distribution Activity duration (in days)

11/7/2017 Temesgen G. 47 11/7/2017 Temesgen G. 48

 11/7/2017

There has been a lot of criticism on the approach of

The average or expected time it is given by
obtaining three ‘‘valid’’ time estimates to put into the PERT
te= (to+4tm+tp)/6 formulas.
For the case of ‘design foundation’, te can be
worked out to be 19 days [(14 + 4 x18 + It is often difficult to arrive at one activity-time estimate;
28)/6]. three subjective definitions of such estimates do not help
the matter (how optimistic and pessimistic should one be).
The fact that te > tm in this case, is a reflection
of the extreme position of tp and the Nevertheless, the three time estimate also provides the
asymmetry in the Beta distribution, even advantages of ascertaining the variability or uncertainty
though computationally the weights given to associated with a particular set of estimate.
to and tp is the same.

11/7/2017 Temesgen G. 49 11/7/2017 Temesgen G. 50

For example, suppose we have two sets of estimate In order to measure the uncertainty
provided by the different estimator for the same associated with the estimate of duration of an
‘design foundation’ activity. activity, the standard deviation (St) and the
In order of (te, tm,tp), let the first set of estimate (14, variance Vt are determined, which in PERT are
18, 28) and the other set of estimates be (17, 18, 25). defined as:
There is large variability in the estimates of first St = (tp-to)/6 and
estimator compared to the second one, even though Vt = (St)2
the expected or average activity duration turns out to The formula for St indicates that it is one sixth
be 19 in both the cases (verify!!!). of the difference between the two extreme
time estimates.

11/7/2017 Temesgen G. 51 11/7/2017 Temesgen G. 52

 11/7/2017

Further, the greater the uncertainty in time estimates, For the two sets of estimate used in ‘design activity’,
the greater the value of (tp-to), and the more spread the St and Vt would be 2.33 days and 5.44 respectively
out will be the distribution curve. for first set of estimates while 1.33 days and 1.77 are
A high St represents a high degree of uncertainty the corresponding values of St and Vt for the second
regarding activity times. In other words there is a set of estimates.
greater chance that the actual time required to The expected length or duration of project Te is
complete the activity will differ significantly from the calculated by summing up the expected duration te’s
expected time te. of activities on the critical path.

11/7/2017 Temesgen G. 53 11/7/2017 Temesgen G. 54

The critical path is determined following the In case, there is more than one critical path in a
forward pass and backward pass explained project network, then the path with the largest
earlier. variance is chosen to determine the VT and ST.
The variance associated with the critical path Mathematically,
is the sum of variances associated with the • Te= ∑te
activities on the critical path. • VT = ∑Vt and
• ST= ∑St

11/7/2017 Temesgen G. 55 11/7/2017 Temesgen G. 56

 11/7/2017

VT and ST represent variability in the expected

Now that, te is the weighted sum of to, tm, and
project duration. The higher the VT and ST values,
tp, it is also treated as a random variable. Since,
the more likely it is that the time required to
Te is the sum of te’s it indeed is a random
complete the project will differ from the expected
variable.
project length Te.
The distribution of Te follows normal
As was pointed earlier, to, tm, and tp are assumed
distribution according to the Central Limit
to be a random variable following Beta
Theorem of statistics.
distribution in the PERT technique.

11/7/2017 Temesgen G. 57 11/7/2017 Temesgen G. 58

Z= (TD- Te)/ ST
Suppose, it is required to compute the probability of
completing the project within a target duration of TD Here Z is the number of standard deviations by which TD exceeds Te. Note
days. that TD might be less than Te, in which case Z is negative. Now the
probability measure the originally sought may be obtained by referring to
Now given the Te of the project it is possible to the following table, extracted from a standard normal table:
calculate the deviation of TD from Te in units of
standard deviation. Z Probability of meeting
Due Date
Z Probability of meeting
Due Date
This is calculated from the normal distribution table. To 3.0 .999 1.0 .841
adopt the table, a ratio called the standardised 2.8 .997 0.8 .788
2.6 .995
deviation or more often the normal deviate, Z, is 2.4 .992
0.6 .726
0.4 .655
derived. Z is defined as the ratio of the difference in TD 2.2 .986
0.2 .579
and Te to ST. Mathematically, Z= (TD- Te)/ ST, 2.0 .977
0.0 .500
1.8 .964
1.6 .945
1.4 .919
11/7/2017 Temesgen G. 59 11/7/2017 Temesgen G. 60
1.2 .885
 11/7/2017

Z= (TD- Te)/ ST
Example: PERT Diagram
Z Probability of meeting Z Probability of meeting
Due Date Due Date

-0.2 .421 E (5,14, 17)

-2.2 .014 30 50
-.4 .345
13
-.6 .274 -2.4 .008 B (2,5,14) H(1, 4,7)
2 F (2,5,14) 2
-.8 .212
6 D (1,2,3) 6
-2.6 .005
-1.0 .159

-1.2 .115 -2.8 .003 A

A (3,12,21) C (6, 15, 30) 40 G (4, 5, 12) 60
10 20
-1.4 .081
-3.0 .001
4
12 16 6
-1.6 .055
-1.8 .036

-2.0 .023

11/7/2017 Temesgen G. 61 11/7/2017 Temesgen G. 62

Expected duration, standard deviations and

variances for activities Computation of early occurrence and late occurrence times
Duration (days) Expected Standard Variance Vt
Activity Optimisti Most Pessimisti duration deviation St = (St)2
Id (days) te= (t -t )/6 No Early occurrence time Late occurrence time Slack
c likely c duration = p o
(to+4tm+tp)/6 de
duration duration tp
to tm
Col 1 Col 2 Col 3 Col 4 Col 5 Col6 Col 7 10 0 12-12=0 0

10-20 3 12 21 12 3 9 20 0+12=12 Min of [(21-6)=15 and (28 - 0

16)=12]=12
20-30 2 5 14 6 2 4
30 12+6=18 Min of [(34-13)=21 and (28 - 3
20-40 6 15 30 16 4 16 2)=26]=21
30-40 1 2 3 2 1/3 1/9 40 Max of [(12+16)=28 and Min of [(36-6)=30 and (34 - 0
(18+2)=20]=28 6)=28]=28
30-50 5 14 17 13 2 4
50 Max of [(18+13)=31 and 36-2=34 0
40-50 2 5 14 6 2 4 (28+6)=34]=34
40-60 4 5 12 6 4/3 16/9 60 Max of [(34+2)=36 and 36 0
11/7/2017 Temesgen G. 63 11/7/2017 (28+6)=34]=36 Temesgen G. 64
50-60 1 4 7 2 1 1
 11/7/2017

Now, the problem of computing the probability In other words, the target duration TD is 1.09
of meeting target duration (TD), such as 42 days standard deviations greater than the expected
shown in the figure is quite simple. Since the time TE=36 days.
total area under the normal curve is exactly one, The equivalent probability P(Z=1.09) can be
the cross hatched area under the normal curve read off a normal probability distribution. This
is directly the probability that the actual corresponds to a probability of 0.862
completion time, will be equal to, or less than,
(86.2%)which implies that there is a 86.2%
42 days.
chance that the project will get completed
In this case Z= (TD- TE)/ ST, = (42-36)/ 5.48 =
within 42 days.
1.09 standard deviations.
11/7/2017 Temesgen G. 65 11/7/2017 Temesgen G. 66

Meeting a Target Duration TD Meeting a Target Duration TD

1.09 Standard Deviations 0 Standard Deviations

P(t  42 days) P(t  36 days)

 82.6%  50%

36 42 36
36
Time - days Time - days
11/7/2017 Temesgen G. 67 11/7/2017 Temesgen G. 68
 11/7/2017

Meeting a Target Duration TD Assuming that time now is zero, one may expect this
project to end at time 36 days (corresponding
probability of achieving this target being 50%,
0.55 Standard Deviations verify!!! Hint: TD=36, TE=36 ); and the probability
that it will end on or before the target duration of 42,
P(t  33 days) without expediting the project is approximately
 29.1% 86.2%.
On the other hand, if one were to schedule towards
TD= 33 days; herein TD<TE; i.e. Z=-0.55 (Note the
33 36 negative sign); the corresponding probability would
Time - days
be 0.291, which is really a very bleak situation.
11/7/2017 Temesgen G. 69 11/7/2017 Temesgen G. 70

In the above, the phrase ‘without expediting’ is very

important.
The feature in PERT on the computation of
probability of completing the project in a particular
In certain projects schedules always may be met by some
means or another, duration is quite useful especially for negotiating
the duration with an owner by the executing
for example:
– by changing the schedule,
agency.
– by changing the project requirement, For example, while agreeing on a particular
– by adding further personnel or facilities, etc. duration, the executing agency would like to judge
However, here it is implied that the probability being his chances on completing the project in that
computed hereinabove is the one that the original duration.
schedule will be met without having to expedite the work
in some way or another.
11/7/2017 Temesgen G. 71 11/7/2017 Temesgen G. 72
 11/7/2017

For being reasonably sure of a particular duration, he

would like to attain a probability of more than 95%.
Thus for the same example, suppose the executing
agency is asked to provide the projected duration for
the project, the agency would find out the duration
corresponding to Z(P=0.95)= 1.65, thus the target
duration for this case could be TD= TE + 1.65 x ST= 36 +
1.65 x 5.48= approximately 45 days. In other words,
the executing agency would be quite confident of
completing the project in 45 days.

11/7/2017 Temesgen G. 73