Controlling time
(lecture & seminar)
Why to control?
Monitoring and analysis of project data should enable
the project manager to
• address problems at an early stage and
• take advantages of opportunities;
• prevent problems rather than responding to them;
• communicate quickly and effectively:
thus project members will work with the same and up-todate information, and can quickly deal with problems.
6 essential features of control
systems
1. A plan must be made
2. This plan must be published
3. Once working, the activity being controlled
must be measured
4. The measurements must then be compared
with the plan
5. Any deviations must be reported to the
appropriate person
6. A forecast of the results of any deviations
must then be made, and corrective actions
taken or a new plan must be made
Bar
chart
PNTs
Measurement of activities
1. The measurement should be appropriately precise
2. The measurement should be pertinent
3. The speed of collection of the information must be
rapid compared with the timecycle of the system as
a whole
4. Measurements need to be accurate or of consistent
inaccuracy
5. The number of data processing points should be
kept as small as possible
The measurement should be
appropriately precise
• Precision is positively correlated with
measurement costs
• PNT indicates the necessity of precision
– Critical activities
– Non-critical activities
The measurement should be pertinent
• What use the collected data can be made
The speed of data collection…
• Measurement must be taken frequently
enough to allow useful action to be taken.
• The longer the time for corrective actions to
be taken, the lower frequency of
measurement is acceptable.
• Thus: the speed of data collection will
increase toward the project’s ending.
Measurements need to be accurate or of
consistent inaccuracy
• Accuracy can be bought with increased cost.
• Inaccurate but consistent measuring
technique can be acceptable.
• PNT: where is the low accuracy tolerable.
• Accuracy and precision is NOT the same:
Accuracy =
degree of closeness to the true value (validity)
Precision =
degree to which repeated measurements leads to the same results
(reliability).
The number of data processing points
should be kept as small as possible
Too much processing points would lead to:
• delay in information use
• distortions in the information
3 useful methods
(the simplest while appropriate is the best)
• The bar (Gantt) chart or the network itself
• Re-analysis
• Negative float
Gantt chart as a time control method
(best for smaller projects)
Activities
time
1.
2.
3.
4.
5.
6.
today
Example
Market research (survey) project. Estimation of activity durations:
1. Creating the SOW (framework for the research). = 7 days
2. Collecting secondary information:
1. models = 15 days
2. empirics = 15 days
3.
4.
5.
6.
7.
8.
9.
Planning the research = 6 days
Formulating the questionnaire = 14 days
Collecting data in 3 destinations form 100-100 people = 10 days each
Entering data = 2 days per one subsample
Analysing data = 6 days
Writing up the research report = 7 days
Presenting the results = 1 day
Example
After 30 days:
• SOW is made in 7 days
• Secondary research on models was
finished in 14 days
• Secondary research on empirics was
finished in 18 days
• Half of the research plan is written
• No other activity is started
Re-analysis with PNT
(or with Gantt chart)
• In large and/or complex projects.
• Taking the original network and inserting into
the actual times instead of the expected
durations. Sometimes new „delay activites”
need to be built in.
OR
• Re-drawing the network leaving out all those
activities that are complete and re-analyse the
others. (Just like planning a new project)
Example 2 (for networks)
• Manufacturing project:
The firm have to purchase two types of raw
materials (A & B; procurement is centralised thus
it is a single activity), and produce two different
product parts from these (P1 and P2). These can
be done separately. P2 has to put through an
obligatory test, P1 do not. After both P1 and P2
are ready, the firm will assemble them with an
externally bought P3 part (this must be purchased
after previous buying is closed) to get the final
product. Project done.
Create the WBS and identify tasks.
Example 2 (sipmlified network diagram)
The duration times for the activities (with activity labels):
• (a) Raw materials procurement: 7 days
• (b) P1 production: 5 days
• (c) P2 production: 2 days
• (d) P2 test: 2 days
• (e) P3 procurement: 4 days
• (f) Assembly: 2 days
b
5
• Network?
• TPT?
a
7
c
2
d
2
e
4
a-b-f = 14 = TPT
a-c-d-f = 13
a-e-f = 13
f
2
Example 2: problems arise
• After the project started, it became obvious that the testing
facility will not be able to start the testing process in time,
only two days later (they have a priority job for another
project).
• The P3 procurement is expected to be 1 day longer, too.
• Re-analyse the project with the network.
• Identify the new TPT.
b
5
a
7
c
2
delay
2
e
5
a-b-f = 14
a-c-(delay)-d-f = 15 = TPT
a-e-f = 14
d
2
f
2
Negative float
• If the network is too large and complex.
• Fix the end date, insert the actual durations
and re-analyse the network
(backward analysis).
• Late activities will appear with negative float.
There must be corrective actions taken
(priority is according to the order of the
absolute value of the negative floats).
Example 2: calculate floats and
negative floats
0
b
5
0
a
7
1
1
c
2
d
2
0
f
2
1
0
e
4
b
5
-1
-1
-1
-1
a
7
c
2
delay
2
d
2
This will be discussed in a more
sophisticated way on the following
classes, after we learn more about the
project network diagrams.
0
e
5
-1
f
2
Reporting rules
• Avoid recrimination (‘past is dead’). Take steps
to avoid a recurrence of failure not to create a
fuss.
• Progress should be reported in the form:
– not complete
• How much time is required to complete the activity?
– complete
• PNT helps to identify problematic areas and
distinguishes areas of authority but not
remove any responsibility.
Forecasting with PNT
• It enables predictions of resultant actions to
be deduced from present or past actions.
Forecasting example with PNT
•
•
Planned total project time (PTPT) of 1000 days.
After 400 days (ATWP) the planned time for the
work performed (PTWP) is 350 days.
a) What is the estimated time to complete (ETC)
the project?
b) What is the estimated time slip for the whole
project?
c) Calculate the schedule performance index (SPI).
Solution
• Planned time to complete (PTC):
PTPT – PTWP = 1000 - 350 = 650 days
• From the past activities we can calculate how these days
have to be recalculated if the performance is not changing:
ATWP/PTWP = 400/350=8/7
• Current estimated time to complete (ETC): 650*(8/7)=743
• Current estimated TPT: 400+743=1143
• Total current estimated time slip: 1000-1143=-143
• Current schedule performance index (SPI):
PTWP/ATWP=350/400=0.875=87.5%
Plotting the limits of performance
level
Problem solving
There is a project with a planned Total Project Time (TPT) of 120 days.
The project consist of three non-overlapping work packages with
a duration of 40-40 days each.
After 20 days only 37.5% of the first work package is completed.
Calculate the following:
• current schedule performance index (SPI)
• total current estimated time slip
• planned time to complete
• current estimated time to complete (ETC)
• current estimated Total Project Time (TPT)
Reading
• Textbook chapter 9
Thank you for listening
Seminar
Seminar problem 1
• There is a project consisting of 4 tasks:
–
–
–
–
A (duration: 3 days) starts at the 1st day
B (1 day) starts at the 4th day
C (2 days) starts at the 5th day
D (2 days) starts at the 7th day
• After 4 days the completition rates are: A (66.67%), B
(100%), others (0%)
• Identify/Calculate the following: PTPT, ATWP, PTWP,
SPI, PTC, ETC, ETPT, expected time slip
• Plot the original and the re-analyzed Gantt charts.
Solution
•
•
•
•
•
•
•
•
•
Plot the Gantt (done in the seminar).
PTPT = 8 days
ATWP = 4 days
PTWP = work units done expressed in days = 3 days
SPI = (work done)/(work scheduled) = ¾ = 75%
PTC = PTPT – PTWP = 5 days
ETC = PTC/SPI = 6.67 days
ETPT = PTPT/SPI = 10.67 days
expected time slip = PTPT – ETPT = -2.67 days
Seminar problem 2
• There is a project consisting of 5 tasks:
–
–
–
–
–
A (duration: 2 days) starts at the 1st day
B (3 days) starts at the 3rd day (completing A is a prerequisite)
C (2 days) starts at the 6th day (completing B is a prerequisite)
D (1 day) starts at the 8th day (completing C is a prerequisite)
E (2 days) starts at the 9th day (completing D is a prerequisite)
• After 5 days the completition rates are: A (100%), B (100%),
C (50%), others (0%)
• Identify/Calculate the following: PTPT, ATWP, PTWP, SPI,
PTC, ETC, ETPT, expected time slip
• Plot the original and the re-analyzed Gantt & PNT charts.
Solution
•
•
•
•
•
•
•
•
•
Plot the charts (done in the seminar).
PTPT = 10 days
ATWP = 5 days
PTWP = work units done expressed in days = 6 days
SPI = 6/5 = 120%
PTC = PTPT – PTWP = 4 days
ETC = PTC/SPI = 3.33 days
ETPT = PTPT/SPI = 8.33 days
expected time slip = PTPT – ETPT = +1.67 days
Seminar problem 3
• There is a project consisting of 6 tasks:
–
–
–
–
–
–
A (duration: 1 day) starts at the 1st day
B (2 days) starts at the 1st day
C (3 days) starts at the 3rd day (A&B are prerequisites)
D (3 days) starts at the 3rd day (A&B are prerequisites)
E (3 days) starts at the 6th day (C&D are prerequisites)
F (1 day) starts at the 9th day (E is a prerequisite)
• After 5 days the completition rates are: A (100%),
B (100%), C (33.33%), others (0%)
• Plot: PNT, Gantt (before and after)
• Identify/Calculate the following: PTPT, ATWP, PTWP,
SPI, PTC, ETC, ETPT, expected time slip
Solution
•
•
•
•
•
•
•
•
•
Plot the Gantt chart (solved in the class)
PTPT = 9 days
ATWP = the number of calendar days spent on the project = 5 days
SPI = (work units completed)/(work units planned during the 5 days)=
= PTWP/ATWP = 4/9 ≈ 44.44%
PTWP = how many calendar days should it have taken to complete
the work units completed during the ATWP =
= ATWP(SPI) =
= (average planned duration of a work unit)(work units performed) =
= (4/9)(5) = 20/9 ≈ 2.22 days of work with 100% schedule
performance
PTC = PTPT – PTWP ≈ 6.78 days
ETC = PTC/SPI = 15.25 days
ETPT = PTPT/SPI = ATWP + ETC = -20.25 days
expected time slip = PTPT – ETPT = -11.25 days
Seminar problem 4 (’treasure map’)
• The project is the following:
1st a treasure map should be decoded (activity a, 2 weeks). After it is
done, the location of the treasure is identified, thus we can hire the
staff (activity b, 2 weeks), buy the equipment (c, 1 week) and arrange
transportation (d, 1 week). These tasks can be performed
independently of each other. The next step is travelling to the site (e,
1 week). Last part is the excavation (f, 4 weeks).
• Plot the PNT diagram. Identify the paths, the critical path(s), the
critical activities, the TPT, and the activity floats.
• Plot the Gantt chart.
• After 3 weeks, a is ready, c is ready, but nothing else.
• Identify/calculate PTPT, ATWP, SPI, PTWP, PTC, ETC, ETPT, expected
time slip.
• Re-draw the Gantt and the PNT charts.
Solution
•
•
•
•
•
•
•
•
•
Plots are done in the seminar.
PTPT = 9 weeks
ATWP = 3 weeks
SPI = 3/5 = 60%
PTWP = 1.8 weeks (= 12.8 days)
PTC = PTPT – PTWP = 7.2 weeks
ETC = PTC/SPI = 12 weeks
ETPT = PTPT/SPI = 15 weeks
expected time slip = PTPT – ETPT = -6 weeks