1. Analysing the AoA network
Project Management

2. Analysing the AoN network
Project Management

3. Data on the activity node
Activity label & description
Total Float
EST
EFT
LST
Duration
LFT

4. Total Project Time
The shortest time in which the project can be completed. Determined by the critical path.
Calculation: forward pass
Forward pass: The earliest start times (EST) of all activities are calculated.
Trom these the earliest finishing times (EFT) are also calculated

5. Critical path
sequence of activities that has no float time (or has the maximum negative float time in absolute value), and that determines the duration of the project. It is the longest path. Activities on the critical path are the critical activities.
The critical path can be identified by a backward pass, calculating the Latest Finishing Times (LFT), and from these the Latest Starting Times (LST).

6. Floats in AoN
Total float: the time by which an activity can be delayed or extended without affecting the TPT. It can be used to delay the start of an activity or to increase its duration. TF = LFT – EST – Duration
Free float: the time by which an activity can be delayed or extended without affecting the start of any succeding activity. FF = ESTj+1 - EFTj

7. Example: organising a conference
Objectives: to organise a 3 days long open scientific conference with participants, lectures, buffet reception, a conference book of the best studies and TV and radio interviews with some of the most known lecturers.
Create the WBS chart and create the task list with estimated durations and precedence relations (in a table form)
Plot both the AoA and the AoN diagram
Calculate the TPT, identify the critical path, the total, and the free float times.

8. Organising participants Invitation and marketing Organising interviews
Example: WBS
Project
Book
Event management
Marketing
Editing
Publishing
Infra-structure
Organising participants
Arranging event
Invitation and marketing
Organising interviews
Collecting articles
Peer reviewing
Facilities
Staff
Materials

9. Task list with precedence relations
Activity label
Task description
Duration (weeks)
Immediate predecessors a
Invitation 2 – b
Organising participants 4 c
Facilities 3 d
Staffing e
Materials f
Collecting articles 6 g
Peer reviewing h
Organising interviews 1
c, d, e i
Publishing 5 j
Arranging event
h, i
Activity label
Task description
Duration (weeks)
Immediate predecessors a
Invitation 2 b
Organising participants 4 c
Facilities 3 d
Staffing e
Materials h
Collecting articles 6 j
Peer reviewing k
Publishing 5 l
Organising interviews 1 m
Arranging event
Activity label
Task description
Duration
Immediate predecessors a
Invitation b
Organising participants c
Facilities d
Staffing e
Materials h
Collecting articles j
Peer reviewing k
Publishing l
Organising interviews m
Arranging event

10. AoA 4 9 19 3 c 5 10 19
4 d 1 2a 2 3 6 6 10 19 4b 9 20 3e 1h 10 21 1j 6 5 f
TPT = 21 i 7 12 8 15 3g
CP: a-b-f-g-i-j

11. AoN TPT: 21 CP: a-b-f-g-i-j c a b d h j e i f g 1 6 9 10 16 19 3 9 2 29 a 2 b 2 6 4 d 6 10 15 19 4 h 10 11 9 19 20 1 j 20 21 1 1 e 6 9 10 16 19 3 i 15 20 5 f 6 12 g 12 15 3
TPT: 21
CP: a-b-f-g-i-j

12. Activity times for the previous diagram (finalize individually)
Duration
EST
LST
EFT
LFT
Total float
Free float a 2 b 4 6 c 3 9 16 19 10 1 d e f g h i j

13. Example 2 (for individual practice)
a) Draw the AoA and AoN diagram with the data below:
Activity label
Duration (weeks)
Immediate predecessors a 1 – b 2 c 5 d 3 e f g 4
f, c, d
b) Determine the TPT and the critical path and activity floats.
c) Compute the EETs, LETs and slack for every node in the AoA & ESTs, LSTs, EFTs, LFTs in the AoN diagram.

14. Solution: AoA 3 5 1 2 6 7 4 2 b 2 e 2 f 1 7 1 a 5 c 4 g 7 3 d TPT: 11
Float: 0 3 5
Float: 0 2 b 2 e 2 f
Float: 0
Slack: 0
Slack: 0 1 2 1 6 7 7 11 1 a 5 c 4 g
T. float: 1
F. float: 1
Float: 0
Float: 0
Slack: 0
Slack: 0
Slack: 0
Slack: 0 4 7 3 d
T. float: 3
F. float: 3
Slack: 3
TPT: 11
CP: a-b-e-f-g

15. Solution: AoN TPT: 11 CP: a-b-e-f-g b e f g a c d 1 3 2 3 5 2 5 7 2 1b 1 3 2 e 3 5 2 f 5 7 2 1 g 7 11 4 a 1 c 1 6 2 7 5 3 d 1 4 3 7
TPT: 11
CP: a-b-e-f-g

16. Example 3 3 2 b 5 3 e 5 1 2 2 a 2 5 c 5 7 8 1 f 14 2 8 14 4 2 4 g 6 3 d 12 2 h 5 12
Calculate the EETs and LETs.
Create a precedence table (with task, duration, immediate predecessor, total and free floats).

17. Solution Activity Duration Total float Free float a 2 b c 5 1 d 3 e fb c 5 1 d 3 e f g 4 h

18. ‘Crashing’ – reducing task durations by increased costs

19. Definition of crashing
Obtaining reduction in time at an increased cost (increasing the employed resources).
Cost-slope: the cost of reducing duration time by a unit of time.
Let’s see the following example: a 4 b 2 c e d 5 f 3 c 6 8 1 7 9 2 e 8 10 1 9 11 2 a 4 b 4 6 2 f 11 14 3 d 6 11 5

20. Procedure for crashing
Crash one time unit at a time
Only the crashing of critical activities has any effect on TPT
Crash that activity first that is the cheapest to reduce in time
Be aware of multiple critical paths
Stop crashing when:
the crash-time is reached at every ‘crashable’ activity,
benefits of possible crashing are lower than crashing costs.

21. Crashing table
If the costs to reduce times are known, then a table can be set up showing the relative costs for the reduction in time of each activity by a constant amount.
Crash-time is the minimum duration of an activity. It is given by technical factors.
Activity
(label)
Duration
(day)
Float (day)
Crash time
Cost-slope (€/day) a 4 2
100 b
150 c 1
110 d 5 3
200 e
160 f
500
Crash time is the minimum duration of an activity (one cannot crash the duration of an activity below the crash time).
Benefit of reducing TPT by one day:
400 €/day

22. Solution method step: identify the critical activities
step: find the critical activity with cheapest crash cost, and if its cost slope is lower than the daily benefit from crashing, reduce its duration with one day. If there is no activity to crash, or it is too costly, stop crashing and go to step 4.
step: reidentify the critical path, and go back to step two.
step: identify the final critical path(s), TPT and the total net benefit of crashing.

23. Path / activity crashed Path / activity crached
Solution
Path durations
Path / activity crashed
normal
step 1
step 2
step 3
step 4
step 5 – a d
d, c
none
a-b-c-e-f 13 12 11 10
a-b-d-f 14
Cost:
100
200
310
Cumulated net benefit:
300
600
800
890
Path durations
Path / activity crached
normal
step 1
step 2
step 3
step 4
step 5
Cost:
Cumulated net benefit:
After crashing:
there are two critical paths
TPT is 10 days
total benefit of crashing is €890

24. Example 2 (for individual work)b 2 d 2 e 5 a 3 g 3 7 c 3 f 3
Identify the critical path and the TPT.

25. Example 2 (for individual work)b 3 5 2 d 5 7 2 e 7 12 5 a 3 g 12 15 3 7 c 3 6 1 4 7 f 6 9 3 12
Critcal: a-b-d-e-g TPT: 15
Using tbe table on the next slide, calculate the optimal TPT with crashing.

26. What is the total profit on crashing? 10 days €3000
Activity
(label)
Normal duration (day)
Float (day)
Crash time
Cost-slope (€/day) a 3 1
500 b 2
550 c
150 d 5
900 e 4
400 f
100 g
200
Activity
(label)
Normal duration (day)
Float (day)
Crash time
Cost-slope (€/day) a 3 1
500 b 2
550 c
150 d 5
900 e 4
400 f
100 g
200
Benefit of reducing TPT by one day:
1200 €/day
What is the new TPT?
What is the total profit on crashing?
10 days
€3000

27. Reading
Lockyer – Gordon (2005) Chapter 8 pp & Chapter 14

28. Thanks for the attention!