Analysing the AoA network Project Management

Analysing the AoN network Project Management

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

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

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).

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

Example: organising a conference Objectives: to organise a 3 days long open scientific conference with 100-200 participants, 30-50 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.

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

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

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

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 2 9 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

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

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.

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

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 1 b 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

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).

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

‘Crashing’ – reducing task durations by increased costs

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

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.

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

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.

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

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.

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.

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

Reading Lockyer – Gordon (2005) Chapter 8 pp. 61-63. & Chapter 14

Thanks for the attention!