chapter 7 project management - welcome to open...
TRANSCRIPT
Learning Objectives Understand how to plan, monitor, and control
projects using PERT/CPM. Determine earliest start, earliest finish, latest
start, latest finish, and slack times for each activity.
Identify critical activities and critical path. Understand impact of variability in activity
times.
Project Management
Used to manage large complex projects
Has three phases: 1. Project planning 2. Project scheduling 3. Project controlling
Phase 1: Project Planning
1. What is the project goal or objective? 2. What are the activities (or tasks) involved? 3. How are activities linked? 4. How much time required for each activity? 5. What resources are required for each
activity?
Phase 2: Project Scheduling 1. When will the entire project be completed? 2. What is the scheduled start and end time for
each activity? 3. Which are the “critical” activities? 4. Which are the noncritical activities? 5. How late can noncritical activities be without
delaying the project? 6. After accounting for uncertainty, what is the
probability of completing the project by a specific deadline?
Project Controlling Control of large projects involves close
monitoring of schedules, resources, and budgets. Questions to be answered:
At any particular date or time, is project on schedule, behind schedule, or ahead of schedule?
At any particular date or time, is money spent on project equal to, less than, or greater than budgeted amount?
Are there enough resources available to finish project on time?
If project is to be finished in shorter amount of time, what is best way to accomplish this at least cost?
7.2 Project Networks
Once project mission or goal has been clearly specified, first issues to address deal with project planning.
Identify activities that constitute project, precedence relationships between these activities, and time and other resources required for each activity.
These activities and precedence relationships are demonstrated graphically in an activity network
Identifying Activities
Subdivides a large project into smaller units Each activity should have a clearly defined
starting point and ending point Each activity is clearly distinguishable from
every other activity Each activity can be a project in itself
Work Breakdown Structure (WBS) Divides the project into its various
subcomponents and defines hierarchical levels of detail
Level 1 Project 2 Major tasks in project 3 Subtasks in major tasks 4 Activities to be completed
Identify for Each Activity:
Which other activities must be completed previously (predecessors)
Time required for completion Resources required
This completes the project planning phase.
Project Scheduling Phase
Commonly used techniques:
Programme Evaluation and Review Technique (PERT)
Critical Path Method (CPM)
Project Management Example: General Foundry Inc.
Has 16 weeks to install a complex air filter system on its smokestack
May be forced to close if not completed within 16 weeks due to environmental regulations
Has identified 8 activities
Drawing the Project Network
AON – Activity on Node networks show each activity as a node and arcs show the immediate predecessor activities
AOA – Activity on Arc networks show each activity as an arc, and the nodes represent the starting and ending points
We will use the AON method
AON Network for General Foundry Both A and B have no predecessors; thus both start the project and we need a dummy start.
If there was only one activity without an immediate predecessor, it would be the only start and we would not need a dummy start.
Dummy Start
D comes after BOTH A and B so it is drawn next with 2 arrows; one from each predecessor activity preceding it.
G comes after both D and E so it is drawn next with 2 arrows; one from each predecessor preceding it.
H comes after both F and G so it is drawn next with 2 arrows. It is the ONLY terminal activity, so there is no dummy.
Determining the Project Schedule Some activities can be done simultaneously so
project duration should be less than 25 weeks (sum of times of all activities = 25)
Critical path analysis is used to determine project duration
The critical path is the longest time path through the network and the SHORTEST time for project completion.
Once project network has been drawn to show all activities and precedence relationships, determine project schedule. This includes the earliest time an activity can start
(ES) and finish (EF) without violating the precedence relationships. All precedence relationships must be honoured.
After we have determined ALL the ES and EF, we must now determine the latest time an activity can start LS, and finish LF, without delaying the project beyond its earliest possible finish.
We then determine the slack for each activity
7.3 Determining The Project Schedule
Critical Path Time Computations To find the critical path, calculate two distinct starting and
ending times for each activity: ES = Earliest time at which an activity can start,
assuming all predecessors have been completed.
EF = Earliest time at which an activity can be finished.
LS = Latest time at which an activity can start so as to not delay completion time of entire
project. LF = Latest time by which an activity has to finish so as to not delay the completion time of
entire project.
Forward Pass Identifies earliest times (ES and EF)
ES Rule: All immediate predecessors must be done before an activity can begin
If only 1 immediate predecessor, then ES = EF of predecessor
If >1 immediate predecessors, then ES = Max {all predecessor EF’s}
EF = ES + activity time.
Forward Pass: Earliest Start & Finish Times
In doing the forward pass, we go from start to finish: go with the flow of arrows
ES and EF only; ES before EF ES of an activity = Max {EF of all its
immediate predecessors} EF = ES + time
ES EF
LS LF
Activity Time Node Notation:
Forward Pass: Earliest Start & Finish Times
ES of both A and B we set at 0 for both activities start the project.
EFA = 2; and EFB = 3 ESC = 2 which = EFA. EFC = 5 ESD = max {EFA,EFB} = max {2, 3} = 3. EFD
= 7 ESE and ESF = EFC = 4. EFE = 8; EFE = 7 ESG = max {EFD,EFE} = 8. EFD = 13 ESH = max {EFF,EFG} = 13. EFD = 15
Backward Pass Identifies latest times (LS and LF) LF Rule (we do not delay the project):
If an activity is an immediate predecessor for a single activity, its LF equals the LS of the activity that immediately follows it, then
LF = LS of immediate follower If activity is the immediate predecessor
to more than one activity, then LF = Min {LS of all immediate followers} LS = LF – activity time
Backward Pass: Latest Start & Finish Times
LFH = 15. Only terminal activity. LSH = 13 LFF = 13; LFG = 13; LSF = 10; LSG = 8 LFD = 8; LFE = 8; LSD = 4; LSE = 4 LFC = min {LSE, LSF} = min {4, 10} = 4. LSC
= 2 LFB = LSD = 4; LSB = 1 LFA = min {LSC, LSD} = min {2, 4} = 2. LSA =
0 LFSTART = min {LSA, LSB} = 0 = LSSTART
Slack Time and Critical Path(s) Slack is the length of time an activity can
be delayed without delaying the project Slack = LS – ES or LF – EF
Activities with 0 slack are Critical Activities The Critical Path is a continuous path
through the network from start to finish that include only critical activities. There are NO gaps or breaks between activities.
Total Slack Time vs. Free Slack Time Total slack time is shared by more than 1
activity Example: A 1 week delay in activity B will leave 0 slack for activity D
Free slack time is associated with only 1 activity Example: Activity F has 6 week of free slack time
Variability in Activity Times
Activity times are usually estimates that are subject to uncertainty
Approaches to variability: 1. Build “buffers” into activity times 2. PERT – probability based 3. Computer simulation
PERT Analysis Uses 3 time estimates for each activity
Optimistic time (a) Pessimistic time (b) Most likely time (m)
These estimates are used to calculate an expected value and variance for each activity (based on the Beta distribution)
Project Variance and Standard Deviation Project variance (σp
2) is the sum of the variances of the CRITICAL activities only = ∑ (variances of all critical path activities) σp
2 = 0.11 + 0.11 + 1.0 + 1.78 + 0.11 = 3.11
Project standard deviation (σp) = SQRT (Project variance) σp = SQRT ( 3.11) = 1.76
Probability of Project Completion What is the probability of finishing the
project within 16 weeks? Assumptions:
Project duration is normally distributed Activity times are independent
Normal distribution parameters: µp = expected completion time= 15 weeks σp = proj standard deviation = 1.76 weeks
Normal Probability Calculations
This means 16 weeks is 0.57 standard deviations above the mean of 15 weeks.
Z(0.57) = 0.2157. Add 0.5 to get 0.7157
Prob (proj completion < 16 weeks) = 0.7157
Project Duration for a Given Probability • What time must be set to ensure a 99%
chance of completion within that time?
• This is the inverse transformation.
• The Z value that corresponds to an area of 0.49 is 2.33
• Project duration = σZ + µ (σ = 1.76; Μ = 15) • Time = 1.76*2.33 + 15 • Time = 19.1008 weeks
Consider The Following Projects Draw The Project Networks Determine the Expected Times and Variances Determine the activity schedule (ES, EF, LS, LF, Slack) Identify the critical activities State the critical paths What times must be set such that there is a guaranteed
90% (and 99%) time of completion? What is the probability that project 1 will be completed
within 20 (and also 24) days? What is the probability that project 2 will be done
within 70 (and also between 72 and 80; also between 60 and 75) days?
Activity Immediate
Predecessors a m b A -- 2 4 7 B A 3 6 10 C A 1 2 3 D B 2 4 6 E B 2 3 5 F B, C 3 4 6 G B 1 2 4 H F 2 3 4 I F, H 1 1.5 2 J I 2 3 4 K D, E 4 7 10
PRO
JECT 1
Activity Immediate
Predecessors a m b A – 8 10 12 B – 6 7 9 C – 3 3 4 D A 10 20 30 E C 6 7 8 F B, D, E 9 10 11 G B, D, E 6 7 10 H F 14 15 16 I F 10 11 13 J G, H 6 7 8 K I, J 4 7 8 L G, H 1 2 4
PRO
JECT 2
PRO
JECT 4
Activity Immediate Predecessors
(a) (m) (b) A – 3 6 8 B A 5 8 10 C A 5 6 8 D B, C 1 2 4 E D 7 11 17 F D 7 9 12 G D 6 8 9 H F, G 3 4 7 I E, F, H 3 5 7
PRO
JECT 5
Activity Predecessors Normal Time A 5 B 6 C A 4 D A, B 3 E B 5 F C, D 2 G E 4 H E, G 2 I F, H 3 J G 2 K I, J 5
Activity on Node (AON) Network • Arcs represent precedence relationships between
activities. • There is only one activity (A) that does not have any
predecessor. Begin immediately, no dummy start. • Draw one node for A. Also the start of the project
Project 1
A
• Activities B and C have activity A as their only predecessor. Draw two arrows from A to B and C.
A B
C
Activity on Node (AON) Network • Activities D, E, F and G have activity B as their
predecessor. Draw arrows from B to all of these. F also has C has a predecessor so draw an arrow from C to F. Notice where we put F so C can be clearly attached (arrows CAN cross).
Project 1
A B
C
D
E
G
F
Activity on Node (AON) Network • Both activities H and I come after activity F. Activity
I also comes after activity H. Draw arrows from F to H and I and draw an arrow from H (newly drawn) to I
Project 1
A B
C
D
E
G
F
H
I
Activity on Node (AON) Network • Activity J comes after activity I. Draw an arrow from
both I to J.
Project 1
A B
C
D
E
G
F
H
I
J
Activity on Node (AON) Network
• Activities D and E precede Activity K. Draw arrows from both D and E to K.
Project 1
K
A B
C
D
E
G
F
H
I
J
Activity on Node (AON) Network • All activities have been drawn, but we must clearly
identify the finish of the project. Notice that activities G, J and K have no successor activities, therefore they are all terminal (finish) activities. All these must be represented at the finish (dummy)
Project 1
A B
C
D
E
G
F
H
I
K
J
Finish