network techniques for project management meaning of project & pm basic network techniques
TRANSCRIPT
Introduction to Project Management
• A temporary endeavour undertaken to create a unique
product or service (PMI, 2004, P.5)
• Variety and unique in nature
• Mainly used for large R& D activities
• Project management mainly developed by military
Two basic Network techniques
• PERT an acronym for Program Evaluation Review Technique• Originally developed to facilitate the planning and
scheduling the Polaris Fleet Ballistic Missile project of US• PERT is Probabilistic• CPM an acronym for Critical Path Method• developed in 1957 by the DuPont company in US to solve
scheduling problems in industrial setting.• CPM is concerned with the trade off between cost and time.
Definitions used in PERT CPM
• Activity: A time consuming task. An effort that require
resources and take a certain time for completion i.e.
designing a part, connecting a bridge girder
• Event: A milestone. Is a specific point in time indicating
beginning or ending of one or more activities. Does not
consume time or resources.
Rules for Network Construction
• Each activity must have a preceding and a succeeding
event
• Each event should have a distinct number
• There should be no loops in project network
• Each activity is represented by a uniquely numbered
arrow.
Example of PERTActivity Optimist
icMost likely
Pessimistic
Te=ta+4tm+tb/6
VarianceV=(b-a)/6
1-2 9 12 21 13 2
1-3 6 12 18 12 2
2-4 1 1.5 5 2 1
3-4 4 8.5 10 8 1
2-5 10 14 24 15 2.33
4-5 1 2 3 2 0.33i. Draw the network diagram ii. Determine critical pathiii. Calculate event slack and activity floativ. Find the standard deviation of CP durationv. Compute the probability of completing the project in 25
weeks
PERT Analysis
Uses 3 time estimates for each activityOptimistic 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)
PERT Analysis
Expected activity time (t)t = (a + 4m + b)
6Variance = [ (b – a) / 6 ]2
Standard deviation = SQRT(variance) = (b – a) 6
Project Variance and STD Deviation
• Project variance (σp2)
= ∑ (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
PERT Analysis
Z = (Target time – expected time)σp
Z = (16 - 15) = 0.57 1.76
This means 16 weeks is 0.57 standard deviations above the mean of 15 weeks.
CPM
• Cost -Time Relationship
• Expediting projects
• Expediting activities requires additional resources,
which means increasing the cost of the project.
• Slope gives us the cost increase associated with a
reduction of one unit of the activity duration
• Slope =Cc-Cn/Tn-Tc
Example 1
i. Find the all normal schedule and costii. Find the all-crash schedule and costiii.Find the least cost plan for all crash time scheduleiv.Find the least cost plan for an intermediate time schedule of 22
weeks.
Activity Activity Normal Crash Cost SlopeCc-Nc/Tn-Tc
week Cost Week
Cost
A 0-1 5 100 4 140 40
B 0-2 9 200 7 300 50
C 1-2 7 250 4 340 30
D 1-3 9 280 7 340 30
E 2-4 5 250 2 460 70
F 2-5 11 400 7 720 80
G 3-5 6 300 4 420 60
I 4-5 8 80 6 140 30
CPM Contd..all normal solution
• 0-1-2-4-5 is the critical path• Duration=5+7+5+8=25 weeks.• Cost=1860
00
11
44
22
33
55
A5
E5
C7
B9
D9
F11
G6H0
I8
CPM Contd.. All crash solution
• 0-1-3-4-5 is the critical path • Duration is 17 days• Cost=2860
0
1
4
2
3
5
A4
E2C4
B7
D7
F7
G4H0
I6
Find the minimum cost for the Crash time
• 0-1-3-4-5 is the critical path • Duration is 17 days• Cost=2860
0
1
4
2
3
5
A4
E2C4
B7
D7
F7
G4H0
I6
Find the minimum cost for the Crash time
• Step 1- Get the non critical activities with their slope• Step-2 Select the activity which has largest slope & expand as much as
possible to achieve as a large a cost reduction is possible• Step-3 Expand the next largest activity slope and continue until all path
become critical.
0
1
4
2
3
5
A 4
E 2
C 4
B 7
D 7
F 7
G 4H 0
I 6
Find the minimum cost for the Crash time
• Step 1- Get the non critical activities with their slope =B-50 C-30,E-70,G-60,F-80• Step-2 Select the activity which has largest slope & expand as much as possible to
achieve as a large a cost reduction is possible= Select F for 2 days expansion cost 160
• Step-3 Expand the next largest activity slope and continue until all path become critical.=E one day is possible at$70 in similar manner G by 2 days cost 120 finally activity B can be expanded by one day “ $50 saving.
• So total cost savings are :160+70+120+50=400.Thus the revised cost for 17 days are 2860-400=$2460
00
11
44
22
33
55
A4
E2C4
B7
D7
F7
G4H0
I6
Determine the least cost plan for any desired number of days ie Least cost plan
for 22 days• First step is to list all critical activities and their slope
• The activity with the least slope will be compressed first, because decreasing the
project time by one day will result in smallest increase in cost.in this case activity C
& I can be selected, arbitrarily C is selected but how many days -the most is up to
is crash time 4 days here. However such a reduction may create one or more
additional critical path. The minute an additional CP is created, the compression
should be stopped. Thus here C is reduced from 7 days to 4 days at a cost of 90.So
the least cost plan for 22 days is 1860+90=1950
Critical activities Activities by events Slope
A 0-1 40
C 1-2 30
E 2-4 70
I 4-5 30
Example2• The following data were obtained from a study of the time
required to overhaul a small power plant.
i. Find the all normal schedule and costii. Find the all-crash schedule and costiii. Find the least cost plan for all crash time scheduleiv. Find the least cost plan for an intermediate time schedule of 17
weeks.
Activity Crash Schedule Normal Schedule
Week cost Week cost
1-2 3 6 5 4
1-3 1 5 5 3
2-4 5 7 10 4
3-4 2 6 7 4
2-6 2 5 6 3
4-6 5 9 11 6
4-5 4 6 6 3
6-7 1 4 5 2
5-7 1 5 4 2
Example2• The following data were obtained from a study of the time
required to overhaul a small power plant.
i. Find the all normal schedule and cost 1-2-6-7=16 ,1-2-4-6-7=31, 1-2-4-5-7=25,1-3-4-6-7=28,1-3-4-7=22 and cost= 4+4+6+2=16
ii. Find the least cost plan for all crash time scheduleiii.Find the least cost plan for an intermediate time schedule of 17
weeks.
Activity Crash Schedule Normal
1-2 3 6 5 4
1-3 1 5 5 3
2-4 5 7 10 4
3-4 2 6 7 4
2-6 2 5 6 3
4-6 5 9 11 6
4-5 4 6 6 3
6-7 1 4 5 2
5-7 1 5 4 2
11
33
66
22
55
77446
1110
5
7
6
5
4
5
Example2• The following data were obtained from a study of the time
required to overhaul a small power plant.
i. Find the all crash schedule and cost 1-2-6-7=6 ,1-2-4-6-7=14, 1-2-4-5-7=13,1-3-4-6-7=9,1-3-4-5-7=8 and cost= 6+7+9+4= 26
ii. Find the least cost plan for all crash time scheduleiii.Find the least cost plan for an intermediate time schedule of 17
weeks.
Activity Crash Schedule Normal
1-2 3 6 5 4
1-3 1 5 5 3
2-4 5 7 10 4
3-4 2 6 7 4
2-6 2 5 6 3
4-6 5 9 11 6
4-5 4 6 6 3
6-7 1 4 5 2
5-7 1 5 4 2
11
33
66
22
55
77444
55
1
2
2
3
1
1
Example2The following data were obtained from a study of the time
required to overhaul a small power plant.
i. Find the least cost plan for all crash time schedule.ii. Find the least cost plan for an intermediate time schedule of
17 weeks.
Activity Crash Schedule
Normal Schedule
SlopeCc-nc/nt-ct
1-2 3 6 5 4 1
1-3 1 5 5 3 0.5
2-4 5 7 10 4 0.6
3-4 2 6 7 4 0.4
2-6 2 5 6 3 0.5
4-6 5 9 11 6 0.5
4-5 4 6 6 3 1.5
6-7 1 4 5 2 0.5
5-7 1 5 4 2 1
11
33
6622
55
77446
1110
5
7
6
5
4
5