1. Project Scheduling: PERT-CPM

2. PERT (Program evaluation and review technique) and CPM (Critical Path Method) makes a managerial technique to help planning and displaying the coordination of all the activities.

3. Activity Description Immediate Predecessors Estimated Time ActivityB C D E F G H I J K L M N
Excavate
Lay the foundation
Put up the rough wall
Put up the roof
Install the exterior plumbing
Install the interior plumbing
Put up the exterior siding
Do the exterior painting
Do the electrical work
Put up the wallboard
Install the flooring
Do the interior painting
Install the exterior fixtures
Install the interior fixtures - A B C E D
E,G
F,I J H
K,L
2 weeks
4 weeks
10 weeks
6 weeks
5 weeks
7 weeks
9 weeks
8 weeks

4. Immediate predecessors:
For any given activity, its immediate predecessors are the activities that must be completed by no later than the starting time of the given activity.

5. AOA (Activity-on-Arc):
Each activity is represented by an arc.
The arcs are used to show the precedence relationships between the activities.

6. START
(Estimated)
Time 2 A
arc
node B 4 C 10 D I 7 6 4 E 5 7 F G J 8 H K L 9 4 5 M 2 N 6
FINISH

7. Path and Length START A B C D G H M FINISH
= 40 weeks
START A B C E F J K N FINISH
= 43 weeks
START A B C E F J L N FINISH
= 44 weeks
Critical Path

8. Critical Path:
A project time equals the length of the longest path through a project network. The longest path is called “critical path”.
Activities on a critical path are the critical bottleneck activities where any delay in their completion must be avoided to prevent delaying project completion.

9. ES :
Earliest Start time for a particular activity
EF :
Earliest Finish time for a particular activity

10. START 2
ES=0
EF=2 A B 4
ES=2
EF=6 C 10
ES=6
EF=16 D I 7 6
ES=16
EF=23
ES=16
EF=22 4
ES=16
EF=20 E 7 5
ES=22
EF=29
ES=20
EF=25 G F J 8 H 9 K 4 L 5 M 2 N 6
FINISH

11. Earliest Start Time Rule:
If an activity has only a single immediate predecessor, then ES = EF for the immediate predecessor.
Earliest Start Time Rule:
The earliest start time of an activity is equal to the largest of the earliest finish times of its immediate predecessors.
ES = largest EF of the immediate predecessors.

12. START
ES=0
EF=2 2 A B 4
ES=2
EF=6 C 10
ES=6
EF=1 D I 7 6
ES=16
EF=23
ES=16
EF=22 4
ES=16
EF=20 E 7 5
ES=22
EF=29
ES=20
EF=25 G F J 8
ES=25
EF=33 H 9
ES=29
EF=38 K 4
ES=33
EF=37 L 5
ES=33
EF=38 M 2
ES=38
EF=40 N 6
ES=38
EF=44
FINISH
ES=44
EF=44

13. Latest Finish Time Rule:
LS:
Latest Start time for a particular activity
LF:
Latest Finish time for a particular activity
Latest Finish Time Rule:
The latest finish time of an activity is equal to the smallest of the latest finish times of its immediate successors.
LF = the smallest LS of immediate successors.

14. START
LS=0
LF=0 2
LS=0
LF=2 A B 4
LS=2
LF=6 C 10
LS=6
LF=16 D I 7 6 4
LS=18
LF=25
LS=20
LF=26
LS=16
LF=20 E 7 5
LS=26
LF=33
LS=20
LF=25 G F J 8
LS=25
LF=33 H 9
LS=33
LF=42 K 4
LS=34
LF=38 L 5
LS=33
LF=38 M 2
LS=42
LF=44 N 6
LS=38
LF=44
FINISH
LS=44
LF=44

15. Earliest
Start Time
Latest
Start Time
S=( 2, 2 )
F=( 6, 6 )
Earliest
Finish Time
Latest
Finish Time

16. START
S=(0,0)
F=(0,0) 2
S=(0,0)
F=(2,2) A
Critical Path B 4
S=(2,2)
F=(6,6) C 10
S=(6,6)
F=(16,16) D I 7
S=(16,18)
F=(23,25) 4
S=(16,16)
F=(20,20) 6
S=(16,20)
F=(22,26) E 7 5
S=(22,26)
F=(29,33)
S=(20,20)
F=(25,25) G F 8
S=(25,25)
F=(33,33) J H 9
S=(29,33)
F=(38,42) K 4
S=(33,34)
F=(37,38) L 5
S=(33,33)
F=(38,38) M 2
S=(38,42)
F=(40,44) N 6
S=(38,38)
F=(44,44)
FINISH
S=(44,44)
F=(44,44)

17. Slack:
A difference between the latest finish time and the earliest finish time.
Slack = LF - EF
Each activity with zero slack is on a critical path.
Any delay along this path delays a whole project completion.

18. Three-Estimates Most likely Estimate (m)
= an estimate of the most likely value of time.
Optimistic Estimate (o)
= an estimate of time under the most favorable
conditions.
Pessimistic Estimate (p)
= an estimate of time under the most
unfavorable conditions.

19. Beta distribution o p m
Mean :
Variance:

20. Mean critical path:
A path through the project network becomes the critical path if each activity time equals its mean.
Activity OE M PE
Mean
Variance 1 2 3 2 A B C 2 8 4 1 6 9 18 10 4
OE: Optimistic Estimate
M : Most Likely Estimate
PE: Pessimistic Estimate

21. Activities on Mean Critical Path
Variance A B C E F J L N 2 4 10 5 8 6 1 4
Project Time

22. Approximating Probability of Meeting Deadline
Assumption:
A probability distribution of project time is a normal distribution.
T = a project time has a normal distribution
with mean and ,
d = a deadline for the project = 47 weeks.

23. Using a table for a standard normal distribution,
the probability of meeting the deadline is
P ( T d ) = P ( standard normal )
= 1 - P( standard normal ) =
0.84.

24. Time - Cost Trade - Offs Activity cost Crash Crash cost Normal
Crashing an activity refers to taking special costly measures to reduce the time of an activity below its normal value.
Activity
cost
Crash
Crash cost
Normal
Normal cost
Crash
time
Normal
time
Activity
time

25. Activity J: Normal point: time = 8 weeks, cost = $430,000.
Crash point: time = 6 weeks, cost = $490,000.
Maximum reduction in time = = 2 weeks.
Crash cost per week saved =
= $30,000.

26. Maximum
Reduction
in Time
(week)
Time
(week)
Cost
($1,000)
Crash Cost
per Week
Saved
Activity N C N C A B J 2 4 8 1 2 6
$180
$320
$430
$280
$420
$490 1 2
$100
$ 50
$ 30
N: Normal C: Crash

27. Using LP to Make Crashing Decisions
Let Z be the total cost of crashing activities.
A problem is to minimize Z, subject to the constraint that its project duration must be less than or equal to the time desired by a project manager.

28. = the reduction in the time of activity j
by crashing it
= the project time for the FINISH node

29. = the start time of activity j
Duration of activity j = its normal time
Immediate predecessor of activity F:
Activity E, which has duration =
Relationship between these activities:

30. Immediate predecessor of activity J:
Activity F, which has time =
Activity I, which has time =
Relationship between these activities:

31. The Complete linear programming model
Minimize

32. One Immediate
Predecessor
Two Immediate
Predecessors
Finish Time = 40
Total Cost = $4,690,000