1. Project Management with PERT/CPM

2. PERT/CPM PERT : program evaluation and review technique
CPM : critical path method
Use a project network, Activity-on-Node (AON):
Nodes: activities, or tasks, to be performed
Arcs: show immediate predecessors to an activity
Times: duration times of activities are written next to the node

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

4. The project network for the Reliable Construction Co. project
START
Activity Code
A. Excavate
B. Foundation
C. Rough wall
D. Roof
E. Exterior plumbing
F. Interior plumbing
G. Exterior siding
H. Exterior painting
I. Electrical work
J. Wallboard
K. Flooring
L. Interior painting
M. Exterior fixtures
N. Interior fixtures A 2 B 4 C 10 I D 6 E 4 7 G F 5 7 J 8 H 9 K 4 L 5 M 2 N 6
FINISH

5. Bake a Cake Example Task Immediate predecessors Task Time
A: Buy frosting ingredients —
½ hr
B: Clean up kitchen
1 hr
C: Buy cake ingredients
D: Prepare frosting
A,B
¼ hr
E: Prepare batter & bake
B,C
2 hrs
F: Frost cake
D,E
1/2 A B D E F 1
1/4 2
Start C
Finish

6. Critical Path
A path through a project network is a route from START to FINISH.
The length of path is the sum of the task times (durations) of the nodes (activities) on the path.
The critical path is the longest path.
The project duration is the length of the longest path.

7. The paths and lengths through Reliable’s project network
START →A →B →C →D →G →H →M→FINISH
START →A →B →C →E →H →M →FINISH
START →A →B →C →E →F →J →K →N →FINISH
START →A →B →C →E →F →J →L →N →FINISH
START →A →B →C →I →J →K →N →FINISH
START →A →B →C →I →J →L →N →FINISH
=40 weeks
=31 weeks
=43 weeks
=44 weeks
=41 weeks
=42 weeks

8. To Find the Critical Path and Slacks
Work from top to bottom in the network, calculating
ES = earliest start time for an activity
EF = earliest finish time for an activity
ES for activity i = largest EF of the immediate predecessors
ES = 0 if no immediate predecessors
EF = ES + activity duration time
Work from bottom to top in the network, calculating
LS = latest start time for an activity
LF = latest finish time for an activity
LS = LF – activity duration time
LF for activity i = smallest LS of the immediate successors
LF at Finish = EF at Finish if no immediate successors
Slack = LF - EF = LS - ES
If slack is zero, the activity is on the critical path.

9. The complete project network showing ES, LS, EF and LF for each activity of the baking example
S = (ES, LS)
F = (EF, LF)
Slack = LS – ES = LF - EF
S=( ) F=( ) Slack=
Start
S=( ) F=( ) Slack=
S=( ) F=( ) Slack= B 1 A
S=( ) F=( ) Slack= C
S=( ) F=( ) Slack= D E 2
S=( ) F=( ) Slack= F
S=( ) F=( ) Slack=
Finish
S=( ) F=( ) Slack=
Work down the network calculating ES and EF (ES at Start = 0, EF at Start = 0)
Work backward up the network calculating LS and LF (LF and LS at Finish is EF and ES at Finish)

10. The complete project network showing ES, LS, EF and LF for each activity of the baking example
S=(0, 0) F=(0, 0) Slack=0
S = (ES, LS)
F = (EF, LF)
Slack = LS – ES = LF - EF
Start
S=(0, 0) F=(1,1) Slack=0
S=(0, 2 ¼ ) F=(½, 2 ¾ ) Slack=2 ¼ B 1 A
S=(0, ½) F=(½, 1) Slack= ½ C
S=(1, 2 ¾) F=(1 ¼, 3) Slack=1 ¾ D E 2
S=(1, 1) F=(3, 3) Slack=0 F
S=(3, 3) F=(3 ½, 3 ½) Slack=0
Finish
S=(3 ½, 3 ½) F=(3 ½, 3 ½) Slack=0
Critical Path is Start →B→E →F →Finish
Activity D has slack of 1¾ hours (Start of D could be delayed without affecting total project duration)
Also, activities A and C have slack of 2 ¼ and ½ respectively.

11. The complete project network showing ES, LS, EF and LF for each activity of the Reliable Construction Co. project
S= (0, 0) F= (0, 0)
START
S = (ES, LS)
F = (EF, LF) A
S= (0, 0) F= (2, 2) 2 B 4
S= (2, 2) F= (6, 6)
S= (6, 6) F= (16, 16) C 10
S= (16,20) F= (22,26) D
S= (16, 16) F= (20, 20) 6 E 4 I
S= (16,18) F= (23,25) 7
S= (22,26) F= (29,33)
S= (20,20) F= (25,25) G 5 7 F
S= (25,25) F= (33,33) J 8
S= (29,33) F= (38,42) H 9
S= (33,34) F= (37,38) K 4 L
S= (33,33) F= (38,38) 5
S= (38,42) F= (40,44) M 2 N
S= (38,38) F= (44,44) 6
S= (44,44) F= (44,44)
FINISH

12. Slack for Reliable’s activities
Activity
Slack (LF - EF)
On Critical Path? A
Yes B C D 4 No E F G H I 2 J K 1 L M N

13. The spreadsheet used by MS project for entering the activity list for the Reliable Construction Co. project

14. Incorporate Uncertain Activity Duration Times (Probabilistic)
PERT Three-Estimate Approach
m = most likely estimate of activity duration time
o = optimistic estimate of activity duration time
p = pessimistic estimate of activity duration time
Assume Beta distribution of activity time
Approximately:

15. Expected value and variance of the duration of each activity for Reliable’s project
Optimistic
Estimate o
Most Likely Estimate m
Pessimistic Estimate p A 1 2 3
1/9 B
3 ½ 8 4 C 6 9 18 10 D
5 ½ E
4 ½ 5
4/9 F G
6 ½ 11 7 H 17 I
7 ½ J K L M N

16. The paths and path lengths through Reliable’s project network when the duration of each activity equals its pessimistic estimate
Path
Length
START →A →B →C →D →G →H →M→FINISH
START →A →B →C →E →H →M →FINISH
START →A →B →C →E →F →J →K →N →FINISH
START →A →B →C →E →F →J →L →N →FINISH
START →A →B →C →I →J →K →N →FINISH
START →A →B →C →I →J →L →N →FINISH
=70 weeks
=54 weeks
=66 weeks
=69 weeks
=60 weeks
=63 weeks

17. mean length of path µp = sum of mean activity times on the path
For a path (typically the critical path), find the mean length (time) µp and the variance σp2
mean length of path µp = sum of mean activity times on the path
(because E[X+Y] = E[X] + E[Y])
Variance of length of path σp2 = sum of variances of activity times on path
(because we assume independence:
Var [X+Y] = Var [X] + Var [Y] if X,Y independent)
Example:
critical path Start →A →B →C →E →F →J →L →N →Finish
µp = = 44
σp2 = 1/ / /9 =9

18. Find the probability the project is completed in 47 weeks
using an assumption of a Normal distribution

19. Time-Cost Trade-offs
If one can expedite the project, use money/resources to reduce task times, what is the best way to allocate money?
For an activity, could pay extra to reduce time (crash)
How much would it cost to reduce total project duration from 44 weeks to 40 weeks? Which activities should be “crashed”?
Could calculate crash cost and crash time for all possible paths – but can also apply LP!
Crash cost
Normal cost
Crash time
Normal time
Activity duration time $
assume linear

20. Crash Cost per Week Saved
Time-cost trade-off data for the activities of Reliable’s project
Activity
Time
Cost
Maximum Reduction
in Time
Crash Cost per Week Saved
Normal
Crash A
2 weeks
1 week
$180,000
$280,000
$100,000 B
4 weeks
$320,000
$420,000
$50,000 C
10 weeks
7 weeks
$620,000
$860,000
3 weeks
$80,000 D
6 weeks
$260,000
$340,000
$40,000 E
$410,000
$570,000
$160,000 F
5 weeks G
$900,000
$1,020,000 H
9 weeks
$200,000
$380,000
$60,000 I
$210,000
$270,000
$30,000 J
8 weeks
$430,000
$490,000 K L
$250,000
$350,000 M N
$330,000
$510,000
If do all tasks normal, 44 weeks, cost is 4.55 million.
If do all tasks crash, 28 weeks, cost is 6.15 million.

21. Marginal Cost Analysis
For small networks, may reduce the project 1 week at a time, and observe the changes.
Activity to Crash
Crash Cost
Length of Path
ABCDGHM
ABCEHM
ABCEFJKN
ABCEFJLN
ABCIJKN
ABCIJLN 40 31 43 44 41 42 J
$ 30,000 39 F
$ 40,000

22. Linear Programming to Make Crashing Decisions
Let
Z = total cost of crashing on any activity
xj = reduction in the duration of activity j due to crashing
j = A,B,C,…,N
xj ≤ maximum reduction time = normal time – crash time
yFINISH= project duration, time at which FINISH node is reached
yj = start time of activity j
yj ≥ yi + normal timei – xi i is an immediate predecessor of j F I J yJ
yJ ≥ yF xF
yJ ≥ yI xI yF 5 yI 7

23. Linear Programming Model

24. The schedule of cumulative project costs when all activities begin at their earliest start times or at their latest start times

25. Time-cost trade-off data for the activities of the baking project
Activity
Normal Time
Crash Time
Normal Cost
Crash Cost
A: Buy frosting
0.5
0.25 5
B: Clean kitchen
1.0 10
C: Buy cake
D: Prepare frosting
E: Prepare batter bake
2.0
1.5
F: Frost cake

26. The project network showing Normal Time and Crash Time for each activity of the baking project
Start
N: Normal time
C: Crash time
If all are normal:
ADF 1 ¼
BDF 1 ¾
BEF 3 ½
CEF 3
Total time is 3 ½.
If all are crashed:
ADF ¾
BDF ¾
BEF
CEF
Total time is 2.
N C
1 , 1/4
N C
1/2 , 1/4 A C
N C
1/2 , 1/4 B D E
N C
1/4 , 1/4
N C
2 , 1 1/2 F
N C
1/2, 1/4
Finish

27. Could solve the crash LP for finish times between 3
Could solve the crash LP for finish times between 3.5 and 2 to evaluate alternatives 25
16.67
Cost
11.67
8.33 5
2.5 2
2.25
2.5
2.75 3
3.25
3.5
T= y
Finish

28. Lasagna Dinner Example
Task
Tasks that must precede
Time
A: Buy the mozzarella cheese
30 mins
B: Slice the mozzarella A
5 mins
C: Beat 2 eggs
2 mins
D: Mix eggs and ricotta cheese C
3 mins
E: Cut up onions and mushrooms
7 mins
F: Cook the tomato sauce E
25 mins
G: Boil large quantity of water
15 mins
H: Boil the lasagna noodles G
10 mins
I: Drain the lasagna noodles H
J: Assemble all the ingredients
I, F, D, B
K: Preheat the oven
L: Bake the lasagna
J, K

29. Construct project network
Start E 7 G 15 A C 2 K 30 15 H 10 F 25 5 B D 3 I 2
A→B →J →L *
C→D →J →L
E→F →J →L
G→H →I →J →L 45 J 10 L 30
Finish

30. Start E 7 G 15 30 A C 2 K 15 H 10 D 3 F 25 5 B I 2 J 10 L 30 Finish
EF = ES + activity time (or duration)
if no predecessors, ES = 0;
otherwise ES = max (EF) (immediate predecessors)
work forward through network
Start
ES=0EF=30
ES=0EF=2
ES=0EF=7
ES=0EF=15 E 7 G 15 30 A C
ES=0EF=15 2 K 15 H 10
ES=15EF=25
ES=7EF=32
ES=30EF=35 D 3
ES=2EF=5 F 25 5 B
ES=25EF=27 I 2
ES=35EF=45 J 10
ES=45EF=75 L 30
Finish

31. The complete project network showing ES, LS, EF, LF and Slack for each activity of the Lasagna dinner example
S = (ES, LS)
F = (EF, LF)
Slack = LF-EF=LS-ES
Start
S=(0,0) F=(30,30) Slack=0 30 A C
S=(0,30)F=(2,32) slack=30
S=(0,3) F=(7,10) Slack=3
S=(0,10) F=(15,25) Slack=10
S=(0,30) F=(15,45) Slack=30 2 E 7 G 15 K 15
S=(30,30) F=(35,35) Slack=0
S=(15,25) F=(25,35) Slack=10 5 B
S=(2,32) F=(5,35) slack=30
S=(7,10) F=(32,35) Slack=3 D F 25 H 10 3
S=(25,33) F=(27,35) Slack=8 I 2
S=(35,35) F=(45,45) Slack=0 J 10
S=(45,45) F=(75,75) Slack=0 L 30
Critical path has zero slack: A→B →J →L
Finish

32. Because of a phone call, you will delayed by 6 minutes to
cut onions and mushrooms (Task E) . By how much will
dinner be delayed?
slack is 3, delay of 6 minutes will delay dinner by 6-3=3
If you use your food processor instead to reduce cutting
time from 7 minutes to 2 minutes, will dinner still be
delayed?
ES=0 LS=8
LS=2 LF=10 E 2
slack = 8, so phone of 6 won’t delay dinner

33. All of the critical path, ES, LS, EF, LF, are based on estimates of the activity times.
How can you incorporate uncertainty into planning?
PERT 3 – estimate approach
Most Likely Estimate (m) most probable event
Optimistic Estimate (σ) if everything goes perfectly
Pessimistic Estimate (p) if everything goes wrong

34. What is the probability of meeting your deadline?
Assume the distribution for activity time is a Beta distribution
density
f(t)
t: time
µ - 3σ
µ + 3σ
µ ± 3σ interval captures 99.73% of distribution

35. Consider Critical Path
μ= Buy mozzarella cheese
σ2 = (6 2/3)2
o= m= p=50 B
μ= Slice cheese
σ2 = (11/3)2
o= m= p=11 J
μ= Assemble
σ2 = (2)2 = 4 L
μ= Bake
σ2 = (3 2/3)2
o= m= p=40

36. Could calculate pessimistic length:
= 115
Longest pessimistic path may not be the critical mean path
The mean length is sum of means:
= 75 = µp
Assume all task times are independent, variance for path is sum of variances:

37. Assume distribution of path time is normal (central limit theorem if lots of tasks on a path)