The critical path method is used to analyze a project schedule network diagram to determine the longest path of activities. Key steps are: 1) Draw the network diagram of activities and dependencies 2) Perform forward and backward passes to calculate early and late start/finish times 3) Identify the critical path(s) with the longest total duration as the schedule cannot be shortened without changing these activities 4) Determine the float of non-critical activities which can be delayed without affecting the project completion date.

1. Time Management
Problem-Set-1
Dr. Syed Amir Iqbal.

2. Produce a network diagram using the data below, identify the critical
path and record the project duration.
Task Duration Predecessor
A 5 Start
B 8 A
C 3 B
D 4 A
E 5 C
F 7 C
G 6 D,E
H 8 F
I 6 G
J 5 H,I

4. Task Duration Predecessor
A 1
B 8 A
C 4 A
P 7 A
L 2 B
M 4 C
Q 4 P,C
N 9 P
Y 5 L,Q
F 10 M
J 2 Q
S 2 N
V 5 Y,F,J
Z 1 V,S
• Construct a
precedence diagram.
• On the diagram,
compute the four
schedule times for
each activity.
• Identify the critical
path

5. An automobile
manufacturer is
developing a new
model of car, and the
following table shows
some of the
information relating to
the major activities
involved in this project:

6. For the following table of information,
1. Draw the network diagram
2. List the network paths
3. Determine the critical path(s)
4. Determine the float for each activity

8. For the following table of information,
1. Draw the network diagram 3. Determine the critical path(s)
2. List the network paths 4. Determine the float for each activity
Task Duration Predecessor
Start 0
A 5 Start
B 2 Start
C 3 A, B
D 5 Start
E 6 Start
F 4 D, E
G 2 C, F
H 5 G
I 7 G
J 3 H
Finish 0 I, J

10. Perform a Forward Pass and a Backward Pass of the network
diagram given and identify critical path?

11. For the information given find:
a) Critical Path
b) Cost of completing the project 3 weeks earlier.
Activity Precedence Time (Weeks) Cost ($)
Normal Crash Normal Crash
A - 4 2 10000 11000
B A 3 2 6000 9000
C A 2 1 4000 6000
D B 5 3 14000 18000
E B,C 1 1 9000 9000
F C 3 2 7000 8000
G E,F 4 2 13000 25000
H D,E 4 1 11000 18000
I H,G 6 5 20000 29000

12. Activity Precedence Time (Weeks) Cost ($) Cost/Week Weeks to Compress
Normal Crash Normal Crash
A - 4 2 10000 11000 500 2
B A 3 2 6000 9000 3000 1
C A 2 1 4000 6000 2000 1
D B 5 3 14000 18000 2000 2
E B,C 1 1 9000 9000 0 0
F C 3 2 7000 8000 1000 1
G E,F 4 2 13000 25000 6000 2
H D,E 4 1 11000 18000 2333 3
I H,G 6 5 20000 29000 9000 1

13. A=4
B=3
E=1
F=3
G=4
I=6
H=4
D=5
C=2
A-B-D-H-I = 22
A-B-E-H-I = 18
A-B-E-G-I = 18
A-C-E-H-I = 17
A-C-E-G-I = 17
A-C-F-G-I = 19

14. Activity Precedence Time (Weeks) Cost ($) Cost ($) /Week Weeks to Compress
Normal Crash Normal Crash
A - 4 2 10000 11000 500 2
B A 3 2 6000 9000 3000 1
C A 2 1 4000 6000 2000 1
D B 5 3 14000 18000 2000 2
E B,C 1 1 9000 9000 0 0
F C 3 2 7000 8000 1000 1
G E,F 4 2 13000 25000 6000 2
H D,E 4 1 11000 18000 2333 3
I H,G 6 5 20000 29000 9000 1

15. A-B-D-H-I = 22
A-B-E-H-I = 18
A-B-E-G-I = 18
A-C-E-H-I = 17
A-C-E-G-I = 17
A-C-F-G-I = 19
A=4
B=3
E=1
F=3
G=4
I=6
H=4
D=5
C=2
For Week-1
• Reduce activity A to 3 weeks and additional cost will be $500.
• Now path A-B-D-H-I will be 21 weeks and is still the critical path.
For Week-2
• Activity A still have 1 week cushion for time reduction. So Reduce
activity A to one more week. So current duration of activity A is 2
weeks with additional cost of $500.
• New duration of path A-B-D-H-I is 20 weeks and it is still the
critical path. Activity A is now exhausted and further reduction in
this activity is not possible.
For Week-3
• Now among activities B-D-H-I, activity D has the lowest reduction
cost/week. So reduce activity D by one week with additional cost
of $ 2000. New duration of activity D is 4 weeks. The total
duration of A-B-D-H-I is 19 weeks.
• Now the project duration is reduced from 22 weeks to 19 weeks
and the project has two critical paths A-B-D-H-I & A-C-F-G-I.

16. A None 7
B A 3
C A 4
D B,C 5
E D 2
F D 4
G E,F 5
Act. Imed. Pred. Time
A(7)
B(3)
C(4)
D(5)
E(2)
F(4)
G(5)

17. Time – Cost Analysis
Crashing the Project (Crash 4 days)
Activity Normal
Time
Crash
Time
Normal
Cost
Crash
Cost
Cost/Day
A 7 6 $7,000 $8,000 $1,000
B 3 2 5,000 7,000 2,000
C 4 3 9,000 10,200 1,200
D 5 4 3,000 4,500 1,500
E 2 1 2,000 3,000 1,000
F 4 2 4,000 7,000 1,500
G 5 4 5,000 8,000 3,000
17

18. For the data given in the table, Compute critical
path using PERT technique.

19. Activity Predecessor
Time
Te S.D.
To Tm Tp
A - 2 4 6 4 0.667
B A 8 11 20 12 2.000
C A 10 15 20 15 1.667
D B 12 18 24 18 2.000
E C 8 13 24 14 2.667
F C 4 7 16 8 2.000
G D,F 14 18 28 19 2.333
H E 10 12 14 12 0.667
I G,H 7 10 19 11 2.000

20. For the data given in the table,
Compute critical path using PERT
technique.

21. Activity Predecessor
Time
Te S.D.
To Tm Tp
A - 12 14 22 15 1.667
B - 16 17 24 18 1.333
C A 14 15 16 15 0.333
D A 13 18 23 18 1.667
E B 16 18 20 18 0.667
F D,E 13 14 21 15 1.333
G C,F 6 8 10 8 0.667