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Problem1:
An assemble is to be made from „2‟ parts „x‟ a d ‟y‟. Botha parts must be turned a lathe. „y‟ must be polished
whereas „x‟ need not be polished. The sequence of actives, to gather with their predecessor s given below.
Activity Description Predecessor activity
A Open work order -
B Get Material for X A
C Get material for Y A
D Then X on lathe B
E Then y on lathe B,C
F Polish Y E
G Assemble x and y D,F
H Pack G
Draw a net work diagram of activities for project.
Solution:
Problem-2: Listed in the table are the activities and sequencing necessary for a maintenance job on the heat exchanging.
Activity Description Predecessor
A Dismantle pipe connections -
B Dismantle heater, closure, and
floating front
A
C Remove Tube Bundle B
D Clean bolts B
E Clean heater and floating head B
1 6
5 3
4
2 8 7
F
A
C
E
B D
G H
Dummy
front
F Clean tube bundle C
G Clean shell C
H Replace tube bundle F,G
I Prepare shell pressure test D,E,H
J Prepare tube pressure test
And reassemble
I
Draw a net work diagram of a activities for the projects.
Solution:
Problem-3: Listed in the table are activities and sequencing necessary for the completion of a recruitment
procedure for management trainees (µt) in a firm.
Solution:
Activity Description Predecessor Activity
A Asking units for recruitments -
B Ascertaining management
trains(MTs) requirement for
commercial function
A
C Asserting MTs requirement for
Account/finance functions
A
D Formulating advertisement for
MT(commercial)
C
E Calling applications from the B
5
10 9
6
8
7
3
4
2 1
C
A B
I
E
G
D
F
H
J
succedfull conditions passing
through the institute
F Calling application s from the
successful conditions passing
through the institute of chartered
Accounts (ICA)
C
G Releasing the advertisement D,E
H Completing applications
received
G
I Screening of applications against
advertisement
H
J Screening of applications
received from ICA
F
K Sending of personal forms I,J
L Issuing interview/regret letters K
M Preliminary interviews L
N Preliminary interviews of
outstanding candidates from ICA
J
O Final inter view M,N
Draw a network of activities for the project.
Solution: the net work diagram for the given project.
PROBLEMS ON CRITICAL PATH METHOD (CPM):
1
8 6 4
3 14 13 12 11 10 9 7 5
2
B
A
C
D R
O
M
L
K
H
G
E
N
F J
I
PROBLEM-1: A project has the following times schedule
Activity 1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-9 8-10 9-10
Time in
weeks
4 1 1 1 6 5 4 8 1 2 1 8 7
Construct CPM Network and compute.
I. TE and TL for each event.
II. Float for each activity
III. Critical path and its duration.
SOLUTION:
The net work is constructed as given in diagram.
FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:
4
6
5
7
8
9
3
2
1
10
4
1
1
1
5
6
8
4 1
2
8
7
1
=18 =18
=17 =17
=15 =15
=1 =1
=7 =7
=4 12
=11 =16
=25 =25
=5 =13
{ } { ) )} , i=2, 3 = { ) )} =Max {5, 2} =5 { } { ) )} , i=6, 7 = { ) )} =Max {12, 17} =17 { } { ) )} , i=4, 8 = { ) )} =Max {10,18} =18 { } { )
{ } { )
)} , J=9, 10 = { ) )} =MIN {17, 17} =17
{ } { )
)} , J=6, 9 = { ) )} =Min {7, 12} =12
{ } { )
)} , J=2, 3
=1
=11
=15
=17
=0
I. The tail event ( )and head event ( ) computed on the network as follows
EVENT 1 2 3 4 5 6 7 8 9 10
0 4 1 5 7 11 15 17 18 25
0 12 1 13 7 16 15 17 18 25
CRITICAL PATH:
Activity
(i.j)
(1)
Duration
of time
( )
(2)
Earliest time
Starting
( )
Finishing time
( )
(3) (4)=(3)+(2)
Latest time
Starting
Finishing
)
(5)=(6)-(2) (6)
Total float
time
T-F= )
(7)=(5)-(3)
1-2 4 0 4 8
12
8
1-3 1 0 1 10 11 10
2-4 1 4 5 12 13 8
3-4 1 1 2 12 13 12
3-5 6 1 7 6 12 5
4-9 5 5 10 13 18 8
5-6 4 7 11 12 16 5
5-7 8 7 15 7 15 0
6-8 1 11 12 16 17 5
7-8 2 15 17 15 17 0
8-9 1 17 18 17 18 0
8-10 8 17 25 8 0 9
9-10 7 18 25 8 0 10
=25
=18
CRITICAL PATH ACTIVITIES:
From the above table we observe that the activities 5-7, 7-8 and 8-9 are critical activities
as their total float is zero. Hence we have the following critical path.
5-7, 7-8 and 8-9 8+8+7=25
Problem-2: The following table gives activities of duration of construction project work.
ACTIVITY 1-2 1-3 2-3 2-4 3-4 4-3
DURATION 20 25 10 12 6 10
a) Draw the network for the project.
b) Find the critical path.
Solution:
FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:
4 5
3
2
1
=30 =30
=46 =46
=20 20
=36 =36
25
20
6
12
10
10
{ } { ) )} , i=2, 3 = { ) )} =Max {32, 36} =36
{ } { )
)} , J= 3,4 = { ) )} =Min {20,14} =20
{ } { )
)} , J=2, 3 = { ) )} =Min {0,5} =0
=30
=0
Activity
(i.j)
(1)
Duration
of time
( )
(2)
Earliest time
Starting
( )
Finishing time
( )
(3) (4)=(3)+(2)
Latest time
Starting
Finishing
)
(5)=(6)-(2) (6)
Total float
time
T-F= )
(7)=(5)-(3)
1-2 20 0 20 0 20 0
1-3 25 0 25 5 30 5
2-3 10 20 30 20 30 0
2-4 12 20 32 24 36 4
3-4 6 30 36 30 36 0
4-5 10 36 46 36 46 0
From the above table we observe that the activities 1-2, 2-3, 3-4, and4-5 are critical activities as
their total float is „0‟. Hence we have the following critical path
1-2-3-4-5 with the total project duration is „46‟ days.
Problem-3:
Find the critical path and calculate the slack time for each event for the following PERT diagram.
Solution:
FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:
4
6
5
7 8 9 3
2
1
2
1
4
8
5
3
2
5
4
5
1
3
{ } { )
J= 1 = { ) ) =Min {12}
{ ) = { ))} = {10} =10
{ ) = { ))} = {11} =11
=0
These values can be represented in the following network diagram.
Activity
(i.j)
(1)
Duration
of time
( )
(2)
Earliest time
Starting
( )
Finishing time
( )
(3) (4)=(3)+(2)
Latest time
Starting
Finishing
)
(5)=(6)-(2) (6)
Total float
time
T-F= )
(7)=(5)-(3)
1-2 2 0 2 5 7 5
{ } { ) ) )} , J= 2, 3, 4 = { ) ) )} =Min {5, 0, 6} =0
4
6
5
7 8 9
2
1
2
1
4
8
5
3
2
5
4
5
1
3 3
=15 =15
=11 =12
=7 =8
=2 =2
=10 =10
=7
=11 =16
=1 =7
=1
=11
=11
=15
1-3 2 0 2 0 2 0
1-4 1 0 1 6 7 6
2-6 4 2 6 7 11 5
3-7 5 2 7 3 8 1
3-5 8 2 10 2 10 0
4-5 3 1 4 7 10 6
5-9 5 10 15 10 15 0
6-8 1 6 7 11 12 5
7-8 4 7 11 8 12 1
8-9 3 11 14 12 15 1
[PERT]
In the net work analysis it is implicitly assumed that the time values are deterministic or
variations in time are insignificant. This assumption is valid in regular jobs such as
i. Maintenance of machine.
ii. Construction of a building or a power
iii. Planning for production.
As these are done from time and various activities could be timed very well
However in reach projects or design of a gear box of a new machine various
Activities .are based on judgment. A reliable time estimate is difficult to get because the technologies is
changing the job? the pert approach taxes into account the uncertainties associated with in that activity.
DEFINATIONS:
1. OPTIMISTIC TIME: the optimistic time is the shortcut possible time in which the activity can be
finished. It assumes that everything goes very well. This is denoted by „ ”
2. MOST LIKELY TIME; the most likely time is the estimate of the activity would take. This assumes
normal delays. If a graph is plotted in the time of completion and the frequency of completion in that time
period than the „most likely time „will represent the highest frequency of a occurrence. This is denoted by
“ ”.
3. PESIMISTIC TIME; the „pessimistic time‟ represents the longest time the activity could take if
everything goes wrong. As in optimistic estimate. This value may be such that value. This is denoted by
„ ”
These 3 types‟ values are, shown in the following diagram in order to obtain these values; one could use time
values available similar jobs. But most of the time the estimator may not be so fortunate to have this data.
Secondary values are the functions of manpower, machines and supporting facility .A better approaches would
be to seek opinion of experts in the field keeping in views the resources available, this estimate does not take
into account such natural catastrophes as fire etc.
In pert calculation all values are used to attain the percent exportation value.
4.EXPECTED TIME; the „expected time‟ is average time an activity will take if it were to be reported on
large number of times and it is base on the assumption that the activity time follows „Bets distribution‟. This is
given by the formula
=(
)
5. VARIANCE: the variance for the activity is given by the formula
*( )
+
Where = the pessimistic time. =the expected time.
Frequency
O
Time
Problem-1: for the project represented by the network diagram, find the earliest time and latest times to reach
each node given the following data.
TASK A B C D E F G H I J K
LEAST TIME 4 5 8 2 4 6 8 5 3 5 6
GREATEST
TIME
8 10 12 7 10 15 16 9 7 11 13
MOST LIKELY
TIME
5 7 11 3 7 9 12 6 5 8 9
Optimistic time
Most likely time
Pessimistic Time
4 6
5 7
8
9
2
1
3
A B
D F
E
H
K C
SOLUTION;
First we calculate the expected time „ „by the formula “ = (
) as follows
Task Optimistic
time( ) Pessimistic
time( )
Most likely time
( ) Expected time
( )
A 4 8 5 = )
= 5.3
B 5 10 7 = )
7.2
C 8 12 11 = )
10.7
D 2 7 3 = )
3.5
E 4 10 7 = )
7
F 6 15 9 = )
9.5
G 8 16 12 = )
12
H 5 9 6 = )
6.3
I 3 7 5 = )
5
J 5 11 8 = )
8
K 6 13 9 = )
9.1
Now the earliest time expected time for each note are obtained by taking sum of the expected times for all
the activities leading to node i, when more than one activity leads to a node i, the maximum of is called.
FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:
G G J
{ } { ) =0+5.3=5.3 { } { ) ) { ) )} =max {12.5,10.4} =12.5 { } { ) ) { ) )} =max {13,19.5} =19.5 { } { ) ) { ) )} =max {15.5,24.5} =24.5
{ ) = { ))} = 24.5
{ ) = { ))} = 25.9
{ } { )
)} , J= 3,7 = { ) )} =Min {19.5,19.5} =2
{ ) = { ))} = 12.5
{ } { )
)} , J= 5,7 = { ) )} =Min {10,12.5} =10
{ ) = { ))} = 7
{ } { )
)} , J= 2,4
= 19.5
2.5
=25.8
=0
TASK A B C D E F G H I J K
LEAST TIME 4 5 8 2 4 6 8 5 3 5 6
GREATEST
TIME
8 10 12 7 10 15 16 9 7 11 13
MOST LIKELY
TIME
5 7 11 3 7 9 12 6 5 8 9
Estimation Time 5.3 7.2 10.7 3.5 7 9.5 12 6.3 5 8 9.2
These calculations are may be arranged in the following table.
NODE SLACK( )
1 5.3 5.3 5.3 0
2 3.5 3.5 10.0 6.5
3 7.2 12.5 12.5 0
4 7 19.5 19.5 0
5 6.3 25.8 25.8 0
6 5 24.5 24.5 0
7 9.1 34.9 34.9 0
8 8.0 32.5 32.5 0
=34.9
=32.5
4 6
5 7
8
9
2
1
3
A(5.3) B(7.2)
D(3.5)
G(12)
E(7)
H(6.3)
I (5) J(8)
K(9.1) C(10.4)
F=9.5
=32.5 =32.5
=34.9 =34.9
=24.5 =24.5
=3.5 =10.0
=19.5 =19.5
=5.3
=25.8 =25.8
=12.5 =12.5
PROBLEM-2: A project has the following characteristics
ACTIVITY 1-2 2-3 2-4 3-5 4-5 4-6 5-7 6-7 7-8 7-9 8-10 9-10
MOST
OPTEMESTI
TIME(a)
1 1 1 3 2 3 4 6 2 5 1 3
MOAST
PESSIMESTIC
TIME(b)
5 3 5 5 4 7 6 8 6 8 3 7
MOST LIKELY
TIME(m)
1.5 2 3 4 3 5 5 7 4 6 2 5
Construct a PERT network. Find critical path and variance for each event find the project duration as
95% probability.
SOLUTION: Activity expected times & their variances are computed by the following formula
Expected time ( ) = (
). V=
)
Activity a b 4m V
1-2 1 5 6 = )
=2 4/9
2-3 1 3 8 = )
2 1/9
2-4 1 5 12 = )
3 4/9
3-5 3 5 16 = )
4 1/9
4-5 2 4 12 = )
1/9
4-6 3 7 20 = )
5 4/9
5-7 4 6 20 = )
5 1/9
6-7 6 8 28 = )
7 1/9
7-8 2 6 16 = )
4 4/9
7-9 5 8 24 = )
6.16 1/9
8-10 1 3 8 = )
1 1/9
9-10 3 7 20 )
5 4/9
5
8
9
2
3
1
10
1
1
3
1 2
2
4
5
6
1
3
1
=139/6 =139/6
=17 =17
=4 =8
=8 =12
=169/9 =169/9
7
FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:
The longest path is 1-2-4-6-7-9-10 can be traced. This is known as critical path.
PROBLEM ON PROJECT CRASHING: The following table gives data on normal time, and cost crash time and cost for a project
Activity Normal
Time (weeks) Cost(Rs/-)
Crash
Time (weeks) Cost(Rs/-)
1-2 3 300 2 400
2-3 3 30 3 30
2-4 7 420 5 580
2-5 9 720 7 810
3-5 5 250 4 300
4-5 0 0 0 0
5-6 6 320 4 410
6-7 4 400 3 470
6-8 13 780 10 900
7-8 10 1000 9 1200
Indirect cost is 50/- per week.
Draw the network diagram for the
4
6
3
=17 =127/6
=10 =10
=5 =5
=2 =2
{ } { ) =0+2=2 { } { ) ) { ) )} ) { } { ) ) { ) )} =max {13,17} =17
{ } { )
) { )
)}
=max {21,
} =
-2=
{ } { )
)} , J= 8,9
= ,(
)
)} =Min {19.5,17} =17
{ ) = { ))} = 10
{ ) = { ))} = 12
{ ) = { ))} = 5
{ } { )
{ } { )
)} , J= 3,4 = { ) )} =Min {6,2} =2
{ } { )
=0
4 6 5 7 8 2 1 3
10
7
3 9 6
5 13 4 3 =32
=32
=22 =22
=6 =7
=12 =12
=3
=18 =18
=10 =12
=0
0
⁄
⁄
