Cpm pert problems

Problems on C.P.M &PERTProblem1:

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 activityA Open work order -B Get Material for X AC Get material for Y AD Then X on lathe BE Then y on lathe B,CF Polish Y EG Assemble x and y D,FH 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

1

1 6

53

4

2 87

F

A

C

E

B D

G H

E1 Dummy

A Dismantle pipe connections

-

B Dismantle heater, closure, and floating front

A

C Remove Tube Bundle BD Clean bolts BE Clean heater and floating

head frontB

F Clean tube bundle CG Clean shell CH Replace tube bundle F,GI Prepare shell pressure

testD,E,H

J Prepare tube pressure testAnd 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:

2

5

109

6

8

7

3

4

21

CA

B

I

D1E

G

D

FD2

HJ

Activity Description Predecessor ActivityA 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 succedfull conditions passing through the institute

B

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,JL Issuing interview/regret

lettersK

M Preliminary interviews LN 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.

3

PROBLEMS ON CRITICAL PATH METHOD (CPM):

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.

4

1

864

3 1413121110975

2

B

A

C

D R

O

M

L

K

H

G

E

N

F J

I

4

6

5

7

8

9

3

2

1

10

4

1

1

1

5

6

8

4 1

28

7

1

E9 =18

E1=0 , L1=0

E8 =17

E7 =15

L7

E3 E5

E2 =4

E6 =11

E10 =25

E4 =5

L4

FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:

I. The tail event (T e¿and head event (T L) computed on the network as follows

EVENT

1 2 3 4 5 6 7 8 9 10

T e 0 4 1 5 7 11 15 17 18 25T L 0 12 1 13 7 16 15 17 18 25

CRITICAL PATH:

5

E1=0

E2=E1+D12=0+4=4

E3=E1+D13=0+1=1

E4=MAX {Ei+Di4 }=MAX ¿ (E3+D34)} , i=2, 3=MAX ¿ =Max {5, 2} =5

E5=E3+D35=1+6=7

E6=E5+D56=7+4=11

E7=E5+D57=7+8=15

E8=MAX {Ei+Di8 }=MAX ¿ (E7+D78)} , i=6, 7=MAX ¿ =Max {12, 17} =17

E9=MAX {Ei+Di9 }=MAX ¿ (E8+D 89)} , i=4, 8=MAX ¿ =Max {10,18} =18

E10=MAX {E i+Di10 }=MAX ¿ (E8+D 810)} , i=9, 8=MAX ¿ =Max {15, 25} =25

L10=E10=0

L9=L10−D 910=25−7=18

L8=¿ MIN {LJ−D 8J }=MIN ¿ (L10−D810)} , J=9, 10=MIN ¿ =MIN {17, 17} =17

L7=L8−D87=17−2=15

L6=L8−D86=17−1=16

L5=MIN {LJ−D5J }=MIN ¿ (L27−D57)} , J=6, 9=MIN ¿ =Min {7, 12} =12

L4=L9−D49=18−5=13

L3=L5−D 35=7−6=11

L2=L4−D24=13−1=12

L1=MIN {LJ−D 1J }=MIN ¿ (L3−D13)} , J=2, 3

E2=4

E3=1

E4=5

E5=7

E6 =11

E7=15

E8 =17

E10

E9=18

E1 =0 L10=0

L9=18

L8=17

L7=15

L6=16

L5=12

L4=13

L3=11

L2=12

L1=O

Activity(i.j)

(1)

Duration of time (Dij)(2)

Earliest timeStarting(Ei)

Finishing time (E J=Ei+Dij)

(3) (4)=(3)+(2)

Latest time StartingLi=L j−Dij

Finishing(L¿¿ J )¿

(5)=(6)-(2) (6)

Total float timeT-F=(L¿¿ i−Ei)¿

(7)=(5)-(3)1-2 4 0 4 8

128

1-3 1 0 1 10 11 102-4 1 4 5 12 13 83-4 1 1 2 12 13 123-5 6 1 7 6 12 54-9 5 5 10 13 18 85-6 4 7 11 12 16 55-7 8 7 15 7 15 06-8 1 11 12 16 17 57-8 2 15 17 15 17 08-9 1 17 18 17 18 08-10 8 17 25 8 0 99-10 7 18 25 8 0 10

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 4-3DURATION 20 25 10 12 6 10

a) Draw the network for the project.b) Find the critical path.

6

Solution:

FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:

Activity(i.j)

(1)

Duration of time (Dij)(2)

Earliest timeStarting(Ei)

Finishing time (E J=Ei+Dij)

(3) (4)=(3)+(2)

Latest time StartingLi=L j−Dij

Finishing(L¿¿ J )¿

(5)=(6)-(2) (6)

Total float timeT-F=(L¿¿ i−Ei)¿

(7)=(5)-(3)1-2 20 0 20 0 20 01-3 25 0 25 5 30 52-3 10 20 30 20 30 02-4 12 20 32 24 36 43-4 6 30 36 30 36 04-5 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.

7

4 5

3

2

1

E1=0 , L1=0

E3 =30

L3

E5 =46

E2 =20

E4 =36

25

20

6

12

1010

E1=0

E2=E1+D12=0+20=20

E3=E2+D23=20+1=30

E4=MAX {Ei+Di4 }=MAX ¿ (E3+D34)} , i=2, 3=MAX ¿ =Max {32, 36} =36

E5=E4+D 45=36+10=46

L5=E5=46

L4=L5−D45=46−10=36

L3=L4−D45=36−6=30

L2=MIN {LJ−D 2J }=MIN ¿ (L4−D24)} , J= 3,4=MIN ¿ =Min {20,14} =20

L1=MIN {LJ−D 1J }=MIN ¿ (L3−D13)} , J=2, 3

E2=20

E3=30

E4=36

E5=46

E1 =0 L5=46

L4=36

L3=30

L2=20

L1=O

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:

8

4

6

5

7 8 93

2

1

2

1

4

8

5

3

2

5

4

5

1

3

E1=0

E2=E1+D12=0+2=2

E3=E1+D13=0+2=2

E4=E1+D14=0+1=1

E6=E2+D26=2+4=6

E7=E3+D37=2+5=7

E5=MAX {Ei+Di5 }=MAX ¿ (E4+D45)} , i= 4,3=MAX ¿ =Max {10, 4} =10

E8=MAX {Ei+Di8 }=MAX ¿ (E7+D78)} , i=6, 7=MAX ¿ =Max {7, 11} =11

E9=MAX {Ei+Di9 }=MAX ¿ (E5+D59)} , i=8, 5

L9=E9=15

L8=MIN {LJ−D8J }=MIN ¿ J= 1=MIN ¿ =Min {12}

L5=¿ ={(15−5 ) ) } = {10} =10

L6=¿ ={(12−1 ) )} = {11} =11

L7=¿ ={(12−4 ) )} = {8} =8

L3=MIN {LJ−D3J }=MIN ¿ (L7−D37)} , J= 3,7=MIN ¿ =Min {2,3} =2

L4=¿ ={(10−3 ) ) } = {7} =7

L2=¿ ={(11−4 ) ) } = {7} =7L1=MIN {LJ−D 1J }=MIN ¿

(L3−D13 )(L4−D14)} , J= 2, 3, 4=MIN ¿ =Min {5, 0, 6} =0

E2=2

E3=1

E4=1

E7=7

E6 =11

E8=11

E1 =0 L9=0

L8=12

L5=10

L6=11

L7=8

L3=2

E9=15

L4=7

L2=7L1=¿0

These values can be represented in the following network diagram.

Activity(i.j)

(1)

Duration of time (Dij)(2)

Earliest timeStarting(Ei)

Finishing time (E J=Ei+Dij)

(3) (4)=(3)+(2)

Latest time StartingLi=L j−Dij

Finishing(L¿¿ J )¿

(5)=(6)-(2) (6)

Total float timeT-F=(L¿¿ i−Ei)¿

(7)=(5)-(3)1-2 2 0 2 5 7 51-3 2 0 2 0 2 01-4 1 0 1 6 7 62-6 4 2 6 7 11 53-7 5 2 7 3 8 13-5 8 2 10 2 10 04-5 3 1 4 7 10 65-9 5 10 15 10 15 06-8 1 6 7 11 12 57-8 4 7 11 8 12 18-9 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.

9

4

6

5

7 8 9

2

1

2

1

4

8

5

3

2

5

4

5

1

33

E9 =15

E1=0 , L1=0

E8 =11

E7 =7

L7 =8

E3

E5 =10

E2 =7 E6 =11

E4 =1

L4

ii. Construction of a building or a poweriii. 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 ‘t o”

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 “tm”.

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 ‘t p”

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

t e =(to+t p+4 tm

6)

5. VARIANCE: the variance for the activity is given by the formula

σ 2=[ (t p−t o )6 ]

2

Where t o=the optimistic time. σ 2=the variance t p = the pessimistic time. t e =the expected time.

tm=themost likely time .

10

Frequency

OTime

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 KLEAST TIME t o 4 5 8 2 4 6 8 5 3 5 6GREATEST TIME

tP 8 10

12

7 10

15

16

9 7 11 13

MOST LIKELY TIME

tM 5 7 11

3 7 9 12

6 5 8 9

SOLUTION;

11

Optimistic time

Most likely time

Pessimistic Time

4 6

57

8

9

2

1

3

AB

DF

G

E

H

G J

KC

First we calculate the expected time ‘t e ‘by the formula “t e = (to+t p+4 tm

6) as follows

Task Optimistic time(t o ¿

Pessimistic time(t p¿

Most likely time(tm¿

Expected time(t e¿

A 4 8 5 =¿¿ = 5.3B 5 10 7 =¿¿7.2C 8 12 11 =¿¿10.7D 2 7 3 =¿¿3.5E 4 10 7 =¿¿7F 6 15 9 =¿¿9.5G 8 16 12 =¿¿12H 5 9 6 =¿¿6.3I 3 7 5 =¿¿5J 5 11 8 =¿¿8K 6 13 9 =¿¿9.1

Now the earliest time expected time Ei 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 Ei is called.

FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:

TASK A B C D E F G H I J KLEAST TIME t o 4 5 8 2 4 6 8 5 3 5 6GREATEST TIME

tP 8 10 12 7 10 15 16 9 7 11 13

12

E1=0

E2=MAX {E i+Di2 }=MAX ¿ =0+5.3=5.3

E3=E1+D13=0+3.5=3.5 E4=MAX {Ei+Di4 }=i=2,1MAX ¿ (E¿¿1+D14)=max {5.3+7.2 )(0+10.4)}¿ =max {12.5,10.4} =12.5

E5=MAX {Ei+Di5 }=i=3,4MAX ¿ (E¿¿ 4+D 45)=max {3.5+9.5 )(12.5+7)}¿ =max {13,19.5} =19.5

E6=E5+D65=19.5+6.3=25.8

E7=MAX {Ei+Di7 }=i=5,3MAX ¿ (E¿¿3+D37)=max {19.5+5 )(12+3.5)}¿ =max {15.5,24.5} =24.5

E8=E6+D68=25.8+9.1=34.9

L9=E9=32.5

L8=E8=34.9

L7=¿ ={(32.5−8 ) ) } = 24.5

L6=¿ ={(34.9−9.11 )) } = 25.9

L5=MIN {LJ−D5J }=MIN ¿ (L7−D57)} , J= 3,7=MIN ¿ =Min {19.5,19.5} =2

L4=¿ ={(19.5−7 ) ) } = 12.5

L3=MIN {LJ−D3J }=MIN ¿ (L7−D37)} , J= 5,7=MIN ¿ =Min {10,12.5} =10

L2=¿ ={(11−4 ) ) } = 7L1=MIN {LJ−D iJ }=MIN ¿ (L4−D24)} , J= 2,4

E2=5.3

E5= 19.5

E4=12.5

E7=24.5

E6 =25.8

E8=34.9

E1 =0 L9=32.5

L8=34.9

L5=25.8

L6=11

L7=24.5E3=3.5

E9=32.5

L4=12.5

L3=10

L2=7

L1=0

MOST LIKELY TIME

tM 5 7 11 3 7 9 12 6 5 8 9

Estimation Time

t e 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 t e Ei Li SLACK(Li−Ei ¿

1 5.3 5.3 5.3 02 3.5 3.5 10.0 6.53 7.2 12.5 12.5 04 7 19.5 19.5 05 6.3 25.8 25.8 06 5 24.5 24.5 07 9.1 34.9 34.9 08 8.0 32.5 32.5 0

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

13

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

E9 =32.5

E1=0 , L1=0

E8 =34.9

L8

E7 =24.5

E3 =3.5

L3 E5 =19.5

E2 =5.3

L2=5.3

E6 =25.8 E4 =12.5

L4

MOST OPTEMESTI TIME(a)

t o 1 1 1 3 2 3 4 6 2 5 1 3

MOAST PESSIMESTIC TIME(b)

t p 5 3 5 5 4 7 6 8 6 8 3 7

MOST LIKELY TIME(m)

t e 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 (t e) = (to+t p+4 tm

6). V=(b−a)

6

2

Activity a b 4m t e 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

FORWARD PASS CALUCLATION: BACKWARD PASS CALUCLATION:

14

4

5

6

8

9

2

3

1

10

1

1

3

12

2

3

4

5

61

3

1

E9 =139/6

L9 =139/6E1=0 , L1=0

E8 =17

L8

E7 =17

L7

E3 E5

E6 =10

E10 =169/9

L10 =169/9

E4 =5

L4

7

E2 =2

L2

E1=0

E2= {Ei+Di2 }=MAX ¿ =0+2=2

E3=E2+D23=2+2=4

E4=E2+D24=2+3=5

E5=MAX {Ei+Di5 }=i=3,4MAX ¿

L10=E10=¿ 169

6

L9=L10−D 910=169

6−5=139

6

L8=L10−D 810=139

6 -2=

1276

L7=MIN {LJ−D7J }=MIN ¿ (L9−D 79)} , J= 8,9

E1 =0 L9=32.5

E2=2

E3=4

L9=139

6

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 NormalTime (weeks) Cost(Rs/-)

CrashTime (weeks)

Cost(Rs/-)

1-2 3 300 2 4002-3 3 30 3 302-4 7 420 5 5802-5 9 720 7 8103-5 5 250 4 3004-5 0 0 0 05-6 6 320 4 4106-7 4 400 3 4706-8 13 780 10 9007-8 10 1000 9 1200

Indirect cost is 50/- per week.Draw the network diagram for the project and identify the critical path.What is the normal project duration associated cost?Find out the total float associated with each activity.Crash the relevant activities systematically and determine the optimal project

completion time cost.

15

E1=0

E2= {Ei+Di2 }=MAX ¿ =0+2=2

E3=E2+D23=2+2=4

E4=E2+D24=2+3=5

E5=MAX {Ei+Di5 }=i=3,4MAX ¿

L10=E10=¿ 169

6

L9=L10−D 910=169

6−5=139

6

L8=L10−D 810=139

6 -2=

1276

L7=MIN {LJ−D7J }=MIN ¿ (L9−D 79)} , J= 8,9

E4=5

E5=8

E6=10

E7=17

E8=17

E2=139

6

E10=1696

L9=1276

L7=17

L6=10

L5=12

L4=5

L3=8

L2=2

L1=0

Solution:

16

4

65

7

8

21

3

10

7

3

96

5 13

4

3 E8 =32

L8 =32

E7 =22

L7

E3 =6

L3 =7

E5 =12

L5 E2 =3

L2=3

E6 =18

L6

E4 =10

L4

E1 =0

L1=0

0

17