Question

You have been appointed project manager for a company critical project. After a detailed WBS analysis you have identified 15 key activities that make up the project. The following table details for each activity its duration and its preceding activities.

Activity	Preceding Activity	Duration Days
A	-	8
B	-	6
C	A	5
D	A	9
E	B	5
F	B	12
G	D,E	4
H	C	5
I	H	12
J	G,F	7
K	J	10
L	J	6
M	G,F	5
N	M	7
O	I,K	3

Senior Management has asked you to produce a comprehensive project plan. As part of this plan complete a CPM (critical path method) analysis of the above project. In your analysis produce a network diagram, define and identify the critical activities. In addition, calculate: Earliest Start, Latest Start, Earliest Finish, and Latest Finish, and slack for each activity.

Why are slack times important information for the project manager?

Answer 1

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Critical path, Total float, Free float, Independent float

A	-	8
B	-	6
C	A	5
D	A	9
E	B	5
F	B	12
G	D,E	4
H	C	5
I	H	12
J	G,H	7
K	J	10
L	J	6
M	G,F	5
N	M	7
O	I,K	3

Solution:

Activity	Immediate Predecessors	Duration
A	-	8
B	-	6
C	A	5
D	A	9
E	B	5
F	B	12
G	D,E	4
H	C	5
I	H	12
J	G,H	7
K	J	10
L	J	6
M	G,F	5
N	M	7
O	I,K	3

Edge and it's preceded and succeeded node

Edge	Node1 → Node2
A	1→2
B	1→3
C	2→4
D	2→5
E	3→5
F	3→6
H	4→8
G	5→7
M	6→11
d	7→6
J	7→9
d	8→7
I	8→10
K	9→10
L	9→12
O	10→12
N	11→12

The network diagram for the project, along with activity time, is

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=8]=0+8=8

E3=E1+t1,3 [t1,3=B=6]=0+6=6

E4=E2+t2,4 [t2,4=C=5]=8+5=13

E5=Max{Ei+ti,5}[i=2,3]

=Max{E2+t2,5;E3+t3,5}

=Max{8+9;6+5}

=Max{17;11}

=17

E6=Max{Ei+ti,6}[i=3,7]

=Max{E3+t3,6;E7+t7,6}

=Max{6+12;21+0}

=Max{18;21}

=21

E7=Max{Ei+ti,7}[i=5,8]

=Max{E5+t5,7;E8+t8,7}

=Max{17+4;18+0}

=Max{21;18}

=21

E8=E4+t4,8 [t4,8=H=5]=13+5=18

E9=E7+t7,9 [t7,9=J=7]=21+7=28

E10=Max{Ei+ti,10}[i=8,9]

=Max{E8+t8,10;E9+t9,10}

=Max{18+12;28+10}

=Max{30;38}

=38

E11=E6+t6,11 [t6,11=M=5]=21+5=26

E12=Max{Ei+ti,12}[i=9,10,11]

=Max{E9+t9,12;E10+t10,12;E11+t11,12}

=Max{28+6;38+3;26+7}

=Max{34;41;33}

=41


Backward Pass Method
L12=E12=41

L11=L12-t11,12 [t11,12=N=7]=41-7=34

L10=L12-t10,12 [t10,12=O=3]=41-3=38

L9=Min{Lj-t9,j}[j=12,10]

=Min{L12-t9,12;L10-t9,10}

=Min{41-6;38-10}

=Min{35;28}

=28

L8=Min{Lj-t8,j}[j=10,7]

=Min{L10-t8,10;L7-t8,7}

=Min{38-12;21-0}

=Min{26;21}

=21

L7=Min{Lj-t7,j}[j=9,6]

=Min{L9-t7,9;L6-t7,6}

=Min{28-7;29-0}

=Min{21;29}

=21

L6=L11-t6,11 [t6,11=M=5]=34-5=29

L5=L7-t5,7 [t5,7=G=4]=21-4=17

L4=L8-t4,8 [t4,8=H=5]=21-5=16

L3=Min{Lj-t3,j}[j=6,5]

=Min{L6-t3,6;L5-t3,5}

=Min{29-12;17-5}

=Min{17;12}

=12

L2=Min{Lj-t2,j}[j=5,4]

=Min{L5-t2,5;L4-t2,4}

=Min{17-9;16-5}

=Min{8;11}

=8

L1=Min{Lj-t1,j}[j=3,2]

=Min{L3-t1,3;L2-t1,2}

=Min{12-6;8-8}

=Min{6;0}

=0

(b) The critical path in the network diagram has been shown. This has been done by double lines by joining all those events where E-values and L-values are equal.
The critical path of the project is : 1-2-5-7-9-10-12 and critical activities are A,D,G,J,K,O

The total project time is 41
The network diagram for the project, along with E-values and L-values, is

For each non-critical activity, the total float, free float and independent float calculations are shown in Table

Activity (i,j) (1)	Duration (tij) (2)	Earliest time Start (Ei) (3)	(Ej) (4)	(Li) (5)	Latest time Finish (Lj) (6)	Earliest time Finish (Ei+tij) (7)=(3)+(2)	Latest time Start (Lj-tij) (8)=(6)-(2)	Total Float (Lj-tij)-Ei (9)=(8)-(3)	Free Float (Ej-Ei)-tij (10)=((4)-(3))-(2)	Independent Float (Ej-Li)-tij (11)=((4)-(5))-(2)
1-3	6	0	6	0	12	6	6	6	0	0
2-4	5	8	13	8	16	13	11	3	0	0
3-5	5	6	17	12	17	11	12	6	6	0
3-6	12	6	21	12	29	18	17	11	3	-3
4-8	5	13	18	16	21	18	16	3	0	-3
6-11	5	21	26	29	34	26	29	8	0	-8
7-6	0	21	21	21	29	21	29	8	0	0
8-7	0	18	21	21	21	18	21	3	3	0
8-10	12	18	38	21	38	30	26	8	8	5
9-12	6	28	41	28	41	34	35	7	7	7
11-12	7	26	41	34	41	33	34	8	8	0

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