Question

QUESTION 2: (20 marks) 2. Critical Path Method (CPM) is a network diagramming technique used to predict total project duration and to determine the critical path. a) Based on the table provided, create the CPM network diagram. Analyse and show the Early Start (ES), Early Finish (EF), Late Start (LS), Late Finish (LF) and the Float/Slack duration (14 marks) b) Analyse the diagram, state and circle the Critical Path clearly. (6 Marks) Activity Predecessor Duration A 6 B А 7 с 00 A D B, C 5 I E С 6 F DE 6

Answer 1

Solution:

Activity	Immediate Predecessors	Duration
A	-	6
B	A	7
C	A	8
D	B,C	5
E	C	6
F	D,E	6

Edge and it's preceded and succeeded node

Edge	Node1 → Node2
A	1→2
B	2→3
C	2→4
D	3→5
d	4→3
E	4→5
F	5→6

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

B7) 3 D(5) 1 A(6) 2 d(0) 5F(6) 6 E(6) C(8) 4

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=6]=0+6=6

E3=Max{Ei+ti,3}[i=2,4]

=Max{E2+t2,3;E4+t4,3}

=Max{6+7;14+0}

=Max{13;14}

=14

E4=E2+t2,4 [t2,4=C=8]=6+8=14

E5=Max{Ei+ti,5}[i=3,4]

=Max{E3+t3,5;E4+t4,5}

=Max{14+5;14+6}

=Max{19;20}

=20

E6=E5+t5,6 [t5,6=F=6]=20+6=26


Backward Pass Method
L6=E6=26

L5=L6-t5,6 [t5,6=F=6]=26-6=20

L4=Min{Lj-t4,j}[j=5,3]

=Min{L5-t4,5;L3-t4,3}

=Min{20-6;15-0}

=Min{14;15}

=14

L3=L5-t3,5 [t3,5=D=5]=20-5=15

L2=Min{Lj-t2,j}[j=4,3]

=Min{L4-t2,4;L3-t2,3}

=Min{14-8;15-7}

=Min{6;8}

=6

L1=L2-t1,2 [t1,2=A=6]=6-6=0


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 : 1-2-4-5-6 and critical activities are A,C,E,F

The total project time is 26
The network diagram for the project, along with E-values and L-values, is

B7) 3 Ez = 14 13 15 D(5) 1 1 A(6) 2 d(0) 5 F(6) 6 E(6) E1 = 0 11 = 0 E2 = 6 12 6 C(8) 4 E4 - 14 E5 = 20 24 = 14 Ls = 20 Eg = 26 46 = 26 2

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)
2-3	7	6	14	6	15	13	8	2	1	1
3-5	5	14	20	15	20	19	15	1	1	0
4-3	0	14	14	14	15	14	15	1	0	0

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