Question

Q1: Use the network below to answer the following questions: D 7 G 8 E 5 C H 10 F 3 B 3 i. ii. Assume the duration is given in weeks(numbers shown above), how long will it take to finish this project? Show your work. (4 marks) Determine the critical path. (1 mark) List the activities that could be delayed without affecting the submission of the project and determine the duration of this delay ( 1 Marks) iii.

Answer 1

The given problem is

Activity	Activity	Duration (in weeks)
1-2	A	8
1-3	B	10
2-3	C	3
2-4	D	7
3-4	E	6
3-5	F	7
4-6	G	5
5-6	H	3

Edge and it's preceded and succeeded node

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

Forward Pass Method
E1=0

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

E3=Max{Ei+ti,3}[i=1,2]

=Max{E1+t1,3;E2+t2,3}

=Max{0+10;8+3}

=Max{10;11}

=11

E4=Max{Ei+ti,4}[i=2,3]

=Max{E2+t2,4;E3+t3,4}

=Max{8+7;11+6}

=Max{15;17}

=17

E5=E3+t3,5 [t3,5=F=7]=11+7=18

E6=Max{Ei+ti,6}[i=4,5]

=Max{E4+t4,6;E5+t5,6}

=Max{17+5;18+3}

=Max{22;21}

=22

Backward Pass Method
L6=E6=22

L5=L6-t5,6 [t5,6=H=3]=22-3=19

L4=L6-t4,6 [t4,6=G=5]=22-5=17

L3=Min{Lj-t3,j}[j=5,4]

=Min{L5-t3,5;L4-t3,4}

=Min{19-7;17-6}

=Min{12;11}

=11

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

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

=Min{17-7;11-3}

=Min{10;8}

=8

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

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

=Min{11-10;8-8}

=Min{1;0}

=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: 1-2-3-4-6 and critical activities are A, C, E, G

The total project time taken ti finish is 22 weeks.
The network diagram for the project, along with E-values and L-values, is

E2=8 L2=8 D(7) 2 4 G(5) A(8) E4=17 L4=17 C(3) E1=0 L1=0 1 E(6) 6 H(3) E6=22 L6=22 B(10) F(7) 3 5 E3=11 L3=11 E5=18 L5=18

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	10	0	11	0	11	10	1	1	1	1
2-4	7	8	17	8	17	15	10	2	2	2
3-5	7	11	18	11	19	18	12	1	0	0
5-6	3	18	22	19	22	21	19	1	1	0