Question

fe 1 se in C d 110-51 k Pr (e er 0 lea a s879 2 ase dvoa. 8 u32435324 si, b g el ti y nh abl (ci 78666847 sdOpe sdhi dd etap ona SSS-2235 els s n it h io io ta on re -iv ABCDEFGH vrr 2 crro FF+S 3

Answer 1

Activity	Immediate Predecessors	Duration
A	-	7
B	-	8
C	-	6
D	A	6
E	B	6
F	B	8
G	C	4
H	E, D	7

L5 = 14 L, = 14 2 2 2 8 A(7) J..,.: 5 E(6 F(8) G(4) E,-81 2 6 L, = 0 2 E 21 し6-21 2 4-1가 4

Forward Pass Method
E1=0
E2=0+7=7
E3=0+8=8
E4=0+6=6

E5 =Max{13;14} =14
E6= Max{16;10;21} =21

Backward Pass Method
L6=E6=21
L5=21-7=14
L4=21-4=17
L3= Min{13;8} =8
L2=14-6=8
L1=Min{11;0;1} =0
The critical path of the project is: 1-3-5-6 and critical activities are B, E, H
The total project time is 21.

Earliest time Start Latest time Earliest time Latest time Start Duration Finish Finish Total Float Free Float Independent Float Activity (ii)) . (4) (4) (5)(4) (4) (4%) (4%) (4-4)-Ei (4-E)-'ij (4-Li)-'ij E.-E)-t 1-4 2-5 3-6 6 6 8 6 14 8 218 21 17 0 6 13 16 17 8 21 13 21 17

Among B, E and H activity, E can be reduced as it is having the lowest cost i.e. 6