Critical path, Total float, Free float, Independent float : Activity i-j, Name of Activity, Duration Example-1

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6. Critical path, Total float, Free float, Independent float : Activity i-j, Name of Activity, Duration example ( Enter your problem )

( Enter your problem )

Example-1

Other related methods

5. Critical path, Total float, Free float, Independent float : Activity i-j, Duration
(Previous method)

7. Project scheduling : Activity, Predecessors, to, tm, tp
(Next method)

1. Example-1

1. Critical path, Total float, Free float, Independent float
1-2 A 4
2-3 B 6
2-4 C 2
3-4 d 0
3-6 D 2
4-5 E 7
5-6 F 4
6-7 G 8
7-8 H 3

Solution:
The given problem is

Activity	Activity	Duration
1-2	A	4
2-3	B	6
2-4	C	2
3-4	d	0
3-6	D	2
4-5	E	7
5-6	F	4
6-7	G	8
7-8	H	3

Edge and it's preceded and succeeded node

Edge	Node1 $->$ Node2
A	1 $->$ 2
B	2 $->$ 3
C	2 $->$ 4
d	3 $->$ 4
D	3 $->$ 6
E	4 $->$ 5
F	5 $->$ 6
G	6 $->$ 7
H	7 $->$ 8

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

B(6)

$B : 2->3$

3

D(2)

$D : 3->6$

6

G(8)

$G : 6->7$

7

H(3)

$H : 7->8$

8

1

A(4)

$A : 1->2$

2

d(0)

$d : 3->4$

F(4)

$F : 5->6$

C(2)

$C : 2->4$

4

E(7)

$E : 4->5$

5

Forward Pass Method

$E_1=0$

$E_2=E_1 + t_(1,2)$ [

$t_(1,2) = A = 4$ ]

$=0 + 4$

$=4$

$E_3=E_2 + t_(2,3)$ [

$t_(2,3) = B = 6$ ]

$=4 + 6$

$=10$

$E_4=Max{E_i + t_(i,4)} [i=2, 3]$

$=Max{E_2 + t_(2,4); E_3 + t_(3,4)}$

$=Max{4 + 2; 10 + 0}$

$=Max{6; 10}$

$=10$

$E_5=E_4 + t_(4,5)$ [

$t_(4,5) = E = 7$ ]

$=10 + 7$

$=17$

$E_6=Max{E_i + t_(i,6)} [i=3, 5]$

$=Max{E_3 + t_(3,6); E_5 + t_(5,6)}$

$=Max{10 + 2; 17 + 4}$

$=Max{12; 21}$

$=21$

$E_7=E_6 + t_(6,7)$ [

$t_(6,7) = G = 8$ ]

$=21 + 8$

$=29$

$E_8=E_7 + t_(7,8)$ [

$t_(7,8) = H = 3$ ]

$=29 + 3$

$=32$

Backward Pass Method

$L_8=E_8=32$

$L_7=L_8 - t_(7,8)$ [

$t_(7,8) = H = 3$ ]

$=32 - 3$

$=29$

$L_6=L_7 - t_(6,7)$ [

$t_(6,7) = G = 8$ ]

$=29 - 8$

$=21$

$L_5=L_6 - t_(5,6)$ [

$t_(5,6) = F = 4$ ]

$=21 - 4$

$=17$

$L_4=L_5 - t_(4,5)$ [

$t_(4,5) = E = 7$ ]

$=17 - 7$

$=10$

$L_3=text{Min}{L_j - t_(3,j)} [j=6, 4]$

$=text{Min}{L_6 - t_(3,6); L_4 - t_(3,4)}$

$=text{Min}{21 - 2; 10 - 0}$

$=text{Min}{19; 10}$

$=10$

$L_2=text{Min}{L_j - t_(2,j)} [j=4, 3]$

$=text{Min}{L_4 - t_(2,4); L_3 - t_(2,3)}$

$=text{Min}{10 - 2; 10 - 6}$

$=text{Min}{8; 4}$

$=4$

$L_1=L_2 - t_(1,2)$ [

$t_(1,2) = A = 4$ ]

$=4 - 4$

$=0$

(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 :

$1-2-3-4-5-6-7-8$ and critical activities are

$A,B,d,E,F,G,H$

The total project time is 32
The network diagram for the project, along with E-values and L-values, is

B(6)

$B : 2->3$

3

$E_(3)=10$

$L_(3)=10$

D(2)

$D : 3->6$

6

$E_(6)=21$

$L_(6)=21$

G(8)

$G : 6->7$

7

$E_(7)=29$

$L_(7)=29$

H(3)

$H : 7->8$

8

$E_(8)=32$

$L_(8)=32$

1

$E_(1)=0$

$L_(1)=0$

A(4)

$A : 1->2$

2

$E_(2)=4$

$L_(2)=4$

$E_(3)=10$

$L_(3)=10$

d(0)

$d : 3->4$

$E_(4)=10$

$L_(4)=10$

F(4)

$F : 5->6$

$E_(6)=21$

$L_(6)=21$

$E_(7)=29$

$L_(7)=29$

$E_(8)=32$

$L_(8)=32$

$E_(1)=0$

$L_(1)=0$

$E_(2)=4$

$L_(2)=4$

C(2)

$C : 2->4$

4

$E_(4)=10$

$L_(4)=10$

E(7)

$E : 4->5$

5

$E_(5)=17$

$L_(5)=17$

$E_(5)=17$

$L_(5)=17$

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

Activity $(i,j)$ $(1)$	Duration $(t_(ij))$ $(2)$	Earliest time Start $(E_i)$ $(3)$	$(E_j)$ $(4)$	$(L_i)$ $(5)$	Latest time Finish $(L_j)$ $(6)$	Earliest time Finish $(E_i+t_(ij))$ $(7)=(3)+(2)$	Latest time Start $(L_j-t_(ij))$ $(8)=(6)-(2)$	Total Float $(L_j-t_(ij))-E_i$ $(9)=(8)-(3)$	Free Float $(E_j-E_i)-t_(ij)$ $(10)=((4)-(3))-(2)$	Independent Float $(E_j-L_i)-t_(ij)$ $(11)=((4)-(5))-(2)$
2-4	2 $t_(2,4)=2$	4 $E_2=4$	10 $E_4=10$	4 $L_2=4$	10 $L_4=10$	6 $6=4+2$ $(E_i+t_(ij))$	8 $8=10-2$ $(L_j-t_(ij))$	4 $4=8-4$ $(L_j-t_(ij))-E_i$	4 $4=(10-4)-2$ $(E_j-E_i)-t_(ij)$	4 $4=(10-4)-2$ $(E_j-L_i)-t_(ij)$
3-6	2 $t_(3,6)=2$	10 $E_3=10$	21 $E_6=21$	10 $L_3=10$	21 $L_6=21$	12 $12=10+2$ $(E_i+t_(ij))$	19 $19=21-2$ $(L_j-t_(ij))$	9 $9=19-10$ $(L_j-t_(ij))-E_i$	9 $9=(21-10)-2$ $(E_j-E_i)-t_(ij)$	9 $9=(21-10)-2$ $(E_j-L_i)-t_(ij)$

This material is intended as a summary. Use your textbook for detail explanation.
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5. Critical path, Total float, Free float, Independent float : Activity i-j, Duration
(Previous method)

7. Project scheduling : Activity, Predecessors, to, tm, tp
(Next method)

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