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

 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)=10L_(3)=10
 D(2) D : 3->6
 6 E_(6)=21L_(6)=21
 G(8) G : 6->7
 7 E_(7)=29L_(7)=29
 H(3) H : 7->8
 8 E_(8)=32L_(8)=32
 1 E_(1)=0L_(1)=0
 A(4) A : 1->2
 2 E_(2)=4L_(2)=4
 E_(3)=10L_(3)=10
 d(0) d : 3->4
 E_(4)=10L_(4)=10
 F(4) F : 5->6
 E_(6)=21L_(6)=21
 E_(7)=29L_(7)=29
 E_(8)=32L_(8)=32
 E_(1)=0L_(1)=0
 E_(2)=4L_(2)=4
 C(2) C : 2->4
 4 E_(4)=10L_(4)=10
 E(7) E : 4->5
 5 E_(5)=17L_(5)=17
 E_(5)=17L_(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 timeStart(E_i)(3) (E_j)(4) (L_i)(5) Latest timeFinish(L_j)(6) Earliest timeFinish(E_i+t_(ij))(7)=(3)+(2) Latest timeStart(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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