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4. Critical path, Total float, Free float, Independent float : Activity, Predecessors, Duration example ( Enter your problem )

 3. Network diagram : Activity i-j, Name of Activity (Previous method) 5. Critical path, Total float, Free float, Independent float : Activity i-j, Duration (Next method)

### 1. Example-1

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

Solution:
 Activity Immediate Predecessors Duration A - 2 B - 4 C - 3 D A 1 E B 6 F C 5 G D,E 7 H F,G 2

Edge and it's preceded and succeeded node
 Edge Node1 -> Node2 A 1->2 B 1->3 C 1->4 D 2->5 E 3->5 F 4->6 G 5->6 H 6->7

The network diagram for the project, along with activity time, is
 2
 A(2) A : 1->2
 D(1) D : 2->5 E(6) E : 3->5
 5
 1
 B(4) B : 1->3
 3
 G(7) G : 5->6
 C(3) C : 1->4
 G(7) G : 5->6
 4
 F(5) F : 4->6
 6
 H(2) H : 6->7
 7

Forward Pass Method
E_1=0

E_2=E_1 + t_(1,2) [t_(1,2) = A = 2]=0 + 2=2

E_3=E_1 + t_(1,3) [t_(1,3) = B = 4]=0 + 4=4

E_4=E_1 + t_(1,4) [t_(1,4) = C = 3]=0 + 3=3

E_5=Max{E_i + t_(i,5)} [i=2, 3]

=Max{E_2 + t_(2,5); E_3 + t_(3,5)}

=Max{2 + 1; 4 + 6}

=Max{3; 10}

=10

E_6=Max{E_i + t_(i,6)} [i=4, 5]

=Max{E_4 + t_(4,6); E_5 + t_(5,6)}

=Max{3 + 5; 10 + 7}

=Max{8; 17}

=17

E_7=E_6 + t_(6,7) [t_(6,7) = H = 2]=17 + 2=19

Backward Pass Method
L_7=E_7=19

L_6=L_7 - t_(6,7) [t_(6,7) = H = 2]=19 - 2=17

L_5=L_6 - t_(5,6) [t_(5,6) = G = 7]=17 - 7=10

L_4=L_6 - t_(4,6) [t_(4,6) = F = 5]=17 - 5=12

L_3=L_5 - t_(3,5) [t_(3,5) = E = 6]=10 - 6=4

L_2=L_5 - t_(2,5) [t_(2,5) = D = 1]=10 - 1=9

L_1=text{Min}{L_j - t_(1,j)} [j=4, 3, 2]

=text{Min}{L_4 - t_(1,4); L_3 - t_(1,3); L_2 - t_(1,2)}

=text{Min}{12 - 3; 4 - 4; 9 - 2}

=text{Min}{9; 0; 7}

=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 : 1-3-5-6-7 and critical activities are B,E,G,H

The total project time is 19
The network diagram for the project, along with E-values and L-values, is
 2 E_(2)=2L_(2)=9
 E_(5)=10L_(5)=10
 A(2) A : 1->2
 E_(2)=2L_(2)=9
 D(1) D : 2->5 E(6) E : 3->5
 5 E_(5)=10L_(5)=10
 1 E_(1)=0L_(1)=0
 B(4) B : 1->3
 3 E_(3)=4L_(3)=4
 G(7) G : 5->6
 E_(1)=0L_(1)=0
 C(3) C : 1->4
 E_(3)=4L_(3)=4
 E_(6)=17L_(6)=17
 G(7) G : 5->6
 E_(7)=19L_(7)=19
 E_(4)=3L_(4)=12
 4 E_(4)=3L_(4)=12
 F(5) F : 4->6
 6 E_(6)=17L_(6)=17
 H(2) H : 6->7
 7 E_(7)=19L_(7)=19

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) 1-2 2 t_(1,2)=2 0 E_1=0 2 E_2=2 0 L_1=0 9 L_2=9 2 2=0+2(E_i+t_(ij)) 7 7=9-2(L_j-t_(ij)) 7 7=7-0(L_j-t_(ij))-E_i 0 0=(2-0)-2(E_j-E_i)-t_(ij) 0 0=(2-0)-2(E_j-L_i)-t_(ij) 1-4 3 t_(1,4)=3 0 E_1=0 3 E_4=3 0 L_1=0 12 L_4=12 3 3=0+3(E_i+t_(ij)) 9 9=12-3(L_j-t_(ij)) 9 9=9-0(L_j-t_(ij))-E_i 0 0=(3-0)-3(E_j-E_i)-t_(ij) 0 0=(3-0)-3(E_j-L_i)-t_(ij) 2-5 1 t_(2,5)=1 2 E_2=2 10 E_5=10 9 L_2=9 10 L_5=10 3 3=2+1(E_i+t_(ij)) 9 9=10-1(L_j-t_(ij)) 7 7=9-2(L_j-t_(ij))-E_i 7 7=(10-2)-1(E_j-E_i)-t_(ij) 0 0=(10-9)-1(E_j-L_i)-t_(ij) 4-6 5 t_(4,6)=5 3 E_4=3 17 E_6=17 12 L_4=12 17 L_6=17 8 8=3+5(E_i+t_(ij)) 12 12=17-5(L_j-t_(ij)) 9 9=12-3(L_j-t_(ij))-E_i 9 9=(17-3)-5(E_j-E_i)-t_(ij) 0 0=(17-12)-5(E_j-L_i)-t_(ij)

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

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