Question

Lecture Exercise #12 Critical Path Method Determine the early start, early finish, late start, late finish times for each act
Lecture Exercise #13 Project Crashing Activity Predecessor(s) Time (weeks) Normal Crash 3 2 2 1 А B None None None с 1 1 Cost
0.00 0.01 0.05 0.02 0.06 0.08 2 0.07 Lecture Exercise #14 0.03 5120 5517 5910 0.04 5160 5557 5948 6331 6700 .7054 .7389 .7704
Lecture Exercise #15 O Consider the following payoff matrix: State of High Demand Low Demand Nature (Prob. = 0.7) (Prob. = 0.
0 0
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Answer #1

EXERCISE - 12:

Solution:

Activity Immediate Predecessors Duration
A - 6
B - 7
C A 3
D A 2
E B 4
F B 6
G C,E 10
H D,F 7


Edge and it's preceded and succeeded node

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

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

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=6]=0+6=6

E3=E1+t1,3 [t1,3=B=7]=0+7=7

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

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

=Max{6+3;7+4}

=Max{9;11}

=11

E5=Max{Ei+ti,5}[i=2,3]

=Max{E2+t2,5;E3+t3,5}

=Max{6+2;7+6}

=Max{8;13}

=13

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

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

=Max{11+10;13+7}

=Max{21;20}

=21


Backward Pass Method
L6=E6=21

L5=L6-t5,6 [t5,6=H=7]=21-7=14

L4=L6-t4,6 [t4,6=G=10]=21-10=11

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

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

=Min{14-6;11-4}

=Min{8;7}

=7

L2=Min{Lj-t2,j}[j=5,4]

=Min{L5-t2,5;L4-t2,4}

=Min{14-2;11-3}

=Min{12;8}

=8

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

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

=Min{7-7;8-6}

=Min{0;2}

=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 : 1-3-4-6 and critical activities are B,E,G

The total project time is 21

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

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)
1-2 6 0 6 0 8 6 2 2 0 0
2-4 3 6 11 8 11 9 8 2 2 0
2-5 2 6 13 8 14 8 12 6 5 3
3-5 6 7 13 7 14 13 8 1 0 0
5-6 7 13 21 14 21 20 14 1 1 0
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