Question

Consider the following information about a small project:

Problem 15.6 Task Duration Estimate (in Days) Immediate Predecessors None А B A B and D E and C

b-1. Identify the critical path.

• A. A-D-E-F

• b. A-B-C-F

• C. A-B-E-F

Identify the earliest start and finish, and slack for each task. (Leave no cells blank - be certain to enter "0" wherever required.)

c. Which of these activities should the project manager track most closely?

• A. A-B-E-F

• b. A-D-E-F

• c. A-B-C-F

Answer 1

Solution:

Activity	Immediate Predecessors	Duration
A	-	2
B	A	5
C	B	1
D	A	2
E	B AND D	3
F	E AND C	12

Edge and it's preceded and succeeded node

Edge	Node1 → Node2
A	1→2
A	1→5
A	1→6
B	2→3
D	2→4
d	2→6
C	3→6
d	4→6

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

B(5 3 C(1) d(0) C(1) A(2) D(2) 4 C(1) A(2) 5 d(0) d(0) C(1) A(2) d(0) d(0) A( 2 A (2) 6

Forward Pass Method
E1=0

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

E3=E2+t2,3 [t2,3=B=5]=2+5=7

E4=E2+t2,4 [t2,4=D=2]=2+2=4

E5=E1+t1,5 [t1,5=A=2]=0+2=2

E6=Max{Ei+ti,6}[i=1,2,3,4]

=Max{E1+t1,6;E2+t2,6;E3+t3,6;E4+t4,6}

=Max{0+2;2+0;7+1;4+0}

=Max{2;2;8;4}

=8


Backward Pass Method
L6=E6=8

L4=L6-t4,6 [t4,6=d=0]=8-0=8

L3=L6-t3,6 [t3,6=C=1]=8-1=7

L2=Min{Lj-t2,j}[j=6,4,3]

=Min{L6-t2,6;L4-t2,4;L3-t2,3}

=Min{8-0;8-2;7-5}

=Min{8;6;2}

=2

L1=Min{Lj-t1,j}[j=6,5,2]

=Min{L6-t1,6;L5-t1,5;L2-t1,2}

=Min{8-2;8-2;2-2}

=Min{6;6;0}

=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 : 1-2-3-6 and critical activities are A,B,C

The total project time is 8
The network diagram for the project, along with E-values and L-values, is

B(5) L 4 = 8 1 A(2) 5 A( 2 A (2) 6 E = 8 Lo = 8

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)
1-5	2	0	2	0	8	2	6	6	0	0
1-6	2	0	8	0	8	2	6	6	6	6
2-4	2	2	4	2	8	4	6	4	0	0
2-6	0	2	8	2	8	2	8	6	6	6
4-6	0	4	8	8	8	4	8	4	4	0

Please make a note that Revised Chegg Guidelines seriously restrict us in answering only one question! However, I've answered first few questions so that you would rate this answer positively. DO NOT VIOLATE Chegg guidelines as they may have a huge impact on your account. So please understand and rate this answer with thumbs up kindly. It is a gesture that shows you support me. Thank you