Question

Question 2 (20%): A precedence table is given below for a process with 6 tasks, their immediate predecessors and their task times: Immediate Task time label predecessors minutes) 2.0 1.5 2.0 3.0 15 2.0 Daily working time is 14 hours. The daily output need is 210 piece a) Draw the relationship structure of these tasks? 8) Calculate the cycle sims? Use the line balancing technique to determine the number of stations and arrange these tasks to each station? Calculate the efficiency of this system?

Answer 1

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a	-	2.0
b	-	1.5
c	-	2.0
d	a	3.0
e	b,c	1.5
f	d,e	2.0

Solution:

Activity	Immediate Predecessors	Duration
A	-	2.0
B	-	1.5
C	-	2.0
D	A	3.0
E	B,C	1.5
F	D,E	2.0

Edge and it's preceded and succeeded node

Edge	Node1 → Node2
A	1→2
B	1→3
C	1→4
D	2→5
d	3→4
E	4→5
F	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=2]=0+2=2

E3=E1+t1,3 [t1,3=B=1.5]=0+1.5=1.5

E4=Max{Ei+ti,4}[i=1,3]

=Max{E1+t1,4;E3+t3,4}

=Max{0+2;1.5+0}

=Max{2;1.5}

=2

E5=Max{Ei+ti,5}[i=2,4]

=Max{E2+t2,5;E4+t4,5}

=Max{2+3;2+1.5}

=Max{5;3.5}

=5

E6=E5+t5,6 [t5,6=F=2]=5+2=7


Backward Pass Method
L6=E6=7

L5=L6-t5,6 [t5,6=F=2]=7-2=5

L4=L5-t4,5 [t4,5=E=1.5]=5-1.5=3.5

L3=L4-t3,4 [t3,4=d=0]=3.5-0=3.5

L2=L5-t2,5 [t2,5=D=3]=5-3=2

L1=Min{Lj-t1,j}[j=4,3,2]

=Min{L4-t1,4;L3-t1,3;L2-t1,2}

=Min{3.5-2;3.5-1.5;2-2}

=Min{1.5;2;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 : 1-2-5-6 and critical activities are A,D,F

The total project time is 7
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)
1-3	1.5	0	1.5	0	3.5	1.5	2	2	0	0
1-4	2	0	2	0	3.5	2	1.5	1.5	0	0
3-4	0	1.5	2	3.5	3.5	1.5	3.5	2	0.5	-1.5
4-5	1.5	2	5	3.5	5	3.5	3.5	1.5	1.5	0