Question

Question 24 (20 points) Use the following table of related activities with their activity times (weeks). Q24-1) Type your answers by the order of ES, EL, LS, LF, Slack for each activity. For example, A. Х, Х, Х, Х, Х Q24-2) Determine the critical path. Immediate Activity Activity Predecessor Time ESEFLSLqslack (weeks 4 C A DIB E A FIB G C,D H |D,F 7 6 5 4 6 G, H, I J

Answer 1

Computation of earliest starting and finishing times is aided by two simple rules:
1. The earliest finish time for any activity is equal to its earliest start time plus its expected
duration, t:
EF = ES + t
2. ES for activities at nodes with one entering arrow is equal to EF of the entering arrow. ES
for activities leaving nodes with multiple entering arrows is equal to the largest EF of the
entering arrow.
Computation of the latest starting and finishing times is aided by the use of two rules:
1. The latest starting time for each activity is equal to its latest finishing time minus its
expected duration:
LS = LF - t
2. For nodes with one leaving arrow, LF for arrows entering that node equals the LS of the
leaving arrow. For nodes with multiple leaving arrows, LF for arrows entering that node
equals the smallest LS of leaving arrows.
Finding ES and EF times involves a forward pass through the network; finding LS and LF
times involves a backward pass through the network. Hence, we must begin with the EF of the
last activity and use that time as the LF for the last activity. Then we obtain the LS for the last
activity by subtracting its expected duration from its LF
Slack = LS-ES = LF-EF
Activities for which Slack = 0 are in critical path

	Expected time	ES	EF	LS	LF	Slack
A	2	0	2	6	8	6
B	4	0	4	0	4	0
C	3	2	5	8	11	6
D	7	4	11	4	11	0
E	6	2	8	8	14	6
F	5	4	9	11	16	7
G	9	11	20	11	20	0
H	4	11	15	16	20	5
I	6	8	14	14	20	6
J	7	20	27	20	27	0

Critical path is B-D-G-J

Completion time is 27 weeks