Activities A to K are required to complete a project. The time estimates and the immediate predecessors of these activities are given in the table. If the project is to be completed in the minimum possible time, the latest finish time for the activity G is ____________ hours

Activity	Time (hours)	Immediate predecessors
A	2	-
B	3	-
C	2	-
D	4	A
E	5	B
F	4	B
G	3	C
H	10	D, E
I	5	F
J	8	G
K	3	H, I, J

The correct answer is 10.

Step-by-Step Explanation:

To find the latest finish time for activity G, we need to construct the project network and perform both forward and backward pass calculations using the Critical Path Method (CPM).

1. Earliest Start (ES) and Earliest Finish (EF) Times (Forward Pass):
The forward pass starts from the beginning of the project (time = 0) and moves forward to find the earliest possible start and finish times for each activity. The formula for the Earliest Finish time is:
$$\mathrm{EF}=\mathrm{ES}+\mathrm{Duration}$$
If an activity has multiple predecessors, its ES is the maximum of the EF times of all its immediate predecessors:
$${\mathrm{ES}}_{\mathrm{activity}}=max\left({\mathrm{EF}}_{\mathrm{predecessors}}\right)$$
Let's compute these values for all activities:
• For start activities with no predecessors (A, B, and C):
ES(A) = 0 → EF(A) = 0 + 2 = 2 hours
ES(B) = 0 → EF(B) = 0 + 3 = 3 hours
ES(C) = 0 → EF(C) = 0 + 2 = 2 hours

• For activities depending on the start activities:
ES(D) = EF(A) = 2 → EF(D) = 2 + 4 = 6 hours
ES(E) = EF(B) = 3 → EF(E) = 3 + 5 = 8 hours
ES(F) = EF(B) = 3 → EF(F) = 3 + 4 = 7 hours
ES(G) = EF(C) = 2 → EF(G) = 2 + 3 = 5 hours

• For subsequent activities:
ES(H) = max(EF(D), EF(E)) = max(6, 8) = 8 → EF(H) = 8 + 10 = 18 hours
ES(I) = EF(F) = 7 → EF(I) = 7 + 5 = 12 hours
ES(J) = EF(G) = 5 → EF(J) = 5 + 8 = 13 hours

• For the final activity K:
ES(K) = max(EF(H), EF(I), EF(J)) = max(18, 12, 13) = 18 → EF(K) = 18 + 3 = 21 hours

The minimum total project duration is therefore 21 hours.

2. Latest Start (LS) and Latest Finish (LF) Times (Backward Pass):
The backward pass starts from the final activity K (setting its LF equal to the project duration of 21 hours) and moves backward to calculate the latest times activities can finish without delaying the project. The formula for the Latest Start time is:
$$\mathrm{LS}=\mathrm{LF}\&\#x2212;\mathrm{Duration}$$
Let's compute the latest finish and latest start times starting from the end:
• For activity K:
LF(K) = 21 → LS(K) = 21 − 3 = 18 hours

• For predecessor activities of K (H, I, and J):
Since K is the only immediate successor of H, I, and J, their LF values are equal to LS(K):
LF(H) = LS(K) = 18 → LS(H) = 18 − 10 = 8 hours
LF(I) = LS(K) = 18 → LS(I) = 18 − 5 = 13 hours
LF(J) = LS(K) = 18 → LS(J) = 18 − 8 = 10 hours

• For activity G:
Since activity J is the immediate successor of activity G, the latest finish time of G is equal to the latest start time of J:
LF(G) = LS(J) = 10 hours

Thus, the latest finish time for activity G is 10 hours.