Answer:
By Using the Greedy- Activity- Selection algorithm
Explanation:
The Greedy- Activity- Selection algorithm in this case involves
First finding a maximum size set S1, of compatible activities from S for the first lecture hall.
Then using it again to find a maximum size set S2 of compatible activities from S - S1 for the second hall.
This is repeated till all the activities are assigned.
It requires θ(n2) time in its worse .
Packet loss is typically resolved by time-outs and retries. For applications where a duplicate operation doesn't matter this is acceptable.
