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.