2: Total understanding
1: Partial understanding
0: No understanding


Consider a reliable sliding window protocol with:
	 Maximum packet size: 1KB.
	 Maximum network BW: 1MB/s
	 Latency: 100ms.
	 Sequence numbers describe bytes.
	 
Assume that the network loses, but does not reorder, packets.	 
(the high speed of the network means that packets can be sent very
quickly, and received very quickly)


0. 12 eggs/hour * 8 hours /day = 96 eggs/day

	 
1. In steady state, assuming no loss, what is the maximum BW that such
   a  protocol can deliver?

1MB/sec.  This is the maximum network bandwidth. No protocol could do better than this.

   

2. To achieve this, how big (in bits) must the sequence number be?
   Sequence number should cover bandwidth-delay (RTT) product
      B: 1MB/s
      D: 2L
      BD = 200KB
      
     My sequence number needs to cover 200KB.
     	2^17 = 128
	2^18 = 256
     
     2^18.
     
     18 bits.


     OR, another way to see this:
     
             
     I need to keep the pipe full, and it takes me 200ms to get an
     ack. 
     200ms worth of data is 200KB.
     
	Send[200ms worth of data]
	Ack for first packet arrives
	Send next pkt
	Ack for second pkt arrives
	...
	
     
     So, I need to describe 200ms worth of sent data by a sequence
     number.
     
     
   
3. How big (# bits) must the advertised window size be in order to
   ensure a BW of at least .1MB/s
   
   This question is the same as "must the window size be" since it
   makes no sense to have an advertised window size larger than the
   window size.
   
   So, how many bits do I need to "cover" a BD of .1MB/s * .2 seconds?
   
   .1 * .2 = .02 MB = 20KB
   = 2^15
   
   = 15 bits.