A. Suppose you have a connection with an RTT of 30ms and negligible packetizatio

Question

A. Suppose you have a connection with an RTT of 30ms and negligible packetization delay. Your maximum TCP segment size is 1.4kB. You want to download a web page that is 29.4kB in size. Approximately how much longer will a connection with an initial window of 2.8kB take than one with an initial congestion window of 14kB? For example, if the large window takes 10 RTTs and the small window takes 25 RTTs, it takes 2.5 times as long, and write 2.5.

B. Consider a simplified form of 64-QAM modulation which supports a bit rate of 150 Mbps. Assuming it uses a coding gain of 5/6, how many symbols does it put on the wire per second (in millions)?

C. Suppose you have a reasonably large BitTorrent swarm where half of all nodes are behind NATs for which no traversal techniques work. What percentage (integer 0 to 100) of peer pairs cannot share data?

D. Suppose you have a swarm where 80% of all nodes are behind NATs for which no traversal techniques work. Your client wishes to maintain open connections to at least 100 peers. What is the smallest swarm that will on average allow your client, which is behind a NAT, to have 100 peers?

Explanation / Answer

1. Given:

RTT = 30 ms, TCP segment size(MSS)=1.4 KB, Data size (web page) (Throughput)= 29.4 KB

i) If congestion window (W) = 2.8 KB,

No. of segments per window = w = W/MSS = 2

Sending rate = w*MSS/RTT = 2*1.4/30 = 0.093 KBs/ms

Hence time required to send = 29.4/0.093 = 316.13 ms

Hence no. of RTTs = 316.13/RTT = 316.13/30 = 10.54 RTTs

ii) If congestion window (W) = 14 KB,

No. of segments per window = w = W/MSS = 14/1.4 =10

Sending rate = w*MSS/RTT = 10*1.4/30 = 0.467 KBs/ms

Hence time required to send = 29.4/0.467 = 62.96 ms

Hence no. of RTTs = 62.96/RTT = 316.13/30 = 2.099 RTTs

Hence, longer time taken by small window in comparison to larger window = 10.54/2.099 = 5.02

2. Given:

Bit rate = 150 MBps = 1.5e+8 bps

Coding gain = 5/6

For 64-QAM, bits/smbol =6

For a normal signal, symbol rate is given as:

Symbol rate = bit rate/bits/symbol = 1.5e+8/6 = 0.25e+8 symbols/s = 2.5e+7 symbols/s = 25.0e+6 symbols/s = 25.0e+6/1.0e+6 million symbol/s = 25.0 million symbols/s

On applying the coding gain, we get

Symbol rate = (5/6)*25.0 = 20.83 million symbols/s

3. NATs may affect the performance of P2P networks like BitTorrent. If traversal of NAT is not possible, then applications running behind NAT cannot receive incoming connection requests from other peers.Hence, if almost half of nodes are behind NATs with no traversal possible, then a large chunk of peer pairs (50% in theory) will not be able to share data. Although there may be ways to connect them.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.