"Transport Layer: UDP, TCP, Flow & Congestion Control, Sockets" — Practice

Question 1

What four values does TCP use to demultiplex an incoming segment, and what does UDP use?

Accepted Answer

TCP uses source address, source port, destination address and destination port. UDP uses the destination port alone.

Question 2

How long does TIME_WAIT last and what does it protect against?

Accepted Answer

Twice the maximum segment lifetime. It ensures the final acknowledgement reaches the peer and prevents any delayed segment from the old connection appearing inside a new one on the same port pair.

Question 3

Does a timeout or three duplicate acknowledgements indicate more severe congestion, and how does TCP respond differently?

Accepted Answer

A timeout is more severe, since nothing is getting through, so the window drops to one segment and slow start restarts. Three duplicates mean later segments are arriving, so Reno halves the window and stays in congestion avoidance.

Question 4

With MSS 1 KB and initial threshold 8, trace the congestion window for 6 round trips under slow start and congestion avoidance.

Accepted Answer

Rounds 1 to 4 give 1, 2, 4, 8 by doubling. On reaching the threshold of 8, congestion avoidance begins, so rounds 5 and 6 give 9 and 10.

Question 5

Continuing that trace, a timeout occurs after round 6 and three duplicate acknowledgements after round 10. Give the window for rounds 7 to 12 under Reno.

Accepted Answer

The timeout sets the threshold to 5 and the window to 1, so rounds 7 to 9 give 1, 2, 4 and round 10 gives 5, capped at the threshold, entering congestion avoidance. Fast recovery then halves 5 to 2, so rounds 11 and 12 give 2 and 3.

Question 6

Samples of 100, 120, 90 and 110 ms arrive with initial estimate 100 and deviation 10. Compute the final timeout.

Accepted Answer

The estimates progress 100, 102.5, 100.94, 102.07 and the deviations 7.5, 10.625, 11.09, 10.58. The timeout is 102.07 plus 4 times 10.58, which is about 144.4 milliseconds.

Question 7

Explain silly window syndrome and give both fixes.

Accepted Answer

A receiver whose application consumes a few bytes at a time advertises tiny windows, so the sender transmits minimal segments carrying 40 bytes of header each. Clark's receiver-side fix advertises zero until half the buffer or one full segment is free, and Nagle's sender-side fix buffers small writes while any data is unacknowledged.

Question 8

Show why a two-way handshake is insufficient, using a delayed duplicate SYN.

Accepted Answer

The server would allocate state and reply on receiving the old SYN, and the client, not opening a connection, would discard the reply, leaving a half-open connection consuming resources. Old data segments from that earlier connection could then be accepted as current. The third message makes the server commit only after the client acknowledges the server's own sequence number.

Question 9

At 10 Gbps, how long until the 32-bit sequence space wraps, and what breaks?

Accepted Answer

The space is about 4.29 gigabytes and the byte rate is 1.25 gigabytes per second, so it wraps in about 3.4 seconds. Since the maximum segment lifetime is about 2 minutes, a delayed segment can reappear inside the current valid window and be accepted as new data, which the timestamp option prevents.

Question 10

With RTT 50 ms, MSS 1460 bytes and loss 0.01 percent, estimate throughput, then recompute at 1 percent loss.

Accepted Answer

MSS over RTT is 29200 bytes per second. At p equal to 0.0001, one over the square root of p is 100, giving about 2.92 MB/s or 23 Mbps. At p equal to 0.01 the factor is 10, giving about 292 KB/s or 2.3 Mbps, so a hundredfold loss increase caused a tenfold throughput drop.

Practice — "Transport Layer: UDP, TCP, Flow & Congestion Control, Sockets"

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