Transport Layer Flashcards

Question

Sender maintains timer for ___ unacked packet. What happens when timer expires?

Answer 1

each. when timer expires, retransmit only that unacked packet

Answer 2

Duplicate data possible

Answer 3

B sends Seq = 79, ACK = 43, Data = 'C' to B | A sends Seq = 43, ACK = 80 to B

Answer 4

SampleRTT: measured time from segment transmission until ACK receipt. DevRTT: SampleRTT deviation ``` EstimatedRTT = (1- a)*EstimatedRTT + a*SampleRTT, usual value of A = 1/8 DevRTT = (1-b)*DevRTT + b * |SampleRTT - EstimatedRTT| ``` TimeoutInterval = EstimatedRTT + 4*DevRTT

Answer 5

receiver controls sender, so sender won’t overflow receiver’s buffer by transmitting too much, too fast

Answer 6

rwnd - free buffer space | rcvBuffer - buffered data

Answer 7

will not overflow

Answer 8

Receiver computes rwnd = RcvBuffer-[LastByteRcvd - LastByteRead Sender computes x = LastByteSent - LastByteAcked If x<= rwnd then sender can send

Answer 9

To solve this problem, sender sends one data byte when rwnd = 0, if the sender receives corresponding ACK => rwnd != 0

Answer 10

Handshake is the process of establishing the connection Sender sends SynBit = 1, Seq = x, where x is the sequence number Receiver sends the acknowledgment: SynBit = 1, Seq = y, ACKBit = 1, ACKNum = x + 1 Sender sends to receiver back: ACKBIT = 1, ACKNum = y + 1

Answer 11

Sender sends FINBit = 1, seq = x to receiver Receiver sends ACKBit = 1, ACKNum = x + 1 After finishing sending the data, receiver sends FINbit=1, seq=y. After that sender sends ACKbit=1; ACKnum=y+1

Answer 12

Routers can drop the packet, due to the buffer being overflowed.

Answer 13

Duplicates

Answer 14

Controls the sender rate based on the congestion level in the network No congestion -> increase rate Congestion -> decrease rate

Answer 15

Roughly: rate = cwnd / RTT bytes/sec

Answer 16

LastByteSent- LastByteAcked <= min{cwnd, rwnd}

Answer 17

congestion window. dynamic, function of perceived network congestion.

Answer 18

sender increases transmission rate (cwnd = window size), probing for usable bandwidth, until loss occurs

Answer 19

increase cwnd by 1 MSS every RTT until loss detected

Answer 20

cut cwnd in half after loss

Answer 21

Slow Start, Congestion Avoidance, Fast Recovery

Answer 22

when connection begins, increase rate exponentially until first loss event: • initially cwnd = 1 MSS • double cwnd every RTT • done by incrementing cwnd for every ACK received initial rate is slow but ramps up exponentially fast

Answer 23

the congestion avoidance algorithm – increment cwnd by 1 MSS

Answer 24

The slow start algorithm is used when cwnd < ssthresh, otherwise, the congestion avoidance algorithm

Answer 25

loss by timeout: cwnd set to 1MSS loss by 3 duplicates: 1. TCP Reno - cwnd is cut in half window and adds 3 MSS 2. TCP Tahoe - set to 1

Answer 26

When cwnd gets to 1/2 of its value before timeout. Implementation: 1. variable ssthresh 2. on loss event, ssthresh is set to 1/2 of cwnd just before loss event

Answer 27

• avg. window size (# in-flight bytes) is ¾ W | avg. thruput is 3/4W per RTT

Answer 28

if K TCP sessions share same bottleneck link of bandwidth R, each should have average rate of R/K. Yes. Because two competing sessions: 1. additive increase gives slope of 1, as throughout increases 2. multiplicative decrease decreases throughput proportionally