Chapter 3: The Transport Layer

Question 1

In a postal exchange, what is analogous to:
a) the transport-layer protocol
b) the network-layer protocol

Which layer is communication between processes, and which is between hosts?

Answer 1

a) The person who picks up the mail from the letterbox.
b) The postal service.

Transport is between processes, and network is between hosts.

Question 2

Why is IP said to be an unreliable service?

Answer 2

It does not guarantee that segments are delivered, segments are delivered in sequence or that the segments are unaffected.

Question 3

What is multiplexing in the context of transport-layer applications?

Answer 3

When information from the application layer is received by the transport layer from different sockets, it is mixed together and passed to the network layer.

Question 4

How is demultiplexing performed in UDP?

Concretely, if Host A sends a UDP message from port 19157 to Host B at port 46428, how is this achieved?

Answer 4

Host A creates a transport layer segment: (19157, 46428, _, _, data) and passes it to the network layer.

If the segment reaches Host B, the transport layer examines the destination port number in the segment, and delivers it to that socket.

Question 5

How is demultiplexing performed in TCP?

Answer 5

A TCP socket has (sip, dip, sport, dport), and this uniquely determines where it is directed to.

Question 6

Why is UDP described as connectionless?

Answer 6

There is no handshake.

Question 7

Give an example of an application layer protocol that uses UDP.

Answer 7

DNS, HTTP/3

Question 8

Give some benefits of UDP over TCP.

Answer 8

Finer application-level control.
No connection establishment.
No connection state, so more clients possible.
Small packet header overhead.

Question 9

What is the packet structure of UDP?

Answer 9

16 bits for:
sport
dport
length
checksum

Application data

Question 10

How is the UDP checksum implemented?

Answer 10

The 1s complement of the sum of the binary words is taken and stored.

At the recipient, the sum of all the words and the checksum should give all 1s.

Question 11

In a completely reliable and integral transfer medium, what is the sending protocol and receiving protocol?

Answer 11

Sending: Receives data from the application is layer, and then encapsulates the packet before passing to the network layer.

Receiver: Takes a packet from the network layer, extracts the data and delivers it to the application layer.

Question 12

In a completely reliable transfer medium with possible bit errors, what is the sending protocol and receiving protocol?

Answer 12

Sender: Receives data from application layer, encapsulates it and sends it to network, with a SEQUENCE NUMBER swapping between 0 and 1.
It then waits for a valid ACK packet with the sequence number positive acknowledgement. If it gets this, it reads the next data from the application layer. If it gets anything else, it resends.

Receiver: Receives packet from the network layer and extracts it. It sends an ACK with the sequence number of the last correctly received packet. If that was the packet just received, it extracts it and sends it to the application layer.
It returns to waiting for a packet.

Question 13

A sender/receiver is set up to deal with a reliable transmission but with possible integrity issues.
What editing is needed to ensure it can deal with unreliable transmission?

Answer 13

When a sender sends a packet, it must start a timer. If the time runs out, it must resend the packet. If a correct acknowledgement is received or a retransmission is sent, it restarts the timer.

Question 14

Explain why a stop-and-wait protocol, while functional, may not be ideal. What is the solution?

Answer 14

As each packet is sent and checked in turn, it takes substantially longer to transmit data across a network than it should.
Solved by pipelining.

Question 15

Explain the Go-Back-N protocol.

Answer 15

The sequence number range is split into 4 sections:
- Packets sent and ack’d
- Packets sent and not ack’d
- Packets not sent but usable
- Packets not sent and not usable

The window constitutes the middle two sections.

If a packet is in the window, it can be sent. The window remains the same size at all times, so when a packet is ack’d, a new once can be added.

If an out-of-order packet is received, then the packet is dropped. The sender, on a timeout, will resend all the packets in the 2nd section.

Question 16

Explain the Selective Repeat protocol.

Answer 16

The same window is set as GBN, but the window may contain ACK’d packets.

When the packet in the 1st position is ACK’d, the window is shifted along until a non-ack’d packet is in position 1.

When the receiver receives a packet in or before the window, it marks it as received and ACKs it.
When the 1st position is received, the window is slid over until a non-received packet into the 1st position.

Each packet must have it’s own timer.

Question 17

What may occur if the window size is too large?

Answer 17

The last element of the sender may correspond to the sequence number of the start of the receiver, so new data may be received instead of the old data.

Question 18

What precaution is taken for random reordering in a network?

Answer 18

Sequence numbers are not reused until it is sure no packet with that number may still be in the network.

Question 19

What is the MSS and MTU in TCP?

Answer 19

MSS (Maximum Segment Size) is the amount of APPLICATION data that can be placed in a segment.

The MTU (Maximum Transmission Unit) is the length of the largest link layer frame that can be sent.

The MSS is normally calculated from determining the MTU.

Question 20

What is in the TCP header?

Answer 20

16 bits for:
- source port
- destination port

32 bits for:
- sequence number
- ack number

4 bits for:
- header length

4 zero bits.

8 bits for:
- flags (ACK, RST, SYN, FIN, CWR, ECE, PSH, URG)

16 bits for:
- receive window
- checksum
- URG data pointer (if URG is set)

Options (variable length)

Question 21

How is the segment sequence number decided in TCP?

Answer 21

By the first byte sequence number of that segment. The first byte of the whole sequence is given a random sequence number.

Question 22

How is the segment acknowledgement number decided in TCP?

Answer 22

The sequence number of the next expected segment.

If the 1st and 3rd packets have been received, then the 2nd seq number will be sent as an ack to the most recent receival.

Question 23

How is the estimated RTT computed in TCP?

Answer 23

At any point, one sample RTT is being calculated for a packet.

Then,
EstimatedRTT = (1-a) * EstimatedRTT + a * sampleRTT

a is recommended as 0.125

Question 24

How is the RTT variability measured?

Answer 24

DevRTT = (1-b) * DevRTT + b * |SampleRTT - EstimatedRTT|

Recommended value: b = 0.25

Question 25

What is the retransmission timeout interval for TCP?

Answer 25

TimeoutInterval = EstimatedRTT + 4 * DevRTT, with initial value of 1 sec. When a timeout occurs, the timeout interval is doubled until a segment is received.

Question 26

What is a TCP fast transmit?

Answer 26

If there is 3 duplicate ACKs received, it is highly likely of a missing segment, so send that sequence number again immediately.

Question 27

What is flow control in TCP?

Answer 27

The reduction of data influx rate to combat overflowing receiving buffers.

Question 28

How is flow controlled in TCP?

Answer 28

Let A send B a large file. The receive window is defined as: rwnd = rcv_buffer - [last_byte_rcvd - last_byte_read] (basically, the spare room in the receive buffer). B will add the value of rwnd to every acknowledgement it sends to A. Host A then keeps track of last_byte_sent and last_byte_acked to ensure it isn't overflowing the buffer. When B's recv window is 0, A must send 1 data byte until the buffer is more than zero.

Question 29

What is the TCP handshake protocol?

Answer 29

1. Client sends a SYN packet with a random sequence number to the server. 2. The server sends a SYN-ACK packet with a random sequence number and the client's seq number + 1 as its acknowledgement. 3. The client sends an ACK packet with sequence number being the client_seq + 1 and acknowledgement of server+1

Question 30

What is the TCP teardown protocol?

Answer 30

1. Client sends a FIN packet to the server 2. Server sends an ACK packet to the client, waits for a bit, and sends a FIN packet. 3. Client sends an ACK packet. There is a timed wait, and the connection is terminated.

Question 31

If a client sends a TCP packet to a server on an invalid port, what happens?

Answer 31

The server sends a packet back with a RST flag of 1.

Question 32

What are the costs of congestion?

Answer 32

- Large queuing delays are experienced. - A sender must send retransmissions in order to compensate for dropped packets. - When a packet is dropped, the transmission capacity that was used at each of the upstream links is wasted.

Question 33

In network assisted congestion control, how might feedback be sent from the network to a sender?

Answer 33

Choke packets from a network router can be sent to a sender. ----- The router can mark a packet as it goes to the receiver, which marks this in the acknowledgement.

Question 34

How does a TCP sender limit the rate at which it sends traffic into its connection?

Answer 34

The sender keeps track of a congestion window, cwnd. Then, last_byte_sent - last_byte_acked <= min(cwnd, rwnd) And thus, the send rate is roughly cwnd/RTT bytes/sec.

Question 35

How does a TCP sender perceive that there is congestion on the network?

Answer 35

The presence of a dropped TCP segment, which is detected by a timeout or triplicate ACK, gives high congestion.

Question 36

What is the TCP congestion-control slow start algorithm?

Answer 36

When a TCP connection begins, the cwnd is initialised to 1 MSS, so sending rate is approx MSS/RTT (small value). It is doubled each time a transmitted segment is first acknowledged. If there is a loss event indicated by a timeout, the algorithm resets. The slow_start_threshold is set to cwnd/2. When ssthresh is reached, slow start algorithm ends. If at any point a triplicate is detected, a fast retransmit is performed and the fast recovery algorithm is started.

Question 37

What is the TCP congestion-control congestion avoidance algorithm?

Answer 37

Each RTT, the cwnd is increased by 1 MSS. This is done by each ACK increasing the cwnd by MSS/cwnd. When a timeout occurs, cwnd is set to 1 MSS and ssthresh is set to half of cwnd when the loss occurred. When a triplicate ACK is received, cwnd is halved and increased by 3, and ssthresh is set to half of cwn when it was received. Fast recovery is then entered.

Question 38

What is the TCP congestion-control fast recovery algorithm?

Answer 38

The value of cwnd is increased by 1 MSS for every duplicate ACK received missing segment that caused TCP to enter fast-recovery. When ACK arrives for the missing segment, it returns to the congestion-avoidance state. If a timeout event occurs, it transitions to the slow start state after performing the standard set cwnd to 1 MSS and ssthresh is set to half of cwnd when it occurred.

Question 39

What is the average throughput of a TCP Reno connection?

Answer 39

0.75 * W/RTT where W is the window size in bytes when a loss event occurs.

Question 40

How does Explicit Congestion Notification work?

Answer 40

A router sets one of the ECN bits in the IP header as it passed to the recipient. The recipient then sets the ECHO bit in the ACK packet. The sender then sends a CWR bit (Congestion Window Reduced) in the header of the next transmitted TCP sender-to-receiver segment.

Question 41

How does Delay-based Congestion Control work?

Answer 41

The uncongested throughput rate is cwnd/RTT_min. If the actual rate is substantially less than this, the TCP sender decreases its sending rate.

Question 42

Is TCP fair? Is UDP fair? What occurs when UDP and TCP packets are sent at the same time, and how is this solved?

Answer 42

TCP is fair. UDP is not. UDP will crowd out the TCP traffic, and a congestion manager is needed.

Question 43

If TCP is fair, how is it possible a TCP application is unfair?

Answer 43

It uses multiple parallel connections.

Question 44

Give some benefits of QUIC.

Answer 44

- Connection-oriented and secure - Encryption and handshake in one, so half the RTTs needed for this. - Parallel streams for each object in a website, so no HOL-blocking. - Reliable and congestion-controlled transfer. NOTE: QUIC is an APPLICATION LAYER protocol, so it can be modified much faster.