- Computer Networking: A Top Down Approach Featuring the Internet, Third Edition, J. Kurose and Keith Ross, Addison-Wesley, 2004
- Computer Networks - A Systems Approach, Fourth Edition, by Larry L. Peterson and Bruce S. Davie, Morgan Kaufmann, 2003
- Computer Networks, Fourth Edition, by Andrew S. Tanenabum, Printice Hall, 2003
Computing devices are separated into desktop computers, mobile devices such as phones and laptops, IoT devices, data centers and supercomputers.
- A network is a collection of computer nodes connected together.
- A distributed system is a collection of computing nodes that's interconnected and appears to the user as a single monolithic system.
- They're related in the sense that every distributed system is a network with special abstracting software on it.
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Protocols ensure successful communication by defining the format and the order of messages, as well as the actions taken on message transmission or receipt.
- Circuit switching:
- dedicated lines between nodes
- data is sent as a continuous stream
- data is sent at a constant rate
- Packet switching:
- different nodes use the same lines
- data is transferred in small messages called packets
- data is sent at a variable rate
Packet switching allows for more efficient use of communication lines.
Layered network models separate the process of of communication into a set of small discrete tasks, each assigned to a particular layer that works to perform that task only where:
- each layer performs a single well-defined function
- each layer provides the services of its function to the next layer in the hierarchy
- the chosen layer boundaries result in minimal information flow between layers
- the number of layers is large enough to avoid assigning separate functionalities to one layer, and small enough that minimal message passing occurs so the architecture stays simple and easy to manage.
The layered architecture is an abstraction of the network that helps us understand the network and its components, making it easier to understand the network, manage it and make changes to it.
b) Explain the relationship and communication between (1.) layers at the same node. (2.) layers at two different nodes.
- layers at the same node communicate through interfaces.
- layers at different nodes communicate through protocols.
It transmits raw bits over the communication channel, manages connection initiation and termination, and defines the electrical specifications of the physical medium.
d) Suppose that node A wants to send data to node B. Outline how encapsulation is done as the data is moved between different layers at nodes A and B
[A] [B]
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| Application 🡫 | | 🡫 Application |
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| Transport 🡫 🡩 | | 🡩 🡫 Transport |
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| Network 🡫 🡩 | | 🡩 🡫 Network |
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| Link 🡫 🡩 | | 🡩 🡫 Link |
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| Physical 🡩 |<--->| 🡩 Physical |
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In the network interface card (NIC).
b) Draw a block diagram to show the relationship of a network adapter to other host components and the protocol stack functionality.
- It's needed to avoid overwhelming one of the communicating nodes with data, especially if they're operating at different speeds.
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- Feedback based: the receiver sends back info that lets the transmitter send more or less data.
- Rate based: a built-in mechanism at the transmitter limits its transmission rate.
d) What is the difference between error detection and error correction? Explain when each strategy should be used.
- Error detection: mechanisms for a receiver node to detect errors.
- Error correction: mechanisms for a receiver node to detect errors and determine their locations so they can be corrected.
- Polling protocols:
- advantages:
- simple
- has no chance of collisions
- easy to implement
- disadvantages:
- inefficient and slow
- requires a central controller which is a single point of failure
- advantages:
- Token passing protocols:
- advantages:
- faster than polling
- decentralized and more robust than polling
- results in lower latency
- disadvantages:
- has a single point of failure, where if the node holding the token fails, the whole ring fails.
- advantages:
b) (1.) Explain the Slotted-Aloha protocol. (2.) If there are 4 nodes willing to transmit their frames, draw a sketch that illustrates the disadvantages of Slotted-Aloha.
- It's a data link layer protocol for frame transmission where:
- frames are split into slots smaller than teh frame size
- all the nodes attempt to transmit data at the beginning of a slot
- if a collision is detected, all the colliding nodes wait a random period of time then attempt to transmit again
- The notation C, E and S represent a collision, an empty slot and a successful transmission, respectively.
Switches are high performance multi-interface bridges.
- They are layer 2 devices that connect multiple LANs/network segments with different technologies together.
- Each LAN segment on a switch port is a separate collision domain.
- There's no limit on how big a network segment can be.
- Many switches operate in full-duplex mode, sending and receiving data at the same time.
LANs are physical IP networks separated by routers, while subnets are logical IP networks separated in software ans can either be a superset of multiple LANs or a subset of on LAN.
They're 6 byte data link layer addresses that are unique and permanent to each interface on a NIC, and are used as to identify devices in a network in order to allow routing in local network segments. They are managed and allocated by IEEE and are burned into the ROMs of each NIC.
It's a protocol that translates between IP Addresses and MAC Addresses on local networks.
It's a table that maps IP addresses to MAC addresses, and is stored in the ARP module of every node in a LAN. It contains entries for:
- IP addresses of each node in the LAN
- MAC addresses of each node in the LAN
- Time To Live (TTL) for each entry
It helps efficiency and performance by aborting the transmission of corrupted, useless frames.
b) The Ethernet frame starts by an 8-byte preamble: (1.) What is the function of this preamble? (2.) Is the preamble included in the calculation of the frame's CRC? Why or why not?
- It's used:
- for clock synchronization between the transmitter and the receiver.
- as a wake up call to the receiver
- It's not included in the CRC calculation because it's redundant data, that's not part of the frame's payload and contains no useful information.
c) Suppose you want to connect 95 nodes with 10Base2 Ethernet. How many repeaters do you need? and why?
Four repeaters, because the max number of nodes supported on each bus is 30, and ceil(95/30) = 3
It's a connectionless service that provides best effort delivery of frames, and doesn't guarantee delivery or order of delivery.
- No ACKs or NACKs are passed by senders or receivers.
- The receiver just discards frames with failed CRC checks.
- It's blind to the data it transmits:
- It doesn't detect gaps in the data
- Or recognize retransmissions
e) (1.) Why does Ethernet use Manchester encoding? (2.) If the bit stream is [11000110], draw a sketch to show how this stream is encoded.
- It ensures that sender and receiver clocks stay in sync.
- Baseband:
Manchester encoding:
- Infrastructure networks are 802.11 networks where all devices connect to each other (and maybe the internet) through a single access point
- Ad-hoc networks are a group of 802.11 stations that group themselves together to form a network with no central control or connections to other networks.
- all network segments must use the same ethernet technology
- independent collision domains are combined into one
- total number of hosts and tiers is limited
- Store and Forward: all frames are stored in a buffer until they are completely received, then they are transmitted. If the output buffer is empty but a frame wasn't fully received, a delay is generated by the switch.
- Cut-through switching: frames are transmitted as they come; If the output buffer is empty the switch transmits the start of the frame before receiving its end.
c) What is the maximum store and forward delay for a switch if the frame length is 900 bytes, the transmission rate of the inbound link is 24*10^5 bit/sec, and teh transmission rate of outbound link is 36*10^5 bit/sec?
max delay = (900*8 bits) / (24*10^5 bits/sec) = 0.003 sec
- A send the frame to interface 1 of the router R1
- the router R1 reads the frame's destination IP and checks whether its directly connected to it or not
- R1 finds that the destination node isn't directly connected to it, but is reachable through router R2, so it transmits the frame to R2 through R1 interface 2
- R2 checks the destination IP and finds that the node is in one of its LANs
- R2 forwards the frame to node E through its second interface.
The switch table is built automatically as follows:
- It's initially empty
- When a frame arrives on one of the interfaces and its destination isn't in the table, the switch forwards copies of the frame on all its interfaces except the one it arrived on.
- for each frame received, the switch stores in its table the source LAN address, the interface on which the frame arrived and the current time.
a) How many collision domains are there in the network? List the collision domains and explain your answer.
There are 3 collision domains: {r,n}, {s,n}, and {p,q,k,t,u,j,h,n,G} All devices are in subnetworks of the same router, and are all connected by hubs except for r, and which are connected by a switch; Hubs combine collision domains, and switches keep them separate.
b) If host p uses 10BaseT Ethernet, do hosts {q,r,s,t,u} also have to use 10BaseT Ethernet? Explain your answer.
No, only hosts {q,t,u} have to use 10BaseT Ethernet, because nodes {r,s} are separated from p by a switch, which is a layer-2 device capable of translating between different ethernet technologies.
c) (1.) Explain how the switch table of n is built. (2.) Write down the table assuming recent communication between all other nodes.
- The table is initially empty
- When a frame arrives on one of the interfaces and its destination isn't in the table, the switch forwards copies of the frame on all its interfaces except the one it arrived on.
- for each frame received, the switch stores in its table the source LAN address, the interface on which the frame arrived and the current time.
- The control plane defines how packets are forwarded; Routing table creation is part of the control plane.
- The data plane is the processes that actually forward the packets in accordance with the control plane.
Switching connects multiple devices on the same network, while routing connects multiple networks together. Switches just connect devices together, but routers work at the network layer to find the shortest path for a packet across the networks.
It defines the properties of the channel connecting the transport layers of two communicating hosts, in terms of:
- Service reliability
- Received packets ordering compared to transmitted
- Time between two consecutive sent or received packets
- Network congestion feedback
| fabric type | advantages | disadvantages |
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| bus | No processor intervention |
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c) Describe how the routing table entries are stored, and explain its effect on the processing time.
The are stored as entries pointing the next hop required to reach a destination IP, as well as the No. of hops till this IP is reached. This allows them to be stored as binary tress where closer together routers (with similar IP addresses or in subnetworks of the same WAN) are grouped together in a subtree and are searchable in O(n) time.
- Input queueing can occur when:
- the switching fabric is slower than the rate at which input packets are received.
- multiple packets at the input ports are destined to the same output port
- a packet at an input id destined to an empty output port, but is preceded by another packet destined to a busy one, which is known as Head-of-Line blocking.
- Output queueing can occur when:
- multiple input packets are destined to a single output port
- the switching fabric is slower than the incoming traffic.
It's when a packet is queued at an input port despite being destined to a an empty output port. It happens only in the case of input buffering when a packet is preceded by a another on the same input port that is waiting on a busy output port.
Class A: 27, Class B: 214, Class C: 221, Class D: 228
Unicast, anycast and multicast.
c) What is the advantage of the tunneling approach over the dual stack approach when transitioning from IPv4 to IPv6? Give an example to explain your answer.
Tunneling maintains all the information in an IPv6 packet header such as the flow label, which is a new header field that dual stack transmission would strip away.
- static routing is when administrators manually configure the routing tables with one entry for each destination.
- dynamic routing is a means of adaptive routing where multiple possible routes are computed ans the best path for traffic is determined through the network.
- centralized routing algorithms are routing algorithms that use the complete information about link costs and connectivity to determine the minimum cost paths between two nodes.
- decentralized algorithms are an iterative process of calculation and information exchange where no node has complete information about all link costs, but only the links attached to it.
It's the union of all the shortest paths towards a destination in a graph. Since the sink tree defines all the optimal paths between the nodes in a network, it's the goal of routing algorithms to compute the sink trees for all routers and use them.
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A believes that the optimal route to C is going through B, while B believes that the optimal route to C is through A. So when one of the two hosts tries to send a packet to C, the other will transmit it back to sender and update the path cost adding 1 to it, resulting in a single packet being forwarded at least 4 times in a loop before the path is updated with the correct link costs, and the packet reaches its destination.
Using poisoned reverse: when A attempts to route through B to get to C, then B attempts to route through A to get to C, A will advertise to B that the cost to C through it is infinity, forcing B to send directly to C advertising the new C cost leading to the appropriate distance vector updates in each node.
Using packet drop feedback, where dummy packets are sent with incremental TTLs so that the first packet with TTL=1 is dropped by the first router, making that router send back a message identifying it as the first in the path, then the second packet with TTL=2 is dropped by the second router which similarly returns feedback identifying it as the second, and so on until the packet reaches its destination.
Suppose that you have routers running a BGP protocol, if router A receives ana updated path of BTWXYN from router B, explain the cases when (1.) the path will be added to the routing table, and (2.) the path will not be added to the routing table.
The route will:
- be added to the table if A isn;t one of the pre-existing {T,W,X,Y,N} routers and the route is the most preferable one according to router A's defined routing and security policies.
- not be added if
- a preferable path is determined by A
- A has a policy not to route packets through on of the routers in the path
- A is the same as one of the pre-existing routers {T,W,X,Y,N}
- Membership query - general: sent on an interfaces to determine the set of all multicast groups that have members on that interface.
- Membership query - specific: contains the multicast address of a group and is sent on an interface th determine if that specific group has members on that interface.
- Membership report: sent by a host as a response to a membership query to indicate that it's a member of a group, or when the host first joins a group, and is received by the router as well as all the hosts on the same interface.
- Leave group message: an optional message sent by hosts when they leave a multicast group to allow the router to determine that no hosts are remaining in the group without using a delay or a timeout. The router can infer that no hosts are remaining in a group when it receives no responses to a membership query message with that group's address, but this typically takes a an entire second.
It's the consolidation of responses arising from a group of hosts as feedback to a single query message, where if a host observes a response from another host in the same group, it discards its own pending membership report to avoid or reduce network congestion resulting from redundant responses, especially since multicast groups are set to scale very large.
| Network multicast |
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| Multicast via unicast |
The sender uses a separate unicast connection to each of the receivers, resulting in a lot of message duplication over the network links. |
A single class D IP address is used as an identifier to a group of hosts, and a copy of a datagram addressed to this identifier is delivered to all of the receivers associated with the group.
- Group-shared tree: Finding a tree within the network that has the minimum cost and connects all the routers with attached hosts belonging to the mcast group.
- Source-based tree: Finding the least unicast-cost tree from every host in the group to every other one. This means that routing occurs in a source-specific manner and an individual routing tree is constructed for each host.
b) Assume hosts n and m are both in a multicast group with an ID 226.3.16.4. Assume router A doesn't know that there are hosts attached to it that have joined the multicast group. Explain how router A will use the IGMP protocol to determine if the group 226.3.16.4 has been joined by hosts attached to it.
- Router A will send a group-specific membership query that'll traverse along all links of the router's unicast tree and get forwarded over all router interfaces along the tree links.
Message content:0: membership query - specific 8: max response time 16: checksum 32: 226.3.16.4 - hosts {n, m} will each wait a random amount of time <= MRT then respond back with a membership report
- one of the two hosts will send out their membership report before the other due to delay randomness
Report content:
0: membership report 8: MRT 16: checksum 32: 226.3.16.4 - Since the hosts n and m aren't in the same LAN, router A will receive the membership report of the first respondent directly then forward it as-is on all its interfaces.
- the second host, that hasn't yet sent its membership report, will detect this message and discard its own.
c) Suppose that the routers are multicast capable. Suppose that a group-shared center-based protocol is used to build the routing tree, with L as the center of the tree. Using your answer in part (b): (1.) Explain how P and T will join the multicast group. (2.) Draw the tree after P and T join the multicast group.
- After selecting L as the center node:
- P sends a membership report as a unicast message to the center node.
- The message arrives at L, and the route it takes L-O-Q-P is considered the initial routing tree.
- T sends its join message
- the message is detected by router O, which adds the branch O-R-T to the multicast tree, resulting in the tree shown below.
flowchart RL
L---O
P---Q
T---R
Q---O
R---O
a) Which of the following multicast routing protocols uses group-shared trees and which uses source-based trees? DVRMP, MOSPF, CBT.
| group-shared | source based |
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| CBT | DVRMP & MOSPF |
b) What is the advantage of using two different multicast distribution scenarios in the PIM protocol?
Sparse and dense modes are optimized for the assumptions that:
| sparse | the group members will be distributed throughout the network with most LANs and subnets not wanting any multicast traffic. |
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| dense | the group members will be packed densely throughout the network with most LANs and subnets wanting any multicast packet transmitted. |
c) What is third-party dependency in BGMP protocol? What is the disadvantage of the third-party dependency?
BGMP could choose a center router with no multicast hosts in its AS, causing an unnecessary burden on that AS, as well as subjecting the mcast group to performance dependencies on other AS's outside of the group's control.
To create virtual networks of multicast capable routers on top of physical networks that contain a mix of multicast and unicast devices, because a small fraction of internet routers are multicast capable.
- unreliable
- connectionless
- stateless
- unregulated
- uses checksums for error detection
Since it's unreliable, it needs error detection to avoid delivering corrupted data to the application layer.
suppose the data field is 1111 1110 1110 1101 1101 1110 1010 1101 1011 1110 1110 1111
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the data is split into 16-bit words, each of which is added together:
1111 1110 1110 1101+
1101 1110 1010 1101+
1011 1110 1110 1111+
_________________________
1001 1100 1000 1001 -
one's complement of the sum is taken:
1001 1100 1000 1001
________________________
0110 0011 0111 0110
At the receiver's side, the data is split into 16-bit words, all of which are added together then added to the checksum. If the result isn't 1111 1111 1111 1111 then the packet was corrupted during transmission.
1111 1110 1110 1101 +
1101 1110 1010 1101 +
1011 1110 1110 1111 +
0110 0011 0111 0110 +
_________________________
1111 1111 1111 1111
| TCP | Connection-oriented | Congestion controlled (regulated) | Advanced error control | large headers | 3-way handshake |
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| UDP | Connectionless | unregulated | uses checksums for error detection only | small 8-byte headers | no handshake |
rdt_send(data) - rdt_rcv(packet) - mk_pkt(pkt, data) -
extract(pkt, data) - udt_send(pkt) - deliver_data(data) -
is_ack(pkt) - is_nack(pkt)
The transport protocol gathers application data and splits it into segments each enveloped with the source and destination port numbers, then passes them down to the network layer multiplexing them in the process.
At the receiver, the transport layer headers are examined to determine the source port, and the source IP - source port combination uniquely identifies the origin of each segment allowing for data demultiplexing.
b) Can the three clients use the same source port number? How will the server differentiate between the data coming from the clients?
Yes. The receiver identifies the segments by the source-IP:source-port pair, where IP addresses are unique to each client.
c) If client B wants to initiate a second FTP connection to the same server, is there a case where the client will be allowed to use the same source port it used in the first connection? Explain.
Only one running connection can use a single port, so the second connection can only use the same port in the case that the first connection was terminated.
a) (1.) What is the drawback of the stop and wait protocols of RDT? (2.) What type od protocols is used to overcome this drawback?
- The send and receive processes take much less time than the time of travel of a packet, leading to poor channel utilization and poor throughput.
- Pipelined RDT protocols as well as unreliable protocols can be used to overcome this.
It's when a sender sends multiple packets without waiting for pending ACKs one-by-one. The sender and receiver sides may have to buffer more than one packet, and there may be multiple in-transit unacknowledged packets.
c) Describe the selective repeat protocol, including actions taken by the sender and receiver when receiving a packet.
Selective Repeat protocols avoid unnecessary retransmissions by having:
- The receiver individually acknowledge correctly received packets whether or not they’re in-order
- Out of order packets are buffered until any missing packets are received, at which point a batch of packets can be delivered in-order to the upper layer
- Data received from above: if the sequence number is within the window, the data is packetized and sent; otherwise it is either buffered or returned
- Timeout: timers are used to protect against lost packets, where each packet has its own timer. A packet is retransmitted if its timer runs out and it has yet to be acknowledged.
- ACK received: the sender marks that packet as received if it’s in the window.
- If the packet's sequence number is equal to
sendbase, the window base is moved forward to the unacknowledged packet with the smallest sequence number - If the window moves and there are untransmitted packets with sequence numbers falling within the window, these packets are transmitted
- If the packet's sequence number is equal to
- Packet with sequence number in range
[rcvbase, rcvbase+N-1]is correctly received: a selective ACK is returned to the sender if:- The packet was not previously received, it is buffered
- The packet has a sequence number equal to the base of the receive window, then it is moved along with all consecutively numbered packets getting delivered to the upper layer, then the receive window is moved forward by the number of packets delivered to the upper layer.
- Packet with sequence number in range
[rcvbase-N, rcvbase-1]is received: a selective ACK is returned to the sender as well, because if there is no ACK for packetsendbasefrom the receiver to the sender:- The sender will retransmit packet
sendbase - The sender's window will never move forward
- The sender will retransmit packet
- Otherwise: Ignore the packet
d) What is the difference between selective repeat and go-back-N with regard to handling acknowledgments and packet loss?
In GBN, If a packet with sequence number N is received correctly and is in-order, the receiver sends an ACK for packet N and in all other cases discards the packet and resends an ACK for the most recently received in-order packet, forcing the sender to retransmit all the packets it sent out starting from N after it times out.
While in RDT, the receiver individually acknowledges correctly received packets whether or not they’re in-order, and out of order packets are buffered until missing packets are received. Since each packet has its own timer, only missing packets are retransmitted individually.
- If TCP receives a segment with a sequence number larger than the next expected in-order sequence number, it detects a gap in the data
- Since TCP does not use NACKs, it re-acknowledges the last in-order byte of data it has received
- If the TCP sender receives 3 ACKs for the same segment, it concludes that the segment following it has been lost
- If a segment has been lost, TCP performs a fast retransmission, retransmitting the missing segment before that segment's timer expires
e) Explain how the RDT of the TCP protocol can do the following: (1.) detect a gap in the data, (2.) inform the sender that a packet is lost, and (3.) conclude that a packet has been lost.
- A gap in the data is detected at the receiver if TCP receives a segment with a sequence number larger than the next expected sequence number
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- At the receiver: if the received packet is out of order, an ACK of the last in-order byte of data is transmitted.
- At the sender: if 3 ACKs for the same segment are received, it is concluded that the segment following it has been lost
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- At the receiver: when there's a gap in the data, this implies the loss of a previously awaited packet
- At the sender: a packet loss is concluded when the timer's up
| TCP | source port, destination port, seq number, ACK number, window size, flag, checksum |
|---|---|
| UDP | source port, destination port, checksum |
| IP | source IP, destination IP, fragmentation ID, fragmentation offset |
c) Assume that hosts X and Y have a TCP connection established. Assume that the two hosts are separated by one router. Why does host X not directly use the MAC address of host Y when constructing its packet to send to Y?
If the router’s interface in X’s LAN isn’t targeted by the packet’s DLL header, the router will discard the packet and not forward it to the other LAN. The TCP connection wouldn’t affect data-link-layer behavior, since TCP is a transport layer protocol separated from the DLL by the network layer.
| client-server architecture | There's an always-on host, called the server, which services requests from many other hosts, called clients. | HTTP, SMTP, DNS |
|---|---|---|
| P2P | Has minimal (or no) reliance on dedicated/central servers, instead the application exploits direct communication between pairs of intermittently connected hosts | BitTorrent, IPFS |
- A socket is the interface between the transport and application layers within one system
- A user agent is the interface between the network application and the user, such as web browsers and mail clients
- Data loss prevention:
- some apps require fully reliable data transmission such as file transfer
- others -like multimedia- are loss tolerant and don't require DLP
- Bandwidth:
- some apps must be able to transmit at a specific rate to be effective, such as media streaming
- other elastic apps can make use of however much bandwidth is available
- Timing: Interactive realtime apps, such as online games and VoIP, require timing constraints on data delivery
- A push protocol is a protocol that sends out resources to their recipients directly without them needing to request them, such as SMTP
- A pull protocol works by a request-response model, where clients request resources from servers first, then the servers respond back with the requested resource, such as HTTP
- They reduce response times for users, since they're selected geographically to be closer than the origin server.
- They reduce the load on the origin (usually popular/central) servers by offloading the requests going to them.
- They provide the infrastructure for rapid content distribution networks.
b) Explain the difference between sending and receiving an email using a mail user agent and using a browser-based email service. In your answer include the protocol used in each case.
| mail user agent |
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| browser user agent |
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c) (1.) What does it mean that HTTP is a stateless protocol? (2.) What is the advantage of a stateless protocol? (3.) Mention a protocol that is NOT stateless, and explain why it's not stateless.
- It doesn't maintain any state information about the client or the server between requests.
- A stateless protocol avoids the overhead that comes with storing and processing state information, allowing for a the maintenance of a large number of concurrent connections.
- FTP is a stateful protocol, because the client establishes an authenticated control connection that spans multiple requests and data transfers throughout the session.
– The browser extracts the hostname from the URL and passes it to the DNS client – The DNS client sends a query containing the hostname to a DNS server – The DNS client eventually receives a reply that includes the IP address for the hostname
e) (1.) What services does the DNS provide? (2.) Explain how DNS is used for load distribution among duplicate servers.
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- Mapping between hostnames and IP addresses
- Host aliasing: some hosts can have one or more alias hostnames
- Load distribution among duplicate servers
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- Sites are replicated over multiple servers with each server running on a different end system, and having a different IP address
- When clients make a DNS query for a name mapped to a set of addresses, the server responds with the entire set of IP addresses, but rotates the ordering of the addresses with each reply.
- A client typically sends its HTTP request message to the first listed IP address, resulting in the load being distributed among the servers.