CS计算机代考程序代写 dns DHCP assembly FTP algorithm 7.Network_Data_Plane

7.Network_Data_Plane

COMP 3331/9331:
Computer Networks and

Applications
Week 7

Network Layer: Data Plane
Reading Guide: Chapter 4: Sections 4.1, 4.3

Our goals:
§ understand principles behind network layer

services, focusing on data plane:
• network layer service models
• forwarding versus routing

§ instantiation, implementation in the Internet

Network Layer: outline

4.1 Overview of Network layer
– data plane
– control plane

4.2 What’s inside a router
— Not Covered

4.3 IP: Internet Protocol
– datagram format
– fragmentation
– IPv4 addressing
– network address

translation
– IPv6

4.4 Generalized forwarding
and Software Defined
Networking (SDN)
– Not Covered

Network Layer, data plane: outline

• 1968: DARPAnet/ARPAnet (precursor to Internet)
– (Defense) Advanced Research Projects Agency Network

• Mid 1970’s: new networks emerge
– SATNet, Packet Radio, Ethernet
– All “islands” to themselves – didn’t work together

• Big question: How to connect these networks?

Some Background

• Cerf & Kahn in 1974,
– “A Protocol for Packet Network

Intercommunication”
– Foundation for the modern Internet

• Routers forward packets from source to
destination
– May cross many separate networks along the way

• All packets use a common Internet Protocol
– Any underlying data link protocol
– Any higher layer transport protocol

Internet Protocol Stack

• Application: Email, Web, …
• Transport: TCP, UDP, …
• Network: IP
• Link: Ethernet, WiFi, ATM, …
• Physical: copper, fiber, air, …

• “Hourglass” model, “thin waist”, “narrow waist”

UDP
TCP

Ethernet

ATM

…

Email
Web

Internetworking

• transport segment from
sending to receiving host

• on sending side encapsulates
segments into datagrams

• on receiving side, delivers
segments to transport layer

• network layer protocols in
every host, router

• router examines header
fields in all IP datagrams
passing through it

application
transport
network
data link
physical

network
data link
physical network

data link
physical

network
data link
physical

network
data link
physicalnetwork

data link
physical

Network Layer

• forwarding: move
packets from router’s
input to appropriate
router output

• routing: determine
route taken by packets
from source to dest.

– routing algorithms

analogy:

v routing: process of
planning trip from source
to dest

v forwarding: process of
getting through single
interchange

Two key nework-layer functions

0111

value in arriving
packet’s header

routing algorithm

local forwarding table
header value output link

0100
0101
0111
1001

3
2
2
1

Interplay between routing and forwarding

routing algorithm determines
end-end-path through network

forwarding table determines
local forwarding at this router

Data plane
§ local, per-router function
§ determines how datagram

arriving on router input
port is forwarded to
router output port

§ forwarding function

Control plane
§ network-wide logic
§ determines how datagram is

routed among routers along
end-end path from source host
to destination host

§ two control-plane approaches:
• traditional routing algorithms:

implemented in routers
• software-defined networking

(SDN): centralised (remote)
servers

2
3

0111

values in arriving
packet header

Network Layer: data vs control plane

Per-router control plane

Routing
Algorithm

Individual routing algorithm components in each and every
router interact in the control plane

data
plane

control
plane

4.1 • OVERVIEW OF NETWORK LAYER 309

tables. In this example, a routing algorithm runs in each and every router and both
forwarding and routing functions are contained within a router. As we’ll see in Sec-
tions 5.3 and 5.4, the routing algorithm function in one router communicates with
the routing algorithm function in other routers to compute the values for its forward-
ing table. How is this communication performed? By exchanging routing messages
containing routing information according to a routing protocol! We’ll cover routing
algorithms and protocols in Sections 5.2 through 5.4.

The distinct and different purposes of the forwarding and routing functions can
be further illustrated by considering the hypothetical (and unrealistic, but technically
feasible) case of a network in which all forwarding tables are configured directly by
human network operators physically present at the routers. In this case, no routing
protocols would be required! Of course, the human operators would need to interact
with each other to ensure that the forwarding tables were configured in such a way
that packets reached their intended destinations. It’s also likely that human configu-
ration would be more error-prone and much slower to respond to changes in the net-
work topology than a routing protocol. We’re thus fortunate that all networks have
both a forwarding and a routing function!

Values in arriving
packet’s header

2
3

Local forwarding
table

header

0100
0110
0111
1001

1101

3
2
2
1

output

Control plane

Data plane

Routing algorithm

Figure 4.2 ! Routing algorithms determine values in forward tables

M04_KURO4140_07_SE_C04.indd 309 11/02/16 3:14 PM

0111

values in arriving
packet header

data
plane

control
plane

Logically centralized control plane (SDN)

A distinct (typically remote) controller interacts with local
control agents (CAs)

Remote Controller

CA CA CA CA

0111

values in arriving
packet header

Q: What service model for “channel” transporting
datagrams from sender to receiver?

A. No guarantee whatsoever is provided by IP
layer in TCP/IP protocol stack. It’s “best effort
service”.

Network Layer: service model

4.1 Overview of Network layer
– data plane
– control plane

4.2 What’s inside a router
— Not Covered

4.3 IP: Internet Protocol
– datagram format
– fragmentation
– IPv4 addressing
– network address

translation
– IPv6

Network Layer, data plane: outline

4.4 Generalized forwarding
and Software Defined
Networking (SDN)
– Not Covered

The Internet network layer

forwarding
table

host, router network layer functions:

routing protocols
• path selection
• RIP, OSPF, BGP

IP protocol
• addressing conventions
• datagram format
• packet handling conventions

ICMP protocol
• error reporting
• router “signaling”

transport layer: TCP, UDP

link layer

physical layer

network
layer

IP Packet Structure

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Fields for Reading Packet Correctly

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Reading Packet Correctly

• Version number (4 bits)
– Indicates the version of the IP protocol
– Necessary to know what other fields to expect
– Typically, “4” (for IPv4)

• Header length (4 bits)
– Number of 32-bit words in the header
– Typically, “5” (for a 20-byte IPv4 header)
– Can be more when IP options are used

• Total length (16 bits)
– Number of bytes in the packet
– Maximum size is 65,535 bytes (216 -1)
– … though underlying links may impose smaller limits

Fields for Reaching Destination and Back

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Telling End-Host How to Handle Packet

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Telling End-Host How to Handle Packet

• Protocol (8 bits)
– Identifies the higher-level Transport protocol
– Important for demultiplexing at receiving host

Application

Transport

Network

Data link

PhysicalL1

L7 SMTP HTTP DNS NTP

TCP UDP

Ethernet FDDI PPP

optical copper radio PSTN

Checksum, TTL and Fragmentation Fields

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Potential Problems

• Loop: TTL

• Header Corrupted: Checksum

• Packet too large: Fragmentation

Preventing Loops (TTL)
• Forwarding loops cause packets to cycle for a long time

– As these accumulate, eventually consume all capacity

• Time-to-Live (TTL) Field (8 bits)
– Decremented at each hop, packet discarded if reaches 0
– …and “time exceeded” message is sent to the source
– Recommended default value is 64

Header Corruption (Checksum)

• Checksum (16 bits)
– Particular form of checksum over packet header

• If not correct, router discards packets
– So, it doesn’t act on bogus information

• Checksum recalculated at every router
– Why?
– Why include TTL?
– Why only header?

IP fragmentation, reassembly

• network links have MTU
(max.transfer size) – largest
possible link-level frame
– different link types,

different MTUs
• large IP datagram divided

(“fragmented”) within net
– one datagram becomes

several datagrams
– “reassembled” only at

final destination
– IP header bits used to

identify, order related
fragments

fragmentation:
in: one large datagram
out: 3 smaller datagrams

reassembly

…

ID
=x

offset
=0

MF flag
=0

length
=4000

ID
=x

offset
=0

MF flag
=1

length
=1500

ID
=x

offset
=185

MF flag
=1

length
=1500

ID
=x

offset
=370

MF flag
=0

length
=1040

one large datagram becomes
several smaller datagrams

example:
v 4000-byte datagram
v MTU = 1500 bytes

1480 bytes in
data field

offset =
1480/8

IP fragmentation, reassembly

Applet:
http://media.pearsoncmg.com/aw/aw_kurose_network_2/applets/ip/ipfragmentation.html

Data = 4000 –
20 (IP header)
=3980
F1=1480
F2=1480
F3=1020

Note: Offset is expressed as multiple of 8 bytes

IPv4 fragmentation procedure

Ø Fragmentation
• Router breaks up datagram in size that output link can

support
• Copies IP header to pieces
• Adjust length on pieces
• Set offset to indicate position
• Set MF (More fragments) flag on pieces except the last
• Re-compute checksum

ØRe-assembly
• Receiving host uses identification field with MF and

offsets to complete the datagram.
Ø Fragmentation of fragments also supported

Tak
en f

rom
[TC

P/IP

Pro
toco

l Su
ite

Beh
roze

For
ouz

an]

MTU=1420

MTU=820

2.1

2.2

IP Fragmentation Attacks … IP Fragmentation Attacks…
https://en.wikipedia.org/wiki/IP_fragmentation_attack

Not on Exam

Path MTU Discovery procedure

Ø Host
• Sends a big packet to test whether all routers in path to the

destination can support or not
• Set DF (Do not fragment) flag

Ø Routers
• Drops the packet if it is too large (as DF is set)
• Provides feedback to Host with ICMP message telling the

maximum supported size

Fields for Special Handling

4-bit
Version

4-bit
Header
Length

8-bit
Type of Service
(TOS)

16-bit Total Length (Bytes)

16-bit Identification
3-bit
Flags 13-bit Fragment Offset

8-bit Time to
Live (TTL) 8-bit Protocol 16-bit Header Checksum

32-bit Source IP Address

32-bit Destination IP Address

Options (if any)

Payload

Special Handling
• “Type of Service”, or “Differentiated Services

Code Point (DSCP)” (8 bits)
– Allow packets to be treated differently based on needs
– E.g., low delay for audio, high bandwidth for bulk transfer
– Has been redefined several times
– Not widely used

• Options (not often used)

ver length

32 bits

data
(variable length,
typically a TCP
or UDP segment)

16-bit identifier
header
checksum

time to
live

32 bit source IP address

head.
len

type of
service

flgs fragment
offset

upper
layer

32 bit destination IP address

options (if any)

RECAP: IP datagram format
IP protocol version

number
header length

(bytes)

upper layer protocol
to deliver payload to

total datagram
length (bytes)

“type” of data
for
fragmentation/
reassemblymax number

remaining hops
(decremented at

each router)

e.g. timestamp,
record route
taken, specify
list of routers
to visit.

how much overhead?
v 20 bytes of TCP
v 20 bytes of IP
v = 40 bytes + app

layer overhead

4.1 Overview of Network layer
– data plane
– control plane

4.2 What’s inside a router
4.3 IP: Internet Protocol

– datagram format
– fragmentation
– IPv4 addressing
– network address

translation
– IPv6

Network Layer, data plane: outline

IP addressing: introduction

• IP address: 32-bit
identifier for host, router
interface

• interface: connection
between host/router and
physical link
– router’s typically have

multiple interfaces
– host typically has one or

two interfaces (e.g., wired
Ethernet, wireless 802.11)

• IP addresses associated
with each interface

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

223.1.1.1 = 11011111 00000001 00000001 00000001

223 1 11

IP addressing: introduction

Q: how are interfaces
actually connected?
A: we’ll learn about that
in the link layer

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2223.1.3.1

223.1.3.27

A: wired Ethernet interfaces
connected by Ethernet switches

A: wireless WiFi interfaces
connected by WiFi base station

Networks
• IP address:

–network part – high
order bits

–host part – low order
bits

• what’s a network ?
–device interfaces with

same network part of IP
address

–can physically reach
each other without
intervening router

inter-network consisting of 3 nets

223.1.1.1

223.1.1.3

223.1.1.4 223.1.2.9

223.1.3.2223.1.3.1

subnet

223.1.1.2

223.1.3.27
223.1.2.2

223.1.2.1

how many? 223.1.1.1
223.1.1.3

223.1.1.4

223.1.2.2223.1.2.1

223.1.2.6

223.1.3.2223.1.3.1

223.1.3.27

223.1.1.2

223.1.7.0

223.1.7.1
223.1.8.0223.1.8.1

223.1.9.1

223.1.9.2

Networks

Answer: 6 as shown

Masking
• Mask

– Used in conjunction with the
network address to indicate how
many higher order bits are used
for the network part of the
address

• Bit-wise AND
– 223.1.1.0 with mask

255.255.255.0
• Broadcast Address

– host part is all 111’s
– E.g., 223.1.1.255

• Network Address
– Host part is all 0000’s
– E.g., 223.1.1.0

• Both are typically not assigned
to any host

223.1.1.0

223.1.2.0

223.1.3.0

Host B Dot-decimal
address

Binary

IP address 223.1.1.2 11111101.00000001.00000001.00000010

Mask 255.255.255.0 11111111.11111111.11111111.00000000

Network Part 223.1.1.0 11111101.00000001.00000001.00000000

Host Part 0.0.0.2 00000000.00000000.00000000.00000010

B: 223.1.1.2

Original Internet Addresses

• First eight bits: network address (/8)
• Last 24 bits: host address, ~16.7 million

Assumed 256 networks were more than
enough!

Next design: Class-ful Addresses

110

1110

1111

Class A

Class B

Class C

Class D

Class E

8 16 24 310

netid hostid

multicast address

reserved for future use

1.0.0.0 to
127.255.255.255

128.0.0.0 to
191.255.255.255

192.0.0.0 to
223.255.255.255

224.0.0.0 to
239.255.255.255

240.0.0.0 to
255.255.255.255

Used till
the introd

uction of
CIDR 19

Problem: Networks only come in three sizes!

27 nets,
224hosts

214 nets,
216 hosts

221 nets,
28 hosts

Finding the address class

What are the issues?
ØAn organization requires 6 nets each of size 30.

Does it have to buy 6 class C address blocks?

ØAn organization requires 512 addresses? How
many IP addresses should it buy?

Subnetting
ØSubnetting is the process of dividing the class

A, B or C network into more manageable
chunks that are suited to your network’s size
and structure.

ØSubnetting allows 3 levels of hierarchy
• netid, subnetid, hostid

ØOriginal netid remains the same and
designates the site

ØSubnetting remains transparent outside the
site

Subnetting
ØThe process of subnetting simply extends the point

where the 1’s of Mask stop and 0’s start
ØYou are sacrificing some host ID bits to gain Network

ID bits

A company is granted the site address 201.70.64.0 (class
C). The company needs six subnets. Design the subnets.

The company needs six subnets. 6 is not a power of 2.
The next number that is a power of 2 is 8 (23). We need
3 more 1s in the subnet mask. The total number of 1s
in the subnet mask is 27 (24 + 3). The mask is

11111111 11111111 11111111 11100000
or 255.255.255.224

Number of addresses in each subnet = 25

= 32

Quiz?

The number of addresses in each subnet is 25 or 32.

Quiz: IP Addressing

• How many IP addresses belong to the subnet
128.119.254.0/25 ? What are the IP addresses
at the two end-points of this range ?

Answer: 27 = 128 addresses (126 are usable)

10000000 01110111 11111110 00000000

subnet
part

host
part

128.119.254.0/25

128.119.254.127

128.119.254.0

/25

Today’s addressing: CIDR

CIDR: Classless InterDomain Routing
§ network portion of address of arbitrary length
§ address format: a.b.c.d/x, where x is # bits in

network portion of address

11001000 00010111 00010000 00000000

Network (Prefix)
part

Host (Suffix)
part

200.23.16.0/23

200.23.17.255

200.23.16.0

/23

IP address

How many IP addresses belong to the subnet
134.45.22.0/23?

A) 32

B) 64A

C) 128

D) 256

E) 512

Quiz: IP Addressing

www.zeeting.com/salil

ANSWER: E (2^9 = 512)
Answer: E

A small organization is given a block with the beginning
address and the prefix length 205.16.37.24/29 (in slash
notation). What are the IP addresses at the two end
points?

The beginning address is 205.16.37.24. To find the
last address we keep the first 29 bits and change the
last 3 bits to 1s.

Beginning:11001101 00010000 00100101 00011000
Ending : 11001101 00010000 00100101 00011111

There are only 8 addresses in this block.
205.16.37.24 to 205.16.37.31

Quiz: IP Addressing

www.zeeting.com/salil

An ISP is granted a block of addresses starting with
190.100.0.0/16. The ISP needs to distribute these
addresses to three groups of customers as follows:

1. The first group has 64 customers; each needs 256 addresses.

2. The second group has 128 customers; each needs 128 addresses.

3. The third group has 128 customers; each needs 64 addresses.

Design the sub-blocks and give the slash notation for
each sub-block. Find out how many addresses are still
available after these allocations.

Quiz: IP Addressing

Group 1

For this group, each customer needs 256 addresses.
This means the suffix length is 8 (28 = 256). The
prefix length is then 32 – 8 = 24.

01: 190.100.0.0/24 è190.100.0.255/24

02: 190.100.1.0/24 è190.100.1.255/24

…………………………………..

64: 190.100.63.0/24è190.100.63.255/24

Total = 64 ´ 256 = 16,384
53

Group 2

For this group, each customer needs 128 addresses.
This means the suffix length is 7 (27 = 128). The
prefix length is then 32 – 7 = 25. The addresses
are:

001: 190.100.64.0/25 è190.100.64.127/25

002: 190.100.64.128/25 è190.100.64.255/25

………………………………………………….

128: 190.100.127.128/25 è190.100.127.255/25

Total = 128 ´ 128 = 16,384
54

Group 3

For this group, each customer needs 64 addresses.
This means the suffix length is 6 (26 = 64). The
prefix length is then 32 – 6 = 26.

001:190.100.128.0/26 è190.100.128.63/26

002:190.100.128.64/26 è190.100.128.127/26

…………………………

128:190.100.159.192/26 è190.100.159.255/26

Total = 128 ´ 64 = 8,192
55

Number of granted addresses: 65,536

Number of allocated addresses: 40,960

Number of available addresses: 24,576

IP addresses: how to get one?
Q: How does a host get IP address?

• hard-coded by system admin in a file
– Windows: control-panel->network->configuration->tcp/ip-

>properties
– UNIX: /etc/rc.config

• DHCP: Dynamic Host Configuration Protocol: dynamically
get address from as server
– “plug-and-play”

DHCP
goal: allow host to dynamically obtain its IP address from

network server when it joins network
– can renew its lease on address in use
– allows reuse of addresses (only hold address while

connected/“on”)
– support for mobile users who want to join network

DHCP overview:
– host broadcasts “DHCP discover”msg
– DHCP server responds with “DHCP offer”msg
– host requests IP address: “DHCP request”msg
– DHCP server sends address: “DHCP ack”msg

DHCP client-server scenario

223.1.1.0/24

223.1.2.0/24

223.1.3.0/24

223.1.1.1

223.1.1.3

223.1.1.4 223.1.2.9

223.1.3.2223.1.3.1

223.1.1.2

223.1.3.27
223.1.2.2

223.1.2.1

DHCP
server

arriving DHCP
client needs
address in this
network

DHCP server: 223.1.2.5 arriving
client

DHCP discover

src : 0.0.0.0, 68
dest.: 255.255.255.255,67
yiaddr: 0.0.0.0
transaction ID: 654

DHCP offer
src: 223.1.2.5, 67
dest: 255.255.255.255, 68
yiaddrr: 223.1.2.4
transaction ID: 654
lifetime: 3600 secs

DHCP request
src: 0.0.0.0, 68
dest:: 255.255.255.255, 67
yiaddrr: 223.1.2.4
transaction ID: 655
lifetime: 3600 secs

DHCP ACK
src: 223.1.2.5, 67
dest: 255.255.255.255, 68
yiaddrr: 223.1.2.4
transaction ID: 655
lifetime: 3600 secs

DHCP client-server scenario

DHCP: more than IP addresses

DHCP can return more than just allocated IP
address on subnet:
§ address of first-hop router for client
§ name and IP address of DNS sever
§ network mask (indicating network versus host

portion of address)

v connecting laptop needs its
IP address, addr of first-hop
router, addr of DNS server:
use DHCP

router with DHCP
server built into
router

v DHCP request encapsulated in
UDP, encapsulated in IP,
encapsulated in 802.1 Ethernet

v Ethernet frame broadcast (dest:
FFFFFFFFFFFF) on LAN,
received at router running DHCP
server

v Ethernet demuxed to IP
demuxed, UDP demuxed to
DHCP

168.1.1.1

DHCP
UDP
IP
Eth
Phy

DHCP

DHCP
UDP
IP
Eth
Phy

DHCP

DHCPDHCP

DHCP: example

• DHCP server formulates
DHCP ACK containing
client ’ s IP address, IP
address of first-hop
router for client, name &
IP address of DNS server

• Encapsulation of DHCP
server, frame forwarded
to client, demuxing up to
DHCP at client

DHCP: example

router with DHCP
server built into
router

DHCP

DHCP
UDP
IP
Eth
Phy

DHCP

DHCP
UDP
IP
Eth
Phy

DHCP

• Client now knows its IP
address, name and IP
address of DSN server, IP
address of its first-hop
router

DHCP: Wireshark
output (home LAN)

Message type: Boot Reply (2)
Hardware type: Ethernet
Hardware address length: 6
Hops: 0
Transaction ID: 0x6b3a11b7
Seconds elapsed: 0
Bootp flags: 0x0000 (Unicast)
Client IP address: 192.168.1.101 (192.168.1.101)
Your (client) IP address: 0.0.0.0 (0.0.0.0)
Next server IP address: 192.168.1.1 (192.168.1.1)
Relay agent IP address: 0.0.0.0 (0.0.0.0)
Client MAC address: Wistron_23:68:8a (00:16:d3:23:68:8a)
Server host name not given
Boot file name not given
Magic cookie: (OK)
Option: (t=53,l=1) DHCP Message Type = DHCP ACK
Option: (t=54,l=4) Server Identifier = 192.168.1.1
Option: (t=1,l=4) Subnet Mask = 255.255.255.0
Option: (t=3,l=4) Router = 192.168.1.1
Option: (6) Domain Name Server

Length: 12; Value: 445747E2445749F244574092;
IP Address: 68.87.71.226;
IP Address: 68.87.73.242;
IP Address: 68.87.64.146

Option: (t=15,l=20) Domain Name = “hsd1.ma.comcast.net.”

Message type: Boot Request (1)
Hardware type: Ethernet
Hardware address length: 6
Hops: 0
Transaction ID: 0x6b3a11b7
Seconds elapsed: 0
Bootp flags: 0x0000 (Unicast)
Client IP address: 0.0.0.0 (0.0.0.0)
Your (client) IP address: 0.0.0.0 (0.0.0.0)
Next server IP address: 0.0.0.0 (0.0.0.0)
Relay agent IP address: 0.0.0.0 (0.0.0.0)
Client MAC address: Wistron_23:68:8a (00:16:d3:23:68:8a)
Server host name not given
Boot file name not given
Magic cookie: (OK)
Option: (t=53,l=1) DHCP Message Type = DHCP Request
Option: (61) Client identifier

Length: 7; Value: 010016D323688A;
Hardware type: Ethernet
Client MAC address: Wistron_23:68:8a (00:16:d3:23:68:8a)

Option: (t=50,l=4) Requested IP Address = 192.168.1.101
Option: (t=12,l=5) Host Name = “nomad”
Option: (55) Parameter Request List

Length: 11; Value: 010F03062C2E2F1F21F92B
1 = Subnet Mask; 15 = Domain Name
3 = Router; 6 = Domain Name Server
44 = NetBIOS over TCP/IP Name Server
……

request

DHCP: further details
• DHCP uses UDP and port numbers 67 (server side) and 68

(client side)
• Usually the MAC address is used to identify clients

– DHCP server can be configured with a “registered list” of acceptable
MAC addresses

• DHCP offer message includes ip address, length of lease,
subnet mask, DNS servers, default gateway

• DHCP security holes
– DoS attack by exhausting pool of IP addresses
– Masquerading as a DHCP server
– Authentication for DHCP – RFC 3118

IP addresses: how to get one?

Q: how does network get subnet part of IP addr?
A: gets allocated portion of its provider ISP’s

address space

ISP’s block 11001000 00010111 00010000 00000000 200.23.16.0/20

Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/23
Organization 1 11001000 00010111 00010010 00000000 200.23.18.0/23
Organization 2 11001000 00010111 00010100 00000000 200.23.20.0/23

… ….. …. ….
Organization 7 11001000 00010111 00011110 00000000 200.23.30.0/23

CIDR: Addresses allocated in contiguous
prefix chunks

Recursively break down chunks as get closer to host

12.0.0.0/8

12.0.0.0/15

12.253.0.0/16

12.2.0.0/16
12.3.0.0/16

:
:

12.3.0.0/22
12.3.4.0/24

:
:

12.3.254.0/23

12.253.0.0/19
12.253.32.0/19
12.253.64.0/19
12.253.64.108/30
12.253.96.0/18
12.253.128.0/17

:
:
:

Hierarchical addressing: route aggregation

“Send me anything
with addresses
beginning
200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7
Internet

Organization 1

ISPs-R-Us “Send me anything
with addresses
beginning
199.31.0.0/16”

200.23.20.0/23
Organization 2

…

hierarchical addressing allows efficient advertisement of routing
information:

Quiz: What should we do if organization 1 decides
to switch to ISPs-R-Us

A: Move 200.23.18.0/23 to ISPs-R-Us (and break up Fly-By-Night’s/20 block).
B: Give new addresses to Organization 1 (and force them to change all their addresses)
C: Some other solution

“Send me anything
with addresses
beginning
200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7
Internet

Organization 1

ISPs-R-Us “Send me anythingwith addresses
beginning
199.31.0.0/16”

200.23.20.0/23
Organization 2

…

Both A and B are valid. A is explained on the next 2 slides

ISPs-R-Us has a more specific route to Organization 1

“Send me anything
with addresses
beginning
200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7
Internet

Organization 1

ISPs-R-Us “Send me anything
with addresses
beginning 199.31.0.0/16
or 200.23.18.0/23”

200.23.20.0/23
Organization 2

…

Hierarchical addressing: more specific routes

Longest prefix matching

Example: continued

• But how will this work?
• Routers in the Internet will have two entries in

their tables
– 200.23.16.0/20 (Fly-by-Night-ISP)
– 200.23.18.0/23 (ISPs-R-Us)

• Longest prefix match

IP address

200.23.31.255

200.23.16.0

/20

200.23.19.255

200.23.18.0

/23

Longest prefix matching

Destination Address Range

11001000 00010111 00010*** *********

11001000 00010111 00011000 *********

11001000 00010111 00011*** *********

otherwise

DA: 11001000 00010111 00011000 10101010

examples:
DA: 11001000 00010111 00010110 10100001 which interface?

which interface?

when looking for forwarding table entry for given
destination address, use longest address prefix that
matches destination address.

longest prefix matching

Link interface

Quiz: Longest prefix matching

• On which outgoing interface will a packet
destined to 11011001 be forwarded?

Prefix Interface

1* A

11* B

111* C

Default D

www.zeetings.com/salil Answer: B

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