Internet Protocol and Applications
Addressing
‘Routing is part of IP, router needs to
interpret addresses’
Format: server. institution. domain
=> Not an actual Internet address
=> Identifies a host computer / server owned by
an institution that is connected to an internet
=> Domain does not have geographic
significance
Address identifies the following using 32-bits:
- Participating network
- Node ID within the network
IPv4 Address Formats
IP Addresses - Class A
 start
with binary 0
 all 0 (in first octet) is reserved
 01111111 (127) (in first octet) is reserved
 range 1.x.x.x to 126.x.x.x
 all allocated
 Very large networks
IP Addresses - Class B
 start
with binary 10
 range 128.x.x.x to 191.x.x.x
 second octet also included in network
address
 214 = 16,384 class B networks
 all allocated
 Fairly large
IP Addresses - Class C
 start
with binary 110
 range 192.x.x.x to 223.x.x.x
 second and third octet also part of network
address
 221 = 2,097,152 networks
 nearly all allocated
 Relatively small
IP Addresses - Class D
Multicast addressing: one address for one
group.
Example Uses:
- Software updates to installed machines
- video stream to selected users
‘Routing is different from uni-cast routing’
Classless Addresses
32-bit addressing has limitations.
Example: Class B assigned, network has 1000
machines
=> Huge unused addresses
Alternative: assign a group of class C networks
Problems:
- Burden to plan for growth (256-increments)
- Additional routing table information
(performance suffers)
Classless Addresses
Classless Inter-domain Routing (CIDR):
- Does not fall into above classes
- Number of bits defining the network
number varies
=> Allows networks of varying sizes
Commonly used for multiple class C
networks.
For 1000 nodes: 211.195.8.0 to
211.195.11.0
Classless Addresses
Classless Inter-domain Routing (CIDR):
=> First 22-bits are the same
=> 22-bit network number, 10-bit local
identifier
‘Router can extract the network number by
AND operation using a subnet mask’
‘several smaller groups into a larger group –
super-netting’
=> Fewer network numbers in routers
Classless Addresses
Classless Inter-domain Routing (CIDR):
‘Router needs to know the number of bits in
the network ID’
Network address w. x. y. z is replaced by
w. x. y. z / m
m- number of bits in the network ID
Subnet Mask Calculation
Binary Representation
Dotted Decimal
IP address
11000000.11100100.00010001 .00111001
192.228.17 .57
Subnet mask
11111111.11111111.11111111 .11100000
255.255.255 .224
Bitwise AND of
address and mask
(result ant
networ k/subn et
number)
11000000.11100100.00010001 .00100000
192.228.17 .32
Subnet numb er
11000000.11100100.00010001 .001
1
Host numb er
00000000.00000000.00000000 .00011001
25
Problems
# Classify the following:
183.104.200.32, 210.20.34.100,
115.193.23.32
# IP address = 140.100.120.02, subnet
mask = 255.255.224.0, network number?
# Can 255.255.224.7 be a mask?
# For 8000 IP addresses, find the number of
class C networks needed and describe
CIDR addressing scheme (subnet mask).
Obtaining an Addresses
Typical approach:
client machine requests an IP address from the
server, server runs a protocol: Dynamic Host
Configuration Protocol (DHCP) that assigns
an IP address from the list it maintains.
Server is maintained by ISP or LAN manager.
Internet Corporation for Assigned Names and
Numbers (ICANN) – allocates IP addresses.
Obtaining an Address
Difficult to memorize IP addresses => get a
host name registered
Host name is stored in a distributed directory
referenced by client programs
Registration is done through an accredited
registrar (ICANN accredits)
www.google.com = 216.239.53.99
Domain Name System
Translates text form of an address to the 32bit address.
It is a distributed database. Why?
Challenge is to manage millions of
addresses among servers and to provide a
quick translation.
‘Concept of domain eases this task’
DNS has hierarchical arrangement of
organization.
Domain Name System
Root server
edu
com
org
mit tamu
microsoft
ieee acm
…
Host sends a request to the local name
server - A. Process is complete if A can
provide the translation.
Domain Name System
Else, A sends the request to another server
B at a higher level.
The process continues until the request is
resolved.
The translated address is stored in the local
cache of all intermediate servers =>
avoids duplication of the complex
translation process
IP Header
IP packets and Fragmentation
Different fields of the IP packet – we
covered
Different network architectures allow
different Maximum Transfer Units (MTU)
IP packet length < MTU => smooth entire
transfer.
IP packet length > MTU => fragmentation.
Identification, flags, offset fields aid the
process of reassembly.
Fragmentation
Flag: More fragment bit (MFB) = ‘1’ for each
fragment except the last.
Do not fragment bit = ‘1’, server responds
with error if IP packet length > MTU, used
repeatedly to determine MTU
Fragment offset field measures offset in
units of 8 bytes.
# IP packet of 4000 data bytes, MTU = 1400
bytes. Show fragments with offset, MFB.
IP routing
‘Is based on routing tables stored at routers
and the interpretation of IP addresses’
Physical address: one used by underlying
physical network.
Example: Ethernet addresses enclosed in
IEEE 802.3 frames, these are 48-bit
numbers assigned to the network interface
cards
Physical addresses have local significance
but none at the global IP scale.
IP routing
IP packets are enclosed in frames if they
travel through LAN,
If the frame goes to a router, IP layer
extracts the packet, examines address,
determines where to send.
If the packet needs to go to a machine in the
attached LAN, IP puts it in a frame and
sends it to the destination – direct routing.
How to obtain the physical address?
IP routing
Dynamic Binding (Address Resolution
Protocol):
Router sends a broadcast-request with an IP
address for a response of physical
address.
Appropriate device responds with physical
address => stored in router-cache
The cache needs periodic update to
accommodate changes.
Routers
# Router 1 is connected to a network (LAN)
with IP addresses 143.200.x.y
Router 2 is connected to a network (LAN)
with IP addresses 143.100.x.y
Router 1 and Router 2 are connected.
* Router 1 gets a packet with destination
143.200.10.5 => direct delivery => get
physical address (using ARP or cache) =>
send a frame
Routers
* Router 1 gets a packet with destination
143.100.20.4 => not connected to the LAN
=> get the next hop (from routing table) =>
send the packet to router 2.
After receiving the packet, router 2 follows
the same process as described in the
earlier case.
‘traceroute / tracert command can be used
to display intermediate routers between
the host and a specified destination’
Routers
Issues:
Finding the next hop: the router looks it up in
the table, challenge is to look it up fast,
otherwise, packets accumulate => cause
delay => buffer overflow (in extreme case)
Hash structure works fast for searching the
table (Content Addressable Memory)
Next hop entry is stored at a location
generated from a hash function of the
destination.
Routers
Issues:
Finding the next hop:
In that case there is no searching and next
hop is found quickly.
Moving packet from input to output port:
Bus => simple, one at a time (slow)
Switch => connects two end points, flexible
connection, complex control
Routers
Issues:
Scheduling packets for transmission => refers
to order in which packets are transmitted,
FIFO – simple, but Quality of Service dictates
the priority of packets (mail versus real time
audio)
# Routers A, B, C connects 7 networks (total),
each of A, B, C connects to three networks,
only two networks connect to two routers,
others to one each. How many common nexthop are in routing table of the middle router?
Problems
# What percentage of total IP addresses
each class represents?
# Network address = 192.168.100.0
Subnet mask = 255.255.255.192
How many subnets possible? How many
hosts in each?
# A company has six departments each
having ten computers (or networked
devices). Find the subnet mask for the
network of each department.
Problems
# A CIDR address is specified as
192.168.100.0/25. How many host
addresses are on the network?
# Given the network address of
192.168.100.0 and the subnet mask of
255.255.255.192, find the number of
subnets created and the number of hosts
per subnet.
Routing Using Subnets
Multicast routing
‘One to selected many’
Internet Group Management Protocol
(IGMP) – operates between a host and the
local router
- Allows the host to join and leave various
multicast groups
IP packets are used with protocol field = 2 to
exchange messages.
Router sends a query to find out group
members.
Multicast routing
Host responds indicating it belongs / no
more belongs to a particular multicast
group (Identified by a class D address).
The challenge is to convey this message to
all routers and to implement some
multicast routing algorithm.
# Example network with multicast group
hosts in selected networks
- Only a few routers are of importance
Multicast routing
If only uni-cast packets are sent, number
of copies are more,
- through multicasting, source sends only a
copy, then it is replicated only one for each
network having multicast group members.
=> much less packets in the networks,
significant difference in large groups.
Router needs to know what to do with such
a packet.
-
Multicast routing
A spanning tree of routers can be formed
that reaches all the hosts in a group –
multicast tree.
Different trees for different multicast groups
– becomes very huge at global scale.
=> Very few IP routers support multicasting.
Distance Vector Multicast Routing Protocol
(DVMRP) uses Reverse Path
Broadcasting (RPB).
Multicast routing
RPB assumes that a router knows the next
link along the shortest path to a given
node.
Router action (after receiving a multicast
packet):
- Identify the source and the port where
received
- Look up the source in the routing table and
find the next hop in a path to the source
Multicast routing
If the next hop corresponds to the port
where the packet arrived, then send the
packet over all other ports.
- Otherwise, drop the packet.
=> Avoids packet-travel in loops (forward
only to lead away from source)
But, may still add redundancy in packettravel.
Pruning is needed to limit the forwarding
from a router.
-
Multicast routing
When a router gets a multicast packet but
has no group member attached, it uses
IGMP with a prune message to the
sender.
=> Sender no longer sends such packets.
If a host subsequently joins the group, that
router sends a Graft message for
resumption of multicast packets.
Example: Multicasting
Other routing
Resource Reservation Protocol (RSVP)
- Deals with QoS over Internet
- Embeds messages in IP packets (protocol
field value 46)
- Messages contain requests that certain
resources be reserved to meet QoS, for
example: buffer space.
- A router chooses the maximum to satisfy
several different requests.
Other routing
Internet Control Message Protocol (ICMP)
- Is used for reporting errors and for
providing router-updates on conditions
that can develop in the Internet.
- Protocol field = 1
Typical messages: Destination unreachable,
Echo request, Echo reply.
ICMP Message Formats
Checksum
 One’s
complement of 16-bit (one’s
complement) addition of all 16-bit words in
the header.
 one’s complement addition – carry added
 with LSB of result
 # Header with ten octets, checksum in the
last two octets: 01 00 F6 F7 F4 F5 F2 03 00
00
 - Find checksum and resulting header
 - Verify the checksum (final value = FFFFH)
Why Change IP?

Address space exhaustion






two level addressing (network and host) wastes
space
network addresses used even if not connected
growth of networks and the Internet
extended use of TCP/IP
multiple vs. single address per host
requirements for new types of service
IPv6 Enhancements
 expanded
128 bit address space
 improved option mechanism

most options not examined by intermediate
routers
 dynamic
address assignment
 increased addressing flexibility

Any-cast & multicast
 support

for resource allocation
labeled packet flows
IPSec
 RFC
1636 (1994) identified security need
 encryption & authentication to be in IPv6
 but designed also for use with current IPv4
 applications needing security include:




branch office connectivity
remote access over Internet
extranet & intranet connectivity for partners
electronic commerce security
IPSec Scenario
IPSec Benefits
 provides
strong security for external traffic
 resistant to bypass
 below transport layer, hence transparent
to applications
 can be transparent to end users
 can provide security for individual users if
needed
IPSec Functions
 Authentication

for authentication only
 Encapsulating

a

Header(AH)
Security Payload (ESP)
for combined authentication/encryption
key exchange function
manual or automated
 VPNs
usually need combined function
Transport Protocol
 ‘Defines
what one device can say to
another on behalf of the user – defines
end user protocol’
 ‘Reliable communication between sites
with logical connection’
 Connection management: establishing,
maintaining, and ending a connection.
 Establishes parameters: sequence
numbers used for bytes, number of bytes
an entity can receive.
Transport Protocol
 Entities
exchange segments, do error
checking, acknowledging, and flow control,
leaving transmission details to lower
levels.
 Establish connection: Two way handshake
 A – send connection request
 B – receive connection request, establish
connection, send acknowledgement
 A – receive acknowledgement, establish
connection
Transport Protocol
A
and B exchange data and eventually
disconnect.
 Problem: if the first request is delayed and
eventually shows up at a much later time
 B thinks it as another connection.
 Worse: if some of A’s data were seriously
delayed and finally arrived after the
retransmitted one.
Transport Protocol
 Three
way handshake is used with initial
sequence numbers each entity uses.
 A – send TCP segment with SYN = 1 and
sequence = x
 B - send TCP segment with SYN = 1,
acknowledgement = x+1 and sequence = y
 A – send TCP segment with
acknowledgement = y+1
Transport Protocol
– send TCP data segments, sequence
beginning with x+1 and
acknowledgements beginning with y+1
 B – send TCP data segments, sequence
beginning with y+1 and
acknowledgements beginning with x+1
A
TCP Header
Transport Protocol
 Disconnect
Protocol:
 A – receive a CLOSE primitive from the
application, send TCP segment with FIN =
1 and sequence = p (current sequence
count)
 B – receive disconnect request and notify
the application of finished data, send TCP
segment with acknowledgement = p+1
Transport Protocol
– receive a CLOSE primitive from the
application, send TCP segment with FIN =
1, and acknowledgement = p+1, sequence
= q (current sequence count)
 A – send TCP segment with
acknowledgement = q+1
B
Transport Protocol
 Flow
Control:
 - In TCP, sequence number refers to byte
sequence, not packet / segment sequence
 - Each entity can alter the size of the
other’s sending window dynamically using
Window field.
 Flow control is implemented using credit
mechanism (window advertisement)
Transport Protocol
A
credit specifies the maximum number of
bytes this entity can receive (and buffer)
from the other entity.
 This is in addition to those already
received and buffered.
 A – initial sequence = 100, can buffer upto
200 bytes (credit)
 B – initial sequence = 700, can buffer upto
200 bytes (credit)
Transport Protocol
 Each
segment contains 100 bytes.
 (All these have been agreed upon by three
way handshake)
A



starts by sending two segments
Data = …, s = 101, a = 701
Data = …, s = 201, a = 701
Waits for more credit.
Transport Protocol
B


A


responds by sending two segments
Data = …, s = 701, a = 301, c = 0
Data = …, s = 801, a = 301, c = 200
responds by sending two segments
Data = …, s = 301, a = 901
Data = …, s = 401, a = 901
 (Credit
restrictions also apply to B, omitted
here for simplicity)
Transport Protocol
 ‘Allows
the protocol to be robust taking
advantage of changing conditions to adjust
credit’
 # Identify applications requiring real-time
quality of service.
 - Downloading audio files
 - Accessing a remote host
 - Watching a live training session
 - Watching a broadcast news
Transport Protocol
-
using FTP to download small file
 - using FTP to download very large file
 # Assume:
 -TCP entities A, B have initial sequence
numbers 400, 900 respectively
 - segment size = 100 data bytes, initial
credit of each = 200 bytes.
 - each entity delivers a segment to the
application, as soon as it receives =>
freeing up buffer
Transport Protocol
-
A is capable of sending segments at
interval of time T (starting at Time = 0)
(permitting flow control) , B is capable of
sending segments at interval of time 3T
(starting at T = 1.5T) (permitting flow
control)
 - Transmission time between A and B is
negligible.
 Diagram showing segment exchange (D,
S, A, C) upto time 12T.
Transport Protocol
#
In TCP’s flow control logic, an entity
uses the credit field to determine when it
can send new segments. What is the
purpose of acknowledgement then?
# A network has an IP packet size of 128
bytes and maximum packet lifetime of 30s.
If 8-bit packet sequence number is used,
what is the maximum data rate achieved?
Transport Protocol
#
Round trip time (RTT) between two
hosts is 100 ms and both hosts use a TCP
window of 32 Kbytes. What is the
maximum throughput that can be achieved
by TCP in this scenario?
 # Two hosts are connected by a 100 Mbps
link and RTT between them is 1 ms. What
is the minimum TCP window size in order
to achieve maximum throughput between
two hosts? (Assume no overhead)
Transport Protocol
 # A host is receiving data from a remote peer by
TCP segments with 1460 bytes of payload.
Assume only TCP and IP header overheads for
acknowledgements. If TCP acknowledges every
other segment, what is the minimum uplink
bandwidth needed to achieve data throughput of
1 Mbytes/sec?
Transport Protocol
 TCP
uses four timers
 Retransmission timer: Handles
retransmission time - waiting time for an
acknowledgement of a segment, controls
a lost or discarded segment.
 Calculation of retransmission time: It
should be variable and dynamically
adjusted for each connection and may
change during the same connection
Transport Protocol
 It
is based on RTT, most common is the
following: retransmission time = 2 (RTT)
 RTT is calculated by measuring the time
between sending a segment and receiving
acknowledgement.
 Updated RTT = α (previous RTT) + (1- α)
(current RTT).
 α is usually 0.9, for two consecutive RTT =
250 and 70µs, Updated RTT = 232µs,
timer = 464µs
Transport Protocol
 Problem:
When receiving ACK for a
retransmitted segment, sender does not
know if it is for the original or for the
second one
 Karn’s solution: do not consider RTT for a
retransmitted segment to update RTT,
consider only the ones with no
retransmission.
Transport Protocol
 Persist
timer: deals with zero window-size
advertisement
 - receiver sends a zero window ACK,
sender stops
 - receiver sends an ACK with non-zero
window but it is lost
 - both continue to wait (deadlock)
 Recovery: sender starts persist timer after
receiving zero window ACK.
Transport Protocol
-
when it times out, a probe segment is
sent alerting the receiver that the ACK is
lost and be resent.
 Keep-Alive timer: used in most
implementations to prevent long idle
connection, each time one end receive
segment from other side, timer is reset,
otherwise (after 2H) a probe is sent, after
several probes terminates the connection.
Transport Protocol
 Silly
Window Syndrome: A serious
problem occurs if :
 - the sending application creates data
slowly
 - the receiving application consumes data
slowly
 1-byte data => 41-byte datagram (20-byte
TCP header, 20-byte IP header)
 => huge overhead
Transport Protocol
 Solution
at the sender’s end:
 Nagle’s Algorithm:
 - Sending TCP sends the first piece of
data (even if 1-byte)
 - Sending TCP accumulates data in the
buffer and waits for either an ACK or to fill
a maximum size segment. Now, it can
send the segment.
 - Above step is repeated for rest of the
transmission.
Transport Protocol
at the receiver’s end:
 Clark’s solution: To send an ACK as the
data arrive but to announce window size of
zero until there is enough space for
maximum size segment or half-buffer is
empty.
 Delayed ACK: Receiver waits until there is
a decent amount of space in the buffer
before sending ACK. It reduces traffic.
 Solution
Transport Protocol
#
TCP is sending data at 1 Mbytes/sec. If
the sequence number starts with 7000,
how long it takes before the sequence
number goes back to zero?
Congestion Control
 flow
control is also used for congestion
control


recognize increased transit times & dropped
packets
react by reducing flow of data
 RFC’s
1122 & 2581 detail extensions
 two categories of extensions:


retransmission timer management
window management
Problems on Multicasting
#
Consider four interconnected (Ring
topology) routers, each attached to only
one network. Each network has 50 hosts
belonging to a multicast group. If one host
sends a packet to this group, find the
following:
 - Number of packets (copies of the original
one) travelling between routers if multiple
‘Unicasting’ is followed.
Problems on Multicasting
-
Number of packets (copies of the original
one) travelling between routers if
‘Multicasting’ is followed.
Problems on Multicasting
#
Flooding implies that a router retransmits
the packet to all outgoing interfaces except
the one in which it was received.
 Packets have unique identifier so that a
router does not flood the same packet
more than once. For the network shown
on slide 40, find the packets travelling on
the links and the networks if flooding is
used.
Example: Multicasting
Problem on Encryption
#
An encrypted message is: 20 5 21 3 49
4 49 3 4 15, k = 7 and n = 55. A through Z
were initially coded as 1 to 26 and a blank
(space) as 27. Decrypt this message using
RSA.
Transport Protocol
 Two
/ Three way handshake
 Flow / Congestion control
 Different Timers
 Silly Window