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THE OSI MODEL
Application
Presentation
Where We’ve Been
Session
Transport
Network
Data-Link
Physical
Chapter 1—Review
By: Allan Johnson
Table of Contents
• Review the OSI Model
• Encapsulation
• LAN Devices & Technologies
• Transport Layer
• IP Addressing
Why A Layered Model?
Application
Presentation
Session
Transport
Network
Data-Link
Physical
• Reduces complexity
• Standardizes interfaces
• Facilitates modular
engineering
• Ensures interoperable
technology
• Accelerates evolution
• Simplifies teaching &
learning
Application Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides network services
(processes) to applications.
 For example, a computer on
a LAN can save files to a
server using a network
redirector supplied by NOSs
like Novell.
 Network redirectors allow
applications like Word and
Excel to “see” the network.
Presentation Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides data representation
and code formatting.
 Code formatting includes
compression and encryption
 Basically, the presentation
layer is responsible for
representing data so that
the source and destination
can communicate at the
application layer.
Session Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides inter-host
communication by establishing,
maintaining, and terminating
sessions.
 Session uses dialog control and
dialog separation to manage the
session
 Some Session protocols:






NFS (Network File System)
SQL (Structured Query Language)
RCP (Remote Call Procedure)
ASP (AppleTalk Session Protocol)
SCP (Session Control Protocol)
X-window
Transport Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides reliability, flow control,
and error correction through the
use of TCP.
 TCP segments the data, adding a
header with control information
for sequencing and
acknowledging packets received.
 The segment header also
includes source and destination
ports for upper-layer applications
 TCP is connection-oriented and
uses windowing.
 UDP is connectionless. UDP does
not acknowledge the receipt of
packets.
Network Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Responsible for logically
addressing the packet and
path determination.
 Addressing is done through
routed protocols such as IP,
IPX, AppleTalk, and DECnet.
 Path Selection is done by
using routing protocols such
as RIP, IGRP, EIGRP, OSPF,
and BGP.
 Routers operate at the
Network Layer
Data-Link Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides access to the media
 Handles error notification,
network topology issues, and
physically addressing the
frame.
 Media Access Control through
either...
 Deterministic—token passing
 Non-deterministic—broadcast
topology (collision domains)
 Important concept: CSMA/CD
Physical Layer
Application
Presentation
Session
Transport
Network
Data-Link
Physical
 Provides electrical,
mechanical, procedural and
functional means for
activating and maintaining
links between systems.
 Includes the medium through
which bits flow. Media can
be...




CAT 5 cable
Coaxial cable
Fiber Optics cable
The atmosphere
THE OSI MODEL
Application
Presentation
Encapsulation
Session
Transport
Network
Data-Link
Peer-to-Peer
Communications
Physical
Table of Contents
Peer-to-Peer Communications
• Peers communicate using the PDU of their
layer. For example, the network layers of the
source and destination are peers and use
packets to communicate with each other.
Application
Data
Application
Presentation
Data
Presentation
Session
Session
Transport
Data
Segments
Transport
Network
Packets
Network
Data-Link
Frames
Data-Link
Physical
Bits
Physical
THE OSI MODEL
Application
Presentation
Session
Transport
Network
Data-Link
LAN Devices &
Technologies
The Data-Link &
Physical Layers
Physical
Table of Contents
Devices
What layer device?
• What does it do?
 Connects LAN
segments;
 Filters traffic based
on MAC addresses;
and
 Separates collision
domains based upon
MAC addresses.
Devices
• What does it do?
What layer device?
 Since it is a multiport bridge, it can
also
 Connect LAN
segments;
 Filter traffic based on
MAC addresses; and
 Separate collision
domains
 However, switches
also offer full-duplex,
dedicated bandwidth
to segments or
desktops.
Devices
What layer device?
• What does it do?
 Concentrates LAN
connections from
multiple devices into
one location
 Repeats the signal (a
hub is a multi-port
repeater)
Devices
• What does it do?
What layer device?
 Interconnects networks
and provides broadcast
control
 Determines the path
using a routing protocol
or static route
 Re-encapsulates the
packet in the appropriate
frame format and
switches it out the
interface
 Uses logical addressing
(i.e. IP addresses) to
determine the path
Media Types
LAN Technologies
Three Most
Common Used
Today in
Networking
Ethernet/802.3
• Cable Specifications:
 10Base2
 Called Thinnet; uses coax
 Max. distance = 185 meters (almost 200)
 10Base5
 Called Thicknet; uses coax
 Max. distance = 500 meters
 10BaseT
 Uses Twisted-pair
 Max. distance = 100 meters
 10 means 10 Mbps
Ethernet/802.3
• Ethernet is broadcast topology.
 What does that mean?
 Every devices on the Ethernet segment sees
every frame.
 Frames are addressed with source and
destination ______ addresses.
 When a source does not know the destination
or wants to communicate with every device, it
encapsulates the frame with a broadcast MAC
address: FFFF.FFFF.FFFF
 What is the main network traffic problem
caused by Ethernet broadcast topologies?
Ethernet/802.3
• Ethernet topologies are also shared
media.
• That means media access is controlled
on a “first come, first serve” basis.
• This results in collisions between the
data of two simultaneously transmitting
devices.
• Collisions are resolved using what
method?
Ethernet/802.3
• CSMA/CD (Carrier Sense Multiple Access with
Collision Detection)
• Describe how CSMA/CD works:
 A node needing to transmit listens for activity on
the media. If there is none, it transmits.
 The node continue to listen. A collision is detected
by a spike in voltage (a bit can only be a 0 or a 1-it cannot be a 2)
 The node generates a jam signal to tell all devices
to stop transmitting for a random amount of time
(back-off algorithm).
 When media is clear of any transmissions, the
node can attempt to retransmit.
Address Resolution Protocol
• In broadcast topologies, we need a way to
resolve unknown destination MAC addresses.
• ARP is protocol where the sending device
sends out a broadcast ARP request which
says, “What’s you MAC address?”
• If the destination exists on the same LAN
segment as the source, then the destination
replies with its MAC address.
• However, if the destination and source are
separated by a router, the router will not
forward the broadcast (an important function
of routers). Instead the router replies with its
own MAC address.
THE OSI MODEL
Application
Presentation
Transport Layer
Session
Transport
Network
A Quick Review
Data-Link
Physical
Table of Contents
Transport Layer Functions
• Synchronization of the connection
 Three-way handshake
• Flow Control
 “Slow down, you’re overloading my
memory buffer!!”
• Reliability & Error Recovery
 Windowing: “How much data can I send
before getting an acknowledgement?”
 Retransmission of lost or unacknowledged
segments
Transport’s Two Protocols
• TCP
 Transmission Control
Protocol
 Connection-oriented
 Acknowledgment &
Retransmission of
segments
 Windowing
 Applications:
 Email
 File Transfer
 E-Commerce
• UDP
 User Datagram
Protocol
 Connectionless
 No
Acknowledgements
 Applications:




Routing Protocols
Streaming Audio
Gaming
Video Conferencing
THE OSI MODEL
Application
Presentation
IP Addressing
Session
Transport
Network
Subnetting Review
Data-Link
Physical
Table of Contents
Logical Addressing
• At the network layer, we use logical,
hierarchical addressing.
• With Internet Protocol (IP), this address is a
32-bit addressing scheme divided into four
octets.
• Do you remember the classes 1st octet’s
value?





Class
Class
Class
Class
Class
A: 1 - 126
B: 128 - 191
C: 192 - 223
D: 224 - 239 (multicasting)
E: 240 - 255 (experimental)
Network vs. Host
Class A:
27 = 126 networks; 224 > 16 million hosts
N
Class B :
H
H
214 = 16,384 networks; 216 > 65,534 hosts
N
Class C :
H
N
H
H
221 > 2 million networks; 28 = 254 hosts
N
N
N
H
Why Subnet?
• Remember: we are usually dealing with
a broadcast topology.
• Can you imagine what the network
traffic overhead would be like on a
network with 254 hosts trying to
discover each others MAC addresses?
• Subnetting allows us to segment LANs
into logical broadcast domains called
subnets, thereby improving network
performance.
Stealing Bits
• In order to subnet, we must steal or “borrow”
bits from the host portion on the IP address.
• First, we must to determine how many
subnets we need and how many hosts per
subnet.
• We do this through the power of 2
 For example, I need 8 subnets from a Class C:
 24 = 16 - 2 = 14 subnets
 Remember: we subtract 2 because these subnets are not
used
 How many host do we have?
 It’s a Class C, so 4 bits are left: 24 = 16 - 2 = 14 hosts
 Remember: we subtract 2 because one address is the
subnet address and one is the broadcast address
Subnet Mask
• We determine the subnet mask by adding up
the decimal value of the bits we borrowed.
• In the previous Class C example, we borrowed
4 bits. Below is the host octet showing the
bits we borrowed and their decimal values.
1
1
1
1
128
64
32
16
8
4
2
1
We add up the decimal value of these bits and get 240.
That’s the last non-zero octet of our subnet mask.
So our subnet mask is 255.255.255.240
Last Non-Zero Octet
• Memorize this table. You should be able to:
 Quickly calculate the last non-zero octet when
given the number of bits borrowed.
 Determine the number of bits borrowed given the
last non-zero octet.
 Determine the amount of bits left over for hosts
and the number of host addresses available.
Bits
Non-Zero
Borrowed Octet
Hosts
2
192
62
3
224
30
4
240
14
5
248
6
6
252
2
CIDR Notation
• Classless Interdomain Routing is a method of
representing an IP address and its subnet
mask with a prefix.
• For example: 192.168.50.0/27
• What do you think the 27 tells you?
 27 is the number of 1 bits in the subnet mask.
Therefore, 255.255.255.224
 Also, you know 192 is a Class C, so we borrowed 3
bits!!
 Finally, you know the magic number is 256 - 224 =
32, so the first useable subnet address is
197.168.50.32!!
• Let’s see the power of CIDR notation.
202.151.37.0/26
• Subnet mask?
 255.255.255.192
• Bits borrowed?
 Class C so 2 bits borrowed
• Magic Number?
 256 - 192 = 64
• First useable subnet address?
 202.151.37.64
• Third useable subnet address?
 64 + 64 + 64 = 192, so 202.151.37.192
198.53.67.0/30
• Subnet mask?
 255.255.255.252
• Bits borrowed?
 Class C so 6 bits borrowed
• Magic Number?
 256 - 252 = 4
• Third useable subnet address?
 4 + 4 + 4 = 12, so 198.53.67.12
• Second subnet’s broadcast address?
 4 + 4 + 4 - 1 = 11, so 198.53.67.11
200.39.89.0/28
• What kind of address is 200.39.89.32?




Class C, so 4 bits borrowed
Last non-zero octet is 240
Magic number is 256 - 240 = 16
32 is a multiple of 16 so 200.39.89.32 is a
subnet address--the second subnet
address!!
• What’s the broadcast address of
200.39.89.32?
 32 + 16 -1 = 47, so 200.39.89.47
194.53.45.0/29
• What kind of address is 194.53.45.26?





Class C, so 5 bits borrowed
Last non-zero octet is 248
Magic number is 256 - 248 = 8
Subnets are .8, .16, .24, .32, ect.
So 194.53.45.26 belongs to the third subnet
address (194.53.45.24) and is a host address.
• What broadcast address would this host use
to communicate with other devices on the
same subnet?
 It belongs to .24 and the next is .32, so 1 less is
.31 (194.53.45.31)