Transcript Semester 3 Chapter 1 - IIS Windows Server

Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
Semester 3
Chapter 1
Session
Transport
Network
Data-Link
Physical
Review
Table of Contents

Review the OSI Model

LAN Devices &
Technologies

IP Addressing

CIDR Notation

Routing

Transport Layer
Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
Review The Model
Session
Transport
Network
Data-Link
Physical
Open Systems
Interconnected Reference
Model
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
Provides access to the media
Handles error notification,
network topology issues, and
physically addressing the
frame.
Media Access Control
through either...
Data-Link

Physical


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
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
Encapsulation Example
Application

Presentation
Session
Transport

Network
Data-Link
Physical

You type an email message.
SMTP takes the data and
passes it to the Presentation
Layer.
Presentation codes the data
as ASCII.
Session establishes a
connection with the
destination for the purpose
of transporting the data.
Encapsulation Example

Application
Presentation
Session
Transport


Network
Data-Link
Physical

Transport segments the
data using TCP and hands it
to the Network Layer for
addressing
Network addresses the
packet using IP.
Data-Link then encaps. the
packet in a frame and
addresses it for local
delivery (MACs)
The Physical layer sends
the bits down the wire.
Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
LAN Devices &
Technologies
Session
Transport
Network
Data-Link
Physical
The Data-Link & Physical
Layers
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 layer device?

What does it do?

Since it is a multi-port
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 layer device?

What does it do?




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



10Base5



Called Thicknet; uses coax
Max. distance = 500 meters
10BaseT



Called Thinnet; uses coax
Max. distance = 185 meters (almost 200)
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 continues 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 (backoff 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.
Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
IP Addressing
Session
Transport
Network
Data-Link
Physical
Subnetting Review
Logical Addressing



At the network layer, we use logical, hierarchical
addressing.
With Internet Protocol (IP), this address is a 32bit addressing scheme divided into four octets.
Do you remember the classes 1st octet’s value?





Class A: 1 - 126
Class B: 128 - 191
Class C: 192 - 223
Class D: 224 - 239 (multicasting)
Class 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.
Four Subnetting Steps

To correctly subnet a given network address into
subnet addresses, ask yourself the following
questions:
1.
2.
3.
4.

How many bits do I need to borrow?
What’s the subnet mask?
What’s the “magic number” or multiplier?
What are the first three subnetwork addresses?
Let’s look at each of these questions in detail
1. How many bits to borrow?



First, you need to know how many bits you
have to work with.
Second, you must know either how many
subnets you need or how many hosts per
subnet you need.
Finally, you need to figure out the number
of bits to borrow.
1. How many bits to borrow?

How many bits do I have to work with?

Depends on the class of your network address.





Class C: 8 host bits
Class B: 16 host bits
Class A: 24 host bits
Remember: you must borrow at least 2 bits for
subnets and leave at least 2 bits for host addresses.
2 bits borrowed allows 22 - 2 = 2 subnets
1. How many bits to borrow?


How many subnets or hosts do I need?
A simple formula:



Total Bits = Bits Borrowed + Bits Left
TB = BB + BL
I need x subnets:
• I need x hosts:
2 2x
BL
2 2x
BB
• Remember: we need to subtract two to
provide for the subnetwork and
broadcast addresses.
1. How many bits to borrow?





Class C Example: 210.93.45.0
Design goals specify at least 5 subnets so how
many bits do we borrow?
How many bits in the host portion do we have to
work with (TB)?
What’s the BB in our TB = BB + BL formula? (8
= BB + BL)
2 to the what power will give us at least 5
subnets?
3
2 - 2 = 6 subnets
1. How many bits to borrow?


How many bits are left for hosts?
TB = BB + BL
8 = 3 + BL
BL = 5
So how many hosts can we assign to each
subnet?
5
2 - 2 = 30 hosts
1. How many bits to borrow?





Class B Example: 185.75.0.0
Design goals specify no more than 126 hosts
per subnet, so how many bits do we need to
leave (BL)?
How many bits in the host portion do we have to
work with (TB)?
What’s the BL in our TB = BB + BL formula? (16
= BB + BL)
2 to the what power will insure no more than 126
hosts per subnet and give us the most subnets?
7
2 - 2 = 126 hosts
1. How many bits to borrow?


How many bits are left for subnets?
TB = BB + BL
16 = BB + 7
BL = 9
So how many subnets can we have?
9
2 - 2 = 510 subnets
2. What’s the 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 3
bits. Below is the host octet showing the bits we
1
1and their
1 decimal values.
borrowed
128 64 32 16
8
4
2
1
We add up the decimal value of these bits and get 224.
That’s the last non-zero octet of our subnet mask.
So our subnet mask is 255.255.255.224
3. What’s the “magic number?”



To find the “magic number” or the
multiplier we will use to determine the
subnetwork addresses, we subtract the
last non-zero octet from 256.
In our Class C example, our subnet mask
was 255.255.255.224. 224 is our last nonzero octet.
Our magic number is 256 - 224 = 32
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
4. What are the subnets?



We now take our “magic number” and use it as a
multiplier.
Our Class C address was 210.93.45.0.
We borrowed bits in the fourth octet, so that’s
where our multiplier occurs




1st subnet: 210.93.45.32
2nd subnet: 210.93.45.64
3rd subnet: 210.93.45.96
We keep adding 32 in the fourth octet to get all
six available subnet addresses.
Host & Broadcast Addresses



Now you can see why we subtract 2 when
determining the number of host address.
Let’s look at our 1st subnet: 210.93.45.32
What is the total range of addresses up to our
next subnet, 210.93.45.64?

210.93.45.32 to 210.93.45.63 or 32 addresses

.32 cannot be assigned to a host. Why?
.63 cannot be assigned to a host. Why?
So our host addresses are .33 - .62 or 30 host
addresses--just like we figured out earlier.


Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
CIDR Notation
Session
Transport
Network
Data-Link
Physical
A Different Way to Represent
a Subnet Mask
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?


Bits borrowed?


256 - 192 = 64
First useable subnet address?


Class C so 2 bits borrowed
Magic Number?


255.255.255.192
202.151.37.64
Third useable subnet address?

64 + 64 + 64 = 192, so 202.151.37.192
198.53.67.0/30

Subnet mask?


Bits borrowed?


256 - 252 = 4
Third useable subnet address?


Class C so 6 bits borrowed
Magic Number?


255.255.255.252
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 200.39.89.0?





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)
No Worksheet Needed!




After some practice, you should never need a
subnetting worksheet again.
The only information you need is the IP address
and the CIDR notation.
For example, the address 221.39.50/26
You can quickly determine that the first subnet
address is 221.39.50.64. How?



Class C, 2 bits borrowed
256 - 192 = 64, so 221.39.50.64
For the rest of the addresses, just do multiples of
64 (.64, .128, .192).
The Key!!

MEMORIZE THIS TABLE!!!
Bits
Non-Zero
Borrowed Octet
Hosts
2
192
62
3
224
30
4
240
14
5
248
6
6
252
2
Practice On Your Own

Below are some practice problems. Take out a
sheet of paper and calculate...



1.
2.
3.
4.
5.
6.
7.
Bits borrowed
Last non-zero octet
Second subnet address and broadcast address
192.168.15.0/26
220.75.32.0/30
200.39.79.0/29
195.50.120.0/27
202.139.67.0/28
Challenge: 132.59.0.0/19
Challenge: 64.0.0.0/16
Answers
Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
Routing Basics
Session
Transport
Network
Data-Link
Physical
Path Determination & Packet
Switching
A Router’s Functions


A router is responsible for determining the
packet’s path and switching the packet out the
correct port.
A router does this in five steps:
1.
2.
3.
4.
5.
De-encapsulates the packet
Performs the ANDing operation
Looks for entry in routing table
Re-encapsulates packet into a frame
Switches the packet out the correct interface
Routed v. Routing Protocols

What is a routed protocol?




Routed protocols are protocols that enable data to be
transmitted across a collection of networks or
internetworks using a hierarchical addressing
scheme.
Examples include IP, IPX and AppleTalk.
A routable protocol provides both a network and node
number to each device on the network. Routers AND
the address to discover the network portion of the
address.
An example of a protocol that is not routable is
NetBEUI because it does not have a network/node
structure.
Routed v. Routing Protocols

What is a routing protocol?



A routing protocol is a protocol that determines the
path a routed protocol will follow to its destination.
Routers use routing protocols to create a map of the
network. These maps allow path determination and
packet switching. Maps become part of the router’s
routing table.
Examples of routing protocols include: RIP, IGRP,
EIGRP, & OSPF
Multi-protocol Routing


Routers are capable of running multiple routing
protocols (RIP, IGRP, OSPF, etc.) as well as
running multiple routed protocols (IP, IPX,
AppleTalk).
For a router to be able use different routing and
routing protocols, you must enable the protocols
using the appropriate commands.
Dynamic v. Static Routing


Dynamic routing refers to the process of allowing
the router to determine the path to the
destination.
Routing protocols enable dynamic routing where
multiple paths to the same destination exist.
Dynamic v. Static Routing


Static routing means that the network
administrator directly assigns the path router are
to take to the destination.
Static routing is most often used with stub
networks where only one path exists to the
destination.
Default Routes

A default route is usually to a border or gateway
router that all routers on a network can send
packets to if they do not know the route for a
particular network.
Routing Protocol Classes

Routing protocols can be divided into three
classes:



Distance–vector: determines the route based
on the direction (vector) and distance to the
destination
Link-state: opens the shortest path first to the
destination by recreating an exact topology of
the network in its routing table
Hybrid: combines aspects of both
Convergence



Convergence means that all routers share the
same information about the network. In other
words, each router knows its neighbor routers
routing table
Every time there is a topology change, routing
protocols update the routers until the network is
said to have converged again.
The time of convergence varies depending upon
the routing protocol being used.
Distance-vector Routing


Each router receives a routing table periodically
from its directly connected neighboring routers.
For example, in the graphic, Router B receives
information from Router A. Router B adds a
distance-vector number (such as a number of hops),
and then passes this new routing table to its other
neighbor, Router C.
Link-state Routing


Link-state protocols maintain complex databases
that summarize routes to the entire network.
Each time a new route is added or a route goes
down, each router receives a message and then
recalculates a spanning tree algorithm and updates
its topology database.
Comparing the Two
DISTANCE-VECTOR
LINK-STATE
Views network topology from
neighbor’s perspective
Gets common view of entire
network topology
Adds distance vectors from
router to router
Calculates the shortest path to
other routers
Frequent, periodic updates:
slow convergence
Event triggered updates: fast
convergence
Passes copies of routing tables Passes link-state routing updates
to neighbors
to all routers in the system.
Hybrid Routing


Cisco’s proprietary routing protocol,
EIGRP, is considered a hybrid.
EIGRP uses distance-vector metrics.
However, it uses event-triggered topology
changes instead of periodic passing of
routing tables.
Institute of Technology,
Sligo Dept of Computing
THE OSI MODEL
Application
Presentation
Transport Layer
Session
Transport
Network
Data-Link
Physical
A Quick Review
Transport Layer Functions

Synchronization of the connection


Flow Control


Three-way handshake
“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