Transcript Nivel Físico

Cap. 3 (Commer)
3.1 Introduccion
 Reunir distintas tecnologias de red dentro de
un todo coordinado.
 Esquema que esconde los detalles del
hardware subyacentes de red a la vez que
proporciona servicios universales de
comunicacion.
 La interconexion trata de ocultar detalles de
la red.
 Enfoques de Interconectividad:
Nivel Aplicación
Nivel red
Nivel Red en Internet
 Provee transportación mejor-esfuerzo
para transmitir datagramas (paquetes)
de la fuente al destino. La fuente y el
destino puede estar en la misma red o
en diferentes redes.
Environment (Medio Ambiente):
Aplicación
Aplicación
Presentación
Red:
re-dirección
Actualizacion
de Tablas
Presentación
Sesión
Transporte
Enlace Datos
Red:
Físico
re-dirección
Enlace Datos
Físico
Red:
re-dirección
Actualizacion
de Tablas
Enlace Datos
Físico
Red:
re-dirección
Actualizacion
de Tablas
Enlace Datos
Físico
Red:
re-dirección
Actualizacion
de Tablas
Sesión
Transporte
Red:
re-dirección
Enlace Datos
Enlace Datos
Físico
Físico
Red:
re-dirección
Actualizacion
de Tablas
Enlace Datos
Físico
Nivel Red en Internet
 A nivel red Internet es una colección
de subredes, conocidos como
sistemas autónomos (AS de sus siglas
en inglés) conectados.
 El cemento que une estas redes es el
nivel red (IP internetwork Protocol).
Environment
Líneas trasatlánticas
rentadas
Backbone
de ASIA
Backbone
de USA
Red
Regional
Red
Universitaria
Líneas trasatlánticas
rentadas
Red
Regional
Red
Local
Backbone
de Europa
Red
Nacional
Red de una
Companía
Cap. 4 (Forouzan)
IP Addressing:
 La dirección IP tiene 32 bits de longitud escritos como 4




octetos.
Dos dispositivo NUNCA pueden tener la misma dirección
IP, pero un dispositivo puede tener MAS de una dirección
IP (multihomed device).
La dirección IP representa localizaciones, NO nombres de
dispositivos.
La dirección IP tiene dos partes: netid y hostid.
Una dirección IP se expresa como 4 conjuntos de 8 bits
separados por un “.”.
11000000.00000101.00100010.00001011
 Por facilidad, se utiliza la notación “dotted-decimal” para
expresar la dirección
192.5.34.11
Direccionamiento en IP
Clase A (8,24): 256 redes y
224
hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
0
Red
Host
Rango de Direcc.
de los hosts.
1.0.0.0 a
127.255.255.255
Clase B (16,16): 32,527 redes y 32,527 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
128.0.0.0 a
1 0
Red
Host
191.255.255.255
Clase C (24,8): 224 redes y 256 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 1 0
Red
Host
192.0.0.0 a
223.255.255.255
Clase D (4,28)
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 1 1 0
Dirección "Mulicast"
224.0.0.0 a
239.255.255.255
Clase E (4,28)
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 1 1 1
Reserv eed
240.0.0.0 a
255.255.255.255
Sample Internet
129.8.0.1
129.8.45.13
220.3.6.3
129.8.0.0
222.13.16.40
220.3.6.0
220.3.6.23
222.13.16.41
G
G
222.13.16.0
220.3.6.1
x.y.z.t
R
207.42.56.2
To the rest of the
internet
R
124.0.0.0
124.100.33.77
124.42.5.45
124.4.51.66
Number of Networks and hosts in
each class
Class
Number of
Networks
Number of
Hosts
A
2 7  2  126
224  2  16,777,214
B
214  16,384
216  2  65,534
C

221  2,097,152
2 8  2  254
D

E

No aplica
(direcciones
especiales)

No aplica(reservadas)
Direcciones Especiales
Special
Address
Netid
Hostid
Comment
Network address
Specific
All 0’s
Network itself is considered an entity with an IP
address.
Direct Broadcast
address.
Specific
All 1’s
It used by a router to send a packet to all hosts in a
specific network.
Limited
Broadcast
address.
All 1’s
All 1’s
A host which wants to send a packet to every other
host in the current network. (Actually a class E
address). The router block this packet to confine
the broadcasting to the local network.
This Host on this
network
All 0’s
All 0’s
Used by a host at bootstrap time when it does not
know its IP address. At this time, the host send this
address as source address and a limited Broadcast
addr. as destination address. (Actually a class A
address)
Specific host on
this network
All 0’s
Specific
It used by a host to send a packet to a another
host on the same network. (Actually a class A
address)
Loopback
address
127
Any
It used by a host to send a packet to itself.
(Actually a class A address). Used to test software.
Direcciones Multicast por
Categoría
Address
Group
224.0.0.0
Reserved
224.0.0.1
All SYSTEMS on this SUBNET.
224.0.0.2
All Routers on this SUBNET.
224.0.0.4
DVPRM ROUTERS
224.0.0.5
OSPFIGP all ROUTERS
224.0.0.6
OSPFIGP Designeted ROUTERS
224.0.0.7
ST Routers
224.0.0.8
ST Hosts
224.0.0.9
RIP2 Routers
224.0.0.10
IGRP Routers
224.0.0.11
Mobile-Agents
Direcciones Multicast para
Conferencia
Address
Group
224.0.1.7
AUDIONEWS
224.0.1.10
IETF-1-LOW-AUDIO
224.0.1.11
IETF-1-AUDIO
224.0.1.12
IETF-1-VIDEO
224.0.1.13
IETF-2-LOW-AUDIO
224.0.1.14
IETF-2-AUDIO
224.0.1.15
IETF-2-VIDEO
224.0.1.16
MUSIC-SERVICE
224.0.1.17
SEANET-TELEMETRY
224.0.1.18
SEANET-IMAGE
Private Networks

Issues:
1.
2.

Apply for a unique address and use it without being connected
to internet.
1.
Advantage: Future Integration to internet without hassle.
2.
Disadvantage: Almost imposible to obtain a class A o B
addresses these days..
Use any class address without registering it with the internet
authorities.

Advantage: They can be used without permission.

Disadvantage: The address does not have to be unique.
(Confusion).
Solution:

Internet authorities have reserved a block of addresses:

Advantage:

They can be used without permission.

veryboy know that these addresses are for private
networks. They are unique inside of the organization.
Number of Networks and hosts in
each class
Class
Number of Networks
Total
A
10.0.0
1
B
172.16 to 172.31
16
C
192.68.0 to 192.68.255 256
Exercises:

Identify the class of the following IP address: 4.5.6.7
A. class A
B.

class C
Identify the class of the following IP address: 191.1.2.3
A. class A
B.

B. class B
D. Class D
class C
B. class B
D. Class
Identify the class of the following IP address: 169.5.0.0
A. class A
B. class C
B. class B
D. Class D
Exercises:

Identify the class of the following IP address:
241.1.2.3
A. class A.
B. C. class

What of the following is a source IP address:
A.
B.

B. class B.
D. Class
This host on this network.
Loopback address
B. limited broadcast address.
D. specific host on this network.
Using the limited broadcast address, a ______ sends
a packet to ______ on the network:
A. host; all other hosts. B. router; all other routers.
B. host; a specific host D. host; itself
Exercises :

What destination address can be used to send a packet
from a host with IP address 188.1.1.1 to all hosts on the
network.
A.
B.


188.0.0.0
255.255.255.255
B. 0.0.0.0
D. b and c.
A host with address 142.5.0.1 needs to test internal
software. What is the destination address in the packet:
A. 127.0.0.0
B. 127.1.1.1
B. 127.127.127.127
D. all the above
A packet send form a node with IP address 198.123.46.20
to all nodes on network 198.123.46.0 requires a
_____address.
A. unicast
B. multicast
B. broadcast
D. a or b
Exercise 1: Find 7 errors
129.8.0.1
129.8.45.13
220.3.6.3
129.8.0.0
222.13.16.40
220.3.0.0
220.3.6.23
G
G
222.13.16.0
220.3.6.1
R
x.y.z.t
207.42.56.1
206.42.56.2
To the rest of the
internet
R
124.0.0.1
124.100.33.77
124.255.255.255
124.4.51.66
Cap. 5 (Forouzan)
Subnetting and Supernetting
 IP addressing works with two levels of hierachy
(netid, hostid), when the two levels of hierachy are
not enough, then we can use either:
 Subnetting:
 Network is divided into several smaller subnetworks
with each subnetwork having its own subnetwork
address.
 The rest of the Internet is not aware of the change.
 The router knows how to route packets in the subnet.
 Supernettting:
 Class C address are stil available.
 Combination of several class C addresses to create
a larger range of addresses.
 The rest of the Internet is not aware of the change.
Subnet:
Class B
Subnet
b
a
l3
l1
l6
d
l2
l4
e
c
l5
B
A l1
R1
Rest of
Internet
Subnet
l
3
l6
D
l2
l4
C
Subnet
b
a
l3
l1
l6
d
l2
l4
c
l5
G
F
l5
l7
E
Subnet (Subred)
 Example:
Clase B (16,6,10): 32,527 redes, 62 subredes y 1,024 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 0
Red
Subred
Host
Clase B (16,10,6): 32,527 redes, 1024 subredes y 64 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 0
Red
Subred
Host
Rango de Direcc.
de los hosts.
128.0.0.0 a
191.255.255.255
Rango de Direcc.
de los hosts.
128.0.0.0 a
191.255.255.255
Masking
 Operation to obtain the subnettwork address
from an IP address.
 Mask: 32-bit number, diviedn in two parts:
 The bits in the mask containing 1s defines

the netid or combination of netid and
subnetid.
The part of the 0s define the hostid
 To get the subnet address, the router applys
the bit-wise-and operation on the IP address
and the mask.
 Ejemplo: Si se pidieron prestados 8 bits al
campo del “host”, entonces la máscara
seria: 255.255.255.0
Special Address in Subnetting
Special
Address
Netid
Hostid
Comment
Subnetwork
address
Specific
All 0’s
Subnetwork itself is considered an entity with an IP
address.
Direct Broadcast
address.
Specific
All 1’s
It used by a router to send a packet to all hosts in a
specific subnetwork.
Limited
Broadcast
address.
All 1’s
All 1’s
A host which wants to send a packet to every other
host in the current subnetwork. (Actually a class E
address). The router block this packet to confine
the broadcasting to the local subnetwork.
This Host on this
subnetwork
All 0’s
All 0’s
Used by a host at bootstrap time when it does not
know its IP address. At this time, the host send this
address as source address and a limited Broadcast
addr. as destination address. (Actually a class A
address)
Specific host on
this subnetwork
All 0’s
Specific
It used by a host to send a packet to a another
host on the same subnetwork. (Actually a class A
address)
Loopback
address
127
Any
It used by a host to send a packet to itself.
(Actually a class A address). Used to test software.
 Example:
11000000.00000101.00100010.00001011
192.5.34.11
Mask:
255.255.255.0
Result:
192.5.34.0
Subnet:
Supernet
Class
C
b
a
l3
l1
l6
d
l2
l4
e
c
l5
B
A l1
R1
Rest of
Internet
Class
C
l
3
l6
D
l2
l4
C
Class
C
b
a
l3
l1
l6
d
l2
l4
c
l5
G
F
l5
l7
E
Subnets
 Example:
Class C (22,2,8): 222 nets, 4 supernets y 256 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 1 0
Red
SN
Host
192.0.0.0 a
223.255.255.255
Clase C (20,4,8): 220 nets, 16 supernet y 256 hosts.
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
1 1 0
Red
Super
Host
192.0.0.0 a
223.255.255.255
Masking
 Operation to obtain the subnettwork address
from an IP address.
 Mask: 32-bit number, divied in two parts:
 The bits in the mask containing 1s defines

the supernetid.
The part of the 0s define the hostid.
 To get the supernet address, the router
applys the bit-wise-and operation on the IP
address and the mask.
 Ejemplo: Si se pidieron prestados 2 bits al
campo del “host”, entonces la máscara
seria: 255.255.252.0
 Example:
11000000.00000101.00100010.00001011
192.5.34.11
Mask:
255.255.252.0
Supernet:
192.5.32.0
Sample Internet
134.18.0.0
255.255.248.0
129.8.0.0
255.255.192.0
129.8.64.2 129.8.127.254
220.3.6.0
129.8.128.2 129.8.181.246
255.255.255.248
129.8.128.0
129.8.64.0
220.3.6.10
220.3.6.8
220.3.6.14
129.8.64.1 129.8.128.1
220.3.6.9
G
222.13.16.40
222.13.16.0
220.3.6.241
255.255.0.0
220.3.6.242
134.
R
x.y.z.t
220.3.6.16
G
222.13.16.41
To the rest of the
internet
220.3.6.246
207.42.56.2
R
124.100.33.77
124.0.0.0
255.224.0.0
124.
124.
124.
124.
124.
124.
124.
124.
124.
Exercises :

In Fig. 5.2, what is the mask for the network.
A. 255.255.0.0
B. 0.0.255.255

A device has the IP address 190.1.2.3. What is the
subnetid?
A. 1
B. 3

C. 255.255.255.0
D. none of the above
C. 2
D. insufficient information to answer.
Which of the followng is the defaul mask for the address:
98.0.46.201?
A. 255.0.0.0
B. 255.255.255.0
C. 255.255.0.0
D. 255.255.255.255
Exercises:

What class of IP address does the subnet mask
255.255.128.0 operate on?
C. class B.
D. class A,B or C.
A. class A.
B. class C

The subenet mask for a class C network is
255.255.255.192. How many subnetworks are
available? (Disregard special address):
A.
B.

2
8
C. 4
D. 192
A supernet mask is 255.255.248.0. How many class
C networks were combined to make this supernet:
A.
B.
2
6
C. 4
D. 8
Cap. 7 (Forouzan)
IP datagram
0
8
Ver
HLEN
16
Service Type
TOS
P R E
Identification
Time to Live=#Hops
24
31
Total_Length=header+Data_length
Flags
Protocol(TCP, UDP...)
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Option (0-40 bytes)
Data
Types of Service
TOS bits
Description
0000
Normal (Default)
0001
Minimize Cost
0010
0100
1000
Maximize
reliability
Maximize
Throughput
Minimize Delay
Default TOS
Protocol
TOS bits
Description
ICMP
0000
Normal
IGP
0010
Maximize Reliability
SNMP
0010
Maximize Reliability
TELNET
1000
Minimize Delay
FTP (data)
0100
Maximize Throughput
FTP (control)
1000
Minimize Delay
SMTP (data)
0100
Maximize Throughput
SMTP (control)
1000
Minimize Delay
IP datagram: Fragmentation
Control
bits
0
8
16
24
Ver
HLEN
Service Type
Flags
Identification
Time to Live
Total Length
TOS
P R E
Fragmentation Offset
D M
Protocol
Header Checksum
Source IP address
Destination IP address
Option
Data
31
Figure 7-7
Figure 7-8
IP datagram: Options
0
8
Ver
HLEN
16
Total Length
Service Type
Identification
Time to Live
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Code
Cp
Css
Number
Length
(code+length+Hdata)
Data
HData
31
Record Route Option
0
8
Ver
HLEN
16
Total Length
Service Type
Identification
Time to Live
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Code
0
00
00111
Pointer
Length
IP addresses
Data
31
Figure 7-14
Strict Source Route Option
0
8
Ver
HLEN
16
Total Length
Service Type
Identification
Time to Live
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Code
1
00
01001
Pointer
Length
IP addresses
Data
31
Figure 7-16
Loose Source Route Option
0
8
Ver
HLEN
16
Total Length
Service Type
Identification
Time to Live
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Code
1
00
00011
Pointer
Length
IP addresses
Data
31
Timestamp Option
0
8
Ver
HLEN
16
Time to Live
Total Length
Service Type
Identification
31
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Code
0
10
00100
Length
Pointer
O-Flow Flags
IP addresses and timestamp to be stored
Data
Figure 7-19
Figure 7-20
Checksum
0
8
Ver
HLEN
16
Service Type
Total Length
TOS
P R E
Identification
Time to Live
24
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Option
Data
31
Figure 7-23
Fragmentation
 Useful when a datagram travel through
different networks.
 MTU (Maximum Transfer Unit) is the maximum
length of data that can be encapsulated in a
frame.
 If a datagram is fragmented is fragment, it
becomes to be a new datagram.
 When a datagram is fragmented, requires
part of the header to be copied by all
fragments.
MTUs for Different Networks
Protocol
MTU
Hyperchannel
65,535
Token Ring
(16 Mbps)
17,914
Token Ring
(4 Mbps)
4,464
FDDI
X.25
PPP
4,532
576
296
No fragmentation needed.
IP Datagram
Header
MTU
Trailer
No fragmentation needed.
IP Datagram
H
MTU
T
H
MTU
T
H
MTU
T
H
H MTU T
H
T
MTU
T
When a host or router process a
frame
0
8
31
16
24
Ver
HLEN Service Type
Identification
Time to Live
Total Length
Flags
Protocol
Fragmentation Offset
Header Checksum
Source IP address
Destination IP address
Option
Data
Figure 7-24
Figure 7-25
Figure 7-26
Cap. 6 Forouzan
Servicios
Servicio
Orientado a
Conexión
Servicio con
Confirmación
Servicio sin
Confirmación
X.25
TCP,
Ethernet,
Frame Relay
Ethernet
UDP,
IP,
Ethernet
Conexión
Datos
Desconexión
Servicio NO
Orientado a
Conexión
Direct vs. Indirect Delivery
Routing Methods
Static vs. Dynamic Routing
 Static Routing:
 The administrator enter the route for each


destination into the table manually.
Not updated automatically.
Used in small networks that do not change
frequently.
 Dynamic Routing:
 Updated periodically using Dynamic
Routing Protocols: RIP, OSPF, BGP
Routing Module
Routing Table
Receive: an IP packet
1.
For each entry in the routing table
1.
2.
Apply the mask to packet destination address
If (the mask matches the value in the destination field)
1.
If (the G flag is absent)
1.
2.
3.
2.
3.
Use packet destination address as next hop address
Send packet to fragmentation module with next hop address
Return
If no match is found, send an ICMP error message.
Return.
Example: Routing table for router R1
Mask
Destination
Next Hop
F.
R.C.
U.
I.
255.0.0.0
111.0.0.0
-
U
0
0
m0
255.255.255.224
193.14.5.160
-
U
0
0
m2
255.255.255.224
194.17.21.192
-
U
0
0
m1
255.255.255.255
194.17.21.16
111.20.18.14
UGH
0
0
m0
255.255.255.0
192.16.7.0
111.15.17.32
UG
0
0
m0
255.255.255.0
194.17.21.0
111.20.18.14
UG
0
0
m0
0.0.0.0
0.0.0.0
111.30.31.18
UG
0
0
m0
……………
When R1 receives a packet for destination:
192.16.7.14
Mask
Destination
Address
Hop
ANDNext
Mask
Match?
F.
R.C.
U.
I.
255.0.0.0
111.0.0.0
192.0.0.0
NoU
0
0
m0
255.255.255.224
193.14.5.160
192.16.7.0
NoU
0
0
m2
255.255.255.224
194.17.21.192 192.16.7.0
-
NoU
0
0
m1
……………
192.16.7.14
255.255.255.255
194.17.21.16
111.20.18.14 NoUGH
192.16.17.14
0
0
m0
255.255.255.0
192.16.7.0
111.15.17.32 Match!!
UG
192.16.17.0
1
1
m0
255.255.255.0
194.17.21.0
111.20.18.14
UG
0
0
m0
0.0.0.0
0.0.0.0
111.30.31.18
UG
0
0
m0
Increase by 1, if
other packet with the
same destination reach
this router.
When R1 receives a packet for destination:
193.14.5.176
Mask
Address AND
Destination MaskNext Hop
193.0.0.0
-
Match
? F.
R.C.
U.
I.
NoU
0
0
m0
Match!
U
1
1
m2
255.0.0.0
111.0.0.0
255.255.255.224
193.14.5.160 193.14.15.160
-
255.255.255.224
194.17.21.192
-
U
0
0
m1
255.255.255.255
194.17.21.16
111.20.18.14
UGH
0
0
m0
255.255.255.0
192.16.7.0
111.15.17.32
UG
0
0
m0
255.255.255.0
194.17.21.0
111.20.18.14
UG
0
0
m0
0.0.0.0
0.0.0.0
111.30.31.18
UG
0
0
m0
……………
193.14.5.176
Increase by 1, if
other packet with the
same destination reach
this router.
When R1 receives a packet for destination:
200.34.12.34
Mask
Destination
Address
Hop
ANDNext
Mask
Match?
F.
R.C.
U.
I.
255.0.0.0
111.0.0.0
200.0.0.0
NoU
0
0
m0
255.255.255.224
193.14.5.160
200.34.12.32
NoU
0
0
m2
255.255.255.224
194.17.21.192 200.34.12.32
-
NoU
0
0
m1
……………
200.34.12.34
255.255.255.255
194.17.21.16
111.20.18.14 NoUGH
200.34.12.34
0
0
m0
255.255.255.0
192.16.7.0
111.15.17.32 NoUG
200.34.12.0
0
0
m0
255.255.255.0
194.17.21.0
111.20.18.14 NoUG
200.34.12.0
0
0
m0
0.0.0.0
0.0.0.0
111.30.31.18 Match!
UG
0.0.0.0
1
1
m0
Increase by 1, if
other packet with the
same destination reach
this router.
Chapter 8
ARP
and
RARP
Static vs. Dynamic
Mapping
Static
Table
Static
Table
ARP (Address Resolution Protocol)
ARP (Address Resolution Protocol)
NL-1: Sender knows
the IP address of
the target.
NL-2: IP ask ARP to
create a request
ARP message.
DLL: encapsulated
using broadcast
address in the
destination field.
NL-1: Recives the
destination
phyisical address
NL-1: recognise the
IP address.
NL-2: IP ask ARP to
create a reply ARP
message with its
physical address.
DLL: encapsulated
using unicast
address in the
destination field.
ARP packet
(i.e. IPv4=0800H)
(i.e. Ethernet=1)
(Physical
Address
Length)
(Logical
Address
Length)
Encapsulation in an Ethernet frame
Figure 8-5, Part I
Figure 8-5, Part II
Proxy ARP
Figure 8-7
RRQ
(if I am not the
destination, only
updates table)
RRQ/srp
Sleep
IP_RE
(if solved in
the cache)
IP_RE/srq
Pending
RRP
(updates the
cache table)
Original Cache Table used for the
examples
State Queue
Attempt
Time-out
900
Protocol
Address
R
5
P
2
2
129.34.4.8
P
14
5
201.11.56.7
R
8
P
12
450
1
180.3.6.1
114.5.7.89
Hardware
Address
ACAE32457342
457342ACAE32
220.55.5.7
F
R
9
P
18
60
3
19.1.7.82
188.11.8.71
4573E3242ACA
Example: ARP output module receive and IP
datagram with address: 114.15.7.89
State Queue
Attempt
Time-out Protocol
Address
R
5
900
P
2
2
129.34.4.8
P
14
5
201.11.56.7
R
8
P
12
450
1
180.3.6.1
114.5.7.89
Hardware
Address
ACAE32457342
457342ACAE32
220.55.5.7
F
R
9
P
18
60
3
19.1.7.82
188.11.8.71
4573E3242ACA
Example: ARP output module receive and IP
datagram with address: 116.1.7.22
State Queue
Attempt
Time-out Protocol
Address
R
5
900
P
2
2
129.34.4.8
P
14
5
201.11.56.7
R
8
P
12
1
220.55.5.7
P
23
1
116.1.7.22
450
180.3.6.1
114.5.7.89
Hardware
Address
ACAE32457342
457342ACAE32
F
R
9
P
18
60
3
19.1.7.82
188.11.8.71
4573E3242ACA
Example: ARP input module receive and IP
datagram with address: 188.11.8.71 and
Hardware Address: E34573242ACA
State Queue
Attempt
Time-out Protocol
Address
R
5
900
P
2
2
129.34.4.8
P
14
5
201.11.56.7
R
8
P
12
1
220.55.5.7
P
23
1
116.1.7.22
450
180.3.6.1
114.5.7.89
Hardware
Address
ACAE32457342
457342ACAE32
F
R
9
60
19.1.7.82
4573E3242ACA
R
18
900
188.11.8.71 E34573242ACA
Example: The cache-control updates every entry.
The time-out values is decremented by 60.
State Queue
R
5
P
2
Attempt
Time-out Protocol
Address
840
3
180.3.6.1
Hardware
Address
ACAE32457342
129.34.4.8
F
201.11.56.7
R
8
390
114.5.7.89
P
12
2
220.55.5.7
P
23
2
116.1.7.22
457342ACAE32
F
F
R
18
840
188.11.8.71 E34573242ACA
RARP
 Why RARP?
RARP
RARP Packet
(i.e. IPv4=0800H)
(i.e. Ethernet=1)
(Physical
Address
Length)
(Logical
Address
Length)
RARP in an Ethernet Frame
1.
2.
3.
4.
In a _______ protocol associates a logical address with a phyisical
address.
A. Static mapping C. physical mapping
B. Dynamic mapping
D. a and b
The _______ is a dynamic mapping protocol in which a logical address
is found for a given physical address.
A. ARP
C. ICMP
B. RARP
D. none of the above
A router reads the _______ address on a packet to determine the next
hop?
1. Logical
C. Source
2. Physical
D. ARP
A ARP reply is_______ to _______.
1. broadcast;all hosts
C. unicast; all hosts
2. multicast; all hosts
D. unicast; one host
Chapter 9
Internet Control
Message Protocol
(ICMP)
ICMP (Internet Control Message
Protocol)
 Internet does not have
 Error Control mechanism
 Flow Control mechanism
 Assistant Mechanism
 ICMP compensate this two problems.
ICMP Position in the network
layer.
ICMP encapsulation
ICMP Format
Types of Messages
Type
Message
Type
Message
3
Destination Unreachable
8 or 0
Echo request or reply
4
Source quench (“flow Control”)
13 or 14 Timestamp request or reply
11
Time exceeded
17 or 18 Address mask request or reply
12
Parameter Problem
10 or 9
5
Redirection
Router solicitation and
advertisement
Types of Error Report Messages
Contents of data-field for the
error message
Destination Unreachable
0 Nework Unreachable
8
Source Host isolated
1 Host Unreachable
9
Destination network prohibited.
2 Protocol Unreachable
10
Destination host prohibited
3 Port Unreachable
11
Network unreachable for the requested
TOS.
4 Fragmentation required
12
Host unreachable for the requested TOS
5 Source Routing can’t be
accomplished.
13
Destination host filtered out.
6 Destination Network Unknown
14
Host precedence violated.
7 Destination host Unknown
15
IP precedence lower the network
precedence level.
Source quenche (“flow control”)
 A message is send for every datagram
is discarded by a router or the
destination host.
Time Exceeded
0
Number of hops exceeded
1
Final destination does not
receive all the fragments.
Parameter Problem
0
1
Error or ambiguity in one of
the header fields
Required part of an option I
missed.
Redirection
 A host usually start with a small routing table that is
gradually augmented and updated. One of the tools
to accomplish this is the redirection message.
Format
0
Redirection for the network-specific route.
1
Redirection for host-specific route.
2
Redirection for network-specific route based
on the specified TOS.
3
Redirection for the host-specific route based
on the specified TOS.
Types of Query Messages
 Used to diagnose some network
problems.
Echo Request and Reply
 For diagnostic purposes.
 Test if



there is communication at the IP level.
an intermediate router is receiving, processing and forwarding
packets.
another host is reachable (ping-packet internet groper).
 The identification and sequence number field are not formally
defined, and can be used arbitrarily by the sender.
Timestamp Request and Reply
 Used for:
 Round-trip time
 Sincronize clocks
Round-trip calculation
t

Original

tReceive
t Transmit 
dSending  tReceive
  t Original
dReceiving  t Returned
 tTransmit

tRoundTrip  dSending  dReceiving
Example:
t Original  46
t Receive  59
t Transmit  60
t Returned  67

dSending  59  46  13
dReceiving  67  60  7
t RoundTrip  13 7  20
Syncronize Clocks
t

Original

tReceive
t Transmit 
dSending  t Received (tOriginal  done way )
done way
t RoundTrip


2
Example:
t Original  46
t Receive  59
t Transmit  60
t Returned  67

dSending  59  46  13
dReceiving  67  60  7
t RoundTrip  13 7  20
dt  59  (46  20/2)  3
Address Mask Request and Reply
 To identify network address,
subnetwork address and host identifier.
Router Solicitation and Advertisement
 In order that host knows the router
connected to its network.
 The host should know if the router is
alive.
Router Solicitation and Advertisement
Solicitatio
n
Advertisme
nt
Advertisme
nt
Advertisme
nt
Format
 Solicitation:
 Advertisement:
Checksum
Figure 9-20
1.
2.
3.
4.
5.
If a host needs to syncronise its clock with another host, it sends a _____ message.
A.
B.
timestamp-request
C. router-advertisement
Source-quenche D. time-exceeded
The purpose of the echo request and echo reply is to _______.
A.
B.
C.
D.
Report errors
Check node-to-node communication
Check packet lifetime
Find IP address.
Which field is always present in a ICMP package?
A.
B.
Type
Code
C. Checksum
D. all the above
When the hop-count field reaches zero and the destination has not been
reached, a ______ error message is sent?
A.
B.
Destination address
Time-excedeed
C. Parameter problem
D. redirection
Error in the header or option field of an IP datagram require a _____ error
message.
A.
B.
Parameter-problem
C. router-solicitation
Source-quenche D. redirection
6.
7.
8.
9.
The _____ packet contains information about a router.
A.
B.
Router-solicitation C. router-advertisement
Router-Information D. router-reply.
Who can send ICMP error-reporting messages?
A.
B.
Routers
Destination hosts
C. Source hosts
D. a and b
A time-exceeded message is generated if ______?
A.
B.
C.
D.
The round-trip time is close to zero.
The time-to-live field has a zero value.
Fragments of a message do not arrive within a set time
b and c
In calculating the time difference between two clocks, a negative value
indicates _______.
1.
2.
3.
4.
An invalid calculation.
The source clock lags behind the destination clock.
The destination clock lags behind the source clock.
The one-way time has been miscalculated.