[networking] RFC 5737 compliance: service subnets

Replace non-compliant subnets in:

intro-basic-networking.rst

Change-Id: Ie0d4aaaffbcf73b48fd10402c231d3cfe578ac4b
Partial-Bug: #1656378
This commit is contained in:
Akihiro Motoki 2017-05-19 09:30:41 +09:00
parent 901ce995b3
commit c8b761a2d0
1 changed files with 42 additions and 42 deletions

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@ -163,11 +163,11 @@ There are two syntaxes for expressing a netmask:
* dotted quad
* classless inter-domain routing (CIDR)
Consider an IP address of 192.168.1.5, where the first 24 bits of the
Consider an IP address of 192.0.2.5, where the first 24 bits of the
address are the network number. In dotted quad notation, the netmask
would be written as ``255.255.255.0``. CIDR notation includes both the
IP address and netmask, and this example would be written as
``192.168.1.5/24``.
``192.0.2.5/24``.
.. note::
@ -178,14 +178,14 @@ IP address and netmask, and this example would be written as
Sometimes we want to refer to a subnet, but not any particular IP
address on the subnet. A common convention is to set the host
identifier to all zeros to make reference to a subnet. For example, if
a host's IP address is ``10.10.53.24/16``, then we would say the
subnet is ``10.10.0.0/16``.
a host's IP address is ``192.0.2.24/24``, then we would say the
subnet is ``192.0.2.0/24``.
To understand how ARP translates IP addresses to MAC addresses,
consider the following example. Assume host *A* has an IP address of
``192.168.1.5/24`` and a MAC address of ``fc:99:47:49:d4:a0``, and
``192.0.2.5/24`` and a MAC address of ``fc:99:47:49:d4:a0``, and
wants to send a packet to host *B* with an IP address of
``192.168.1.7``. Note that the network number is the same for both
``192.0.2.7``. Note that the network number is the same for both
hosts, so host *A* is able to send frames directly to host *B*.
The first time host *A* attempts to communicate with host *B*, the
@ -194,25 +194,25 @@ the local network. The request is a broadcast with a message like
this:
*To: everybody (ff:ff:ff:ff:ff:ff). I am looking for the computer who
has IP address 192.168.1.7. Signed: MAC address fc:99:47:49:d4:a0*.
has IP address 192.0.2.7. Signed: MAC address fc:99:47:49:d4:a0*.
Host *B* responds with a response like this:
*To: fc:99:47:49:d4:a0. I have IP address 192.168.1.7. Signed: MAC
*To: fc:99:47:49:d4:a0. I have IP address 192.0.2.7. Signed: MAC
address 54:78:1a:86:00:a5.*
Host *A* then sends Ethernet frames to host *B*.
You can initiate an ARP request manually using the :command:`arping` command.
For example, to send an ARP request to IP address ``10.30.0.132``:
For example, to send an ARP request to IP address ``192.0.2.132``:
.. code-block:: console
$ arping -I eth0 10.30.0.132
ARPING 10.30.0.132 from 10.30.0.131 eth0
Unicast reply from 10.30.0.132 [54:78:1A:86:1C:0B] 0.670ms
Unicast reply from 10.30.0.132 [54:78:1A:86:1C:0B] 0.722ms
Unicast reply from 10.30.0.132 [54:78:1A:86:1C:0B] 0.723ms
$ arping -I eth0 192.0.2.132
ARPING 192.0.2.132 from 192.0.2.131 eth0
Unicast reply from 192.0.2.132 [54:78:1A:86:1C:0B] 0.670ms
Unicast reply from 192.0.2.132 [54:78:1A:86:1C:0B] 0.722ms
Unicast reply from 192.0.2.132 [54:78:1A:86:1C:0B] 0.723ms
Sent 3 probes (1 broadcast(s))
Received 3 response(s)
@ -225,8 +225,8 @@ command:
$ arp -n
Address HWtype HWaddress Flags Mask Iface
10.0.2.3 ether 52:54:00:12:35:03 C eth0
10.0.2.2 ether 52:54:00:12:35:02 C eth0
192.0.2.3 ether 52:54:00:12:35:03 C eth0
192.0.2.2 ether 52:54:00:12:35:02 C eth0
.. _DHCP:
@ -251,11 +251,11 @@ client. The exchange looks like this:
1. The client sends a discover ("Im a client at MAC address
``08:00:27:b9:88:74``, I need an IP address")
2. The server sends an offer ("OK ``08:00:27:b9:88:74``, Im offering
IP address ``10.10.0.112``")
3. The client sends a request ("Server ``10.10.0.131``, I would like
to have IP ``10.10.0.112``")
IP address ``192.0.2.112``")
3. The client sends a request ("Server ``192.0.2.131``, I would like
to have IP ``192.0.2.112``")
4. The server sends an acknowledgement ("OK ``08:00:27:b9:88:74``, IP
``10.10.0.112`` is yours")
``192.0.2.112`` is yours")
OpenStack uses a third-party program called
@ -265,9 +265,9 @@ Dnsmasq writes to the syslog, where you can observe the DHCP request
and replies::
Apr 23 15:53:46 c100-1 dhcpd: DHCPDISCOVER from 08:00:27:b9:88:74 via eth2
Apr 23 15:53:46 c100-1 dhcpd: DHCPOFFER on 10.10.0.112 to 08:00:27:b9:88:74 via eth2
Apr 23 15:53:48 c100-1 dhcpd: DHCPREQUEST for 10.10.0.112 (10.10.0.131) from 08:00:27:b9:88:74 via eth2
Apr 23 15:53:48 c100-1 dhcpd: DHCPACK on 10.10.0.112 to 08:00:27:b9:88:74 via eth2
Apr 23 15:53:46 c100-1 dhcpd: DHCPOFFER on 192.0.2.112 to 08:00:27:b9:88:74 via eth2
Apr 23 15:53:48 c100-1 dhcpd: DHCPREQUEST for 192.0.2.112 (192.0.2.131) from 08:00:27:b9:88:74 via eth2
Apr 23 15:53:48 c100-1 dhcpd: DHCPACK on 192.0.2.112 to 08:00:27:b9:88:74 via eth2
When troubleshooting an instance that is not reachable over the network, it can
be helpful to examine this log to verify that all four steps of the DHCP
@ -307,25 +307,25 @@ Here is an example of output from :command:`ip route show`:
.. code-block:: console
$ ip route show
default via 10.0.2.2 dev eth0
10.0.2.0/24 dev eth0 proto kernel scope link src 10.0.2.15
192.168.27.0/24 dev eth1 proto kernel scope link src 192.168.27.100
192.168.122.0/24 dev virbr0 proto kernel scope link src 192.168.122.1
default via 192.0.2.2 dev eth0
192.0.2.0/24 dev eth0 proto kernel scope link src 192.0.2.15
198.51.100.0/25 dev eth1 proto kernel scope link src 198.51.100.100
198.51.100.192/26 dev virbr0 proto kernel scope link src 198.51.100.193
Line 1 of the output specifies the location of the default route,
which is the effective routing rule if none of the other rules match.
The router associated with the default route (``10.0.2.2`` in the
The router associated with the default route (``192.0.2.2`` in the
example above) is sometimes referred to as the *default gateway*. A
DHCP_ server typically transmits the IP address of the default gateway
to the DHCP client along with the client's IP address and a netmask.
Line 2 of the output specifies that IPs in the ``10.0.2.0/24`` subnet are on
Line 2 of the output specifies that IPs in the ``192.0.2.0/24`` subnet are on
the local network associated with the network interface eth0.
Line 3 of the output specifies that IPs in the ``192.168.27.0/24`` subnet
Line 3 of the output specifies that IPs in the ``198.51.100.0/25`` subnet
are on the local network associated with the network interface eth1.
Line 4 of the output specifies that IPs in the ``192.168.122.0/24`` subnet are
Line 4 of the output specifies that IPs in the ``198.51.100.192/26`` subnet are
on the local network associated with the network interface virbr0.
The output of the :command:`route -n` and :command:`netstat -rn` commands are
@ -337,27 +337,27 @@ routes would be formatted using these commands:
$ route -n
Kernel IP routing table
Destination Gateway Genmask Flags MSS Window irtt Iface
0.0.0.0 10.0.2.2 0.0.0.0 UG 0 0 0 eth0
10.0.2.0 0.0.0.0 255.255.255.0 U 0 0 0 eth0
192.168.27.0 0.0.0.0 255.255.255.0 U 0 0 0 eth1
192.168.122.0 0.0.0.0 255.255.255.0 U 0 0 0 virbr0
0.0.0.0 192.0.2.2 0.0.0.0 UG 0 0 0 eth0
192.0.2.0 0.0.0.0 255.255.255.0 U 0 0 0 eth0
198.51.100.0 0.0.0.0 255.255.255.128 U 0 0 0 eth1
198.51.100.192 0.0.0.0 255.255.255.192 U 0 0 0 virbr0
The :command:`ip route get` command outputs the route for a destination
IP address. From the below example, destination IP address ``10.0.2.14`` is on
IP address. From the below example, destination IP address ``192.0.2.14`` is on
the local network of eth0 and would be sent directly:
.. code-block:: console
$ ip route get 10.0.2.14
10.0.2.14 dev eth0 src 10.0.2.15
$ ip route get 192.0.2.14
192.0.2.14 dev eth0 src 192.0.2.15
The destination IP address ``93.184.216.34`` is not on any of the connected
local networks and would be forwarded to the default gateway at ``10.0.2.2``:
The destination IP address ``203.0.113.34`` is not on any of the connected
local networks and would be forwarded to the default gateway at ``192.0.2.2``:
.. code-block:: console
$ ip route get 93.184.216.34
93.184.216.34 via 10.0.2.2 dev eth0 src 10.0.2.15
$ ip route get 203.0.113.34
203.0.113.34 via 192.0.2.2 dev eth0 src 192.0.2.15
It is common for a packet to hop across multiple routers to reach its final
destination. On a Linux machine, the ``traceroute`` and more recent ``mtr``