What is ARP? [ Was Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask. ]
On Thu, Jul 10, 2003 at 02:39:15PM +0300, Amir Sela wrote: An ARP request? What for? Is it to find the MAC of the default gateway? ARP maps the MACs into IPs, doesn't it? An ARP request would send the MAC address and expects to get in reply the IP that correspond that MAC, isn't it? Wrong. it's the other way around. When you ping x.x.x.x, the computer knows the IP already. You just typed it, didn't you? What it doesn't know is the MAC address of the computer with this IP, so proper one-to-one communication can't be established. So, an ARP request is sent to request that the computer with the pertaining IP reply its MAC address. You are right that when I ping x.x.x.x I do know the IP address. Yet according to the DSL-HOWTO/appendix.html ARP Address Resolution Protocol. Converts MAC addresses to IP addresses. The way I read this is that an ARP request would send the MAC and expect the IP in return. That is, what is known is the MAC and what is looked after is the IP. If my understanding is correct then what both of us are missing is how this integrates into Ethernet communication. -- Shaul Karl, [EMAIL PROTECTED] e t = To unsubscribe, send mail to [EMAIL PROTECTED] with the word unsubscribe in the message body, e.g., run the command echo unsubscribe | mail [EMAIL PROTECTED]
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
On Tue, 2003-07-08 at 21:09, Beni Cherniavsky wrote:
Shaul Karl wrote on 2003-07-08:
I still don't get something. Quoting section 7 of the IP Sub-Networking
Mini-Howto:
For the sake of this example, let us assume that you have decided to
subnetwork you C class IP network number 192.168.1.0 into 4 subnets
(each of 62 usable interface/host IP numbers). However, two of these
subnets are being combined into a larger single network, giving three
physical networks.
These are :-
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
As far as I understand, they assume that sub-networking is uniform:
they know the full network mask and know it is subnetted with a given
subnet mask. They assume it doesn't only apply to their own subnet
but to each subnet of the network. No, they wouldn't treat
192.168.*.* in this way, the full network mask seems to be
255.255.255.0 so they only assume things about sibling subnets i.e.
192.168.1.*.
--
Actually, I am not quite sure this is correct. The smaller networks do not do any
assumptions regarding other networks.
The routing decision is done at the router, while the transmitting node is totally
unaware of the subnet mask implemented at the destination network.
If you really want the guts, it depends on the routing protocol implemented on the
router and whether it supports propagating VLSM masks and whether classless routing is
implemented.
As long as routing is done by classless routing protocol, all-zeros and all-ones
subnets become available for use.
As the abbriviation of VLSM applies (Variable Length Subnet Mask), in this case you
have the benefit of not loosing bits for what would look like a broadcast or network
address,
which does not really exists.
In the above axample, with VLSM implemented, the node does the following:
1) The destination address is not in node's network, so it will be transmitted to
default gateway.
2) ARP broadcast to get the MAC address of the default gateway.
3) The node sends the packet to DG
4) DG (the router) performs a lookup in it's routing table and notices that
192.168.1.191 is in 192.168.1.128/25 network which is directly attached to the router.
5) The router queries for the subnet mask associated with the interface and sees
255.255.255.128
5) The packet will be sent through the interface attached to 192.168.1.128/25 (this is
not a broadcast !!!)
6) ARP broadcast to get the destination MAC
7) Packet gets to it's destination.
As you can see, this is a normal unicast address when using VLSM.
References:
1) Routing TCP/IP. Volume 1 by Jeff Doyle, Cisco Press
2) Cisco CCNA Exam 640-607 Certification Guide by Wendell Odom, Cisco Press
Guy
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Re: What is ARP? [ Was Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask. ]
On Fri, Jul 11, 2003 at 07:57:35PM +0300, Shaul Karl wrote: You are right that when I ping x.x.x.x I do know the IP address. Yet according to the DSL-HOWTO/appendix.html ARP Address Resolution Protocol. Converts MAC addresses to IP addresses. The way I read this is that an ARP request would send the MAC and expect the IP in return. That is, what is known is the MAC and what is looked after is the IP. If my understanding is correct then what both of us are missing is how this integrates into Ethernet communication. The way I read this is that the quoted document is in error. Either they were reffering to RARP (reverse ARP, used to be used for things like what DHCP does today, in combination with bootp) or they just mixed IP and MAC. Maybe you want to write the author about this. Cheers, --Amos = To unsubscribe, send mail to [EMAIL PROTECTED] with the word unsubscribe in the message body, e.g., run the command echo unsubscribe | mail [EMAIL PROTECTED]
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
On ?, 2003-07-10 at 00:23, Shaul Karl wrote:
On Wed, Jul 09, 2003 at 03:35:49PM +0300, Amir Sela wrote:
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
I start feeling that someone should tell me straight in the face that
I should do a lot of reading before posting another message after this
one.
You need to do a lot of reading :)
First of all, just to make sure we're on solid ground - you need to
realize that once you subnet, the lowest common denominator is used,
that is, as in the example, if you have 8 bits of hosts, and decided to
subnet that to 1100, you now have 4 subnetworks. Essentially what
they did in the example was a two stage operation:
A) create 4 different subnetworks - subnetmask(in our case) is 26 bits
B) supernet between two of the newly created subnetworks - subnetmask
is 25 bits on two of the subnetworks.
Must I begin with 4 subnetworks and then merge 2 of them in order to
end up with exactly those 3 subnetworks? The answer is probably yes and
the reason for this is what I fail to understand.
Practically speaking, you don't need to actually do that. You simply
need to give a /25 subnetmask to the 124 upper hosts, and /26
subnetmasks to the 2x62 lower ones. But as I've stated before, the
reason is simply the way TCP-IP works. With such subnetmasks, 192 is not
a host ID. It's a network ID.
192.168.0.192/26 - 110.10101000..1100/26
| |||
\\//\-\/-/
Network ID Host ID
So, from the point of view of 192.168.0.15/26, 192 is a network
number, not a complete IP (network number+host ID which is not all zeros
or all ones). This is why it will be unavailable from the two smaller
networks. If the subnetmask is /25, as it is with the 124 upper hosts,
those IP addresses can be used. But since you would obviously want the
hosts on the two smaller networks to be able to communicate with all
computers on the large network, you can't use these addresses.
I'm stating this because your question was defocused. the 2 smaller
networks know that 192.168.1.191 and 192 are special addresses,
because they still are. Lets assume that our networks are A,B,CD.
When the tcp-ip stack in a host in A initializes, it ANDs its own IP
address against the subnetmask to determine its home network number.
Then, every outgoing packet goes through the same process. if the
resulting bit string is different than that host's own result, the
packet is known to be out of this host's network, and the ARP request is
made to the default gateway, for forwarding.
An ARP request? What for? Is it to find the MAC of the default
gateway? ARP maps the MACs into IPs, doesn't it? An ARP request would
send the MAC address and expects to get in reply the IP that correspond
that MAC, isn't it?
Wrong. it's the other way around. When you ping x.x.x.x, the computer
knows the IP already. You just typed it, didn't you? What it doesn't
know is the MAC address of the computer with this IP, so proper
one-to-one communication can't be established. So, an ARP request is
sent to request that the computer with the pertaining IP reply its MAC
address.
What will happen with 191
is that you will be requesting the host on subnet A to send a broadcast
packet to network C
Why would a host on subnet A consider 191 to be a broadcast address?
Why it wouldn't consider 192.168.*.191 to be all broadcast addresses?
See above. 192.168.*.191 is completely irrelvant here. with subnet masks
of /24 and
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
First of all, just to make sure we're on solid ground - you need to
realize that once you subnet, the lowest common denominator is used,
that is, as in the example, if you have 8 bits of hosts, and decided to
subnet that to 1100, you now have 4 subnetworks. Essentially what
they did in the example was a two stage operation:
A) create 4 different subnetworks - subnetmask(in our case) is 26 bits
B) supernet between two of the newly created subnetworks - subnetmask
is 25 bits on two of the subnetworks.
I'm stating this because your question was defocused. the 2 smaller
networks know that 192.168.1.191 and 192 are special addresses,
because they still are. Lets assume that our networks are A,B,CD.
When the tcp-ip stack in a host in A initializes, it ANDs its own IP
address against the subnetmask to determine its home network number.
Then, every outgoing packet goes through the same process. if the
resulting bit string is different than that host's own result, the
packet is known to be out of this host's network, and the ARP request is
made to the default gateway, for forwarding. What will happen with 191
is that you will be requesting the host on subnet A to send a broadcast
packet to network C(again, by definition, supernetting 2 smaller
networks out of 4 is completely transparent to hosts in the 2 small
networks). And with 192, you'll be requesting the host on subnet A to
send a packet to the network address of network D. Both of these cases
are obviously broken.
The funny thing is that the so called CD network does not really exist.
Supernetting is actually a sort of a hack. you simply remove 1 bit on
all the hosts on networks C and D, and it just works.. You couldn't
have supernetted it AC,D,B, for example. You should play a bit with the
binary versions of the IP addresses to get a hang of this..
To summon it up: from the two other networks' point of view, you're
trying to send packets to broadcast and network addresses. And they know
it because they're supposed to know it. Their subnet masks tells them
that. Hosts in A would say something like Hey.. if hosts in C and D
want to ignore proper subnet mask rules, it's their own lookout. I still
know 191 and 192 are special addresses, and that's how I'll treat them
I hope this helps..
Amir.
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Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
On Wed, Jul 09, 2003 at 03:35:49PM +0300, Amir Sela wrote:
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
I start feeling that someone should tell me straight in the face that
I should do a lot of reading before posting another message after this
one.
First of all, just to make sure we're on solid ground - you need to
realize that once you subnet, the lowest common denominator is used,
that is, as in the example, if you have 8 bits of hosts, and decided to
subnet that to 1100, you now have 4 subnetworks. Essentially what
they did in the example was a two stage operation:
A) create 4 different subnetworks - subnetmask(in our case) is 26 bits
B) supernet between two of the newly created subnetworks - subnetmask
is 25 bits on two of the subnetworks.
Must I begin with 4 subnetworks and then merge 2 of them in order to
end up with exactly those 3 subnetworks? The answer is probably yes and
the reason for this is what I fail to understand.
I'm stating this because your question was defocused. the 2 smaller
networks know that 192.168.1.191 and 192 are special addresses,
because they still are. Lets assume that our networks are A,B,CD.
When the tcp-ip stack in a host in A initializes, it ANDs its own IP
address against the subnetmask to determine its home network number.
Then, every outgoing packet goes through the same process. if the
resulting bit string is different than that host's own result, the
packet is known to be out of this host's network, and the ARP request is
made to the default gateway, for forwarding.
An ARP request? What for? Is it to find the MAC of the default
gateway? ARP maps the MACs into IPs, doesn't it? An ARP request would
send the MAC address and expects to get in reply the IP that correspond
that MAC, isn't it?
What will happen with 191
is that you will be requesting the host on subnet A to send a broadcast
packet to network C
Why would a host on subnet A consider 191 to be a broadcast address?
Why it wouldn't consider 192.168.*.191 to be all broadcast addresses?
packet to network C(again, by definition, supernetting 2 smaller
networks out of 4 is completely transparent to hosts in the 2 small
networks).
Why the 2 smaller networks have to consider the fact that CD is a
supernet? Once again, why 192.168.1.191 is considered by them to be a
broadcast address while any other IP from 192.168.*.191 is fine?
And with 192, you'll be requesting the host on subnet A to
send a packet to the network address of network D. Both of these cases
are obviously broken.
The funny thing is that the so called CD network does not really exist.
According to the mini HOWTO it exists in the routing table of the
router (section 7.1). It also exists as far as hosts on CD are
concerned. Nothing mentions C and D separately, the router specifically
mentions the combined CD and you still claim that the separate networks
exists but not the combined one. I can't understand that.
Supernetting is actually a sort of a hack. you simply remove 1 bit on
all the hosts on networks C and D, and it just works..
Except from the fact that for some hosts (the ones on A and B) 2 hosts
out of it don't exists.
Let me put it otherwise. Suppose I was sub netting valid Internet
addresses, say 192.1.1.0 in a similar manner. Would you, with your
current IP, be able to contact 192.1.1.191 or would it be regarded as a
broadcast address as far as you are concerned?
To summon it up: from the two other networks' point of view, you're
trying to send packets to broadcast and network addresses. And they know
it because they're supposed to know it. Their
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
Gilad Ben-Yossef wrote: Shaul Karl wrote: Am I right that in theory one can construct a 2 interfaces Ethernet network with a netmask of 255.255.255.254? No. Are there reasons to skip such an Ethernet network and keep the smallest network with a minimum of 4 hosts and a netmask of 255.255.255.252? You need a network address and a broadcast address for each network, so the minimal sized network is 4 IPs. Gilad. It's not exactly true if you control both hosts on the network (at least on linux). You can configure it using point-to-point addresses: [EMAIL PROTECTED] tmp]$ /sbin/ifconfig eth0.100 10.100.0.1 netmask 255.255.255.255 pointopoint 10.100.0.2 (and ofcourse with the appropriate command on the other host) then you'll get: [EMAIL PROTECTED] tmp]$ /sbin/ip addr ls | grep 0.10 4: eth0.100: BROADCAST,MULTICAST,UP mtu 1500 qdisc noqueue inet 10.100.0.1 peer 10.100.0.2/32 brd 10.255.255.255 scope global eth0.100 [EMAIL PROTECTED] tmp]$ /sbin/ip route ls | grep 0.10 10.100.0.2 dev eth0.100 proto kernel scope link src 10.100.0.1 [EMAIL PROTECTED] tmp]$ ping -c 1 10.100.0.2 PING 10.100.0.2 (10.100.0.2) from 10.100.0.1 : 56(84) bytes of data. 64 bytes from 10.100.0.2: icmp_seq=1 ttl=64 time=0.231 ms Actually the IP addresses of both sides don't even need to have any relation to one another... -- Eran Mann Senior Software Engineer MRV International Tel: 972-4-9936297 Fax: 972-4-9890430 www.mrv.com = To unsubscribe, send mail to [EMAIL PROTECTED] with the word unsubscribe in the message body, e.g., run the command echo unsubscribe | mail [EMAIL PROTECTED]
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
On Tue, Jul 08, 2003 at 08:39:55AM +0300, Oron Peled wrote:
So for your case you need two bits -- i.e: 255.255.255.252
Now calculate the 2 correct host numbers as an exercise (and
reply to the group, so others will benefit as well...)
A possible allocation of IPs with a 255.255.255.252 netmask might be:
network address:192.168.0.8.
1st host: 192.168.0.9.
2nd host: 192.168.0.10.
broadcast address: 192.168.0.11.
Another similar example:
network address:192.168.25.252.
1st host: 192.168.25.253.
2nd host: 192.168.25.254.
broadcast address: 192.168.25.255.
I still don't get something. Quoting section 7 of the IP Sub-Networking
Mini-Howto:
\bgein{quotation}
For the sake of this example, let us assume that you have decided to
subnetwork you C class IP network number 192.168.1.0 into 4 subnets
(each of 62 usable interface/host IP numbers). However, two of these
subnets are being combined into a larger single network, giving three
physical networks.
These are :-
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
This illustrates an important point with subnetworks - the usable
addresses are determined by the SMALLEST subnetwork in that address
space.
7.1. The routing tables
Let us assume that a computer running Linux is acting as a router for
this network. It will have three network interfaces to the local LANs
and possibly a fourth interface to the Internet (which would be its
default route.
Let us assume that the Linux computer uses the lowest available IP
address in each subnetwork on its interface to that network. It would
configure its network interfaces as
__
Interface IP Address Netmask
eth0192.168.1.1 255.255.255.192
eth1192.168.1.65255.255.255.192
eth2192.168.1.129 255.255.255.128
__
The routing it would establish would be
__
Destination Gateway Genmask Iface
192.168.1.0 0.0.0.0 255.255.255.192 eth0
192.168.1.640.0.0.0 255.255.255.192 eth1
192.168.1.128 0.0.0.0 255.255.255.128 eth2
__
On each of the subnetworks, the hosts would be configured with their
own IP number and net mask (appropriate for the particular network).
Each host would declare the Linux PC as its gateway/router, specifying
the Linux PCs IP address for its interface on to that particular
network.
Robert Hart Melbourne, Australia March 1997.
\end{quotation}
I will repeat the question in case you haven't noticed that I have
\begin{interruptRequest} in the middle:
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
--
Shaul Karl, [EMAIL PROTECTED] e t
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Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
Shaul Karl wrote on 2003-07-08:
I still don't get something. Quoting section 7 of the IP Sub-Networking
Mini-Howto:
For the sake of this example, let us assume that you have decided to
subnetwork you C class IP network number 192.168.1.0 into 4 subnets
(each of 62 usable interface/host IP numbers). However, two of these
subnets are being combined into a larger single network, giving three
physical networks.
These are :-
__
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63255.255.255.192 62
192.168.1.64192.168.1.127 255.255.255.192 62
192.168.1.128 192.168.1.255 255.255.255.128 124 (see note)
__
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
\begin{interruptRequest}
How does the 2 smaller networks know that 192.168.1.191 and 192 were
initially a broadcast and network addresses? Would they treat any one of
192.168.*.19[12] in the same way?
\end{interruptRequest}
As far as I understand, they assume that sub-networking is uniform:
they know the full network mask and know it is subnetted with a given
subnet mask. They assume it doesn't only apply to their own subnet
but to each subnet of the network. No, they wouldn't treat
192.168.*.* in this way, the full network mask seems to be
255.255.255.0 so they only assume things about sibling subnets i.e.
192.168.1.*.
--
Beni Cherniavsky [EMAIL PROTECTED]
If I don't hack on it, who will? And if I don't GPL it, what am I?
And if it itches, why not now? [With apologies to Hilel ;]
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Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
Shaul Karl wrote: Am I right that in theory one can construct a 2 interfaces Ethernet network with a netmask of 255.255.255.254? No. Are there reasons to skip such an Ethernet network and keep the smallest network with a minimum of 4 hosts and a netmask of 255.255.255.252? You need a network address and a broadcast address for each network, so the minimal sized network is 4 IPs. Gilad. = To unsubscribe, send mail to [EMAIL PROTECTED] with the word unsubscribe in the message body, e.g., run the command echo unsubscribe | mail [EMAIL PROTECTED]
Re: A 2 hosts Ethernet network with a 255.255.255.254 netmask.
On Tue, 8 Jul 2003 04:01:25 +0300
Shaul Karl [EMAIL PROTECTED] wrote:
Am I right that in theory one can construct a 2 interfaces Ethernet
network with a netmask of 255.255.255.254?
No. You gave only 1 bit for the host part. This is not enough as
in any subnet there are two addresses that are reserved:
A host number all zero -- represent the network address
A host number all ones -- represent the broadcast address
E.g: With a netmask of 255.255.255.224 (three most significant bits of 1)
you have 5 host bits. However, in each of the subnet you may
have only 2^{5} - 2 == 30 host addresses.
So for your case you need two bits -- i.e: 255.255.255.252
Now calculate the 2 correct host numbers as an exercise (and
reply to the group, so others will benefit as well...)
Will this setup work under Linux? Should ifconfig for such a setup be
run with the pointopoint (point-to-point) flag?
You do not need any point to point flag (assuming your connection is
ethernet). If you just cable them back to back, than you need crossed
cable (to save a hub/switch).
Are there reasons to skip such an Ethernet network and keep the
smallest network with a minimum of 4 hosts and a netmask of
255.255.255.252?
Yes, now you know them...
Hope it helps.
--
Oron Peled Voice/Fax: +972-4-8228492
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