What the heck. I'll throw in my two cents.
I think the answer is "none of the above". :-)
Here are the steps I think would happen:
1) .18 host checks if 192.168.1.255 is on local
subnet. It is not. So,
2) .18 host sends packet to proper gateway. Assume
there's only one gateway defined (R1) for .18 and it's
a default for all traffic. .18 forwards the
IP packet to R1's interface on the 192.168.1.16/29
subnet. Note this is done using unicast ethernet forwarding.
.18 has no clue that .255 may represent a broadcast
address on some other remote subnet. Thus, no hosts
on the .16/29 network would receive this as a broadcast frame.
3) R1 receives frame, extracts the IP packet and sees
the destination is for 192.168.1.255. R1 sees that
this IP address falls within the subnet defined on its R1-R2
interface.
4a) If broadcast forwardng is enabled on R1's R1-R2 interface, the
packet is queued onto the R1-R2 interface. Before the packet
is forwarded, the destination IP address (192.168.1.255) in the
packet is changed to 255.255.255.255 (the default physical IP
broadcast address)
4b) All devices (basically just R2) on the R1-R2 subnet receive the
packet. Unless R2 has some form of bridging enabled, the
255.255.255.255 packet does not get forwarded to any other
interfaces/subnets.
5) If broadcast forwarding is not enabled on R1's R1-R2 interface,
the packet would get dropped by R1.
I didn't realize a directed broadcast was converted into an
interface specific IP broadcast address (default is the all ones
broadcast address [255.255.255.255]) when the physical broadcast
translation occurs. I had instead throught the packet
would maintain its IP dst of the subnet broadcast address and appear
on the wire with the broadcast ethernet destination address... But
the mapping to the specified IP broadcast address (255.255.255.255)
is what's suggested in the CCO docs and that's what I'm see in my
lab...
Which, if true, is interesting in that when overlaying multiple
logical subnets on a network, one would lose the subnet broadcast
information and all clients would have to process the now broadcast
IP packet. That is, a client would receive the ethernet broadcast,
extract the IP packet, and now see 255.255.255.255 as the dst. If
the dst IP address remained the directed subnet broadcast, one could
still broadcast to just the target set of clients (those on that subnet).
Brian
-----Original Message-----
From: Bob Vance [mailto:[EMAIL PROTECTED]]
Sent: Saturday, August 05, 2000 10:15 AM
To: CISCO_GroupStudy List (E-mail)
Subject: RE: --- Not own broadcast
I found this question interesting.
Maybe, I missed it, but I'm surprised that there wasn't more discussion
on
this one.
> 192.168.1.16/29 ----- R1 ----- 192.168.1.32/29
> |
> |192.168.1.252/30
> |
> 192.168.1.48/29 ----- R1 ----- 192.168.1.64/29
>
>node at address 192.168.1.18 sends a packet to address
> 192.168.1.255.
>which node or nodes will receive the packet?
> a) All nodes on all subnetworks
> b) Only the node at address 192.168.1.255
> c) Only the nodes on subnetwork 192.168.1.16 that have broadcast
reception
> enabled
> d) All nodes on subnetwork 192.168.1.16
> e) All nodes on subnetworks 192.168.1.16, 192.168.1.32, and
192.168.1.48
My answer is D.
The question says nothing about *accepting* the packet,
so it's reduced to a routing question:
Onto which segments, if any, will the routers send this packet?
I believe that the question, as stated, doesn't have enough information
to
be answered absolutely (but, then which ones do ;>).
Certainly, all nodes on the local segment, 192.168.1.16, will receive
the
packet (whether or not they accept the packet is entirely another
question,
although we hope that the author understands the distinction |>)
So, D) could be a correct answer.
B is out, since it is at a minimum a subnet broadcast address, not a
specific node address.
E and C
(Let's leave out the question of,
"...what the meaning of "is", is."
)
are also right out.
So, we are left with whether A is correct.
Under classful rules, with subnet prefix length /29, the address
192.168.1.255
is host 7 on subnet 192.168.1.248/29.
This 7/-3 is a subnet broadcast address, -1.
In addition, 248/5 is the -1 subnet. So we have the address:
network=192.168.1 subnet=-1 host=-1
.
This is supposed to be *recognized* as an "all-subnets" broadcast by
hosts
and must be accepted (RFC1122) by all hosts in all subnets of
192.168.1.0/24
network.
But, again, but this is irrelevant to the question of who *receives* the
packet.
Under classless rules, the address
192.168.1.255
is simply host 7 on network prefix 192.168.1.248/29.
Thus, this would simply be a directed broadcast.
The question is what will R1 (and R2) do with the packet?
At least two questions remain for the diagram:
. are the routers configured to forward directed subnet broadcasts?
. do they ignore all-subnets broadcast address
Are we to assume
no ip directed-broadcasts
?
Under 11.3, the default is
ip directed-broadcast
.
OTOH, this only has effect when
ip forward-protocol ...
is set.
Presumably, in the absence of facts (I hate those things -- they always
get
me confused), we would assume the lowest possible configuration, so we
would
*not* assume that any broadcasts being forwarded.
Presumably, also the all-subnets broadcast would also be blocked.
So, I have to answer D.
***However***
Now, more interesting, is what if we ***did*** have
ip forward-protocol ...
and ip directed-broadcast
on all interfaces?
*Now*, what would R1 do with the packet?
Will he see it as a directed broadcast to 192.168.1.252/30 only?
Or will he see it as an all-subnets broadcast and put it on
192.168.1.32/29
as well?
I frankly don't know what the behavior of IOS is vis-a-vis the
all-subnets
broadcast issue.
But, the all-subnets broadcast is deprecated as early as RFC1812.
Seeing it as an all-subnets broadcast would require that the routers be
configured for some kind of deprecated classful behavior.
I would therefore think that a classless (for lack of a better term)
router
with no local 192.168.1.0/24 addresses would simply treat the address
192.168.1.255 as any other routable directed broadcast address.
If so, in our case, R1 & R2 know about 192.168.1.252/30, so the packet
would
be put on that segment (ignoring, for the moment, that the connection
between R1 & R2 looks like a point-to-point serial connection).
Now,
All of the above having been said, I tried this on my own network:
(which consists of 4 segments connected by a 3640 and a few other
segments connected by some 25xx. The segments have several Unix and
windows hosts on them.
)
pluto ## ping 10.255.255.255 # network's mask (nonVLSM)= 255.255.252.0
PING 10.255.255.255: 64 byte packets
64 bytes from 10.10.120.12: icmp_seq=0. time=4. ms
64 bytes from 10.10.120.1: icmp_seq=0. time=5. ms
64 bytes from 10.10.120.11: icmp_seq=0. time=6. ms
64 bytes from 10.10.120.10: icmp_seq=0. time=7. ms
64 bytes from 10.10.120.12: icmp_seq=1. time=0. ms
64 bytes from 10.10.120.10: icmp_seq=1. time=2. ms
64 bytes from 10.10.120.11: icmp_seq=1. time=3. ms
64 bytes from 10.10.120.1: icmp_seq=1. time=4. ms
64 bytes from 10.10.124.200: icmp_seq=1. time=58. ms
So all the local hosts responded, as expected (answer D).
But then there was this 10.10.124.200 response. I have no idea what
this
box is (I no longer have control over the network).
Either it's one of the 3640's interfaces, or I have a rogue box on my
10.10.120.0/22 subnet, or the 3640 passed the packet somewhere.
Oooh, my head's starting to hurt, now.
As Linda Richman would say,
"Discuss it amongst yourselves."
-------------------------------------------------
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