From: Elias Chari (elias.chari@gmail.com)
Date: Thu Nov 30 2006 - 18:04:29 ART
Here is an example.
Take the following:
224.10.8.5
224.138.8.5
These two map to the same MAC. How?
Only the 7 LSBits from the second byte get mapped so that would leave you
the second byte in both addresses to be 10. Now an easy way to figure it out
without complicated maths, since the last MSB of the second byte is not
mapped (in value is 128), subtract 138-128 and you have are left with 10,
same as the other address 2nd byte.
Therefore both map to 01:00:5e:0a:08:05
If my maths are correct the 32:1 ratio is worked out as follows:
M/cast addresses = 2^28
Available mappings = 2^23
Devide the two and you get 2^5 i.e 32 m/cast will map to one mac.
HTH
Elias
On 11/30/06, Ryan <ryan95842@gmail.com> wrote:
>
> (use courier to view)
>
> This is the same information I've read in a what seems like a dozen books
> and web pages. They all say that you only use last 23bits of the IP to
> make
> up the MAC address. Nothing new here. Yet, there is is this 32:1 overlap,
> but I've not found any documentation or book that goes beyond that to
> explain.
>
> So if I try this out for myself for example:
>
> Lets take a random Multicast IP address. 228.45.3.2
> (I just typed in some numbers, so that's as random as were going to get)
>
> Let's make this IP into a Multicast MAC address...
>
> 228.45.3.2
>
> 01:00:5e:xx:xx:xx
>
> xxx.45.3.2
>
> 45 . 3 . 2
> 0010 1101.0000 0011.0000 0010
> ^
> 2 D: 0 3: 0 2
> x010 1101.0000 0011.0000 0010
>
> So if I've done my math right, 228.45.3.2 should be: 01:00:5e:2d:03:02
>
> If I add a '128' to it to simulate the 1 in the 24th spot that gets
> excluded
> in the conversion,
>
> 228.173.3.2 should ALSO be: 01:00:5e:2d:03:02
>
> So, based on this, I would have to say that for any given Multicast IP
> address, there is going to be EXACTLY 32 duplicate MAC address's.
>
> 224.y.x.x
> 224.(y+128).x.x
> 225.y.x.x
> 225.(y+128).x.x
> ...
> 238.y.x.x
> 238.(y+128).x.x
> 239.y.x.x
> 239.(y+128).x.x
>
>
>
> So, what happens with well know Multicast IP's like the 224.0.0.xaddress's?
>
> This would seem to mean that you could not use any Multicast address is
> y.0.0.x range or y.128.0.x range as it would conflict with the
> 224.0.0.xaddress's.
>
> This correct?
>
> -Ryan
>
>
>
>
>
> On 11/29/06, sabrina pittarel <sabri_esame@yahoo.com> wrote:
> >
> >
> >
> > Hi Rayn,
> > the way the MAC Destination Address for a multicast packet is built is
> the following:
> >
> > the last 23 bits of the IP multicast address are copied into the last
> 23
> bits of the MAC address, while the beginning of the MAC address is fixes
> and
> set to 0100.5e(xx.xxxx)
> > This leaves 9 bits of the IP multicast address (32-23 = 9) that are not
> reflected anyhow in the L2 MAC. These are the first 9 bits of the IP
> address.
> > Of these 9 bits 4 are fixed (1110b => class D address) but 5 are
> variable
> so all the combinations of these 5 bits (2^5) generates ip multicast
> addresses that matches the same mac.
> >
> > Hope this helps,
> >
> > Sabrina
> >
> >
> >
> >
> >
> > ----- Original Message ----
> > From: Ryan <ryan95842@gmail.com>
> > To: ccielab@groupstudy.com
> > Sent: Wednesday, November 29, 2006 6:02:17 PM
> > Subject: Grasping the Multicast 32:1 duplicate MAC address issue...
> >
> >
> >
> > I'm a little confused by the Multicast duplicate MAC address issue...
> >
> > Is the 32:1 duplicate MAC address issue just for certain address's (32
> > total)?
> >
> > 224.1.1.1
> > 224.129.1.1
> > 225.1.1.1
> > 225.129.1.1
> > ...
> > 238.1.1.1
> > 238.129.1.1
> > 239.1.1.1
> > 239.129.1.1
> >
> > Or does it apply to ANY Multicast address? So for any one Multicast
> address,
> > there are exactly 32 duplicates....
> >
> >
> >
> > -Ryan
> >
> > _______________________________________________________________________
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