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TOPIC: Multicast overlap

Multicast overlap 13 years 10 months ago #8751

  • stefke
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Dear all,

Following is confusing me regarding multicast L2 to L3 mapping:

I've been reading about the 32:1 ratio for mac addresses to ip addresses. As far as I understand is that when you convert a multicast IP address to a multicast MAC address you :

a) Place the lower 23 bits from the IP into the MAC address
b) use the 00:01:5e as first 3 bytes

But what I'm having difficulties to grasp is the 32:1 overlap.

e.g. IP = --> MAC = 01:00:5E:59:E6:45

So the next IP address with the same MAC should be..... ??
I thought that adding 128 bits to the second byte should do the trick but somehow it doesn't add up:

231.(217+128).230.69 --> BUT then the MAC = 01:00:5E:5B:E6:45 and that's not the same MAC !!

Anybody any idea ???



Re: Multicast overlap 13 years 10 months ago #8883

Okay I just went through this chapter at school this past week so I am really new at this myself. But after going over your post and working my way through it I think I might have an answer.

In your example, --> MAC = 01:00:5E:59:E6:45

there is only one possible IP address besides the one above that will match the MAC address. As you stated you take the last 23 bits of the IP address and convert it into the last 24 bits of the MAC address. The first bit from the second octet (128) is always translated as 0 into the MAC address.

Thus, the only two IP addresses that would translate to that MAC are 231.217 and 231.89 (which is 217-128).

11011001 = 217
01011001 = 89

Since the first bit is always translated to the MAC as 0 both of those numbers are converted to 59 in hexidecimal form.

Now, since the range of multicast addresses are from - you would have two addresses in each network that would translate to 01:00:5E:59:E6:45

They would be and, followed by and, and so on up until and

The only problem with that is it only gives you 30 possible IP addresses that translate to the same MAC address. Where do the remaining two come from?
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