Count of MAC addresses misunderstanding

Question

I know that MAC address has to be unique. It has 6 elements - k. And it chooses from 36 chars (A-Z,0-9) - n. It means that there are n^k combinations.

36⁶= 2176782336 If there are 7.5 billions of people, it's about 0.29 devices per person per life. It seems like incredibly low number.

Am I right or not? What and where am I doing mistake?

Mistake found it's not 6 elements but 12 (36¹² isn't still correct)

Answer 1

I know that MAC address has to be unique.

Strictly speaking it only has to be unique on the local network. The way this was supposed to be ensured was by making the default MAC address of devices globally unique. This was acheived through splitting the address into a vendor part and then a part that could be assigned by the vendor.

In practice some vendors are sloppier about this than others. Most VM software and at least one hardware vendor assign addresses randomly from within a vendor ID (which may or may not actually belong to the vendor in question) and just work on the assumption that the number of devices on a network will be small enough that MAC collisions are unlikely.

It has 6 elements - k. And it chooses from 36 chars (A-Z,0-9) - n. It means that there are nk combinations.

Wrong

A MAC address is a 48 bit binary number, conventionally written as six hexadecimal numbers each with two digits.

So that gives 2^48 addresses, however not all of them are usable for a few reasons.

1. Two bits are special, one indicates whether an address is "local" or "global", the other indicates whether it is unicast or multicast.
2. MAC addresses were traditionally allocated to a vendor in blocks of 2²⁴. Most vendors will never use that many leading to lots of wasted addresses. Smaller blocks are now available but only in a very limited range of sizes, so there is still a lot of wastage.

The IEEE has some concerns about the possibility of MAC address exhaustion, mainly driven by applications that assign large numbers of MAC adddresses to the same device. New applications that don't need backwards compatibility are strongly advised to use EUI-64 instead.

Answer 2

There are 2^48 addresses. It means 281474976710656 unique combinations.

To make it clearer:

MAC Addresses are 6 bytes, for example:

00:50:56:29:33:d2

Each byte goes from (in hex) 00 to ff or in binary 0000000 to 11111111.

6 bytes * 8 bits per byte = 48 bits

total combinations of 48 bits = 2^48 = 281474976710656

MAC addresses can be re-used. Manufacturers usually send cards with duplicate addresses to different parts of the world, so there's no chance of two network cards with the same MAC address on the same network.

The MAC addresses must be unique on a local network, but there's not a problem for having duplicate MAC addresses if they are separated by one or more routers since the devices won't see each other directly.

Answer 3

MAC address consists of 48 bits. It is 2**48 or 281 474 976 710 656 addresses, which is more than enough at the moment. Plus there is practice to reuse MAC addresses of obsoleted equipment.

MAC address is usually represented in hex, which is n in your formula. But one "element" has 00 to FF - which is 256 - combinations. So 256**6 should be used for calculation.