Taking your comment literally, I'd say that the number of blocks is more
relevant than the block size, so maybe the formula shouldn't be there.
Just in case you didn't understand what a block is, I illustrate the
ideas below.
The basic idea is that, instead of assigning a single sequence of ports
to a CE, the port space from whatever to 65535 is divided up into
equal-size chunks, or blocks, and the CE is assigned a sequence of m
ports from each block. I assume that this is done to make a
port-guessing attacker's job more difficult, but I'm not sure it does so
if the attacker knows the algorithm. My own view is that the MAP
document could get rid of a bunch of complexity if, like lw4o6, it left
the *port* mapping algorithm to a separate draft. The PSID could stay,
since it is needed, but the only thing important to MAP is the *address*
mapping algorithm.
Anyway, here is your illustration. Assume for example that a = 4,
implying a block size of 4096 (0x1000) port numbers. Thus port number
space is divided up into 16 blocks as shown in the figure below. Hex
notation is used for neatness.
What is also shown below is the set of ports allocated to a particular
CE, represented by P..P. This is a sequence occuring at the same
relative position in every block. The PSID is an index that points to
that relative position. The first port number, in fact, is PSID * m from
the beginning of the block, where m is the total number of ports in the
sequence P..P.
Port
----
0x0000
Block not used -- contains system ports.
0x0FFF
----
0x1000
P..P
0X1FFF
----
0x2000
.
.
.
.
0xEFFF
----
0xF000
P..P
0xFFFF
----
On 29/01/2013 12:54 AM, Qi Sun wrote:
Dear Tom,
IMHO, these are valuable comments. However, I'm kind of lost on this
part:
A Indexes the blocks of port numbers. The lowest block number
(A = 0) and possibly others contain the system port numbers, and
hence should not be used if port numbers 0-1023 are to be avoided.
a Width of the field containing the block index A. Indirectly a
determines the block size, by the formula: Block size = 65536 /
2^a
I'm not sure of the meaning of 'Block size', which is equal to
2^(k+m). As I interpret, each CE can get 2^(a+m) ports and each port
range is 2^m. What can we use the 'Block size' for?
Thanks in advance!
Best Regards, Qi Sun
On 2013-1-29, at 上午12:59, Tom Taylor wrote:
...
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