I believe your calculation of the FSB bandwidth is incorrect.
In order to calculate FSB bandwidth you multiply bus frequncy (266.66) times the transfers per clock as you said (4) and the FSB width (32bit or 4 byte).
Therefore, a system with a 266.66Mhz FSB (stock Core 2 Duo) has a FSB bandwidth of:
266.66 x 4 x 4 = 4266.66MB/s
Your memory has a 64bit (8 byte) width and a capability of 2 transfers per clock (DDR).
Therefore, to flood the FSB bandwidth you get:
4266.66MB/s = X Mhz * 2 * 8
4266.66MB/s = X Mhz * 16
266.66Mhz = X
Therefore a memory bus speed of 266.66Mhz or DDR2-533 will flood the FSB bandwidth.
Math does not lie.
I am totally correct actually
and yes Mathematics do not lie:
Evidence (Pictures, had to split them up into two parts).
Ok, so that states that a Pentium 4 at 800Mhz QDR = a 6400MB/s bandwidth, correct?
Throws all you're calculations out of the window
I always have advanced technical documentation for queries like this, and refer to them if necessary. Part of the job of an engineer, not an enthusiast
Right, so 6400MB/s for 800MHz
Lets re-do 1066MHz.
1066MHz QDR - Transfers per clock is Four
. Therefore 266MHz x 4 = 1066Mhz DQR.
To generate the bus speed you must multiply this value by eight. Therefore: 8528MB/s.
I am very tired, but can continue if you really want.