ashFTW
Senior member
I measured the die. See the picture above. Assuming the XCC die is 400 mm2, which Is what Intel reported, the area taken up by the cores, and mesh, and cache is 200 mm2 for 15 cores. I’m not going to argue the minuscule difference between 13.3 and 12.5.Your Numbers are Off.
Xeon based Golden Cove core die area size is 10.5 mm^2 with L3$. 10.5 x 16 = 168 mm^2, the rest(32 mm^2) is used by the Mesh/Ring Bus interconnect. so with L2$ and Ring Bus the total die area is 12.50 mm^2 per core
I was comparing Intel 7 to Intel 3, to come up with 1.5x density increase. Intel 4 to 3 is expected to be 8-10% denser for the high performance library.I don't expect Intel3 to be as big of a jump that we are seeing from Intel7 to Intel4(based on current information) and I don't expect that Granite Rapids will add much more to the new instruction set provided by Sapphire Rapids
If you are going to use Redwood cove size on Intel 4, then you should use 400mm2 to estimate the number of cores. And you will arrive at similar number as me -- 40 cores per tile, and 160 per chip. You need to reread my post.We can extrapolate that Server Class Redwood Core on was going to be 9.37 mm^2 including L3$ and Mesh/Ring interconnect . so using your number to calculate it will be 600/9.37 = Exactly 64 cores per tile and 256 Core per CPU and Sierra Forest has the die area size to fit exactly 10,24 e cores.



