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Discussion in 'MacBook Pro' started by Serban, Oct 23, 2013.
What is the difference between haswell from 13" and the cpu from 15"? anybody knows? Thank you
ya one's a dual core one's a quadcore
yea but why they changed the name? the 13" is haswell and 15" Crystal Well...why? only because is quad?
Because it has Iris Pro. Intels Codename for Haswell with eDRAM is Crystwell.
Taken from Wiki:
This certainly sounds like it would be a performance boost, but how much? I dont know.
Haswell chips with Iris Pro come in a special configuration that includes an additional L4 cache of 128MB in form of an additional chip on the same package. Intel calls this combination of a Haswell processor and a L4 cache "Crystalwell".
This L4 cache boosts the integrated graphics chip significantly, but it can also be used by the CPU. So if the integrated graphics are disabled in favor of the dGPU, Crystalwell has effectively a pretty remarkable 128MB L4 cache. (L3 caches in Haswell are 4MB-8MB).
Basically the Crystalwell means it has that added 50GB/s bidirectional 128MB L4 cache which especially the GPU needs. The dual channel main memory offers only 25.6GB/s.
So the Iris in the 13" is really rather slow. The GPU is big and in theory can chrunch numbers but it quickly runs against a wall because it starves for data to process. It has only those 25.6GB/s not the 50GB/s x2 added.
Intel plans to add such a codename Crystalwell memory on package on all CPUs they sell. These Quad Core are the first to benefit from it.
This is the first step. The benefit is apperantly speed, latency and power efficiency. Traditional memory in dram modules with GDDR5 or DDR3 are always a tradeoff. GDDR5 has horrible latency and is unsuitable for CPUs, DDR3 doesn't have the bandwidth GPUs want. Some form of HMC memory directly on the CPU package is the holy grail they are aiming for.
It really only matters for the onboard GPU. The CPU part doesn't yet really need it, but in some rare cases it does give some extra boost (Mostly because of the lower latency I think and not the added bandwidth).