5 ms·

Honest question: How does disabling Hyper-Threading cut costs?

dogma1138 8y ago
It doesn’t but it allows them to have more products based of the same exact die just like i7s and i5s share the same die which reduces manufacturering costs. That said it would simply mean that the i7 would become the new i5 and also fill the same price range quite likely. They might run the i5, 7 and 9 in the mainstream line now and bring the i3 to where the Pentium line is currently is however I still think that this is quite likely a configuration issue than a new segmentation since the i9 is their HEDT platform now.
- gameswithgo 8y ago
  they already had too many products! you have to spend like 5 days figuring out which one is 1% faster than the other for your particular use case. choice is bad! :D
  - dsr_ 8y ago
    The good news is that you can ignore the 1% differences. Here's your new decision tree: 1. How much can I afford? 2. In that budget, do I need maximum instructions per second on one core or maximum cores? If maximum cores, AMD is over there.
pizza234 8y ago
Transistors, I guess; HT adds extra registers for each core. However, I have some doubts that removing them would make a tangible impact on the final price.
- kingosticks 8y ago
  Or you could leave it all in there (cheaper and easier) just disabled and instead enjoy better yields: * Parts with broken HT can be shipped as fully working non-HT parts. * High-leakage parts can be shipped as they now pass the power screening.
  - gsnedders 8y ago
    This is the status quo with i5 v. i7 binning, AFAIK.
  - wtallis 8y ago
    > Parts with broken HT can be shipped as fully working non-HT parts. It's unlikely that they ever get parts with broken HT on an otherwise salvageable core. Too much of that functionality is simply partitioning existing resources in half. There aren't that many transistors that are simply HT overhead that the core can do without when HT is not in use.
- theandrewbailey 8y ago
  Intel claimed that HT used about 5% of the die area, but that was with HT Pentium 4. The numbers probably changed in 15 years.
  - monocasa 8y ago
    It's probably less as a percentage now.
- Symmetry 8y ago
  Not registers per se. You want a large number of physical registers for renaming and speculative execution in any event. What you need with HT is another sets of mappings form architectural to physical registers for the committed state.
- dragontamer 8y ago
  Not actually. Remember that architectural registers (such as RAX or EAX) are "fake" and remapped to the real, physical, microcode registers. Code like "xor eax, eax" is translated into "physical-register #51 = 0" with regards to uops, and doesn't even use any execution units on modern Skylake or Zen processors! Hyperthreads only need to contain one more remapping of physical registers to architectural registers.

It doesn’t but it allows them to have more products based of the same exact die just like i7s and i5s share the same die which reduces manufacturering costs. That said it would simply mean that the i7 would become the new i5 and also fill the same price range quite likely. They might run the i5, 7 and 9 in the mainstream line now and bring the i3 to where the Pentium line is currently is however I still think that this is quite likely a configuration issue than a new segmentation since the i9 is their HEDT platform now.

gameswithgo 8y ago
they already had too many products! you have to spend like 5 days figuring out which one is 1% faster than the other for your particular use case. choice is bad! :D
- dsr_ 8y ago
  The good news is that you can ignore the 1% differences. Here's your new decision tree: 1. How much can I afford? 2. In that budget, do I need maximum instructions per second on one core or maximum cores? If maximum cores, AMD is over there.

pizza234 8y ago

Transistors, I guess; HT adds extra registers for each core. However, I have some doubts that removing them would make a tangible impact on the final price.

kingosticks 8y ago
Or you could leave it all in there (cheaper and easier) just disabled and instead enjoy better yields: * Parts with broken HT can be shipped as fully working non-HT parts. * High-leakage parts can be shipped as they now pass the power screening.
- gsnedders 8y ago
  This is the status quo with i5 v. i7 binning, AFAIK.
- wtallis 8y ago
  > Parts with broken HT can be shipped as fully working non-HT parts. It's unlikely that they ever get parts with broken HT on an otherwise salvageable core. Too much of that functionality is simply partitioning existing resources in half. There aren't that many transistors that are simply HT overhead that the core can do without when HT is not in use.
theandrewbailey 8y ago
Intel claimed that HT used about 5% of the die area, but that was with HT Pentium 4. The numbers probably changed in 15 years.
- monocasa 8y ago
  It's probably less as a percentage now.
Symmetry 8y ago
Not registers per se. You want a large number of physical registers for renaming and speculative execution in any event. What you need with HT is another sets of mappings form architectural to physical registers for the committed state.
dragontamer 8y ago
Not actually. Remember that architectural registers (such as RAX or EAX) are "fake" and remapped to the real, physical, microcode registers. Code like "xor eax, eax" is translated into "physical-register #51 = 0" with regards to uops, and doesn't even use any execution units on modern Skylake or Zen processors! Hyperthreads only need to contain one more remapping of physical registers to architectural registers.