5 ms·

The problem is that IPC is also a crude metric. Even leaving aside fundamental algorithmic differences, an implementation of some algorithm with IPC of 0.5 is n

The problem is that IPC is also a crude metric. Even leaving aside fundamental algorithmic differences, an implementation of some algorithm with IPC of 0.5 is not necessarily faster than an implementation that somehow manages to hit every execution port and deliver an IPC of 4. I can improve IPC of almost any algorithm (assuming it is not already very high) by slipping lots of useless or nearly useless cheap integer operations into the code. People always tell you "branch misses are bad" and "cache misses are bad". You should always ask: "compared to what"? If it was going to take you 20 cycles worth of frenzied, 4 instructions per clock, work to calculate something you could keep in a big table in L2 (assuming that you aren't contending for it) you might be better off eating the cache miss. Similarly you could "improve" your IPC by avoiding branch misses (assuming no side effects) by calculating both sides of a unpredictable branch and using CMOV. This will save you branch misses and increase your IPC, but it may not improve the speed of your code (if the cost of the work is bigger than the cost of the branch misses).

faragon 9y ago
A sustained IPC of 2 is already quite high (e.g. for a OoOE CPU having 2 load/store units, and 2 [vector-]ALUs, i.e. having a maximum IPC of 4). You can reach near-full CPU utilization when processing sequential data (read/process/store). When using trees, hashes, etc. on "big enough data", it is very hard to have good CPU IPC (even when having 99% L1 cache hits, e.g. because of code where the branch predictor has no enough information, when many CPUs are accessing to data in the same zone of memory, etc.).
- jwatte 9y ago
  You should say "ALU utilization" when that's what you mean.
  - faragon 9y ago
    Full CPU utilization regarding IPC is not just about the ALUs, it includes all units (ALU units, load/store, and other, e.g. if you have differentiated integer and floating point ALUs, etc.). E.g. with a 4 maximum IPC CPU (2 load/store, 2 ALU), on a synthetic benchmark doing only register-register ALU optimal work, you would get an IPC of 2. Also, depending on what algorithm you're doing, you'll never get the maximum IPC, even with optimal code, e.g. for a hash digest you don't need the store unit very much, so in the above CPU example you would reach an IPC of 3 with optimal code. So not always not having the maximum IPC means you're under-using the CPU for a given algorithm.
hesselink 9y ago
If I understand correctly, the point is not raising IPC necessarily, it's finding what to optimize. If you have low IPC, optimize for memory access. If you have high IPC, optimize for code execution. This is in the article under "Interpretation and actionable items". In the end, what you want to improve is the wall-clock time of your program (or benchmarks, realistically). Slipping in useless operations is not going to do this.

faragon 9y ago

A sustained IPC of 2 is already quite high (e.g. for a OoOE CPU having 2 load/store units, and 2 [vector-]ALUs, i.e. having a maximum IPC of 4). You can reach near-full CPU utilization when processing sequential data (read/process/store). When using trees, hashes, etc. on "big enough data", it is very hard to have good CPU IPC (even when having 99% L1 cache hits, e.g. because of code where the branch predictor has no enough information, when many CPUs are accessing to data in the same zone of memory, etc.).

jwatte 9y ago
You should say "ALU utilization" when that's what you mean.
- faragon 9y ago
  Full CPU utilization regarding IPC is not just about the ALUs, it includes all units (ALU units, load/store, and other, e.g. if you have differentiated integer and floating point ALUs, etc.). E.g. with a 4 maximum IPC CPU (2 load/store, 2 ALU), on a synthetic benchmark doing only register-register ALU optimal work, you would get an IPC of 2. Also, depending on what algorithm you're doing, you'll never get the maximum IPC, even with optimal code, e.g. for a hash digest you don't need the store unit very much, so in the above CPU example you would reach an IPC of 3 with optimal code. So not always not having the maximum IPC means you're under-using the CPU for a given algorithm.

hesselink 9y ago

If I understand correctly, the point is not raising IPC necessarily, it's finding what to optimize. If you have low IPC, optimize for memory access. If you have high IPC, optimize for code execution. This is in the article under "Interpretation and actionable items". In the end, what you want to improve is the wall-clock time of your program (or benchmarks, realistically). Slipping in useless operations is not going to do this.