llvm dev - Jun 2019 - [llvm-mca] What's the difference between Rthroughput and "total cycles" in llvm-mca

What is the difference between the two? I thought "Rthroughput" is
basically the number of cycles required to execute a single iteration at
steady state, but this does not seem to match with the schedule/timeline
generated by llvm-mca.
Thanks in advance,
Tom
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Hi Tom,

Field 'Total Cycles' from the summary view simply reports the elapsed
number of cycles for the entire simulation.

Rthroughput (from the "Instruction Info" view) is the reciprocal of
the
instruction throughput.
Throughput is computed as the maximum number of instructions of a same type
that can be executed per clock cycle in the absence of operand dependencies.

Example (x86 - AMD Jaguar):
   ADD EAX, ESI

The integer unit in Jaguar has two ALU pipelines. An ADD instruction can
issue to any of those pipelines. That means, two independent ADD can be
issue during a same cycle. Therefore, throughput is 2  (instructions per
cycle), and RThrougput (1/throughput) is 0.5.

I hope it helps,
-Andrea

On Thu, Jun 6, 2019 at 10:11 PM Tom Chen via llvm-dev <
llvm-dev at lists.llvm.org> wrote:
> What is the difference between the two? I thought "Rthroughput"
is
> basically the number of cycles required to execute a single iteration at
> steady state, but this does not seem to match with the schedule/timeline
> generated by llvm-mca.
> Thanks in advance,
> Tom
> _______________________________________________
> LLVM Developers mailing list
> llvm-dev at lists.llvm.org
> https://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>
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Hi Andrea,
So does this definition make sense for basic blocks with more than one
instructions? E.g. how should one interpret a basic block with RThroughput
of 2.3?

On Fri, Jun 7, 2019 at 7:39 AM Andrea Di Biagio <andrea.dibiagio at
gmail.com>
wrote:
> Hi Tom,
>
> Field 'Total Cycles' from the summary view simply reports the
elapsed
> number of cycles for the entire simulation.
>
> Rthroughput (from the "Instruction Info" view) is the reciprocal
of the
> instruction throughput.
> Throughput is computed as the maximum number of instructions of a same
> type that can be executed per clock cycle in the absence of operand
> dependencies.
>
> Example (x86 - AMD Jaguar):
>    ADD EAX, ESI
>
> The integer unit in Jaguar has two ALU pipelines. An ADD instruction can
> issue to any of those pipelines. That means, two independent ADD can be
> issue during a same cycle. Therefore, throughput is 2  (instructions per
> cycle), and RThrougput (1/throughput) is 0.5.
>
> I hope it helps,
> -Andrea
>
> On Thu, Jun 6, 2019 at 10:11 PM Tom Chen via llvm-dev <
> llvm-dev at lists.llvm.org> wrote:
>
>> What is the difference between the two? I thought
"Rthroughput" is
>> basically the number of cycles required to execute a single iteration
at
>> steady state, but this does not seem to match with the
schedule/timeline
>> generated by llvm-mca.
>> Thanks in advance,
>> Tom
>> _______________________________________________
>> LLVM Developers mailing list
>> llvm-dev at lists.llvm.org
>> https://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>>
>
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