问题
I have GeForce GTX460 SE, so it is: 6 SM x 48 CUDA Cores = 288 CUDA Cores. It is known that in one Warp contains 32 threads, and that in one block simultaneously (at a time) can be executed only one Warp. That is, in a single multiprocessor (SM) can simultaneously execute only one Block, one Warp and only 32 threads, even if there are 48 cores available?
And in addition, an example to distribute concrete Thread and Block can be used threadIdx.x and blockIdx.x. To allocate them use kernel <<< Blocks, Threads >>> (). But how to allocate a specific number of Warp-s and distribute them, and if it is not possible then why bother to know about Warps?
回答1:
The situation is quite a bit more complicated than what you describe.
The ALUs (cores), load/store (LD/ST) units and Special Function Units (SFU) (green in the image) are pipelined units. They keep the results of many computations or operations at the same time, in various stages of completion. So, in one cycle they can accept a new operation and provide the results of another operation that was started a long time ago (around 20 cycles for the ALUs, if I remember correctly). So, a single SM in theory has resources for processing 48 * 20 cycles = 960 ALU operations at the same time, which is 960 / 32 threads per warp = 30 warps. In addition, it can process LD/ST operations and SFU operations at whatever their latency and throughput are.
The warp schedulers (yellow in the image) can schedule 2 * 32 threads per warp = 64 threads to the pipelines per cycle. So that's the number of results that can be obtained per clock. So, given that there are a mix of computing resources, 48 core, 16 LD/ST, 8 SFU, each which have different latencies, a mix of warps are being processed at the same time. At any given cycle, the warp schedulers try to "pair up" two warps to schedule, to maximize the utilization of the SM.
The warp schedulers can issue warps either from different blocks, or from different places in the same block, if the instructions are independent. So, warps from multiple blocks can be processed at the same time.
Adding to the complexity, warps that are executing instructions for which there are fewer than 32 resources, must be issued multiple times for all the threads to be serviced. For instance, there are 8 SFUs, so that means that a warp containing an instruction that requires the SFUs must be scheduled 4 times.
This description is simplified. There are other restrictions that come into play as well that determine how the GPU schedules the work. You can find more information by searching the web for "fermi architecture".
So, coming to your actual question,
why bother to know about Warps?
Knowing the number of threads in a warp and taking it into consideration becomes important when you try to maximize the performance of your algorithm. If you don't follow these rules, you lose performance:
In the kernel invocation,
<<<Blocks, Threads>>>, try to chose a number of threads that divides evenly with the number of threads in a warp. If you don't, you end up with launching a block that contains inactive threads.In your kernel, try to have each thread in a warp follow the same code path. If you don't, you get what's called warp divergence. This happens because the GPU has to run the entire warp through each of the divergent code paths.
In your kernel, try to have each thread in a warp load and store data in specific patterns. For instance, have the threads in a warp access consecutive 32-bit words in global memory.
回答2:
Are threads grouped into Warps necessarily in order, 1 - 32, 33 - 64 ...?
Yes, the programming model guarantees that the threads are grouped into warps in that specific order.
As a simple example of optimizing of the divergent code paths can be used the separation of all the threads in the block in groups of 32 threads? For example: switch (threadIdx.s/32) { case 0: /* 1 warp*/ break; case 1: /* 2 warp*/ break; /* Etc */ }
Exactly :)
How many bytes must be read at one time for single Warp: 4 bytes * 32 Threads, 8 bytes * 32 Threads or 16 bytes * 32 Threads? As far as I know, the one transaction to the global memory at one time receives 128 bytes.
Yes, transactions to global memory are 128 bytes. So, if each thread reads a 32-bit word from consecutive addresses (they probably need to be 128-byte aligned as well), all the threads in the warp can be serviced with a single transaction (4 bytes * 32 threads = 128 bytes). If each thread reads more bytes, or if the the addresses are not consecutive, more transactions need to be issued (with separate transactions for each separate 128-byte line that is touched).
This is described in the CUDA Programming Manual 4.2, section F.4.2, "Global Memory". There's also a blurb in there saying that the situation is different with data that is cached only in L2, as the L2 cache has 32-byte cache lines. I don't know how to arrange for data to be cached only in L2 or how many transactions one ends up with.
来源：https://stackoverflow.com/questions/11816786/why-bother-to-know-about-cuda-warps