Skip to content
← CUDA, Triton & GPU Performance / 16
hardNewNVIDIATogether AIFireworks AI

Inside one CUDA graph, decode kernel boundaries still cost time. How does programmatic dependent launch hide them?

A graph removes the CPU from the step but not the boundary between two kernels. The answer prices that boundary across 390 kernels, then shows which half of the next kernel may start early and which half must wait.

Updated Oct 2026 · Worked answers on inference mechanisms, with calculations, implementation details, and follow-up questions.

A graph removes the CPU from the step but not the boundary between two kernels. The answer prices that boundary across 390 kernels, then shows which half of the next kernel may start early and which half must wait.

10 answers per topic instead of 5, plus saved progress and bookmarks · costs nothingor unlock all 213 remaining answers · $25

Every worked answer and derivation in all 12 topics, every concept page, the full practice-test bank, and every code drill’s edge-case tests and reference solution.

6 months · One payment · No auto-renewal

UP NEXT ON YOUR JOURNEY
Next in this trackOn a B200, how does an NVFP4 GEMM apply its block scales, and when does FP4 actually make a layer faster than FP8?Next in this trackA cp.async double-buffered kernel fails a handful of CI runs and never fails under -G. Find the race and fix it.Next in this trackA kernel PR claims 12x over eager PyTorch on the SwiGLU activation and passes the tests. How do you decide whether it is real?
DISCUSSION · 0

Nobody has commented on this answer. Add a correction or another way to work it.