LinkedIn Inference Engineer interview questions

Both reuse KV for shared prefixes, and on a 2K-token chat prompt their hit rates differ by under one percent. The answer finds the differences that do matter, from eviction policy to whether a cache key means anything outside one process.

A matrix-vector product has one job: keep enough bytes in flight. The answer counts how many that is per SM, then finds the three places a decode GEMV quietly drops below it.

Two GEMV kernels do the same arithmetic on the same matrix and differ only in which thread reads which element. One streams at HBM speed; the other turns every warp load into 32 separate transactions.

Occupancy is the minimum of four separate budgets, and candidates usually compute one of them. The answer works all four for an H100, finds which one binds, and then asks whether raising occupancy would help at all.

The one-column pad is the textbook fix, and it works for FP32 scalars. The answer derives why, then shows what happens with BF16 and with 16-byte vector stores, where the pad stops working and kernels swizzle instead.

Every dimension of a GEMM tile is pinned by a different piece of hardware. The answer derives the reuse that sets the M and N sides, the swizzle row that sets K, and the register and shared-memory budgets that stop the tile growing.

Eight tiles too many turns a one-wave GEMM into two. The answer prices the waste in microseconds for a real prefill shape, then compares the three fixes, including the one that lives in the scheduler, not the kernel.

RMSNorm does almost no arithmetic, so its cost is bytes, and the eager PyTorch version moves nine times more of them than it needs to. The answer writes the one-pass kernel, tests it on awkward hidden sizes, and says where fusion stops paying.

A Python loop over 256 experts is correct and slow for three separate reasons, only one of which is launch count. The answer builds the expert-sorted layout the fused kernel runs on and counts what it costs in padding.

A 4-bit weight is only cheaper if turning it back into FP16 costs less than the bytes it saved. The answer counts the instruction budget per weight at full HBM bandwidth and shows the bit trick that makes the conversion fit.

Every kernel in the step can run near peak bandwidth while the step itself runs at two thirds of it. The answer reads the three rows of the timeline that tell you why, and turns the gaps into milliseconds.

Both changes move work off the threads: one takes over address generation, the other takes over operand staging. The answer counts the instructions each one removes and shows why a Hopper kernel is organized around barriers instead of thread blocks that do everything.

Hopper's biggest GEMM tiles spend half the register file on the accumulator and feed every operand byte through one SM's shared memory. Blackwell attacks both limits, and the answer counts the bytes each change gives back.

CUDA lets you reserve part of L2 for data you choose, and the KV cache is re-read every step. The answer measures what fits against what a decode step streams, finds the one case where pinning pays, and the kernel change that pays more.

A kernel time is a distribution measured under conditions, and a reviewer will ask about both. The answer builds the timing loop, says what each line defends against, and adds the one sanity check that catches most bad numbers.

On paper an H200 SXM is an H100 with more and faster memory, so nothing should get slower. The answer works through the four ways it can anyway, from the spec sheet you did not read to the tuning table that does not know the new device's name.

Tensor parallelism divides the weight bytes but not the all-reduce count, so each doubling buys less than the one before. The answer prices that fixed cost, then finds the workload where TP=8 wins anyway.