I/O Size vs Latency: Amortizing Seek Time

Topic Overview

Amortize Fixed Seek Costs

On HDDs the seek time is fixed whether you read 4KB or 1MB afterward. Large sequential reads spread that penalty across many bytes, so effective latency per byte drops. This is why database sequential scans and log append prefer big read sizes on spinning rust.

Effective Latency Formula

Total time = seek + rotation + (bytes / throughput). Per-byte cost = total / bytes.

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• SSD random reads have no seek; size matters less for latency
• Read-ahead in the kernel exploits sequential detection
• Storage engines align prefetch to multiples of page size
• Batch small reads in application buffers when HDD bound

Working Through the Exercise

Keep the relevant documentation open while you implement. When your output disagrees with the reference, trace one failing case by hand before changing random lines.

• Trace first: walk one failing input step by step
• Read the manual: skim the official doc for the mechanism named in this task
• Compare baselines: save measurements from the prior step so speedups are honest

Why for this exercise

You will calculate per-operation and per-byte latency for several request sizes on HDD and SSD models. This exercise requires showing how bigger I/O amortizes mechanical costs.

Your Task

Every I/O request pays a fixed cost before any data moves (command processing, queueing, and on an HDD a seek), then a transfer cost proportional to its size. Model one request:

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That is a 0.05 ms fixed overhead plus a transfer at 1 GB/s. At 4 KiB the fixed part is most of the latency; at 1 MiB the transfer dominates. That is why databases batch small writes and read large sequential runs.

Input

One I/O size in bytes.

Output

io_size=S latency_ms=L, with L rounded to three decimals: io_size=4096 latency_ms=0.054.