HDD vs SSD: Modeling Physical Storage Latency

Topic Overview

Mechanical vs Flash Latency

HDD random reads pay seek plus rotational delay before a 4KB sector arrives. SSD reads are electrical with tens of microseconds latency but very different write semantics: NAND must erase before reprogramming, driving write amplification and wear leveling.

Latency Components

Model HDD random read as seek + average rotation + transfer. Model SSD read as fixed microseconds; model SSD random write as read-modify-erase-write on affected pages.

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• Sequential throughput can saturate interface on both media
• SSD FTL hides physical layout; logical LBA still maps to pages
• Databases cache hot pages in RAM regardless of media
• Understanding the gap explains why LSM trees favor sequential writes

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 compute latency and IOPS tables for HDD and SSD constants provided in the starter code. This exercise asks you to print a comparison that motivates buffering and sequential layout.

Your Task

"SSDs are faster" hides the real lesson: a hard disk pays a mechanical positioning cost, a seek plus half a rotation on average, before every random I/O, while its sequential bandwidth is respectable. Build a small latency model for disks and SSDs, run workloads against it, and see exactly where the 100× gap comes from, and where it shrinks to about 10×.

Input

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Up to 8 devices, with unique names of up to 15 characters. run evaluates the workload on every device, in definition order.

Model (all integer nanoseconds; "round" means half up)

• transfer(bytes) = round(bytes × 1000 / MBPS) (MB = 10^6 bytes).
• HDD: half_rotation = round(30,000,000,000 / RPM) and position = SEEK_US × 1000 + half_rotation.
  • random: COUNT × (position + transfer(size));
  • seq: position + transfer(COUNT × size) (one seek, then streaming).
• SSD: lat = READ_US or WRITE_US, × 1000.
  • random: COUNT × (lat + transfer(size));
  • seq: lat + transfer(COUNT × size).

Output

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• Times are in ms with 3 decimals: round the ns to µs, then print µs / 1000. ms/op is round(total / COUNT) ns, printed the same way.
• IOPS = round(COUNT × 10^9 / total_ns), and MB/s = round(total_bytes × 10000 / total_ns) / 10, with one decimal.
• With two or more devices, add fastest: A, R.Rx faster than B, where A has the smallest total, B the largest (the first defined wins ties), and R = round(10 × worst / best) / 10. If all are equal, print all devices take the same time instead.
• Errors:
  • error: usage: hdd NAME RPM SEEK_US MBPS (also for a duplicate name or a 9th device);
  • error: usage: ssd NAME READ_US WRITE_US MBPS;
  • error: usage: run random|seq read|write COUNT SIZE_KIB;
  • error: workload larger than 1 TiB (COUNT × SIZE_KIB > 2^30);
  • error: no devices;
  • error: unknown command CMD.

Example

Input:

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Output:

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Requirements

1. Use 64-bit integer arithmetic and the rounding rules exactly. Floating point would make the last digit disagree.
2. Keep the device list generic: the same run must work for any mix of disks and SSDs.

Hidden tests cover sequential workloads (where the gap shrinks), writes, a 15k-RPM disk against a SATA SSD, a single-device run, equal devices, large transfers, and malformed input.