20 - Storage Systems & Disk Management

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# 20 - Storage Systems & Disk Management ## 🎯 Learning Objectives After reading this topic, you will be able to: - Describe the storage hierarchy (cache → RAM → disk → tape) - Calculate disk access time (seek + rotational + transfer) - Compute MTTF for disk systems - Understand the role of buffer management ## 📖 C...

20 - Storage Systems & Disk Management

🎯 Learning Objectives

After reading this topic, you will be able to:

Describe the storage hierarchy (cache → RAM → disk → tape)
Calculate disk access time (seek + rotational + transfer)
Compute MTTF for disk systems
Understand the role of buffer management

📖 Core Content

20.1 Storage Hierarchy

(Diagram)

Level	Speed	Size	Volatile?
Cache	1 ns	KB	Yes
RAM	10-100 ns	GB	Yes
SSD	0.1 ms	GB-TB	No
HDD	5-10 ms	TB	No
Tape	Seconds	TB-PB	No

20.2 Magnetic Disk Structure

A hard disk consists of:

Platters: Circular disks coated with magnetic material
Tracks: Concentric circles on each platter
Sectors: Divisions of tracks (typically 512 bytes)
Cylinders: Same track across all platters
Read/write head: Moves across the platter to access data

20.3 Disk Access Time

T_{access} = T_{seek} + T_{rotation} + T_{transfer}

Component	Description	Formula
Seek time ( $T_{seek}$ )	Time to move arm to correct track	$T_{seek} = \text{avg seek time}$
Rotational latency ( $T_{rotation}$ )	Time for platter to rotate to correct sector	$T_{rotation} = \frac{1}{2} \times \frac{60}{RPM}$
Transfer time ( $T_{transfer}$ )	Time to read/write data	$T_{transfer} = \frac{\text{data size}}{\text{transfer rate}}$

Transfer rate =

\frac{\text{sectors per track} \times \text{sector size}}{\text{rotation time}}

Worked Example

Problem: Seek time = 3ms, rotational speed = 30000 RPM, 200 sectors/track, sector size = 512 bytes. File size = 1000KB.

Transfer rate = $200 \times 512 \div (60/30000) = 102400 \div 0.002 = 51.2 \text{ MB/s}$
Rotational latency = $\frac{1}{2} \times \frac{60}{30000} = 1 \text{ ms}$
Access time per sector = 3 + 1 = 4 ms If file is contiguous: Transfer time = 1000KB / 50 KB/ms = 20ms. Total = ~20ms. If file is scattered: Need 2000 sectors (1000KB / 512 bytes). Total = 2000 × 4ms + 20ms = 8.02 seconds!

20.4 MTTF (Mean Time To Failure)

MTTF for a single disk: Average time until the disk fails
MTTF for a system of n disks: $\frac{MTTF_{disk}}{n}$ (assuming independent failures)

Worked Example

30 disks tested for 24 hours. 15 failed after 18 hours, 8 failed after 22 hours, 7 survived entire 24 hours. Total operating hours =

15 \times 18 + 8 \times 22 + 7 \times 24 = 270 + 176 + 168 = 614

hours Number of failures = 23 MTTF for one disk =

614 / 23 = 26.7

hours MTTF for 15 disks =

26.7 / 15 = 1.78

hours

📐 Key Formulas

Formula	Description
$T_{access} = T_{seek} + T_{rotation} + T_{transfer}$	Total disk access time
$T_{rotation} = \frac{60}{2 \times RPM}$	Average rotational latency
$MTTF_{system} = \frac{MTTF_{disk}}{n}$	System MTTF for n disks

⚠️ Common Pitfall

The Mistake: Calculating rotational latency as full rotation time instead of half. Why: Average rotational latency is half the rotation time (

\frac{60}{2 \times RPM}

), not the full rotation (

\frac{60}{RPM}

📝 Practice Questions

Q1. A disk has 30000 RPM, seek time 5ms, 400 sectors/track, sector size 512B. Find the transfer rate.

Answer

Data per track = 400 × 512 = 204800 bytes Rotation time = 60/30000 = 0.002s = 2ms Transfer rate = 204800 / 0.002 = 102.4 MB/s

Q2. What is the storage hierarchy? Why is it important for DBMS?

Answer

Cache → RAM → SSD → HDD → Tape. Each level is larger, slower, and cheaper. DBMS uses this hierarchy strategically: frequently used data stays in memory (cache/RAM), bulk data on disk, archives on tape.

🔗 Cross-References

Next Topic: 21 - RAID Systems
Related: BSCS4022 (OS) — Storage management
Textbook: Chapter 10 (Storage and File Structure) Join Discord Previous19 - Application DesignNext21 - RAID Systems