Andrew S. Tanenbaum (OS) Edition 4 Exercise 5 Question 28 (Page No. 431)

Question

Consider a magnetic disk consisting of $16$ heads and $400$ cylinders. This disk has four $100$-cylinder zones with the cylinders in different zones containing $160, 200, 240.$ and $280$ sectors, respectively. Assume that each sector contains $512$ bytes, average seek time between adjacent cylinders is $1\: msec,$ and the disk rotates at $7200$ RPM. Calculate the

1. disk capacity,
2. optimal track skew, and
3. maximum data transfer rate.

Answer 1

a)$720896000$

c)$17,203,200 \frac{\text{Bytes}}{\text{sector}}$

Explanation:

The size of the zone $=\text{platter} \times \text{cylinders}\times \frac{\text{sectors}}{\text{track}}\times \frac{\text{Bytes}}{\text{sector}} $

Capacity of zone 1: $16 \times 100\times 160 \times 512 = 131072000\;\text{Bytes}$

Capacity of zone 2: $16 \times 100\times 200 \times 512 = 163840000\;\text{Bytes}$

Capacity of zone 3: $16 \times 100\times 240 \times 512 = 196608000\;\text{Bytes}$

Capacity of zone 4: $16 \times 100\times 280 \times 512 = 229376000\;\text{Bytes}$

Sum$=131072000 + 163840000 + 196608000 + 229376000 = 720896000$

(c) The rate of data transfer will be maximum when the cylinders are being written or read in the outermost zone.

In the outer zone, in one second, $280$ sectors are being read $120$ times.

Thus the maximum data rate $= 280 \times 120 \times 512 = 17,203,200 \;\frac{\text{Bytes}}{\text{sec}}$