Math Problem Statement

A disk has the following specifications:

Number of cylinders: 18,500 Number of heads: 16 Number of sectors per track: 75 Spin rate: 5400 RPM Seek time between adjacent tracks: 2 ms Sector size: 1000 bytes (a) Compute the disk capacity. (b) Calculate the time required to read the entire disk sequentially. (c) Calculate the time to transfer 350 bytes of data.

Solution

Let's address each part of the problem step by step:


(a) Compute the disk capacity

The disk capacity is calculated using the formula: Disk Capacity=Number of Cylinders×Number of Heads×Sectors per Track×Sector Size\text{Disk Capacity} = \text{Number of Cylinders} \times \text{Number of Heads} \times \text{Sectors per Track} \times \text{Sector Size}

Given:

  • Number of cylinders = 18,50018,500
  • Number of heads = 1616
  • Sectors per track = 7575
  • Sector size = 1,000bytes1,000 \, \text{bytes}

Disk Capacity=18,500×16×75×1,000\text{Disk Capacity} = 18,500 \times 16 \times 75 \times 1,000

Disk Capacity=22,200,000,000bytes=22.2GB\text{Disk Capacity} = 22,200,000,000 \, \text{bytes} \, = 22.2 \, \text{GB}


(b) Calculate the time required to read the entire disk sequentially

The time required to read the entire disk is influenced by the disk's spin rate and the total number of sectors. First, let's calculate the total number of sectors:

Total Sectors=Number of Cylinders×Number of Heads×Sectors per Track\text{Total Sectors} = \text{Number of Cylinders} \times \text{Number of Heads} \times \text{Sectors per Track}

Total Sectors=18,500×16×75=22,200,000sectors\text{Total Sectors} = 18,500 \times 16 \times 75 = 22,200,000 \, \text{sectors}

Each sector is read in a single spin (one revolution). The spin rate is given as 5400 RPM, which translates to: Spin Time (per revolution)=605400seconds=0.0111seconds (11.1 ms)\text{Spin Time (per revolution)} = \frac{60}{5400} \, \text{seconds} = 0.0111 \, \text{seconds (11.1 ms)}

Total Time to Read Entire Disk=Total Sectors×Spin Time (per sector)\text{Total Time to Read Entire Disk} = \text{Total Sectors} \times \text{Spin Time (per sector)}

Total Time to Read Entire Disk=22,200,000×0.0111seconds246,420seconds=68.45hours\text{Total Time to Read Entire Disk} = 22,200,000 \times 0.0111 \, \text{seconds} \approx 246,420 \, \text{seconds} \, = 68.45 \, \text{hours}


(c) Calculate the time to transfer 350 bytes of data

Each sector contains 1,000bytes1,000 \, \text{bytes}, and the transfer rate depends on how long it takes to read a sector. Given the spin time per revolution is 11.1ms11.1 \, \text{ms}, we estimate the time to transfer 350 bytes as a proportion of the time to transfer a full sector:

Time to Transfer 350 Bytes=3501,000×11.1ms\text{Time to Transfer 350 Bytes} = \frac{350}{1,000} \times 11.1 \, \text{ms}

Time to Transfer 350 Bytes=3.885ms\text{Time to Transfer 350 Bytes} = 3.885 \, \text{ms}


Summary

(a) Disk capacity: 22.2 GB
(b) Time to read entire disk sequentially: 68.45 hours
(c) Time to transfer 350 bytes: 3.885 ms

Would you like to explore the calculations further or clarify any steps?


5 Related Questions:

  1. How would increasing the spin rate impact the time to read the entire disk?
  2. What would be the effect of reducing sector size on disk capacity?
  3. Can we optimize the time to transfer 350 bytes by modifying seek time?
  4. How does the number of heads affect the overall performance of the disk?
  5. What is the role of rotational latency in disk performance?

Tip:

For faster data transfer rates, modern systems use SSDs,

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Math Problem Analysis

Mathematical Concepts

Arithmetic operations
Unit conversions
Proportionality
Data storage and retrieval calculations

Formulas

Disk Capacity = Number of Cylinders × Number of Heads × Sectors per Track × Sector Size
Spin Time (per revolution) = 60 ÷ Spin Rate (RPM)
Total Time to Read = Total Sectors × Spin Time
Time to Transfer = (Data Size ÷ Sector Size) × Spin Time

Theorems

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Suitable Grade Level

Grades 10-12