Secondary Storage Practice Questions

Given:
Rotation speed = 6000 RPM (Revolutions Per Minute)
Average seek time = 5 ms (milliseconds)
Sectors per track = 500
Sector size = 512 bytes
File size = 3000 sectors, located randomly on the disk.

1. Calculate Rotation Time:
   1 minute = 60 seconds.
   6000 RPM = 6000 revolutions / 60 seconds = 100 revolutions / second.
   Time for 1 revolution = 1 / 100 seconds = 0.01 seconds = 10 ms.

2. Calculate Average Rotational Latency:
   Average rotational latency = (Time for 1 revolution) / 2 = 10 ms / 2 = 5 ms.

3. Calculate Time to Transfer one Sector:
   Time for 1 revolution (10 ms) transfers 500 sectors.
   Time to transfer 1 sector = (Time for 1 revolution) / (Sectors per track)
   = 10 ms / 500 = 0.02 ms.

4. Calculate Total Time to Read 3000 Sectors:
   Since sectors are located randomly, for each sector, we incur average seek time and average rotational latency.
   Total time = 3000 sectors $\times$ (Average seek time + Average rotational latency + Sector transfer time).
   Total time = $3000 \times (5 \text{ ms} + 5 \text{ ms} + 0.02 \text{ ms})$
   Total time = $3000 \times (10.02 \text{ ms})$
   Total time = $30060 \text{ ms}$.

5. Convert to seconds and round:
   Total time = $30060 \text{ ms} = 30060 / 1000 \text{ seconds} = 30.06 \text{ seconds}$.

The total time is 30.06 seconds.
The range given in the PDF solution is [29.50 to 30.50]. Our calculated value 30.06 falls within this range.
===END_Q53===

Given hard disk specifications:
Total capacity = 512 GB
Number of storage surfaces = 32
Sectors per track = 4096
Sector size = 1024 bytes

The total capacity of a hard disk can be expressed as:
Total Capacity = Number of Cylinders $\times$ Number of Surfaces $\times$ Sectors per Track $\times$ Sector Size

We need to find the number of cylinders. Let this be $C$.
$512 \text{ GB} = C \times 32 \text{ surfaces} \times 4096 \text{ sectors/track} \times 1024 \text{ bytes/sector}$

First, convert GB to bytes:
$512 \text{ GB} = 512 \times 2^{10} \text{ MB} = 512 \times 2^{10} \times 2^{10} \text{ KB} = 512 \times 2^{10} \times 2^{10} \times 2^{10} \text{ bytes}$
$512 = 2^9$.
So, $512 \text{ GB} = 2^9 \times 2^{10} \times 2^{10} \times 2^{10} \text{ bytes} = 2^{39} \text{ bytes}$.

Now, substitute the values into the formula:
$2^{39} = C \times 32 \times 4096 \times 1024$
Express the numbers as powers of 2:
$32 = 2^5$
$4096 = 2^{12}$
$1024 = 2^{10}$

So, $2^{39} = C \times 2^5 \times 2^{12} \times 2^{10}$
$2^{39} = C \times 2^{(5+12+10)}$
$2^{39} = C \times 2^{27}$

Solve for $C$:
$C = \frac{2^{39}}{2^{27}}$
$C = 2^{(39-27)}$
$C = 2^{12}$

$2^{12} = 4096$.
Therefore, the number of cylinders in the hard disk is 4096.

The total average time is the sum of seek time, rotational delay, transfer time, and controller overhead.

1. Average Rotational Delay: Half of one rotation time.
   Time for 1 rotation = $\frac{60 \text{ sec}}{15000 \text{ rotations}} = 4$ ms. Average delay = $0.5 \times 4 = 2$ ms.
2. Average Seek Time: Given as twice the average rotational delay.
   Seek Time = $2 \times 2 = 4$ ms.
3. Transfer Time: Time to transfer 512 bytes.
   Time = $\frac{512 \text{ B}}{50 \times 10^6 \text{ B/s}} \approx 10.24$ microseconds.
4. Controller Overhead: 10 times the transfer time.
   Overhead = $10 \times 10.24 \mu s = 102.4 \mu s \approx 0.1$ ms.
   Total Time = $4 + 2 + 0.01024 + 0.1 \approx 6.1$ ms.

The total time to read the file is the sum of the time for all sector accesses plus the total data transfer time.

1. Time for one rotation: $60 \text{ sec} / 10000 \text{ RPM} = 0.006 \text{ sec} = 6 \text{ ms}$.
2. Average rotational latency: Half of one rotation time, which is $6 \text{ ms} / 2 = 3 \text{ ms}$.
3. Time per sector access: This is the sum of seek time and rotational latency, as a seek is required for each sector. Time = $4 \text{ ms} + 3 \text{ ms} = 7 \text{ ms}$.
4. Total access time for 2000 sectors: $2000 \times 7 \text{ ms} = 14000 \text{ ms}$.
5. Data transfer rate: (Bytes per track) / (Rotation time) = $(600 \times 512 \text{ bytes}) / 6 \text{ ms} = 51200 \text{ B/ms}$.
6. Total transfer time: (Total file size) / (Transfer rate) = $(2000 \times 512 \text{ bytes}) / 51200 \text{ B/ms} = 20 \text{ ms}$.
7. Total time: $14000 \text{ ms} + 20 \text{ ms} = 14020 \text{ ms}$.

To find the cylinder number of the last sector, we first calculate the total number of sectors occupied by the file and the capacity per cylinder.

1. Convert file size to sectors:
   File size = 42797 KB = $42797 \times 1024$ bytes.
   Since each sector is 512 bytes, the total number of sectors occupied by the file is:
   $\text{Total sectors} = \frac{42797 \times 1024}{512} = 42797 \times 2 = 85594 \text{ sectors}$

2. Calculate sectors per cylinder:
   Each cylinder has 16 surfaces and each surface has 64 sectors.
   $\text{Sectors per cylinder} = 16 \text{ surfaces} \times 64 \text{ sectors/surface} = 1024 \text{ sectors/cylinder}$

3. Determine sectors remaining in the starting cylinder:
   The starting location is (1200, 9, 40). This means cylinder 1200, surface 9, sector 40.
   Sectors before the starting sector in cylinder 1200: $9 \text{ surfaces} \times 64 \text{ sectors/surface} + 40 \text{ sectors} = 576 + 40 = 616 \text{ sectors}$.
   Remaining sectors in cylinder 1200: $1024 - 616 = 408 \text{ sectors}$.

4. Calculate remaining sectors of the file:
   $\text{Remaining file sectors} = 85594 - 408 = 85186 \text{ sectors}$

5. Calculate full cylinders needed for the remaining sectors:
   $\text{Full cylinders} = \lfloor \frac{85186}{1024} \rfloor = 83 \text{ cylinders}$
   These full cylinders are $1201, 1202, \dots, 1200 + 83 = 1283$.
   The sectors used in these full cylinders are $83 \times 1024 = 84992 \text{ sectors}$.

6. Calculate sectors remaining after full cylinders:
   $\text{Sectors \in the last cylinder} = 85186 - 84992 = 194 \text{ sectors}$
   Since there are remaining sectors, the file extends into the next cylinder.
   The last cylinder will be $1283 + 1 = 1284$.

To find the total time to load all libraries, we first need to calculate the time for a single disk access.
The disk's rotational speed is 6000 rpm, which means it completes 6000 rotations in 60 seconds.
Thus, one rotation takes $\frac{60 \text{ seconds}}{6000 \text{ rotations}} = 0.01 \text{ seconds} = 10 \text{ ms}$.
The average rotational latency (time to wait for the desired sector to rotate under the head) is half of one rotation time, so $10 \text{ ms} / 2 = 5 \text{ ms}$.
The total time for one disk access is the sum of seek time and average rotational latency: $10 \text{ ms} + 5 \text{ ms} = 15 \text{ ms}$.
Since there are 100 libraries, and each requires one disk access from a random location, the total time to load all libraries is $100 \times 15 \text{ ms} = 1500 \text{ ms} = 1.5 \text{ seconds}$.

Given the parameters:

• Sectors per track ($S_T$): 63
• Recording surfaces per platter: 2
• Total platters: 10
• Total recording surfaces ($H$): $10 \times 2 = 20$
  The address is (c,h,s), where c is cylinder, h is surface, s is sector. Sectors are 0-indexed globally.

The number of sectors per cylinder ($S_C$) is $H \times S_T = 20 \times 63 = 1260$.
We want the address of the $1039^{th}$ sector (global index $G = 1039$).

1. Cylinder number ($c$):
   $c = \left\lfloor \frac{G}{S_C} \right\rfloor = \left\lfloor \frac{1039}{1260} \right\rfloor = 0$
2. Remaining sectors within this cylinder ($G'$):
   $G' = G \pmod{S_C} = 1039 \pmod{1260} = 1039$
3. Surface number ($h$):
   $h = \left\lfloor \frac{G'}{S_T} \right\rfloor = \left\lfloor \frac{1039}{63} \right\rfloor = 16$
4. Sector number ($s$):
   $s = G' \pmod{S_T} = 1039 \pmod{63} = 31$
   Thus, the address is (0, 16, 31).

The total number of recording surfaces (heads) is $10 \text{ platters} \times 2 \text{ surfaces/platter} = 20$.
The total number of sectors per cylinder is $20 \text{ surfaces} \times 63 \text{ sectors/track} = 1260$.

The sector number for an address $(c,h,s)$ is given by the formula:
$\text{Sector Number} = c \times (\text{$total_surfaces$} \times \text{$sectors_per_track$}) + h \times \text{$sectors_per_track$} + s$
Substitute the given values $c=400$, $h=16$, $s=29$, total surfaces $= 20$, and sectors per track $= 63$:
$\text{Sector Number} = 400 \times (20 \times 63) + 16 \times 63 + 29$
$\text{Sector Number} = 400 \times 1260 + 1008 + 29$
$\text{Sector Number} = 504000 + 1008 + 29 = 505037$