Chapter 20, Problem 46E

(a)

To determine

Theaverage latency time of the floppy disk.

Answer to Problem 46E

The average latency time of the floppy diskis $0.1\text{ s}$ .

Explanation of Solution

Write the equation for the average latency time.

  $t_l=\frac{T}{2}$        (I)

Here, $t_l$ is theaverage latency timeand $T$ is therotational period.

Write the equation for the rotational period.

  $T=\frac{1}{f}$

Here, $f$ is the frequency of rotation.

Put the above equation in equation (I).

  $\begin{array}{c}{t}_l=\frac{1/f}{2}\\ =\frac{1}{2f}\end{array}$        (II)

Conclusion:

Frequency is the number of rotations per second. It is given that the floppy disk rotates at $300{\text{ rev min}}^{-1}$.

Find the frequency of rotation of the floppy disk.

  $\begin{array}{c}f=300{\text{ rev min}}^{-1}\frac{1\text{ min}}{60\text{ s}}\\ =5\text{ rev/s}\end{array}$

Substitute $5\text{ rev/s}$ for $f$ in equation (II) to find $t_l$ .

  $\begin{array}{c}{t}_l=\frac{1}{2\left(5\text{ rev/s}\right)}\\ =0.1\text{ s}\end{array}$

Therefore, the average latency time of the floppy disk is $0.1\text{ s}$ .

(b)

To determine

The average seek time of the floppy disk.

Answer to Problem 46E

Theaverage seek time of the floppy diskis $164\text{ ms}$.

Explanation of Solution

Write the equation for the access time.

    $t_a=t_s+t_l$

Here, $t_a$ is the access time and $t_s$ is the seek time.

Rewrite theabove equation for $t_s$.

    $t_s=t_a-t_l$        (III)

Conclusion:

Substitute $264\text{ ms}$ for $t_a$ and $0.1\text{ s}$ for $t_l$ in equation (III) to find $t_s$ .

    $\begin{array}{c}{t}_s=264\text{ ms}-0.1\text{ s}\frac{1000\text{ ms}}{1\text{ s}}\\ =264\text{ ms}-100\text{ ms}\\ =164\text{ ms}\end{array}$

Therefore, the average seek time of the floppy disk is $164\text{ ms}$ .

(c)

To determine

The average latency and seek times of the floppy disk.

Answer to Problem 46E

The average latency time of the disk is $8.26\text{ ms}$ and the average seek time is $23.04\text{ ms}$.

Explanation of Solution

Equation (II) can be used to determine average latency time and equation (III) can be used to determine average seek time.

Conclusion:

It is given that the floppy disk rotates at $3633{\text{ rev min}}^{-1}$ .

Find the frequency of rotation of the floppy disk.

  $\begin{array}{c}f=3633{\text{ rev min}}^{-1}\frac{1\text{ min}}{60\text{ s}}\\ =60.55\text{ rev/s}\end{array}$

Substitute $60.55\text{ rev/s}$ for $f$ in equation (II) to find $t_l$ .

  $\begin{array}{c}{t}_l=\frac{1}{2\left(60.55\text{ rev/s}\right)}\\ =8.26\times {10}^{-3}\text{ s}\\ =8.26\times {10}^{-3}\text{ s}\frac{1\text{ ms}}{{10}^{-3}\text{ s}}\\ =8.26\text{ ms}\end{array}$

Substitute $31.3\text{ ms}$ for $t_a$ and $8.26\text{ ms}$ for $t_l$ in equation (III) to find $t_s$ .

    $\begin{array}{c}{t}_s=31.3\text{ ms}-8.26\text{ ms}\\ =23.04\text{ ms}\end{array}$

Therefore, the average latency time of the disk is $8.26\text{ ms}$ and the average seek time is $23.04\text{ ms}$ .