ESSENTIALS OF COMPUTER ORGAN..-TEXT
4th Edition
ISBN: 9781284033144
Author: NULL
Publisher: JONES+BART
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Chapter 11, Problem 21E
Explanation of Solution
Performance:
- The speed of the CPU is a key factor that often decides the performance of the system.
- Disk access speed is considered to be another factor that is used to determine the performance.
- The read write operation that is performed on the disk (milliseconds) is considered to take significantly longer access time in comparison with that of the operation that is performed in a CPU (nanoseconds).
- To improve the performance of the disk access they are different ways present.
- The
algorithm that determines the order in which the data access made in the disk will be beneficial is calculated.
Given:
It is know that there are 100 tracks that are ranged from 0 to 99.
At the beginning the disk arm is considered to be at the position 50 and it will getting move through and towards the lower numbered tracks.
It will require 500 nanoseconds to pass over the track.
The desired track is accessed in 2milliseconds.
The service request order is as follows 35,53,90,67,79,37,76, and 47.
Disk traversed using FCFS (First come First Serve):
The service request is being handled in the way they are being received:
| Service request | Access made | Number of access |
| 35 | 50-35 | 15 |
| 53 | 35-53 | 18 |
| 90 | 53-90 | 37 |
| 67 | 90-67 | 23 |
| 79 | 67-79 | 12 |
| 37 | 79-37 | 42 |
| 76 | 37-76 | 39 |
| 47 | 76-47 | 29 |
| Total number of disk access | 215 | |
| Total time taken | 8*2 milliseconds+(218-8)*500 nanoseconds | |
Therefore, total number of disk access required is “215”.
Disk traversed using SSTF (Shortest Seek Time First):
- The service request that are from the nearest of the sector to that of the current position of the disk arm are being handled on priority.
- The list is processed to be generated in the way from the current position of the disk arm and the instances of the next nearest access are being obtained at every instance.
| Service request | Access made | Number of access |
| 47 | 50-47 | 3 |
| 53 | 47-53 | 6 |
| 67 | 53-67 | 14 |
| 76 | 67-76 | 9 |
| 79 | 76-79 | 3 |
| 90 | 79-90 | 11 |
| 37 | 90-37 | 53 |
| 35 | 37-35 | 2 |
| Total number of disk access | 101 | |
| Total time taken | 8*2 milliseconds+(101-8)*500 nanoseconds | |
Therefore, total number of disk access required is “101”...
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My code is experincing minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place.
My code:
% Define frequency range for the plot
f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz
w = 2 * pi * f; % Angular frequency
% Parameters for the filters - let's adjust these to get more reasonable cutoffs
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
% For bandpass, we need appropriate L value for desired cutoffs
L = 0.1; % Inductance in henries - adjusted for better bandpass response
% Calculate cutoff frequencies first to verify they're in desired range
f_cutoff_RC = 1 / (2 * pi * R * C);
f_resonance = 1 / (2 * pi * sqrt(L * C));
Q_factor = (1/R) * sqrt(L/C);
f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor));
f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor));
% Transfer functions
% Low-pass filter (RC)
H_low = 1 ./ (1 + 1i * w *…
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Chapter 11 Solutions
ESSENTIALS OF COMPUTER ORGAN..-TEXT
Ch. 11 - Prob. 1RETCCh. 11 - Prob. 2RETCCh. 11 - Prob. 3RETCCh. 11 - Prob. 4RETCCh. 11 - Prob. 5RETCCh. 11 - Prob. 6RETCCh. 11 - Prob. 7RETCCh. 11 - Prob. 8RETCCh. 11 - Prob. 9RETCCh. 11 - Prob. 10RETC
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