The Essentials of Computer Organization and Architecture
4th Edition
ISBN: 9781284045611
Author: Linda Null, Julia Lobur
Publisher: Jones & Bartlett Learning
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Chapter 11, Problem 22E
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 48,14,85,35,84,61,30, and 22.
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 |
| 48 | 50-48 | 2 |
| 14 | 48-14 | 34 |
| 85 | 14-85 | 71 |
| 35 | 85-35 | 50 |
| 84 | 35-84 | 49 |
| 61 | 84-61 | 23 |
| 30 | 61-30 | 31 |
| 22 | 30-22 | 8 |
| Total number of disk access | 268 | |
| Total time taken | 8*2 milliseconds+(268-8)*500 nanoseconds | |
Therefore, total number of disk access required is “268”.
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 |
| 48 | 50-48 | 2 |
| 61 | 48-61 | 13 |
| 84 | 61-84 | 23 |
| 85 | 84-85 | 1 |
| 35 | 85-35 | 50 |
| 30 | 35-30 | 5 |
| 22 | 30-22 | 8 |
| 14 | 22-14 | 8 |
| Total number of disk access | 110 | |
| Total time taken | 8*2 milliseconds+(110-8)*500 nanoseconds | |
Therefore, total number of disk access required is “110”...
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Here is a clear background and explanation of the full method, including what each part is doing and why.
Background & Motivation
Missing values: Some input features (sensor channels) are missing for some samples due to sensor failure or corruption.
Missing labels: Not all samples have a ground-truth RUL value. For example, data collected during normal operation is often unlabeled.
Most traditional deep learning models require complete data and full labels. But in our case, both are incomplete. If we try to train a model directly, it will either fail to learn properly or discard valuable data.
What We Are Doing: Overview
We solve this using a Teacher–Student knowledge distillation framework:
We train a Teacher model on a clean and complete dataset where both inputs and labels are available.
We then use that Teacher to teach two separate Student models:
Student A learns from incomplete input (some sensor values missing).
Student B learns from incomplete labels (RUL labels missing…
here is a diagram code :
graph LR subgraph Inputs [Inputs] A[Input C (Complete Data)] --> TeacherModel B[Input M (Missing Data)] --> StudentA A --> StudentB end subgraph TeacherModel [Teacher Model (Pretrained)] C[Transformer Encoder T] --> D{Teacher Prediction y_t} C --> E[Internal Features f_t] end subgraph StudentA [Student Model A (Trainable - Handles Missing Input)] F[Transformer Encoder S_A] --> G{Student A Prediction y_s^A} B --> F end subgraph StudentB [Student Model B (Trainable - Handles Missing Labels)] H[Transformer Encoder S_B] --> I{Student B Prediction y_s^B} A --> H end subgraph GroundTruth [Ground Truth RUL (Partial Labels)] J[RUL Labels] end subgraph KnowledgeDistillationA [Knowledge Distillation Block for Student A] K[Prediction Distillation Loss (y_s^A vs y_t)] L[Feature Alignment Loss (f_s^A vs f_t)] D -- Prediction Guidance --> K E -- Feature Guidance --> L G --> K F --> L J -- Supervised Guidance (if available) --> G K…
details explanation and background
We solve this using a Teacher–Student knowledge distillation framework:
We train a Teacher model on a clean and complete dataset where both inputs and labels are available.
We then use that Teacher to teach two separate Student models:
Student A learns from incomplete input (some sensor values missing).
Student B learns from incomplete labels (RUL labels missing for some samples).
We use knowledge distillation to guide both students, even when labels are missing.
Why We Use Two Students
Student A handles Missing Input Features: It receives input with some features masked out. Since it cannot see the full input, we help it by transferring internal features (feature distillation) and predictions from the teacher.
Student B handles Missing RUL Labels: It receives full input but does not always have a ground-truth RUL label. We guide it using the predictions of the teacher model (prediction distillation).
Using two students allows each to specialize in…
Chapter 11 Solutions
The Essentials of Computer Organization and Architecture
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
Ch. 11 - Prob. 11RETCCh. 11 - Prob. 12RETCCh. 11 - Prob. 13RETCCh. 11 - Prob. 14RETCCh. 11 - Prob. 15RETCCh. 11 - Prob. 16RETCCh. 11 - Prob. 17RETCCh. 11 - Prob. 18RETCCh. 11 - Prob. 19RETCCh. 11 - Prob. 20RETCCh. 11 - Prob. 1ECh. 11 - Prob. 2ECh. 11 - Prob. 3ECh. 11 - Prob. 4ECh. 11 - Prob. 5ECh. 11 - Prob. 6ECh. 11 - Prob. 7ECh. 11 - Prob. 8ECh. 11 - Prob. 10ECh. 11 - Prob. 11ECh. 11 - Prob. 12ECh. 11 - Prob. 13ECh. 11 - Prob. 14ECh. 11 - Prob. 15ECh. 11 - Prob. 16ECh. 11 - Prob. 17ECh. 11 - Prob. 18ECh. 11 - Prob. 19ECh. 11 - Prob. 20ECh. 11 - Prob. 21ECh. 11 - Prob. 22ECh. 11 - Prob. 23ECh. 11 - Prob. 24ECh. 11 - Prob. 25ECh. 11 - Prob. 26E
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