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 20E
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.
The service request order is as follows 82,97,35,75,53,47,17, and11.
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 |
| 82 | 50-82 | 32 |
| 97 | 82-97 | 15 |
| 35 | 97-35 | 62 |
| 75 | 35-75 | 40 |
| 53 | 75-53 | 22 |
| 47 | 53-47 | 6 |
| 17 | 47-17 | 30 |
| 11 | 17-11 | 6 |
| Total number of disk access | 213 | |
Therefore, total number of disk access required is “213”.
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 that is nearest | Access made | Number of access |
| 50 | 47 | 50-47 | 3 |
| 47 | 53 | 47-53 | 6 |
| 53 | 35 | 53-35 | 18 |
| 35 | 17 | 35-17 | 18 |
| 17 | 11 | 17-11 | 6 |
| 11 | 75 | 11-75 | 64 |
| 75 | 82 | 75-82 | 7 |
| 82 | 97 | 82-97 | 15 |
| Total number of disk access | 137 | ||
Therefore, total number of disk access required is “137”...
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Here are two diagrams. Make them very explicit, similar to Example Diagram 3 (the Architecture of MSCTNN).
graph LR subgraph Teacher_Model_B [Teacher Model (Pretrained)] Input_Teacher_B[Input C (Complete Data)] --> Teacher_Encoder_B[Transformer Encoder T] Teacher_Encoder_B --> Teacher_Prediction_B[Teacher Prediction y_T] Teacher_Encoder_B --> Teacher_Features_B[Internal Features F_T] end subgraph Student_B_Model [Student Model B (Handles Missing Labels)] Input_Student_B[Input C (Complete Data)] --> Student_B_Encoder[Transformer Encoder E_B] Student_B_Encoder --> Student_B_Prediction[Student B Prediction y_B] end subgraph Knowledge_Distillation_B [Knowledge Distillation (Student B)] Teacher_Prediction_B -- Logits Distillation Loss (L_logits_B) --> Total_Loss_B Teacher_Features_B -- Feature Alignment Loss (L_feature_B) --> Total_Loss_B Partial_Labels_B[Partial Labels y_p] -- Prediction Loss (L_pred_B) --> Total_Loss_B Total_Loss_B -- Backpropagation -->…
Please provide me with the output image of both of them . below are the diagrams code
I have two diagram :
first diagram code
graph LR subgraph Teacher Model (Pretrained) Input_Teacher[Input C (Complete Data)] --> Teacher_Encoder[Transformer Encoder T] Teacher_Encoder --> Teacher_Prediction[Teacher Prediction y_T] Teacher_Encoder --> Teacher_Features[Internal Features F_T] end subgraph Student_A_Model[Student Model A (Handles Missing Values)] Input_Student_A[Input M (Data with Missing Values)] --> Student_A_Encoder[Transformer Encoder E_A] Student_A_Encoder --> Student_A_Prediction[Student A Prediction y_A] Student_A_Encoder --> Student_A_Features[Student A Features F_A] end subgraph Knowledge_Distillation_A [Knowledge Distillation (Student A)] Teacher_Prediction -- Logits Distillation Loss (L_logits_A) --> Total_Loss_A Teacher_Features -- Feature Alignment Loss (L_feature_A) --> Total_Loss_A Ground_Truth_A[Ground Truth y_gt] -- Prediction Loss (L_pred_A)…
I'm reposting my question again please make sure to avoid any copy paste from the previous answer because those answer did not satisfy or responded to the need that's why I'm asking again
The knowledge distillation part is not very clear in the diagram. Please create two new diagrams by separating the two student models:
First Diagram (Student A - Missing Values):
Clearly illustrate the student training process.
Show how knowledge distillation happens between the teacher and Student A.
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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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