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Which of the follwings are true? (There might be multiple correct answers) Compare and swap is a hardware based solution correct correct ]Basic Compare and swap ensures bounded waiting Test and set is a hardware solution that can be extended to a more complex alternative ]Implementing semaphore is more sophisticated than implementing test and set

Which of the follwings are true? (There might be multiple correct answers) Compare and swap is a hardware based solution correct correct ]Basic Compare and swap ensures bounded waiting Test and set is a hardware solution that can be extended to a more complex alternative ]Implementing semaphore is more sophisticated than implementing test and set

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Which of the follwings are true? (There might be multiple carrect answers) Compare and swap is a hardware based solution correct correct Basic Compare and swap ensures bounded waiting |Test and set is a hardware solution that can be extended to a more complex alternative |Implementing semaphore is more sophisticated than implementing test and set Which of the follwings are true? (There might be multiple correct answers) |In race condition multiple processes can execute shared information concurrently. Process synchronization ensures data consistency. Race condition does not create data inconsistency in a shared region. Semaphore is only used to solve critical section problem.
Transcribed Image Text:Which of the follwings are true? (There might be multiple carrect answers) Compare and swap is a hardware based solution correct correct Basic Compare and swap ensures bounded waiting |Test and set is a hardware solution that can be extended to a more complex alternative |Implementing semaphore is more sophisticated than implementing test and set Which of the follwings are true? (There might be multiple correct answers) |In race condition multiple processes can execute shared information concurrently. Process synchronization ensures data consistency. Race condition does not create data inconsistency in a shared region. Semaphore is only used to solve critical section problem.
Suppose the 2 processes in peterson's solution are numbered 0 and 1. Which of the following is a valid code statement in the entry section of process 1? O flag[1] = false O turn = 0 O flag[0] = false turn = 1 # Define Flag[1]=TRUE; # Define Flag[0]=FALSE; # Define turn=1; # We have 4 line of codes in critical section. # Each line of the peterson solutions and critical section requires 2sec to execute. Now if process 1 try to execute and enter in critical section after executing the entry section. what will be the situation if context switch happens after 8sec? O Process PO might enter in the critical section. Process P1 will terminate and will not execute the reminder section. O Contex switch can not take place after 8sec of execution. Process PO never can enter the critical section. Numerical Input Suppose counting semaphore S is started from the value of 15. After that 10 times call the wait S operations and 10 times call the signal S. What is the final value of S?
Transcribed Image Text:Suppose the 2 processes in peterson's solution are numbered 0 and 1. Which of the following is a valid code statement in the entry section of process 1? O flag[1] = false O turn = 0 O flag[0] = false turn = 1 # Define Flag[1]=TRUE; # Define Flag[0]=FALSE; # Define turn=1; # We have 4 line of codes in critical section. # Each line of the peterson solutions and critical section requires 2sec to execute. Now if process 1 try to execute and enter in critical section after executing the entry section. what will be the situation if context switch happens after 8sec? O Process PO might enter in the critical section. Process P1 will terminate and will not execute the reminder section. O Contex switch can not take place after 8sec of execution. Process PO never can enter the critical section. Numerical Input Suppose counting semaphore S is started from the value of 15. After that 10 times call the wait S operations and 10 times call the signal S. What is the final value of S?
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