Question

In a one processor system, there is an interrupt clock which is set to a “time slice Q”, that is, every Q an interrupt occurs to stop the process. There is only one I/O device in the system    which is interrupted when the process needs an I/O. There are 3 processes A, B, C.

    

     Process A executes an I/O interrupt every T units of time.

     Process B executes an I/O interrupt every 2T units of time.

     Process C executes an I/O interrupt every 3T units of time.

(a) Which of the above processes will benefit the most and which is not when Q = T.

      Explain your answer

(b) Which of the above processes will benefit the most and which is not when Q = 2T.

      Explain your answer

(a) Which of the above processes will benefit the most and which is not when Q = 3T.

      Explain your answer

What is the best value should Q takes for good performance ? explain your answer in full.

Answer 1

Ans for the given problem

i) Time-sharing:- Time sharing is a type of operating system, it enables multiple user who are located at the various terminals, to use the same computer system at the same time. In a time sharing system the main objective is to minimize the response time. The time- sharing systems were built and developed to provide an interactive use of the computer system. It uses CPU scheduling and multi - programming to provide each user with a small portion of a time - shared computers.

II) a) process which will benefit the most and which is not when Q = T :-

The Process C will benefit the most for Q = T as it is already defined above that Process A executes an I/O interrupt every T units of time. Assuming, for 1 sec second it states that for every one second Process A executes an I/O interrupt and hence, it will interrupts the CPU when new data has arrived and is ready to be retrieved by the system processor. and so Process A having Q = T will interrupt the least. Since, excessive amounts of processing time will be spent handling interrupts

b) process which will benefit the most and which is not when Q = 2T :-

Process C will benefit the most in this case as it will produce less interruption in the I/Odevice for every one second. Whereas, Process A will be the least beneficial .

a) process which will benefit the most and which is not when Q = 3T :-

Process C will be the most beneficial in thiss case having the same amount of interruption for each second whereas Process A will be least beneficial.

Hence, Process C will be the best among all for Q to get good performance as Interrupts provide low overhead and good latency at low load, but degrade significantly at high interrupt rate. excessive amounts of processing time spent in handling interrupts can cause interruption storm and hinder the performace of the system.

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