GATE CSE 2021 | Set 2 | Question: 14

Question

Which of the following statement(s) is/are correct in the context of $\text{CPU}$ scheduling?

• A. Turnaround time includes waiting time
• B. The goal is to only maximize $\text{CPU}$ utilization and minimize throughput
• C. Round-robin policy can be used even when the $\text{CPU}$ time required by each of the processes is not known apriori
• D. Implementing preemptive scheduling needs hardware support

Comments

Self note:

Turnaround time is the time the process spends from it's arrival to ready queue until it's termination
According to this definition, the process can spend it's lifecycle in Running state, Blocked(Waiting for I/O), Ready state(Waiting to get CPU chance again after some I/O operation has been completed for that process)

So, it will include times such as Running + Waiting + I/O operation,  till the termination of the process

Formulas:

1. Turnaround Time = Completion Time - Arrival Time
Completion Time includes --> Running Time + Waiting Time + Time for Process in I/O

So, Turnaround Time = (Running time + Waiting Time + Time for I/O) - Arrival Time

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I think even if the option B gave maximize throughput it would be wrong ,because the goal is not only this two ,its also to min waiting time ,min tat time ,min response time,ensure fairness,reduce starvation etc @Arjun sir pls verify

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narainkaarthik

Obviously, the option says "only"...There are several criteria's like you mentioned which the short time scheduler has to consider during the thread scheduling

Answer 1

• A. Turnaround time includes waiting time
  • TRUE. $\text{Turnaround Time} = \text{Waiting Time} + \text{Burst Time}$
• B. The goal is to only maximize CPU utilization and minimize throughput
  • FALSE. CPU scheduling must aim to maximize CPU utilization as well as throughput. Throughput of CPU scheduling is defined as the number of processes completed in unit time. $\textsf{SJF}$ scheduling gives the highest throughput. 
• C. Round-robin policy can be used even when the CPU time required by each of the processes is not known apriori
  • TRUE. Round-robin scheduling gives a fixed time quantum to each process and for this there is no requirement to know the CPU time of the process apriori (which is not the case say for shortest remaining time first).
• D. Implementing preemptive scheduling needs hardware support
  • TRUE. Preemptive scheduling needs hardware support to manage context switch which includes saving the execution state of the current process and then loading the next process.

Correct Answer: A;C;D

Reference: Stanford Notes

Comments

@Thadymademe. In xv6 it requires hardware support for pushing minimal context but I don't know how linux does it can you explain it in short?

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for optn B can we think of like due to thrashing the cpu utlization is high but since none or less process are completed the throughput is less/minimum?@Arjun @Deepak Poonia sir please clarify.

 

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may be it is useful, 

source: Operating System Concepts" by Silberschatz, Galvin, and Gagne

Answer 2

Ans. A,C and D
A. turnaround time =completion time(burst+waiting)-arrival time.
B. The goal is to maximize throughput not to minimize it.
C. Round Robin is used in modern day systems as it doesn’t requires the information about the CPU time for processes rather it works just by providing same time quanta to every process.
D. Hardware support is essential for preemption.

Comments

Busy waiting

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Hardware support is required for preemptive scheduling. 

Here, hardware support means generating periodic cycles that trigger an interrupt to the OS scheduler. The OS scheduler (or the dispatcher, same thing) checks whether enough time quanta has passed or not. If the time quanta of the current execution process is over, it triggers a context switch and brings another process from the $\texttt{READY QUEUE}$.

Without periodic interrupts from the hardware, there is no way of doing preemptive scheduling. An application can just keep executing without ever giving control back to the kernel.

In the case of non-preemptive scheduling, no interrupts are required as the application currently executes calls $\texttt{sched_yield()}$ when it is done with its operation. 

The StackOverflow link people are posting talks about HOW the context switch is done, like who saves the context of the current process (register values, stack pointer, etc.). There are two ways of doing that: the hardware saves the context, or the OS does. Both are valid and can be selected based on what is required.

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Correct @sandeep007734.

Answer 3

Option A is correct because Turnaround Time is the total time a process spends in the system from start to finish. It is calculated as $\text{Turnaround Time} = \text{Burst Time} + \text{Waiting Time}$, so it naturally includes the waiting time.

Option B is incorrect because a good operating system wants to get as much work done as possible. Therefore, the goal is to maximize throughput (finish more processes), not minimize it.

Option C is correct because Round-Robin doesn't care how long a process will take. It simply gives every process a fixed slice of time (time quantum) and moves to the next one, making it perfect for systems where we don't know the execution times in advance.

Option D is correct because to stop a running process forcefully (preemption), the system needs a built-in hardware timer. This timer counts down and triggers a physical hardware interrupt to hand control back to the OS.

Correct Options: A, C, and D