Operating Systems | Set 3 - GeeksforGeeks

17 Jun, 2021
Following questions have been asked in GATE CS exam. 
1. Suppose the time to service a page fault is on the average 10 milliseconds, while a memory access takes 1 microsecond. Then a 99.99% hit ratio results in average memory access time of (GATE CS 2000) (a) 1.9999 milliseconds (b) 1 millisecond (c) 9.999 microseconds (d) 1.9999 microseconds 
Answer: (d) Explanation:  
Average memory access time =
      [(% of page miss)*(time to service a page fault) +
                  (% of page hit)*(memory access time)]/100
So, average memory access time in microseconds is. (0.01*10*1000 + 99.99*1)/100 = (100+99.99)/100 = 199.99/100 =1.9999 μs 
2. Which of the following need not necessarily be saved on a context switch between processes? (GATE CS 2000) (a) General purpose registers (b) Translation look-aside buffer (c) Program counter (d) All of the above 
Answer: (b) Explanation: In a process context switch, the state of the first process must be saved somehow, so that, when the scheduler gets back to the execution of the first process, it can restore this state and continue. 
The state of the process includes all the registers that the process may be using, especially the program counter, plus any other operating system specific data that may be necessary. 
A Translation lookaside buffer (TLB) is a CPU cache that memory management hardware uses to improve virtual address translation speed. A TLB has a fixed number of slots that contain page table entries, which map virtual addresses to physical addresses. On a context switch, some TLB entries can become invalid, since the virtual-to-physical mapping is different. The simplest strategy to deal with this is to completely flush the TLB. References: http://en.wikipedia.org/wiki/Context_switch http://en.wikipedia.org/wiki/Translation_lookaside_buffer#Context_switch 
3. Where does the swap space reside ? (GATE 2001) (a) RAM (b) Disk (c) ROM (d) On-chip cache Answer: (b) Explanation: Swap space is an area on disk that temporarily holds a process memory image. When physical memory demand is sufficiently low, process memory images are brought back into physical memory from the swap area. Having sufficient swap space enables the system to keep some physical memory free at all times. References: http://docs.hp.com/en/B2355-90672/ch06s02.html 
4. Which of the following does not interrupt a running process? (GATE CS 2001) (a) A device (b) Timer (c) Scheduler process (d) Power failure 
Answer: (c) Explanation: Scheduler process doesn’t interrupt any process, it’s Job is to select the processes for following three purposes. Long-term scheduler(or job scheduler) –selects which processes should be brought into the ready queue Short-term scheduler(or CPU scheduler) –selects which process should be executed next and allocates CPU. Mid-term Scheduler (Swapper)- present in all systems with virtual memory, temporarily removes processes from main memory and places them on secondary memory (such as a disk drive) or vice versa. The mid-term scheduler may decide to swap out a process which has not been active for some time, or a process which has a low priority, or a process which is page faulting frequently, or a process which is taking up a large amount of memory in order to free up main memory for other processes, swapping the process back in later when more memory is available, or when the process has been unblocked and is no longer waiting for a resource. 
5. Which of the following scheduling algorithms is non-preemptive? (GATE CS 2002) 
a) Round Robin b) First-In First-Out c) Multilevel Queue Scheduling d) Multilevel Queue Scheduling with Feedback 
Answer: (b) 
mohit17067
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Operating Systems
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