GATE-1998
Which of the following devices should get higher
priority in assigning interrupts?
(a) Hard disk (b) Printer
(c) Keyboard (d) Floppy disk
Ans:- Option-C
GATE-1999
1.
System calls are usually invoked by using
(a)a software interrupt (b) polling
(c) an indirect jump (d) a privileged instruction
Ans:- Option-A
GATE-1999
2.
A multi-user, multi-processing operating system cannot be implemented on
hardware that does not support
(a) Address-translation
(b) DMA for disk transfer
(c) At least two modes of CPU execution (privileged and non-privileged)
(d) Demand paging
Ans: Option A,B,C
GATE-1999
3. Which
of the following actions is/are typically not performed by the
operating system when switching context from process A to process
B?
(a) Saving current register values and restoring saved register values for process B.
(b) Changing address translation tables.
(c) Swapping out the memory image of process A to the disk.
(d) Invalidating the translation look-aside buffer.
Ans: Option C,D
Explanation:
When the context switch occur processor save the currently running data into register and then load new data of other process.
Swapping out the memory image of
process to the disk occurs only when the process is suspended.
GATE-2000
1. Which of the following need not necessarily be saved
on a context switch
between processes?
(a) General purpose registers (b) Translation
look-aside buffer
(c) Program counter (d) All of the above
Answer: Option-B
Explanation:
When a context switch
occur, running process data stored in PCB, so that, when the scheduler gets
back to the execution of the same process again, it can restore this state and
continue.
A TLB (Translation look-aside buffer) is a cache memory for
CPU it stores the process data which is frequently used by CPU.
GATE-2000
2. 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
(a) 1.9999 milliseconds (b) 1 millisecond
(c) 9.999 microseconds (d) 1.9999 microseconds
Ans:- Option-D
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.
(99.99*1 + 0.01*10*1000)/100 = (99.99+100)/1000 = 199.99/1000 =1.9999 µs
(99.99*1 + 0.01*10*1000)/100 = (99.99+100)/1000 = 199.99/1000 =1.9999 µs
GATE-2001
1. Which of the following statements is false?
(a) Virtual memory implements the translation of a
program’s address space into physical memory address space
(b) Virtual memory allows each program to exceed the
size of the primary memory
(c) Virtual memory increases the degree of
multiprogramming
(d) Virtual memory reduces the context switching
overhead.
Answer: Option-A
GATE-2001
2. Consider a set of n tasks with known run times r1, r2,
….rn to be run on a
uniprocessor machine. Which of the following processor
scheduling algorithms will result in the maximum throughput?
(a) Round-Robin (b) Shortest-Job-First
(c) Highest-Response-Ratio-Next (d)
First-Come-First-Served
Ans:- Option-B
Explanation:-
Throughput:-
Total number of tasks executed per unit time.
Shortest Job First has maximum
throughput because in this scheduling technique shortest jobs are executed
first hence maximum number of tasks are completed.
GATE-2001
3. Where does the swap space reside?
(a) RAM (b) Disk
(c) ROM (d) On-chip cache
Answer: Option-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.
Gate 1998:-
ReplyDeleteWhich of the following devices should get higher priority in assigning interrupts?
Hard disk
Printer
Keyboard
Floppy disk
Answer should be A-hard disk.
plz correct the answer bcoz it confuse others.....Keybord is lowest priority