GATE-1995
1. Which scheduling policy is
most suitable for a time-shared operating systems?
(a) Shortest Job First
(b)
Round Robin
(c) First Come First Server
(d) Elevator
Ans: option-B
Explanation:-
The round robin scheduling gives each
process a fair time for processing.
GATE-1995
GATE-1995
Which of the following page replacement algorithms suffers from Belady’s
anamoly?
(a) Optimal replacement (b) LRU (c) FIFO (d)
Both (a) and (c)
Ans:-Option-C
Explanation:-
Belady’s anamoly:-
It is a phenomenon
in which when increasing number of page frame number increasing page faults. In the FIFO manner when we increase the page frame
the page fault also increase.
GATE-1995
The principle of locality justifies the use of
(a) interrupts (b) DMA
(c) Polling (d) Cache Memory
Ans:- Option-D
Explanation:-
Principle of Locality
Locality means we can predict with some reason that
which instructions or data a process will use in near future based on the
accesses in the past.
GATE-1995
GATE-1995
In a paged segmented scheme of memory
management, the segment table itself must have a page table because
(a) the segment table is often too large to fit in
one page
(b) each segment is spread over a number of pages
(c) segment tables point to page table and not to
the physical
(d) the processor’s description base register
points to a page table
Ans:-Option-B
GATE-1996
1. A critical section is a
program segment
(a) which should run in a certain
specified amount of time
(b) which avoids deadlocks
(c) where shared resources are
accessed
(d) which must be enclosed by a
pair of semaphore operations, P and V
Ans: option-C
GATE-1996
2. A
solution to the Dining Philosophers Problem which avoids deadlock is
(a)
ensure that all philosophers pick up the left fork before the right fork
(b)
ensure that all philosophers pick up the right fork before the left fork
(c)
ensure that one particular philosopher picks up the left fork before the right
fork, and that all other philosophers pick up the right fork before the left
fork
(d)
None of the above
Ans:
option-C
Explanation:-
GATE-1998
1. A counting semaphore was initialized to 10. Then
6 P (wait) operations and 4V (signal) operations were completed on this
semaphore. The resulting value of the semaphore is
(a) 0 (b) 8 (c) 10
(d) 12
Ans: option-B
Explanation:-
The wait operation (P) decrements the semaphore
value.
The Signal operation (v) increments the semaphore value.
6P => decrements the semaphore 6 times. So the
value becomes 4.
4V => increments the semaphore 4 times. So the
value becomes 8.
GATE-1998
1. Which of the following is true?
(a) Unless enabled, a CPU will not be able to process interrupts.
(b) Loop instructions cannot be interrupted till they complete.
(c) A processor checks for interrupts before executing a new instruction.
(d) Only level triggered interrupts are possible on microprocessors
Ans: option-A
Explanation:
Interrupt:-
An interrupt is a signal to
the processor emitted
by hardware or software indicating an event that needs immediate attention. An
interrupt alerts the processor to a high-priority condition requiring the
interruption of the current code the processor is executing (the current thread).
The processor responds by suspending its current activities, saving its state, and executing a small program called an interrupt
handler (or interrupt service routine, ISR) to deal with the event.
Interrupts are unexpected events in a sequence of execution of instructions causing an interruption of the normal program flow. Read more
Interrupts are unexpected events in a sequence of execution of instructions causing an interruption of the normal program flow. Read more
Option(b) is false.
Option (c) depends upon the architecture of the processor.
Option (d) is false because we have level-triggered interrupts and edge-triggered interrupts.
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