1
GATE CSE 2003
MCQ (Single Correct Answer)
+2
-0.6
A processor uses $$2$$-level page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both $$32$$ bits wide. The memory is byte addressable. For virtual to physical address translation, the $$10$$ most significant bits of the virtual address are used as index into the first level page table while the next $$10$$ bits are used as index into the second level page table. The $$12$$ least significant bits of the virtual address are used as offset within thepage. Assume that the page table entries in both levels of page tables are $$4$$ bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of $$96$$%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of $$90$$%. Main memory access time is $$10$$ ns, cache access time is $$1$$ ns, and TLB access time is also $$1$$ ns.

Suppose a process has only the following pages in its virtual address space: two contiguous code pages starting at virtual address $$0 \times 00000000,$$ two contiguous data pages starting at virtual address $$0 \times 00400000,$$ and a stack page starting at virtual address $$0 \times FFFFF000.$$ The amount of memory required for storing the page tables of this process is

A
$$8$$ KB
B
$$12$$ KB
C
$$16$$ KB
D
$$20$$ KB
2
GATE CSE 2002
MCQ (Single Correct Answer)
+2
-0.6
Dynamic linking can cause security concerns because
A
Security is dynamic
B
The path for searching dynamic libraries is not known till runtime
C
Linking is insecure
D
Cryptographic procedures are not available for dynamic linking
3
GATE CSE 2001
MCQ (Single Correct Answer)
+2
-0.6
Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4KB, what is the approximate size of the page table?
A
16 M
B
8 MB
C
2 MB
D
24 MB
4
GATE CSE 2000
MCQ (Single Correct Answer)
+2
-0.6
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
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