Test: Memory Management & Virtual Memory- 3 - Computer Science Engineering (CSE) MCQ

If the property of locality of reference is well pronounced then the page to be accessed will be found in the memory more likely and hence page faults will be less. Also since access time will be only of the upper level in the memory hierarchy bence execution will be faster.

Even in a non-multiprogramming system, memory protection may be used, for example, spooling is being used.

Hit ratio = 0.35
Time (secondary memory) = 100 ns
T(main memory) = 10 ns
Average access time = h(T_m) + (1 - h) (T_s)
= 0.35 x 10 +(0.65) x 100
= 3.5 + 65 
= 68.5 ns

The size of virtual memory depends on the size of the address bus. Processor generates the memory address as per the size of virtual memory.

In case of dynamic loading, memory management unit converts logical address to physical address at the time of execution.
In doing this user programs do not need any support from the O.S. or the hardware.

By using the overlays we can execute much greater processes simultaneously which cannot be execute and reside in the memory at the same time. In this the process to be executed process brought to memory only when it is needed at the time of execution.

Number of block in cash 
Number of block in main memory 
Since the system uses direct mapping hence different memory blocks that can map into the cache block.
= Number of blocks in MM% Number of blocks in cache

Virtual address space = 8 x 2²⁹ bytes
= 2³² bytes
No. of pages within each segment

Offset within the page = 8 bits (for 2⁸ bytes)

Virtul address space = 2⁶⁴ bytes
Page size = 4 KB
= 2¹² Bytes
Physical Memotry = 256 MB.
= 2²⁸ Bytes
Number of entries in inverted page table = Number of frames in physical memory

[(1- p)100 + p x 50 x 10⁶] ns
Since, 1 ns = 10^-9sec
=> 10^-9[(1 -p)100 + p x 50 x 10⁶] sec
=> [(1 - p) x 10^-7 + p x 50 x 10^-3] sec
=> 10^-7 - 10^-7p + p x 50 x 10^-3
=> 10^-7 + l0^-3 x p f [50 - 10^-4]
~ [10^-7 + p x 10^-3 x 49.9] sec

Virtual address space consists of pages.
Given that,
Number of pages = 1 K = 2¹⁰
 Page size = 2KB = 2¹¹B
Hence virtual address space
= 2¹¹ x2¹⁰
= 2²¹ B
∴  21 bits are required to represent the virtual address space.

Virtual address space = 256 terabytes
= 256 x 2⁴⁰ bytes
= 2⁴⁸ bytes [1 terabyte = 2⁴⁰ bytes] ∴ Number of bits to represent the virtual memory
= 48 bits

Page size = 4 K
= 2¹² Bytes
Number of pages in virtual memory 
=> 2⁶.2³⁰ pages = 64 giga pages.