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


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15 Questions MCQ Test Question Bank for GATE Computer Science Engineering - Test: Memory Management & Virtual Memory- 3

Test: Memory Management & Virtual Memory- 3 for Computer Science Engineering (CSE) 2024 is part of Question Bank for GATE Computer Science Engineering preparation. The Test: Memory Management & Virtual Memory- 3 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Memory Management & Virtual Memory- 3 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Memory Management & Virtual Memory- 3 below.
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Test: Memory Management & Virtual Memory- 3 - Question 1

If the property of locality of reference is well pronounced a program:
1. The number of page faults will be more.
2. The number of page faults will be less.
3. The number of page faults will remain the same.
4. Execution will be faster.

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 1

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.

Test: Memory Management & Virtual Memory- 3 - Question 2

Memory protection is of no use in a

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 2

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

Test: Memory Management & Virtual Memory- 3 - Question 3

In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 3

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

Test: Memory Management & Virtual Memory- 3 - Question 4

The size of the virtual memory depends on the size of the

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 4

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.

Test: Memory Management & Virtual Memory- 3 - Question 5

To obtain better memory utilization, dynamic loading is used. With dynamic loading, a routine is not loaded until is called for. For implementing dynamic loading.

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 5

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.

Test: Memory Management & Virtual Memory- 3 - Question 6

Which one of the following is true?

Test: Memory Management & Virtual Memory- 3 - Question 7

Which of the following is true?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 7

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.

Test: Memory Management & Virtual Memory- 3 - Question 8

Consider a computer with 8 Mbytes of main memory and a 128 K cache. The cache block size is 4 K. It uses a direct mapping scheme for cache management. How many different main memory blocks can map onto a given physical cache block?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 8

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

Test: Memory Management & Virtual Memory- 3 - Question 9

Consider a system using paging and segmentation. The virtual address space consist of up to 8 segments and each segment is 229 bytes long. The hardware pages each segment into 28 byte pages.

How many bits in the virtual address specify the offset within page?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 9

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

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

Test: Memory Management & Virtual Memory- 3 - Question 10

Consider a system using paging and segmentation. The virtual address space consist of up to 8 segments and each segment is 229 bytes long. The hardware pages each segment into 28 byte pages.

How many bits in the virtual address specify the entire virtual address?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 10

Test: Memory Management & Virtual Memory- 3 - Question 11

For 64 bit virtual addresses, a 4 KB page size and 256 MB of RAM, an inverted page table requires

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 11

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

Test: Memory Management & Virtual Memory- 3 - Question 12

If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p.  

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 12

[(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

Test: Memory Management & Virtual Memory- 3 - Question 13

Consider a system with 1 K pages and 512 frames and each page is of size 2 KB. How many bits are required to represent the virtual address space Memory is

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 13

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.

Test: Memory Management & Virtual Memory- 3 - Question 14

Consider a virtual memory of 256 terabytes. The page size is 4K. This logical space is mapped into a physical memory of 256 megabytes.

How many bits are there in the virtual memory?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 14

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

Test: Memory Management & Virtual Memory- 3 - Question 15

Consider a virtual memory of 256 terabytes. The page size is 4K. This logical space is mapped into a physical memory of 256 megabytes.

How many pages are there in the virtual memory?

Detailed Solution for Test: Memory Management & Virtual Memory- 3 - Question 15

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

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