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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  Consider a virtual memory of 256 terabytes. T... Start Learning for Free
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?
  • a)
    28
  • b)
    48
  • c)
    38
  • d)
    58
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider a virtual memory of 256 terabytes. The page size is 4K. This ...
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
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Most Upvoted Answer
Consider a virtual memory of 256 terabytes. The page size is 4K. This ...
Solution:

Virtual Memory

Virtual memory is a memory management technique that allows a computer to compensate for shortages of physical memory by temporarily transferring pages of data from random access memory (RAM) to disk storage. It allows a computer to run larger applications or multiple applications concurrently, and it can improve overall system performance.

Given,

Virtual memory = 256 terabytes
Page size = 4K

To calculate the number of bits in virtual memory, we need to find the number of pages in virtual memory and then multiply it by the page size.

Number of Pages in Virtual Memory

We know that the size of virtual memory is 256 terabytes, which is equivalent to 2^48 bytes.

Page size is 4K, which is equivalent to 2^12 bytes.

Therefore, the number of pages in virtual memory can be calculated as:

Number of pages = 2^48 / 2^12
Number of pages = 2^36 pages

Size of Virtual Memory

Now, we can calculate the size of virtual memory by multiplying the number of pages by the page size.

Size of virtual memory = Number of pages x Page size
Size of virtual memory = 2^36 x 2^12
Size of virtual memory = 2^48 bytes

Number of Bits in Virtual Memory

Finally, we can calculate the number of bits in virtual memory by taking the log base 2 of the size of virtual memory.

Number of bits = log2(2^48)
Number of bits = 48

Therefore, the number of bits in virtual memory is 48.

Answer: (B) 48
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Community Answer
Consider a virtual memory of 256 terabytes. The page size is 4K. This ...
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
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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 the page. 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 0x00000000, two contiguous data pages starting at virtual address 000400000, and a stack page starting at virtual address 0FFFFF000. The amount of memory required for storing the page tables of this process is

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 the page. 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. Assuming that no page faults occur, the average time taken to access a virtual address is approximately (to the nearest 0.5 ns)

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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?a)28b)48c)38d)58Correct answer is option 'B'. Can you explain this answer?
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