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A Computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is _______________ bits
- a)36
- b)32
- c)28
- d)40
Correct answer is option 'A'. Can you explain this answer?
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A Computer system implements 8 kilobyte pages and a 32-bit physical ad...
Max size of virtual address can be calculated by calculating maximum number of page table entries.
Maximum Number of page table entries can be calculated using given maximum page table size and size of a page table entry.
Given maximum page table size = 24 MB
Let us calculate size of a page table entry.
A page table entry has following number of bits.
1 (valid bit) +
1 (dirty bit) +
3 (permission bits) +
x bits to store physical address space of a page.
Maximum Number of page table entries can be calculated using given maximum page table size and size of a page table entry.
Given maximum page table size = 24 MB
Let us calculate size of a page table entry.
A page table entry has following number of bits.
1 (valid bit) +
1 (dirty bit) +
3 (permission bits) +
x bits to store physical address space of a page.
Value of x = (Total bits in physical address) - (Total bits for addressing within a page)
Since size of a page is 8 kilobytes, total bits needed within a page is 13.
So value of x = 32 - 13 = 19
Since size of a page is 8 kilobytes, total bits needed within a page is 13.
So value of x = 32 - 13 = 19
Putting value of x, we get size of a page table entry = 1 + 1 + 3 + 19 = 24bits.
Number of page table entries
= (Page Table Size) / (An entry size)
= (24 megabytes / 24 bits)
= 223
Number of page table entries
= (Page Table Size) / (An entry size)
= (24 megabytes / 24 bits)
= 223
Vrtual address Size
= (Number of page table entries) * (Page Size)
= 223 * 8 kilobits
= 236
Therefore, length of virtual address space = 36
= (Number of page table entries) * (Page Size)
= 223 * 8 kilobits
= 236
Therefore, length of virtual address space = 36
Most Upvoted Answer
A Computer system implements 8 kilobyte pages and a 32-bit physical ad...
Given information:
- Page size = 8 kilobytes
- Physical address space = 32-bit
- Page table entry contains valid bit, dirty bit, three permission bits, and translation
- Maximum size of the page table of a process = 24 megabytes
To find:
The length of the virtual address supported by the system
Solution:
First, let's calculate the number of pages that can be supported in the physical address space.
1. Physical address space = 32-bit
2. Page size = 8 kilobytes = 8 * 1024 bytes
3. Number of pages = Physical address space / Page size
= 2^32 / (8 * 1024)
= 2^(32 - 3 - 10)
= 2^19
So, the physical address space can support 2^19 pages.
Next, let's calculate the maximum number of entries in the page table.
1. Maximum size of the page table of a process = 24 megabytes
2. Page size = 8 kilobytes = 8 * 1024 bytes
3. Maximum number of entries = Maximum size of the page table / Page size
= (24 * 1024 * 1024) / (8 * 1024)
= 3 * 1024 * 1024
So, the maximum number of entries in the page table is 3 * 1024 * 1024.
Since each page table entry contains a valid bit, dirty bit, three permission bits, and translation, the total number of bits required for each entry can be calculated as follows:
1. Number of bits required for valid bit = 1
2. Number of bits required for dirty bit = 1
3. Number of bits required for permission bits = 3
4. Number of bits required for translation = log2(Number of pages)
= log2(2^19)
= 19 bits
Total number of bits required for each entry = 1 + 1 + 3 + 19 = 24 bits.
Finally, let's calculate the length of the virtual address supported by the system.
1. Length of virtual address = Number of entries in the page table * Number of bits required for each entry
= 3 * 1024 * 1024 * 24
= 72 * 1024 * 1024
Since the length of the virtual address is 72 * 1024 * 1024 bits, it can be simplified as 72 megabits.
Therefore, the length of the virtual address supported by the system is 72 megabits, which is equivalent to 36 bits.
Hence, the correct answer is option A) 36.
- Page size = 8 kilobytes
- Physical address space = 32-bit
- Page table entry contains valid bit, dirty bit, three permission bits, and translation
- Maximum size of the page table of a process = 24 megabytes
To find:
The length of the virtual address supported by the system
Solution:
First, let's calculate the number of pages that can be supported in the physical address space.
1. Physical address space = 32-bit
2. Page size = 8 kilobytes = 8 * 1024 bytes
3. Number of pages = Physical address space / Page size
= 2^32 / (8 * 1024)
= 2^(32 - 3 - 10)
= 2^19
So, the physical address space can support 2^19 pages.
Next, let's calculate the maximum number of entries in the page table.
1. Maximum size of the page table of a process = 24 megabytes
2. Page size = 8 kilobytes = 8 * 1024 bytes
3. Maximum number of entries = Maximum size of the page table / Page size
= (24 * 1024 * 1024) / (8 * 1024)
= 3 * 1024 * 1024
So, the maximum number of entries in the page table is 3 * 1024 * 1024.
Since each page table entry contains a valid bit, dirty bit, three permission bits, and translation, the total number of bits required for each entry can be calculated as follows:
1. Number of bits required for valid bit = 1
2. Number of bits required for dirty bit = 1
3. Number of bits required for permission bits = 3
4. Number of bits required for translation = log2(Number of pages)
= log2(2^19)
= 19 bits
Total number of bits required for each entry = 1 + 1 + 3 + 19 = 24 bits.
Finally, let's calculate the length of the virtual address supported by the system.
1. Length of virtual address = Number of entries in the page table * Number of bits required for each entry
= 3 * 1024 * 1024 * 24
= 72 * 1024 * 1024
Since the length of the virtual address is 72 * 1024 * 1024 bits, it can be simplified as 72 megabits.
Therefore, the length of the virtual address supported by the system is 72 megabits, which is equivalent to 36 bits.
Hence, the correct answer is option A) 36.
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A Computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is _______________ bitsa)36b)32c)28d)40Correct answer is option 'A'. Can you explain this answer?
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