A computer has 32-bit virtual addresses and the page size is 2^3 KB. Suppose the text, data, and stack segments of a process need 5 page(s) each. If the computer uses two-level page tables, with 3 bits for the second-level index, how many page table entries (PTEs) are in each second-level page table?
Here the page size = 2^3KB => Page offset = log 2^3KB = 13 bits.
Therefore page number bits = 32 - 13 = 19 bits.
Here we are using 2 level page table.
These 19 bit page number will be divided into 2 level page table.
But in the question, total # of page occupied is 15.
Therefore page number bits = ceil ( log (15)) = 4
Here it's mentioned that computer is using 3 bits for second level index.
Therefore # of bits for 1st level page table = 4 - 3 = 1
Thus the bits distribution is :
First Level Page Table (1) |
Second Level Page Table (3 ) |
Page Offset (13 bits) |
Now # of bits for second level page table is 3.
This means # of page table entries in each second level page table is 2^3 = 8.
If you have any questions comment down and please? upvote thanks...
A computer has 32-bit virtual addresses and the page size is 2^3 KB. Suppose the text,...
The RISC-V 32-bit architecture supports virtual memory with 32-bit virtual addresses mapping to 32-bit physical addresses. The page size is 4Kbytes, and page table entries (PTEs) are 4 bytes each. Translation is performed using a 2-level page table structure. Bits 31:22 of a virtual address index the first-level page table. If the selected first-level PTE is valid, it points to a second-level page table. Bits 21:12 of the virtual address then index that second-level page table. If the selected second-level...
it’s from operating system. please provide me correct answer. A computer has 32-bit virtual addresses and 4-KB pages. The program and data together fit in the lowest page (0-4095) The stack fits in the highest page. How many entries are needed in the page table if traditional (one-level) paging is used? How many page table entries are needed for two-level paging, with 10 bits in each part? Submission: Upload a document with your calculations and results
computer architecture Virtual memory 4 physice memar Assume a 64 bit machine with 40 bit addresses, and 16GB of actual memory. Memory blocks are 4K 1. How many bits in the virtual memory INDEX and OFFSET fields? 2. What fields do you need in a page table entry, and how many bits are needed for each? How many bytes do you need for each page table entry? (each entry is allocated in a whole number of bytes, even if there...
Consider a virtual memory system with the following properties: 36 bit virtual byte address, 8 KB pages size, and 32 bit physical byte address. Please explain how you determined your answer. a. What is the size of main memory for this system if all addressable frames are used? b. What is the total size of the page table for each process on this processor, assuming that the valid, protection, dirty, and use bits take a total of 4 bits and...
Consider a computer system with a 32-bit logical address and 8-KB page size. The system supports up to 1GB of physical memory. How many entries are there in a page table?
A computer system has a 36-bit virtual address space with a page size of 8K, and 4 bytes per page table entry. How many pages are in the virtual address space? What is the maximum size of addressable physical memory in this system? If the average process size is 8GB, would you use a one-level, two-level, or three-level page table? Why? Compute the average size of a page table in part c above
A certain byte-addressable computer system has 32-bit words, a virtual address space of 4GB, and a physical address space of 1GB. The page size for this system is 4 KB. Assume each entry in the page table is rounded up to 4 bytes. a) Compute the size of the page table in bytes. b) Assume this virtual memory system is implemented with a 4-way set associative TLB (Translation Lookaside Buffer) with a total of 256 address translations. Compute the size...
17. A computer system implements a paged virtual memory system. Assume a 16-bit virtual address space and a 24-bit physical address space. Assume that the first 6 bits of a virtual address index the page table and the rest of the bits are the page offset. A process has the following indexed page table. Index Page Table Entry (PTE) 0x3800 0x3600 0x3200 0x1000 2 3 Each page table entry qives a hexadecimal page frame addresses. Translate the following two hexadecimal...
Consider a pure paging system that uses 32-bit addresses (each of which specifies one byte of memory), contains 2 GB of main memory, and has a page size (ps) of 8 KB. 1. Given 32-bits for each PTE, how many total bytes of memory are required to store the page table?
4. Assume that we have a machine with the following memory specifications: Virtual addresses are 32 bits wide Physical addresses are 26 bits wide . Page size is 16 Kbytes 4/A) How many pages are in the virtual memory space? 4/B) How many page frames are in the physical memory space? 4/C) If each page table entry consists of a physical frame number, 1 present/absent bit Answers Pages .Page Frames and 1 dirty/clean bit (which shows if the page has...