Suppose we have 212 bytes of virtual memory and 27 bytes of physical main memory. Suppose the page size is 24 bytes.
a) How many pages are there in virtual memory?
b) How many page frames are there in main memory?
c) How many entries are in the page table for a process that uses all of virtual memory?
Virtual memory = 212 bytes
Physical memory = 27 bytes
Page size = 24 bytes
(a) No of pages in virtual memory = 212/24 = 8. I think virtual memory should be the exact multiple of page size.
(b) No of page frames = 27/24 = 1
(c) There is only one page frame in main memory. If a process uses all of the virtual memory, the process will be using 8 pages in virtual memory. So, there will be eight entries in its page table.
Suppose we have 212 bytes of virtual memory and 27 bytes of physical main memory. Suppose the pag...
Suppose we have 2^20 bytes of virtual memory and 216 bytes of physical main memory. Suppose the page size is 2^8 bytes. a) How many pages are there in virtual memory? b) How many page frames are there in main memory? c) How many entries are in the page table for a process that uses all of virtual memory?
Exercise l: Suppose that we have a virtual memory space of 28 bytes for a given process and physical memory of 4 page frames. There is no cache. Suppose that pages are 32 bytes in length. 1) How many bits the virtual address contain? How many bits the physical address contain? bs Suppose now that some pages from the process have been brought into main memory as shown in the following figure: Virtual memory Physical memory Page table Frame #...
A computer uses a byte-addressable virtual memory system with a four-entry TLB and a page table for a process P. Pages are 16 bytes in size. Main memory contains 8 frames and the page table contains 16 entries. a. How many bits are required for a virtual address? b. How many bits are required for a physical address?
Q1. Suppose we have a virtual memory of size 2 Terabytes (2048GB, or 241 bytes), where pages are 8KB (213 bytes) each, and the machine has 4GB (232 bytes) of physical memory. a) Compute the number of page table entries needed if all the pages are being used. b) Compute the size of the page table if each page table entry also required 4 additional bits (valid, protection, dirty, use). Q2. For this problem, you are given a...
Virtual memory address translation: a) Consider a machine with a physical memory of 8 GB, a page size of 4 KB, and a page table entry size of 4 bytes. How many levels of page tables would be required to map a 52-bit virtual address space if every page table fits into a single page? b) Without a cache or TLB, how many memory operations are required to read or write a page in physical memory? c) How much physical...
Suppose you have a byte-addressable virtual address memory system with 8 virtual pages of 64 bytes each, and 4-page frames. Assuming the following page table, answer the questions below: Page #Frame #Valid Bit0111312-03014215-06-07-0a) How many bits are in a virtual address? b) How many bits are in a physical address? c) What physical address corresponds to the following virtual addresses (if the address causes a page fault, simply indicate this is the case)? 1) Ox00 2) 0x44 3) OxC2 4) 0x80
The physical memory is split into physical pages (aka page frames), which have equal size. We know that the number of page frames in the physical memory equals the number of bytes in a page frame. We also know that physical addresses have 12 bits. How many bytes are in a page frame?
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...
A certain computer provides its users with a virtual memory space of 2^32 bytes. The computer has 2^18 bytes of physical memory. The virtual memory is implemented by paging, and the page size is 4,096 bytes. A user process generates the virtual address 11123456 hexadecimal. a. How many entries are there in the page table? b. Explain how the system establishes the corresponding physical location.
You have a processor that supports virtual memory. The page size used for the virtual memory is 16 KiBytes, The virtual address size is 24 bits. Each table entry is 2 bytes. How many pages will the processor support? ___ How big is the page table (assume a single level page table) ___