1. Assume that your machine uses 16-bit virtual address and 14-bit physical address with 4KB page size. To get full credit, show your work.
a. Show bits of the virtual address and physical address.
b. What is the maximum program size that can be run in this machine?
c. What is the maximum physical memory that this machine can have?
d. Show the numbers of virtual pages and page frames.
e. What is the table size if a page entry size is 8 bytes?
f. List fragmentation types and explain them. Which fragmentation can be seen in this machine?
1. Assume that your machine uses 16-bit virtual address and 14-bit physical address with 4KB page...
Assume you have: 32-bit addresses, 4KB Page size, 4MB Physical Memory Space, 4KB Cache with 4-way set associative and LRU replacement, 32 Byte Cache block size, 4-entry fully associative TLB. A program to be run on this machine begins as follows: double A[1024]; int i, j; double sum = 0; for( i = 0; i < 1024; i++ ) // first loop A[i] = i; for( j = 0; j < 1024; j += 16 ) // second loop ...
page addressing (a) The following are virtual 16 Considered bit-addresses, with the upper 8th Bit the number of the page and the lower one 8th Bit represent the offset. The physical memory included in this example 256 Page frames ( frames ) to each 256 Bytes per page, ie physical addresses from 0x0000 to 0xFFFF, where frame 0 at 0x0000 starts. Given the following page table for a process that 10 Pages (page numbers 0 - 9), some of which...
Consider a logical address space of 8 pages; each page is 2048 byte long, mapped onto a physical memory of 64 frames.(i) How many bits are there in the logical address and how many bits are there in the physical address?(ii) A 6284 bytes program is to be loaded in some of the available frames ={10,8,40,25,3, 15,56,18,12,35} . Show the contents of the program's page table.(iii) What is the size of the internal fragmentation?(iv) Convert the following logical addresses 2249...
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...
1. Address Layouts Our OS uses 32-bit virtual addresses, 24-bit physical addresses, and page sizes of 4KB. Draw the layout of both the virtual and physical addresses along with the size in bits. Please just do the drawing, that's all that is needed
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 #...
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...
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 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?
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