Consider a page size of 4 bytes and a physical memory of 32
bytes (8 pages). If logical page 4 maps to physical page 20, then
logical address 18 (logical page 4, offset 2) maps to physical
address
a) 80
b) 84
c) 82
d) 18
ANSWER:
GIVEN THAT:
FOR CALCULATING THE PHYSICAL ADDRESS THE FORMULA IS AS FOLLOWS:
physical_address=(physical_page_number*page-size)+offset
= 20*4+2
= 82
SO THE ANSWER IS OPTION IS 'C' CORRECT ANSWER
Consider a page size of 4 bytes and a physical memory of 32 bytes (8 pages)....
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 #...
Consider a simple paging system with the following parameters: 232 bytes of physical memory; page size of 210 bytes; 216 pages of logical address space. How many entries in the page table? How many bits in each page table entry? Assume each page table entry contains a valid/invalid bit.
1. Consider a simple paging system with the following parameters: 232 bytes of physical memory; page size of 210 bytes; 216 pages of logical address space. How many bits are in a logical address? How many bytes are in a frame! How many bits in the physical address specify the frame? How many entries are in the page table? How many bits are in each page table entry? Assume each page table entry contains a valid/invalid bit. 2. Consider a...
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...
A simple paging system has a memory size of 256 bytes and a page size of 16 bytes. i. What is the size of the page table? ii. How many bits exist for an address, assuming 1-byte incremental addressing? iii. State p and d values (i.e. the page number and the offset). iv. Perform address translation of 64 bytes to physical address space using the page table below. 0 8 1 6 2 3 3 11 4 7
A simple paging system has a memory size of 256 bytes and a page size of 16 bytes. i. What is the size of the page table? ii. How many bits exist for an address, assuming 1-byte incremental addressing? iii. State p and d values (i.e. the page number and the offset). iv. Perform address translation of 64 bytes to physical address space using the page table below. 0 8 1 6 2 3 3 11 4 7
2" | Page # (10 pts)On a machine with 8 bytes page size, given the page table for a process, and 2 of these 4 entries are mapped to page frames. Frame 0 starts at physical 0 address 0. (All numbers given are in decimal.) a. What are the range(s) of logical address (decimal, byte-level) that would cause 2 a page fault? | Frame # 4 Not in main memory Not in main memory b. Convert the following logical address...
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...
Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory of 64 frames. a. How many bits are required in the logical address? b. How many bits are required in the physical address?
NAME: The total number of pages that make up the executable program is shown (no fragmentation) The total size of a frame is equal to 1024 bytes The CPU accesses the next line of code to be executed The first four bits are the op code The size of a word in the system is 32 bits (16 bits) total program logical address pages page table 0 13 3 4 15 12 6 19 A. What is the logical page...