frame size = 512
page size = 512
(1)2345
page number = 2345/512 = 4
page offset = 2345 mod 512 = 297
Page-4 maps to frame number 2
Physical address = frame number * page size + page offset
Physical address = 2 * 512 + 297 = 1321
(2)1024
page number = 1024/512 = 2
page offset = 1024 mod 512 = 0
Page-2 maps to frame number 10
Physical address = frame number * page size + page offset
Physical address = 10 * 512 + 0 = 5120
(3)6780
page number = 6780/512 = 13
page offset = 6780 mod 512 = 124
There is no mapping for page-13 in page table.
It is a page fault
Consider a computer system that uses a paging system. The memory contains 16 frames, each frame...
Exercise 5 (2.5 points) Assume a memory management system built on paging, its physical memory has the total size of 4 GB. It placed over 16 KB pages. The limit of the logical address space for each process is 512 MB. 1. What is the total number of bits in the physical address? 2. What is the number of bits that specifies the page displacement? 3. Determine how many physical frames in the system. Explain the layout for the logical...
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...
Exercise 6.4.1: Parameters of paging and segmentation. A memory system employs both paging and segmentation: The logical address size is 32 bits. Page size is 512 words. The segment table contains 213 entries. (a) What is the size of w? (b) What is the maximum number of pages per segment?
Address Translation Question [8 points] Suppose a computing system uses paging with a logical address of 24 bits and a physical address of 32 bits. The page size is 4KB. Answer each of the following. If an answer is a power of 2, you can leave it in the form of a power of 2. ... 2. [20 points] Memory address translation and TLB performance [8 points] Suppose a computing system uses paging with a logical address of 24 bits...
Suppose a memory manager employs paging with page size of 4 Kbytes. It has a memory of 256 Mbytes. A process of size 25 Kbytes needs to be loaded into memory. Answer the following. (a) How many frames are there in the memory? (b) How many bits are necessary to represent the physical address as <frame#, offset>? (c) How many frames need to be allocated to the process? Suppose a memory manager employs paging with page size of 4 Kbytes....
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?
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...
2. In the paging system, assume that the logic address space is 28 and the size of a page is 32B a) What is the format of the logical/virtual address? b) Given the following page table Page number Frame number 0 4 4 6 0 Please convert the following logical address into physical address OxFF 0x86 0xA3
paging Example • Consider the following page table. The page size is 8 bytes each. Page Frame 0 6 1 2 0 N 3 3 4 5 • What are the physical addresses for the following logical addresses? • (1,5) . (2,0) • (0,6) • (3,8)
Paging Questions 1. A page is 1 KB in size. How many bits are required to store the page offset? 2. A page entry has 10 bits. What is the size of the page table? 3. A logical address is 32 bits long. The page size is 4 KB. Divide the address into its page number and offset. 4. The following hexadecimal addresses are used in a system with a 20-bit logical address where the page size is 256 bytes....