Logical address size = 48 bits => Virtual address space =
Number of segments = 256 = => Segment Number = 8 bits
Size of segment = Logical address space / #segments =
Each segment will be divided into pages where page size = 8 KB = 8*1024 bytes =
Page table entry size = 4 bytes
Size of page table
= Number of entries in page table * page table entry size
=
Consider a system with 48-bit address that supports paging AND segmentation. The page size is 8KB...
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?
6) Paging [26 pts] Suppose you have a computer system with a 38-bit logical address, page size of 16K, and 4 bytes per page table entry a) How many pages are there in the logical address space? Suppose we use two level paging and each page table can fit completely in a frame. [4 pts] How many pages? [2 pts] Show your calculations here: b) For the above-mentioned system, give the breakup of logical address bits clearly indicating number of...
I need help with this problem.I am currenlty struggelling with it. Consider a computer system using paging, where the address space of every process has a size of C = 2c bytes and the page size is S = 2s bytes. Each entry in the page table uses E bytes. Calculate the number of pages of a process, and the size of a page table (in bytes). Assume that the space wasted by a process in main memory is defined...
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
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?
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?
In a system using paged segmentation, the logical address space of each process consists of a maximum of 16 segments, each of which can be up to 64 Kbytes in size. Physical pages are 512 bytes. Determine how many bits are needed to specify each of the quantities below, justifying each of your answers by demonstrating the calculations. a) Segment number (segment address) b) Number of a logical page within the segment c) Displacement within a page e) Complete logical...
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....