Logical address size = 32 bits
page size = 512 words = 512*4 = 2048 bytes
Number of entry in segment table = 213
(a)size of w = 232/213 = 219 bytes
(b)maximum number of pages per segment = 219/211 = 28 = 256 pages
Exercise 6.4.1: Parameters of paging and segmentation. A memory system employs both paging and segmentation: The...
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 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.
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...
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...
Consider a computer system that uses a paging system. The memory contains 16 frames, each frame can accommodate 512 memory locations (size of frame = 512). The page table is as follows: Page Number 0 1 2 3 4 5 6 7 Frame Number 5 3 10 0 2 9 11 14 What are the physical addresses for the following logical addresses? Show your work Logical Address Physical Address 2345 1024 6780
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....
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...
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...
Consider a system with 48-bit address that supports paging AND segmentation. The page size is 8KB and each page table entry (PTE) is 4B. A process can have up to 256 segments and each segment table entry (STE) is 8B. How large in bytes is the largest possible page table?
1- A 64-bit computer system employs a 16Gbyte main memory and a 32 Kilo word cache. Determine the number of bits in each field of the memory address register (MAR) as seen by cache in the following organizations (show your calculations): Fully associative mapping with line size of 2 words. A. Direct mapping with the line size of 8 words. B. C. 4-way associated mapping with the line size of 1 words. 1- A 64-bit computer system employs a 16Gbyte...