Consider a system with 39-bit virtual addresses, 44-bit physical addresses, 4096 byte pages, and the PTE size is 4 bytes. This system is impossible
Q: Given a 32-bit virtual address space and a 24-bit physical address, determine the number of bits in…
A: As given a 32-bit virtual address space and a 24-bit physical address, determination the number of…
Q: Consider a logical address space of 8 pages of 1024 words each, mapped into a physical memory of 32…
A: The correct answer is option d) 13 bits and 15 bits Explanation: Physical Memory = 32 frames = 2^5…
Q: Consider a logical address space of 8 pages of 1024 words mapped into memory of 32 frames. How many…
A: Introduction :Given , Logical address space = 8 pages size of a page = 1024 wordsmemory size = 32…
Q: Consider a computer system that uses virtual memory with paging (no TLB). Suppose main access time…
A: Discontinuum Buffer for Localization: The CPU creates physical memory location in memory systems.…
Q: Consider a computer system with a 24-bit logical address and a 28-bit physical address. Let's…
A: Multi level paging is a paging scheme which consist of two or more levels of paging tables in…
Q: Consider a paging system with the following: Physical memory= 32 bytes. Page size=4 bytes. Page…
A: Page size = 4B. So page offset = 2 bits. The offset for physical and virtual address is same. A)…
Q: Consider a logical address space of 32 pages of 1024 words each, mapped onto a physical memory of 8…
A: Logical Address space: This address space is generated by the process running on the CPU for a…
Q: Consider a computer system that uses virtual memory with paging (no TLB). Suppose main access time…
A: Paging technique means dividing virtual memory into equal size partitions called pages and storing…
Q: In a page addressing system of 15 bits, where eight bits are used for the page number, what would be…
A: The above question is solved in step 2 :-
Q: Consider a system with 39-bit virtual addresses, 44-bit physical addresses, 4096 byte pages, and the…
A: Physical Address: It is a memory address or location of memory cell in the main memory. Pages: Page…
Q: b. Suppose we use two-level paging and arrange for all page tables to fit into a single page frame.…
A:
Q: Consider a memory system that generates 16-bits addresses and the frame size is 32-byte. At time To…
A: According to the information given:- we have to find the maximum number of entries, page table…
Q: Consider a logical address space of eight pages of 1024 words ach, mapped on to a physical memory of…
A: ANSWER:-
Q: Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB. Answer…
A: Here in this question we have given a 32 bit address system. Page size = 2KB. Virtual address =…
Q: a) A paging system with 512 pages of logical address space, a page size of 2³ and number of frames…
A: Here in this question we have given Page size = 256 No of frame = 1024 Page in logical address=…
Q: Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory…
A: The answer is as follows.
Q: consider a computer system using paging to manage memory; suppose it has 64K (2-) bytes of memory…
A:
Q: Unlike a byte that is always 8 bits, word size can vary from machine to machine. Let's say we have a…
A: since we have 4 address lines so no of unique addresses=2^4=16 so we can store 16 unique words
Q: ry split into pages. If each byte in the virtual memory has a virtual address, and the last 8 bits…
A: Pages are typically 512 to 8192 bytes in size, with 4096 being the most common value. For reasons…
Q: Consider a system with N bytes of physical RAM, and M bytes of virtual address and frames are K…
A: Here the virtual memory size = M bytes And page size is K bytes. The virtual address is divided…
Q: Consider a system with 39-bit virtual addresses, 44-bit physical addresses, 4096 byte pages, and
A: The answer is
Q: Suppose that the offset field of a byte-addressed 32-bit paged logical address is 12 bits. Then, a…
A: 1) Byte addressable 32-bit system can accomodate 232 bytes = 4,294,967,296 bytes 2) 12- bit logical…
Q: Consider a system that has 4K pages of 512 bytes in size in thể logical address space. The physical…
A: The Answer is Given data Number of pages = 4k page size ps = 512 B = 2^9 B
Q: Given a 32-bit virtual address space featuring a 10-10-12 split and a 4-byte PTE size, suppose a…
A: The memory is split as 10 - 10 - 12 This means page size is 4KB. Now process size is 9MB Therefore…
Q: Consider a machine where total addressable physical memory size is 32 Kbytes. If we would like to…
A: total addressable physical memory size=32 Kbytes. we like to run 2 processes P and R as 8KB using…
Q: A computer has a cache, main memory, and a disk used for virtual Memory • An access to the cache…
A: Cache hit ratio =0.9 So cache miss ratio = 0.1 Main memory hit ratio = 0.8 So main memory miss ratio…
Q: Suppose we have a byte-addressable computer using fully associative o 20-bit main memory addresses…
A: Actually, 1 byte =8 bits. cache memory is a fast access memory.
Q: Consider a computer system with 32-bit virtual addressing and a page size of sixty-four kilobytes.…
A: In this question, we are given virtual address and physical memory size and also page size. Page…
Q: Il- Consider a virtual memory system with the following properties: • 42-bit virtual byte address •…
A:
Q: Consider a memory system with a cache, a main memory and a virtual memory. The access times and hit…
A: Introduction :Given , Hit rate(H) and access time(T) for cache(1) , main memory(2) , virtual…
Q: Consider a system with 36-bit virtual addresses, 32-bit physical addresses, and 4KB pages. The…
A: Given,The virtual Address space = 236 bytesPage size = 212 bytesPages = 236 / 212 = 224 Pages
Q: A system has 4-kB pages, a 48-bit virtual address space, and a 33-bit physical address space.…
A: Lets see the solution in the next steps
Q: Consider a logical address space of 256 pages of 1024 words each. This is mapped onto a physical…
A: Given: logical address space= of 256 pages of 1024 words each. the physical memory=32 frames. We…
Q: Consider a system with 4-byte pages. A process has the following entries in its page table: logical…
A: Answer (a): An address of 32 corresponds to the byte that has the logical address of two. This is…
Q: Consider a system consisting of thirty-two bits virtual address, page size is 16 KB and a 512 lines…
A: The virtual address is of 32 bits. Whenever CPU wants any data from the memory, it generates 32bit…
Q: Consider a system with 4-byte pages. A process has the following entries in its page table: logical…
A: We are given page size, page table and logical address and asked the physical address for it. First,…
Q: a) A paging system with 512 pages of logical address space, a page size of 2* and number of frames…
A:
Q: A computer with 32 bits virtual address uses a two-level page table. Virtual addresses are split…
A: GIVEN: A computer with 32 bits virtual address uses a two-level page table. Virtual addresses are…
Q: Consider a system with 16-bit virtual addresses, 256 byte pages, and 4 byte page table entries. The…
A: PTE's per process = 216/28=28 PTE's per page = 28/22 = 26.
Q: Consider a logical address space of 1K pages with a 4KB page size, mapped onto a physical memory of…
A: (1) Logical address space (size)=2m then: Logical address space(size)= no of pages ×page size…
Q: Given a process with address space of size 32 bytes and page size of 8 bytes, if the CPU asks to…
A: Actually. OS is a system software which manages computer hardware and software.
Q: Consider a logical address space of 1024 pages with 2 KB page size, mapped onto a physical memory of…
A: As given, we need to find out, how many bits are required in the logical address. Given data -…
Q: Consider a computer system with a 24-bit logical address and a 28-bit physical address. Let's…
A: logical address space = 224 physical address space = 210
Q: Given a virtual memory of size 4 GiB, physical memory of size 512 KiB, and page size equal to 4 KiB.…
A: Ans:) The higher order bits of any virtual address creates a page number while lower order bits form…
Q: Consider a system with 64-bit address that supports multi-level page tables with two levels. The…
A: The address of a process is first divided into equal-sized pages. These pages are kept under a table…
Q: Assume a logical address space consists of 4 pages of 4096 words each and is mapped to a physical…
A: logical address space consists of 4 pages of 4096 words 256 frames
Q: Given a page table and a logical address=0002022H, where is it physically (what is its physical…
A: As per our guidelines we are supposed to answer?️ only one question. Kindly repost other questions…
Q: Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up…
A: (1) it is given that: logical memory space = 30 bit = 2^30 bytes given page size is 4KB = 2^12…
Q: Take a look at a 64-bit logical address space. a. Assuming a page size of 4 KB Discover the truth.…
A: Introduction: the question is about Take a look at a 64-bit logical address space. a. Assuming a…
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- Consider a computer system with a 64-bit logical address and 32-KB page size. Thesystem supports up to 4096 MB of physical memory. How many entries are there ineach of the following?a. A conventional single-level page table?b. An inverted page table?Consider a system with word-addressable memory in which a word is of 1 byte, 32-bit logical addresses memory, 2 kilobyte page size and page table entries of 4 bytes each. What is the size of the page table in the system in Mega word?Computer Science 9. In a paging system , the logical address is formed of 20 bits. the most significant 8 bits denote the page number, the least significant 12 bits denote the offset. Memory size is 256K bits. a. What is the page size (in bits)? b. What is the maximum number of pages per process? c. How many frames does this system have? d . Give the structure of the page table of a process executing in this system. Assume 2 bits are reserved for attributes. e. How many bits are required for each page table entry? f. If physical memory is upgraded to 1024 K bits, how will your answer to c and e change?
- Given a system with 32 bits logical address and paging is used for memory management, the page size is 4 KB (4096 bytes); If 4 Bytes would be needed to store one entry of the page table, then what is the page table size for a process of 4 MB (4096x496 bytes) address space? 1 KB O a. O b. 8 KB О с. O d. 4 KB 16 KBFor an old computing system with 2K bytes physical memory, and the virtual address has 13bits. Suppose that the size of page/frame is 256 bytes. For a process A, it has its codes and data inpage 0, 1, 2, 10, 11, 28, 29, where pages 0, 1, 10, 29 are in frame 1, 3, 4 and 6, respectively.Moreover, frame 0 contains kernel OS code/data and all other frames are free.a. Show the page table and the content of each PTE for process A;b. Use a figure to illustrate the address translation for virtual address 1110000100000 and explainwhat happens during the translation (interaction among page table, physical memory, disk, andoperating system);c. Suppose that there is a TLB with 4 entries and the current content has the mapping informationfor pages 0, 1. Draw a new figure to illustrate the translation of address 101000011000 andexplain what happens during the translation process.Consider a system with 8192 byte pages, a page table entry (PTE) size of 8 bytes, 40-bit virtual addresses, and 56-bit physical addresses. Considering the size you calculated in the last question, is it realistic to use a single page table in the real world? Explain why or why not. A- BI U - How many PTES could fit on a single page? Answer: How many levels of page tables would you need, if you ensure that each (smaller) page table fits on a page. Answer: Explain how you arrived at your answer for the previous question.
- Consider a logical address space of 1024 pages of 2048 words each, mapped onto a physical memory of 4096 frames. a. How many bits are there in the logical address? b. How many bits are there in the physical address? Show the steps of your solution.Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? With this system, what’s the maximum number of pages that a process can have? Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number.Suppose a computer system uses 16 but addresses for both virtual and physical address. In addition, assume each page (and frame) has a size of 256 bytes. -How many bits are used for page number? -How many bits for offset? -what is the max number of pages a process can have?
- Please if solution could be typed Consider a computer system with a 16-bit logical address and 2-KB page size. The system supports up to 1MB of physical memory. How many entries are there in each of the following? A conventional, single-level page table An inverted page tableSuppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? 1024 Bytes Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…This question is about physical addressability in a system that uses paging. Let's say we have a 32-bit virtual address, with a 4 KB page size. Let's assume that the system we're running upon has a maximum of 1 GB of physical memory. a) How many entries will be there in a single-level page table and inverted page table?