Given a 32-bit virtual address space and a 24-bit physical address, determine the number of bits in the VPN, VPO, PPN, and PPO for the following page sizes P. Drag the appropriate labels to their respective targets.
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: I need an answer to question no. 2
A: Assuming the page size to be 4KB. The calculations for physical addresses are given below: (a) VA =…
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: Suppose you have a byte-addressable virtual address memory system with eight virtual pages of 64…
A: Answer: 1. Since it is be addressable, each page consists of 8 bits(i.e 1byte).It consists…
Q: Imagine a virtual memory system with page size equal 256 bytes. Only the first 5 entries in the…
A: Here page size given is 256 bytes. This means number of page offset bits is log 256 = 8 bits. Thus…
Q: In an architecture with 18 bits of "virtual address" width, "page size" is given as 1024 bytes and…
A: We are given TLB as 2 -way set associative with 16 data blocks. # sets= data blocks/…
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 with a page size of 16 KB. How many bits must be used to represent the…
A: The number of bits required to store the page offset in the logical address is 14. Therefore, the…
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 logical address of 128 pages of 1024 words each, mapped onto a physical memory of 64…
A: The Logical address will have 17 bits and the phycial address will have 16 bits. Explaination :…
Q: Consider a paging system with the following: Physical memory= 32 bytes. Page size=4 bytes. Page…
A: First we see what is actual location and what is coherent location :- Consistent Address - Logical…
Q: Given a 32-bit virtual address space and a 24-bit physical address, determine the number of bits in…
A: Given: Given a 32-bit virtual address space and a 24-bit physical address, determine the number of…
Q: Determine the number of page table entries (PTEs) that areneeded for the following combinations of…
A: Number of page table entries = address size / page size
Q: Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses. (a) What is the…
A: Given Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses.(a) What is…
Q: 5. Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses. a) What is the…
A: A virtual address space or address space is the set of ranges of virtual addresses that an operating…
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: For the following problems assume 1 kilobyte (KB) = 1024 bytes and1 megabyte (MB) = 1024 kilobytes.…
A: Below is the answer to abobe question. I'm providing the answr to all su
Q: Consider a system with a 7-bit virtual address space,32-byte pages and 512-byte physical memory. The…
A: In this question, we are given virtual address and physical address with page size. Virtual address…
Q: Consider a 48-bit virtual address space. The system hierarchical paging. Therefore, the page table…
A: 1)P1 x 1 15/16" 3)8-bit
Q: 2. If the contents of the page table are as follows: VPN PPN Valid 021 1 31 20 3 11 4 5 01 10 6-0 7…
A: (a) 0x1AE in decimal is 430 which on modulus by 8 (since there are 8 VPN given) gives 6. Since at…
Q: Suppose that an MMU has page table entries like the ones shown at right (formatted as in the slides…
A: The solution of the above question is:
Q: 1. Consider a logical address of 128 pages of 1024 words each, mapped onto a physical memory of 64…
A: The answer as given below:
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 the following portion of a page table from a system with 4 KiB (i.e., 4096 byte) pages.…
A: We have given portion of page table and page size is 4KiB. Using virtual memory address, we have to…
Q: Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but…
A: Below is the answer to the above question. I hope this will meet your requirement.
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: In a page addressing system of 10 bits, where four bits are used for the page number, what would be…
A: A page is contiguous virtual memory which is smallest unit to store data in memory management…
Q: When a Page Fault is encountered during a Virtual Address translation to Physical Address, either…
A: When a page fault occur , exception is raised by hardware when a running program accesses a memory…
Q: For the following problems assume 1 kilobyte (KB) = 1024 bytes and 1 megabyte (MB) = 1024 kilobytes.…
A: Below is the answer to above question. I hope this will be helpful for you...
Q: In a system, with 10 bit addresses of which 4 bits is for the page number, give following page…
A: Here number of bits for page number=4 Thus number of bits for offset = 10-4 = 6. These are least…
Q: Consider a system with 36-bit addresses that employs both segmentation and paging. Assume each PTE…
A: This is Operating system related question.
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: Suppose the virtual address width is 32 bits. Also suppose that the virtual and physical page size…
A: We have , Virtual Address = 32-bits Virtual and Physical page size = 4KB…
Q: Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory…
A: a. Logical address space (/size) = 2mLogical address space (/size) = number of pages × page…
Q: Let's assume a system has 64 bytes of physical memory, 4 byte pages, and 16-byte virtual address…
A: The information given are : Physical memory size = 64 bytes Size of page = 4 bytes Size of virtual…
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 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: 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: Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but…
A: Solution: The answer to the above questions are: a) 32 bits. b) 19 bits. c) 13 bits. d) 220 page…
Q: Our system is using virtual memory and has 48-bit virtual address space and 32-bit physical address…
A: A. Page size = 8KiB So page offset bits = 13 bits Total # of page table entries required = total…
Q: Q3: If page in logical address space consists of 32 words. And the size of that address is 12 bits,…
A:
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: Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB, and a PTE…
A: Data given in the question, 32-bit address system page size of 2 KB PTE (Page Table Entry) size of 1…
Q: Determine the number of page table entries (PTES) that are needed for the following combinations of…
A: A page table is the data structure used by a virtual memory system in a computer operating system to…
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: We use the concept "paging" to map logical address to physical address. If the process size is 9216…
A: As it is mentioned that paging is used to map the virtual address space to physical address space.…
Q: Determine the number of page table entries (PTES) that are needed for the following combinations of…
A: The connection between page table entries, virtual location size and page size is given as Page…
Q: Consider an 8-bit memory architecture that uses a single level page table with 4-bit page numbers…
A: Given Data : 8-bit memory architecture 4-bit page number. Page numbers are numbered from bottom to…
Q: A system that uses a two-level page table has 212 bytes pages and 32-bit virtual addresses. Assume…
A: Data given, 2^12 bytes pages 32-bit virtual addresses 4-byte each entry 10 bits of address serve as…
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
- Given a 32-bit virtual address space and a 24-bit physical address, determine the number of bits in the VPN, VPO, PPN, and PPO for the following page sizes P. Drag the appropriate labels to their respective targets.Suppose the virtual address width is 32 bits. Also suppose that the virtual and physical page size is 4Kbytes. If the physical address limit is 220, what is the number of bits in the Physical frame part of the physical address?Assume a 32-bit virtual address and 4 MBs of memory (i.e., DRAM). If the page size is 1 KB, then what is the size of the page table (ignoring any permission or other overhead bits)?
- In an architecture with 18 bits of "virtual address" width, "page size" is given as 1024 bytes and "physical address" width is given as 15 bits. TLB has a “2-way set associative” structure and contains a total of 16 data blocks. What is the TLB Tag field width in this architecture? A) 4 B) 5 C) 6 D) 7 E) 8Determine the number of page table entries (PTEs) that areneeded for the following combinations of virtual address size(n) and page size (P):Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB, and a PTE (Page Table Entry) size of 1 B. Answer the following questions, assuming a virtual address 0x00030f40 a. What is the virtual page number (VPN) and the offset in binary for the given virtual address? b. How many virtual pages are there in the system?
- 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?Fill in blank Suppose that linear page table is used where the memory addresses are 12-bit binary numbers and the page size is 256 bytes. If a virtual address in binary format is 101000011100, then the VPN (virtual page number) in binary format will be ---------Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses.(a) What is the main advantage of a multilevel page table over a single-level one?(b) With a two-level page table, 16-KB pages, and 4-byte entries, how many bits should be allocated for the top-level page table field and how many for the next-level page table field? Explain.
- Consider a logical address space of 512 pages with a 2-KB page size, mapped onto a physical memory of 64 frames. How many bits are required in the logical address?Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB. Answer the following questions, assuming a virtual address 0x00030f40 A. What is the virtual page number (VPN) and the offset in binary for the given virtual address? B. How many virtual pages are there in the system?Suppose you have a byte-addressable virtual address memory system with 16 virtual pages of 64 bytes each and 4-page frames. Assuming the following page table, answer the questions below: Page # Frame # Valid bit 0 2 1 1 3 1 2 - 0 3 0 0 4 1 1 5 - 0 6 - 0 7 - 0 A. What physical address corresponds to the virtual address 0x42? Answer should be in hexadecimal number. (if the address causes a page fault, answer as "page fault" with the proper explanation) B. What physical address corresponds to the virtual address 0x72? Answer should be in hexadecimal number. (if the address causes a page fault, answer as "page fault" with the proper explanation) C. What physical address corresponds to the virtual address 0x84? Answer should be in hexadecimal number. (if the address causes a page fault, answer as "page fault" with the proper explanation)