Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but that each page is of size8 bytes, For each of the questions below please answer with only the number. (a) How many bits are in the virtual address in this system? (b) How many bits are in the VPN? (c) How many bits are in the Offset? (d) How many entries would the Page Table contain?
Q: What exactly is a virtual address, and how does one go about converting one into a real one?
A: The virtual address is a logical address created by the CPU that does not exist in the real world.…
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 =…
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Q: A very old computer had 24-bit addresses and 4KB pages. How many virtual address bits were occupied…
A: Solution: Given, Virtual memory system = 24-bit Page size = 4 KB
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: Consider a system with 256Mbytes of main memory with page size of 4Kbytes. It has a logical address…
A: Answer: Given Main Memory 256Mbytes Page Size =4Kbytes and logical address=26 bits
Q: 1. Consider a 32bit address space with 2K bytes page size, assuming that each entry consists of 4…
A: Solution : Address Space consist of 32 bits, program size is 2^32 bytes i.e. 4GB. Page Size = 2K…
Q: Consider a system with 39-bit virtual addresses, 44-bit physical addresses, 4096 byte pages, and the…
A: The answer is
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: Consider a 128-word L2 memory and a 16-word direct mapped L1 cache. a. How many bits are in the…
A: a. The total number of bits needed to address an L2 word is = log (128) => 7 bits. b. Assume that…
Q: Consider a logical address of 128 pages of 1024 words each, mapped onto a physical memory of 64…
A: Hey there, I am writing the required solution based on the above given question. Please do find the…
Q: Suppose we have a computer system with 44-bit logical addresses, page size of 64KiB, and 4 bytes per…
A: The answer is explained in detailed below:
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: 5. Your system has an 8-way set associative cache and addresses and data of length 16 bits. The…
A: Since it is given that set index bits are 9, we have Total sets = 29 = 512
Q: Given a 4-way set associative cache, which has 256 blocks and 64 bytes per block. Assume a 32-bit…
A:
Q: For the following problems assume 1 kilobyte (KB) = 1024 bytes and1 megabyte (MB) = 1024 kilobytes.…
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Q: Suppose a computer using direct mapped cache has 4G Bytes of main memory and a cache of 256 Blocks,…
A: refer to step 2 for the answer.
Q: Q: Consider a system which has Logical Address = 4 GB, Physical Address PA = 64 MB, Page Size = 4…
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Q: Consider a logical address space of 32 pages of 1024 words each, mapped onto a physical memory of 64…
A: Logical space contains 32 pages = 5 bits ( 25 = 32) Each page contains 1024 words = 10 bits (210 =…
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: In a page addressing system of 10 bits, where four bits are used for the page number, what would be…
A: your question is about what would be the number of frames that would be required in the physical…
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: Logical Address = 4 GB, Physical Address PA= 64 MB, Page Size = 4 KB, then calculate Number of…
A: Logical Address = 4 GB, Physical Address PA= 64 MB, Page Size = 4 KB, then Number of pages =…
Q: Given the 128K page size associated with the 1GB address space, calculate the number of pages in the…
A: Here in this question we have given virtual address space =1GB. Page size= 128kb Physical address…
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Q: A system with 64GB memory with a block size of 4 words, and a 256KB direct map cache. The word size…
A: Given: Memory =64 GB that means total bits = log2(64GB)=log2(236)=36 Cache size = 256 KB Block Size=…
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: 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: source address to a destination address, 1 byte at a time. st = 0×5FC0, src = ®x1A10, length =…
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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: 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: Consider a 64K L2 memory and a 4K L1 2-way associative cache with block sizes of 512. a. How many…
A: Given, size of L2 = 64K and size of L1 = 4K associativity = 4 - way and , block size = 512
Q: Assume a 32-bit address system that uses a paged virtual memory, with a page size of 2 KB, and a PTE…
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Q: Consider a system with 64-bit address that supports multi-level page tables with two levels. The…
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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: In the following three questions, assume a 32-bit virtual address space and page size equal to 4096…
A: Given: Virtual address space = 32 bit page size = 4096 bytes To find: Number of page table entry
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…
Q: Assume you have a small virtual address space of size 2048 KB, and that the system uses paging, but…
A:
Q: Consider a logical address space of 1024 pages of 2048 words each, mapped onto a physical memory of…
A: Question :-
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- A system that uses a two-level page table has 2¹2 - byte pages and 32-bit virtual addresses. The first 8 - bits of the address serve as the index into the first level page table. The number of bits specify the second level index isAssume you have a small virtual address space of size 2048 KB, and that the system uses paging, but that each page is of size 8 bytes. For each of the questions below please answer with only the number. (a) How many bits are in the virtual address in this system? (b) How many bits are in the VPN? (c) How many bits are in the Offset? (d) How many entries would the Page Table contain? pageConsider a computer system with a byte- addressable main memory of size 16 Gigabytes. The computer system has a direct-mapped cache of size 64 Kilobytes, and each cache block is of size 32 bytes. What is the size of the tag field in bits?
- For a page size of 200 words, what is the page number and offset for a logical address of 1142On a simple paging system with 224 bytes of physical memory, 256 pages of logical address space, and a page size of 210 bytes. 1. How many bits are needed to store an entry in the page table (how wide is the page table)? Assume a valid/invalid 1-bit is included in each entry. 2. If the page table is stored in the main memory with 250nsec access time, how long does a paged memory reference take? 3. If the page table is implemented using associative registers that takes 95nsec. and main memory that takes 200nsec, what is the total access time if 75% of all memory references find their entries in the associative registers?Glven the below 16-bit Logical Address: 0001000011001101 And the related Process Page Table: Page# Contents 00101 1 00110 01001 3 11001 4 11101 00010 00111 Show the physical address that results from the translation using paging, knowing that the page size is 2K.
- 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.Consider a logical address of 128 pages of 1024 words each, mapped onto a physical memory of 64 frames. i. How many bits are there in the logical address? ii. How many bits are there in the physical address? III. What is the size of the page table?What is an Internet Protocol (IP) address, and how does it work? What exactly is a mnemonic address, and how does it function in practice? The number of different domains that may be represented using the 32 bit format is limited only by the available memory. And how many machines are allowed to be a member of a domain at any one time?
- In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of page contains 2 permission bit,1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system? (answer in bits)Suppose a computer using direct-mapped cache has 232 bytes of byte-addressable main memory and a cache of 1024 blocks, where each cache block contains 32 bytes.Q.) What is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, block, and offset fields?Computer Science Explain in detail the steps in translating a 24-bit Virtual Address to a 32-bit Physical Address in a system with a two-level page table where the page directory is 32 entries and the size of each page table is 512 entries. Include the length of the bit fields that make up the Virtual Address and the Physical address.