For a system employing paging, consider a logical address space of 1024 pages of 4096 bytes, mapping onto a physical memory of 64 frames. The frame size is exactly equal to the page size (i.e. 4096 bytes). Each memory location stores one byte of data. Determine the number of bits required to address the whole logical address space. Show your steps clearly in deriving the answer. Which table is used to translate the logical address to physical address?
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A: The Answer is in step2
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A: ANSWER:-
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A: The answer is explained in detailed below:
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A: Given : Physical memory = 224 bytes Logical address space = 256 pages Page size = 210 bytes
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A: The answer is as follows.
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A: Actually, given information logical address of 128 pages of 1024 words each, mapped onto a physical…
Q: consider a computer system using paging to manage memory; suppose it has 64K (2-) bytes of memory…
A:
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A: Given Suppose that a machine has 38-bit virtual addresses and 32-bit physical addresses.(a) What is…
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A: Given: Page size: 2-KB Frames: 512
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A: Here in this question we have given Page size = 256 No of frame = 1024 Page in logical address=…
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A: The answer is
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A: Logical space contains 32 pages = 5 bits ( 25 = 32) Each page contains 1024 words = 10 bits (210 =…
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A: The answer as given below:
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A: Given that, Logical address of 16 pages of 2048 words each. Number of physical memory frames= 4
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A: Actually, 1 byte =8 bits. cache memory is a fast access memory.
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A: Given,The virtual Address space = 236 bytesPage size = 212 bytesPages = 236 / 212 = 224 Pages
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A: Given,The virtual Address space = 236 bytesPage size = 212 bytesPages = 236 / 212 = 224 Pages
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A: Since, number of memory accesses required by a paged memory reference = 2 Time taken by a memory…
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…
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A:
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A: (1) it is given that: logical memory space = 30 bit = 2^30 bytes given page size is 4KB = 2^12…
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A: Actually, 1 byte = 8 bits.
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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- 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?Consider a logical address space of 32 pages of 1024 words each, mapped onto a physical memory of 8 frames. How many bits are there in logical address and how many bits are there in physical address?Consider a memory-management system based on paging. The total size of the physical address space 64 MB, Pages of size 4 KB, the Logical address space of 4GB. total number of pages are 16384, total number of frames are 16384 and the number of entries in a page table are 1048576.Now Calculate: a)Size of Page Table b) No of bits in Physical Address c) No of Bits in Logical Address
- 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?Suppose that a machine has 42-bit virtual addresses and 32-bit physical addresses.{a} How much RAM can the machine support (each byte of RAM must be addressable)?{b} What is the largest virtual address space that can be supported for a process?{c} If pages are 2 KB, how many entries must be in a single-level page table?{d} If pages are 2 KB and we have a two-level page table where the first level is indexed by 15-bits, then how many entries does the first-level page table have?{e} With the same setup as part {d}, how many entries are in each second-level page table?{f} What is the advantage of using a two-level page table over single-level page table?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?
- 1. Consider a computer system with a 30-bit logical address and 4-KB page size. The systemsupports up to 512 MB of physical memory. How many entries are there in each of the following?Assume that each page table entry is 4 Bytes.c. A conventional single-level page table?d. An inverted page table?e. A two-level hierarchical page table? 2. Consider a virtual memory system with a 50-bit logical address and a 38-bit physical address.Suppose that the page/frame size is 16K bytes. Assume that each page table entry is 4 Bytes.a. How many frames are in the systems? How many pages in the virtual address space for aprocess?b. If a single-level page table is deployed, calculate the size of the page table for each process.c. Design a multilevel page table structure for this system to ensure that each page table can fitinto one frame. How many levels do you need? Draw a figure to show your page systemsConsider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory of 64 frames. (book: 8.23) How many bits are required in the logical address? How many bits are required in the physical address?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 eight pages of 1024 words ach, mapped on to a physical memory of 32 frames. Find the number of bits in the logical address and the physical addressConsider a virtual memory split into pages. If each byte in the virtual memory has a virtual address, and the last 8 bits of each virtual address give the offset within a page. How big is each page?Take a look at a 64-bit logical address space. a. Assuming a page size of 4 KB Discover the truth. How many page table entries are there? Answer. b. Determine if each entry is 16 bytes long. Page Table Dimensions Answer.