irtual address space and a 24-bit physical address, determ PPO for the following page sizes P.
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- 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)