8.18. Consider a paged logical address space (composed of 32 pages of 2 Kbytes each) mapped into a 1-Mbyte physical memory space. a. What is the format of the processor's logical address? b. What is the length and width of the page table (disregarding the "access rights" bits)? c. What is the effect on the page table if the physical memory space is reduced by half?
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- CA_10 Let the virtual address be V bits and the virtual addtess space be byte-addressable, the page size be P KB (and P is a power of 2), and the the main memory size be MM MB(where [MM MB]) is divide into [P KB]). (a) How large is the user's memory spac (b) How large is the main memory page frames? (c) How large is the page table (in number of table entries)? (d)How many of the virtual memory bits need to be translated? (e) How many bits will be produced if the virtual-to-pyysical address translation is "successful" (f) How many bits does a physical address have, and how are each of these bits obtained?) Consider the following sequence of virtual memory references (in decimal) generatedby a single program in a pure paging system:100, 110, 1400, 1700, 703, 3090, 1850, 2405, 2460, 4304, 4580, 3640a. Derive the corresponding reference string of pages (i.e. the pages the virtual addressesare located on), assuming a page size of 1024 bytes. (Assume that page numberingstarts at 0)b. For the page sequence derived above, determine the number of page faults for each ofthe following page replacement strategies, assuming that two (2) page frames areavailable to the program.i. LRUii. FIFOiii. OPT (Optimal)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 systems
- Consider a virtual memory system that can address a total of 32 bytes. You have unlimited hard disk space, but are limited to only 16MB of semiconductor (physical) memory. Assume that virtual and physical pages are each 4 KB in size. What is the total size of the page table in bytes? (Assume that, in addition to the physical page number, each page table entry also contains some status information in the form of a valid bit (V) and a dirty bit (D)).CA_6 We study the properties of cache memory, and for reasons of easier design and efficient circuits, we assume that the cache capacity is 2i Bytes, and cache line size is 2j Bytes, with i and j being natural numbers: (a) How many bits should the tag field have? And can the tag field contain 0 bit (i.e., be empty)? Elaborate (b) Repeat the above for the index field. (c) Repeat the above for the byte-offset field. (d) Finally, depict a figure showing a cache line, indicate what fields it possibly has, state the possible sizes of these fields, and explain the uses of these fields.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?
- I have a 128-bit computer where addresses are 128 bits wide. I want to implement virtual memory with paging for this computer, and I want to use a page size of 4KB. I will design it so that a page table consumes exactly one page. How many page table entries (PTEs) fit within a page?(Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GBSuppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a cache of 1024 blocks, where each cache block contains 32 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields? c) To which cache block will the memory address 0x000063FA map?
- Suppose a computer using fully associative cache has 4 Gbytes of byte-addressable main memory and a cache of 256 blocks, where each cache block contains 32 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address? Provide the names and the sizes of the fields. c) To which cache block will the memory address 0x01752 map?Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last wordAssume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at?