Main memory contains a total of 24 frames and each frame is 2³ bytes. How many bits are in the offset field? How many bits are required for the physical address?
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- 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.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? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 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? 1024 Bytes 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. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…Suppose the physical address of a memory location is 33330 H and the offset is 1111H. What is the base address (segment number) of the segment? What is the highest address in that segment (assume non overlapping segmentation)?
- Suppose the RAM for a certain computer has 4M words, where each word is 16 bits long. What is the capacity of this memory expressed in bytes? If this RAM is byte addressable, how many bits must an address contain? If this RAM is word addressable, how many bits must an address contain?Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses direct mapping with a block size of 4 words. i) How many bits are there in the tag, index, block & words fields of the address formats. ii) How many bits are there in each word of cache? iii) How many blocks can the cache accommodate?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 word
- Write the binary translation of the logical address 0001010010111010 to physical address, under the following hypothetical memory management schemes, A paging system with a 256 pages, The corresponding frame number is 00000101What is the difference between bit address 7CH and byte address 7CH? What is the specific location of bit address 7CH in memory?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?
- Computer Science This question is about paging-based virtual memory A computer has a virtual-momory space of 250MB (megabytes) The computer has 325) of primary memory. The pige som s-4000 by the address is 1011 0001 0101 1110 0010 1010 0010 a. How many frames can it have? b. Which of the bits in the virtual address correspond to the Page number? c. Which of the bits correspond to the page offset?Consider 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?In a microcomputer, the addresses of memory locations are binary numbers that identify each memory circuit where a byte is stored. The number of bits that make up an address depends on how many memory locations there are. Since the number of bits can be very large, the addresses are often specified in hex instead of binary. If a microcomputer uses a 24-bit address, how many different memory locations are there? What is the largest number that can be How many hex digits are needed to represent the address of a memory location? What is the hex address of the 172th memory location? (Note: The first address is always ) The computer program is stored in the lowest 3 kbyte block of memory. Give the start and end address of this What is the BCD code of the hex address of the 172th memory location?