For the following problems assume 1 kilobyte (KB) 1024 kilobytes 1024 bytes and 1 megabyte (MB) For this problem, assume you have address translation hardware with the following properties: 1 Virtual...


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For the following problems assume 1 kilobyte (KB)<br>1024 kilobytes<br>1024 bytes and 1 megabyte (MB)<br>For this problem, assume you have address translation hardware with the following<br>properties:<br>1<br>Virtual addresses, physical addresses, and page table entries are 32 bits wide.<br>The page size in the system is 4 KB<br>A virtual address is a page number followed by a byte offset within the page.<br>(a)<br>How many bits of the virtual address must be used for the offset, so that every<br>byte in the page can have a unique address?<br>(b)<br>How many bits are left over in the virtual address to store the page number?<br>(c)<br>How many different page numbers does an address space in this system<br>support? (You can express this as a power of two)<br>(d)<br>If a page table consists of a page table entry for each page number in an address<br>space, how much space in MB would the page table take up if it were stored in<br>physical memory?<br>2<br>Given the assumptions above and the page table below, translate the following virtual<br>addresses (expressed as base ten integers) to physical addresses (also base ten<br>integers). If the corresponding page is not resident in memory (NR), indicate a page<br>fault<br>Hint: rather than translating these values to binary or hexadecimal, you can use the<br>following formulas<br>pageNum = |virtualAddress/pageSize] where [x] is x rounded down to the<br>nearest integer<br>offset virtualAddress mod pageSize where x mody is the remainder after<br>dividing x by y<br>physicalAddress frameNum x pageSize + offset.<br>Show your work<br>

Extracted text: For the following problems assume 1 kilobyte (KB) 1024 kilobytes 1024 bytes and 1 megabyte (MB) For this problem, assume you have address translation hardware with the following properties: 1 Virtual addresses, physical addresses, and page table entries are 32 bits wide. The page size in the system is 4 KB A virtual address is a page number followed by a byte offset within the page. (a) How many bits of the virtual address must be used for the offset, so that every byte in the page can have a unique address? (b) How many bits are left over in the virtual address to store the page number? (c) How many different page numbers does an address space in this system support? (You can express this as a power of two) (d) If a page table consists of a page table entry for each page number in an address space, how much space in MB would the page table take up if it were stored in physical memory? 2 Given the assumptions above and the page table below, translate the following virtual addresses (expressed as base ten integers) to physical addresses (also base ten integers). If the corresponding page is not resident in memory (NR), indicate a page fault Hint: rather than translating these values to binary or hexadecimal, you can use the following formulas pageNum = |virtualAddress/pageSize] where [x] is x rounded down to the nearest integer offset virtualAddress mod pageSize where x mody is the remainder after dividing x by y physicalAddress frameNum x pageSize + offset. Show your work
(a) VA 0<br>(b) VA 4095<br>Page<br>Frame #<br>0<br>(c) VA 4096<br>NR<br>1<br>(d) VA 10000<br>2<br>0<br>2<br>(e) VA 20000<br>NR<br>(f) VA 30000<br>NR<br>6<br>1<br>7<br>NR<br>LO<br>

Extracted text: (a) VA 0 (b) VA 4095 Page Frame # 0 (c) VA 4096 NR 1 (d) VA 10000 2 0 2 (e) VA 20000 NR (f) VA 30000 NR 6 1 7 NR LO
Jun 04, 2022
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