6. Consider a memory-management system based on paging. Let the total size of th

Question

6. Consider a memory-management system based on paging. Let the total size of the physical memory be 2GB laid out over pages of size 4KB. Let the logical address space of each process be limited to 128MB. Based on the information above, A. Determine the physical address layout in the system (i.e., the total number of bits in a physical address, B. How many physical frames are there in the system? C. How many pages are there in the logical address space of each process? D. Compute the number of entries in each page table and the size of each page table entry.

Explanation / Answer

The overall physical memory size/lenght is 2GB, therefore each physical address calls for 31 bits. The page length is 4KB, expecting 12 bits for the displacement within each cope with/address. Thus, the physical cope with /address the format is 31 bits, incorporating 19 bits for the page frame num and 12 bits of displacement. Note: there are 512K web page frames within the system.

The logical address space for every system is 128MB, expecting a TOt of 27 bits. The page length/size is identical to the physical pages (which is, 4KB). Accordingly, the logical deal with the layout is 27 bits, with 15 bits which are for the page num and 12 bits for the displacement. Note: there are 32K pages within the address space of every process.

Each page table should incorporate as many entries as there are logical pages within the address with space of every technique. Therefore, there are 32K entries in each page desk. Each access in a page desk/table incorporates a valid/invalid bit and a web page frame Num. Since the page frame num has 19 bits, the period of each entry inside the page desk/table is 20 bits.

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