Please show step by step solution Consider virtual addressing of 32 bits. Assume

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Please show step by step solution

Consider virtual addressing of 32 bits. Assume that the page size is 4 KB. You have 2 GB of RAM. How many entries do you need in the page table? How many bits are needed to index the page table? Assume that Table 1 lists all the virtual pages of process A that are currently in physical memory. If a virtual page is not listed in Table 1, it implies that any address in that range will result a page fault and will be a miss. Translate the virtual addresses below to the physical address using Table 1 and mention which will be hit and which will cause a miss: Translate the following virtual addresses to physical address using Table 1 (if possible): 1546, 20716, 85060, 36320, 2455

Explanation / Answer

= (2^32)/4KB

= (2^32)/2^12

= 2^20

The no. of index bits are need to index the page table is (2^31)/2^12 =2^31-12 =2^19.

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