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Consider a simple paging syste

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 Consider a simple paging system with the following parameters: 2 32 bytes of physical memory; page size of 2 10 bytes; 2 16 pages of logical address space. （ 10 pts ）

（ 1 ） How many bits are in a logical address? （ 2 pts ）

（ 2 ） How many bytes in a frame? （ 2 pts


（ 3 ） How many bits in the physical address specify the frame? （ 2 pts ）

（ 4 ） How many entries in the page table? （ 2 pts ）

（ 5 ） How many bits in each page table entry? Assume each page table entry contains a valid/invalid bit. （ 2 pts ）

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四、 Answer ： （ 10 pts ）

（ 1 ） The number of bytes in the logical address space is (2 16 pages)*(2 10 bytes/page) = 2 26 bytes. Therefore, 26 bits are required for the logical address.

（ 2 ） A frame is the same size as a page, 2 10 bytes.

（ 3 ） The number of frames in main memory is (2 32 bytes of main memory)/(2 10 bytes/frame) = 2 22 frames. So 22 bits is needed to specify the frame.

（ 4 ） There is one entry for each page in the logical address space. Therefore there are 2 16

entries.

（ 5 ） In addition to the valid/invalid bit, 22 bits are needed to specify the frame location in main memory, for a total of 23 bits.

发表于 2017-05-16 21:26:23 回复(0)
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软件工程
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