-操作系统精髓与设计原理(第五版)+课后题答案1
7.12 a. The number of bytes in the logical address space is (216 pages) (210
bytes/page) = 226 bytes. Therefore, 26 bits are required for the logical address.
b. A frame is the same size as a page, 210 bytes.
c. The number of frames in main memory is (232 bytes of main memory)/(210 bytes/frame) = 222 frames. So 22 bits is needed to
specify the frame.
d. There is one entry for each page in the logical address space. Therefore there are 216 entries.
e. 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.
40M40M60M40M30M40M40M
d. The three starting addresses are 80M, 230M, and 360M, for the 40M, 20M, and 10M blocks, respectively.