Assume that a task is divided into four equal-sized segments and that the system builds an eight-entry page descriptor table for each segment. Thus, the system has a combination of segmentation and paging. Assume also that the page size is 2 kbytes. What is the maximum size of each segment and the maximum logical address space for the task?
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2
We know the offset is 11 bits since the page size is 2K. The page table for each segment has eight entries, so it needs 3 bits. That leaves 2 bits for the segment number. So the format of the address is 2 bits for segment number, 3 bits for page number, and 11 bits for offset.
The physical address is 32 bits wide total, so the frame number must be 21 bits wide. Thus 00021ABC is represented in binary as:
Frame Offset
0000 0000 0000 0010 0001 1 | 010 1011 1100
The maximum physical address space is 232 = 4 GB.
The physical address is 32 bits wide total, so the frame number must be 21 bits wide. Thus 00021ABC is represented in binary as:
Frame Offset
0000 0000 0000 0010 0001 1 | 010 1011 1100
The maximum physical address space is 232 = 4 GB.
Answered by
1
Answer:
16K
Explanation:
max size of each seg,
8 entries in page table x page size
8 x 2
16k
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