Question

Assume a task is divided into four equal-sized segments, and the system builds an
eight-entry page descriptor table for each segment. Thus, the system has a combination of segmentation and paging. Assume also the page size is 4 KB

4.1 (
1/2 What is the maximum size of each segment? )
4.2 (
1/2 What is the maximum logical address space for the task? )
4.3 Assume an element in physical location 00021ABC16 is accessed by this task. (3)
Provide the format of the logical address that the task generates for it and
explain how this format is produced. What is the maximum physical address
space for the system?

Answer 1

Answer:

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.

Answer 2

Answer:

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Explanation:

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