Virtual Memory-Stallings

Question

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.
a. What is the maximum size of each segment?
b. What is the maximum logical address space for the task?
c. Assume that an element in physical location 00021ABC is accessed by this task.
What is the format of the logical address that the task generates for it? What is the
maximum physical address space for the system?

My answers to the parts were as follows

(a) 16KB

(b)16 bits

(c)4GB physical memory

Are my answers correct?

Comments

i dont know about it,

i just know that last 11 bits of virtual adress will be 010(BC)_H    

how can i tell first 5 bits of virtual address with given data.....

you have any idea ??

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cpu --> (2, 14) --> (segment_x, 3, 11) --> (segment_x, page_y, 11) --> (frame_number, offset).

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I AM NOT GETTING THIS QUETION PLEASE EXPLAIN.

Answer 1

(a)   (8 entries in the page table) x 2K = 16K.

(b)    (16 K) x 4 segments per task = 64K.

(c)    

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  | 1010 1011 1100

   The maximum physical address space is 2^32 = 4 GB.

Comments

How physical address is 32bits? is it by the given address location 00021ABC? Also, frame offset should be 11 bits right? Why have you chosen 12bits for offset in the representation?