Question

1. Four functions, each of length 500 words, are linked together into one process and loaded into memory. Consider four possible combinations of paging and segmentation: (15 Points) 1. Paging (no segmentation): page size: 512 words page table occupies 1 page 2. Segmentation (no paging): segment table size: 512 words 3. Segmentation with paging (each function becomes a separate segment): page and segment size: 512 words page and segment tables occupy 1 page each 4. Two-level paging (page table is paged): page size: 512 words all page tables occupy 1 page each a. For each system, determine the total amount of occupied memory space in words, including all page or segment tables. (8 Points)

Answer 1

1.

Paging (no segmentation)

• There will be 1 page table, occupying 1 * 512 = 512 bytes
• Each function has 500 bytes, but each function will be allocated full 1 page. So, total of 4 pages will be used, occupying 4 * 512 = 2048 bytes

So, total bytes occupied

= 512 + 2048

= 2560 bytes

2.

Segmentation (no paging)

• There will be 1 segment table, occupying 1 * 512 = 512 bytes
• Now, either all 4 functions might combine into 1 segment, or each function might have its own segment. Irrespective, total space occupied will be 4 * 500 = 2000 bytes

So, total bytes occupied

= 512 + 2000

= 2512 bytes

3.

Segmentation and paging (each function becomes a separate segment)

• There will be 1 segment table, occupying 1 * 512 = 512 bytes
• There will be 1 page table for each segment/function. So total space for these 4 page tables = 4 * 512 = 2048 bytes
• Also, each function has 500 words, but will occupy 1 full page. So there will be total 4 pages. So total space occupied by 4 pages = 4 * 512 = 2048 bytes

So, total bytes occupied

= 512 + 2048 + 2048

= 4608 bytes

4.

Two-level paging (page table is paged)

• There will be 1 first-level page table, occupying 1 * 512 = 512 bytes
• Now, we assume that there will be 1 second-level page-table for each function. With this assumption, there will be total 4 second-level page tables, occupying 4 * 512 = 2048 bytes
• As before, each function will occupy 1 page. So total space by all 4 pages = 4 * 512 = 2048 bytes

So, total bytes occupied

= 512 + 2048 + 2048

= 4608 bytes