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Operating System A paging system uses 16-bit address and 4K pages. The following shows the page...

Operating System

A paging system uses 16-bit address and 4K pages. The following shows the page tables of two running processes, Process 1 and Process 2. Translate the logical address 16,000 of Process 1 and the logical address 9,000 of Process 2 to their physical addresses. Fill your answers into the table below (10 pts)

Process 1                        Process 2

0 ----------- 0                      0 ----------- 3

1 ----------- 4                      1 ----------- 1

2 ----------- 5                      2 ----------- 7

3 ----------- 2                     3 ----------- 6

                                      4 ----------- 8

Processes                 Address           Page#              Offset         Physical Address

Process 1                 16,000             ----------             ----------       ---------------------------

Process 2                   9,000         ----------             ----------       ---------------------------

Suppose both processes ask for a shared memory of 4K, and suppose further that the system decides to allocate page frame 10 for this purpose. What virtual (or logical) addresses processes Process 1 and Process 2 will receive and what are the new page tables? You should provide sufficient reasoning.

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Answer #1
  1. Logical address: 6000

First need to find out page number:

6000 % 4096(4K) = 1904

Floor (6000 / 4096) = 1

So 6000 maps to the 1904 address of page 1.

Page 1 virtual address mapped to frame 4.

So physical address is: 4 * 4096 = 16384 So frame 4 have base address of 16384 in the physical memory.

So 1904th byte of frame 4 is at: 16384 + 1904 = 18288

2. Logical address: 19000

First need to find out page number:

19000 % 4096(4K) = 2616

Floor (19000 / 4096) = 4

So 19000 maps to the 2616 address of page 4.

Page 4 virtual address mapped to frame 8.

So physical address is: 8 * 4096 = 32768 So frame 8 have base address of 32768 in the physical memory.

So 2616th byte of frame 8 is at: 32768 + 2616 = 35384

Process Address Page# Offset Physical Address
Process 1 6,000 1 1904   18288
Process 2 19,000 4 2616 35384
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