Virtual Memory

Section 13.3 Virtual Memory

Checkpoint 13.3.1. Program Memory Regions.

Match the components of a program with the section of the virtual address space where they are located.

Instructions
Code
Global variables
Data
Memory allocated using malloc
Heap
Arguments passed to a function
Stack

Checkpoint 13.3.2. Program Memory Regions.

Given the following C code:

Match each of the following expressions with their corresponding memory region.

x
Data
a
Stack
myfunc
Code
*p
Heap

Checkpoint 13.3.3. Virtual Address Space.

True or False: To ensure that processes cannot access each other’s data, the OS divides the virtual address space up between multiple processes.

True
Incorrect.
False
Correct! The virtual-memory abstraction gives each process its own separate entire virtual address space.

Checkpoint 13.3.4. Virtual Addresses.

True or False: The same virtual address in two different processes can map to two different physical addresses.

True
Correct!
False
Incorrect. See Figures 2 and 3 in Section 13.3.

Checkpoint 13.3.5. Virtual-to-Physical Memory Mapping.

True or False: Pages 100 and 101 of virtual memory can never map to frames 500 and 501, respectively, of physical memory.

True
Incorrect. There is no guarantee where two consecutive pages will be located in physical memory, but nothing prevents them from being mapped to consecutive frames.
False
Correct!

Checkpoint 13.3.6. Virtual Memory.

Which of the following is NOT an advantage of using virtual memory?

Eliminates the need for physical memory in the system
Correct!
Allows programs to use more memory than physically available
Incorrect.
Increases system security by isolating process’s memory
Incorrect.
Enables efficient multitasking by providing separate address spaces for processes
Incorrect.

Checkpoint 13.3.7. Address Translation.

Fill in the blanks with the appropriate word:

The MMU translates from addresses to addresses.

Checkpoint 13.3.8. Virtual and Physical Address Lengths.

True or False: To support virtual memory the OS must ensure that the length of the virtual addresses are the same as the length of physical addresses (the same number of bits).

True
Incorrect. Refer to Figures 6 and 7 in Section 13.3.3.
False
Correct!

Checkpoint 13.3.9. Vocabulary.

Match the term that best fits each sentence below.

A ____ is a fixed sized chunk of a process’s virtual address space.
page
A ____ is a fixed sized chunk of physical memory.
frame
A process uses a ____ to refer to the place in memory where a variable is stored.
virtual address
A ____ is a location in RAM.
physical address
The computer hardware component that implements address translation is the ____.

Checkpoint 13.3.10. Virtual Addresses.

In a virtual memory system, what is the primary role of the offset portion of a virtual address?

Specifies the exact byte location within a page
Correct!
Identifies the specific page within the page table
Incorrect.
Determines the frame number in physical memory
Incorrect.
Maps virtual addresses to physical addresses in the cache
Incorrect.

Checkpoint 13.3.11. Virtual to Physical Address Translation.

Which of the following must the OS do during a context switch to support virtual address to physical address translation?

Update the MMU hardware to refer to the swapped-on process’s virtual memory.
Correct!
Select a physical frame for each page of the process’s virtual memory.
Incorrect.
Zero out (erase) the contents of the swapped-out process’s virtual memory.
Incorrect.
Find enough contiguous frames of RAM to fit the swapped-on process’s virtual pages.
Incorrect.

Checkpoint 13.3.12. Memory Address Size.

In a computer system that uses 8 bits for a memory address, what is the range of valid addresses?

0 – 255
Correct!
0 - 7
Incorrect.
0 - 8
Incorrect.
0 - 256
Incorrect.

Checkpoint 13.3.13. Virtual vs. Physical Addresses.

In a computer system with 32-bit virtual addresses, a physical memory size of 1 GB, and a page size of 4 KB, how many bits are used for the virtual page number (VPN) and for the physical frame number (PFN)?

VPN: 20 bits, PFN: 18 bits
Correct!
VPN: 10 bits, PFN: 8 bits
Incorrect.
VPN: 18 bits, PFN: 20 bits
Incorrect.
VPN: 20 bits, PFN: 20 bits
Incorrect.

Hint.

1 GB is 220 bytes, 1 KB is 210 bytes

Checkpoint 13.3.14. Physical Memory Size.

If there are 4192 frames of physical memory, and the page size is 512 bytes, how large is the physical memory space in bytes?

\(2^{9}\) bytes
Incorrect.
\(2^{10}\) bytes
Incorrect.
\(2^{12}\) bytes
Incorrect.
\(2^{20}\) bytes
Incorrect.
\(2^{21}\) bytes
Correct!

Hint.

Physical memory size depends on the number of frames and the size of each frame (the page size).

Checkpoint 13.3.15. Fork and Memory.

Given the following C code:

   pid_t pid1;
   int x;


   x = 1;
   pid1 = fork();
   if (pid1 == 0) {
       x = 2;
   } else {
       x = 3;
   }
   printf("x = %d\n", x);

When run, the print statement will be executed twice, once by the parent and once by the child process.

Choose the answer that describes the two values of x that are printed and the justification.

The two values are different because the variable x in each of their virtual memory spaces is mapped to is mapped to different physical memory addresses.
Correct!
Both values of x are the same because the variable x in each of their virtual memory spaces is mapped to the same physical memory address.
Incorrect.
The two values are different because the child process executes before the parent process.
Incorrect.
The two values are the same because the value written by the child process overwrites the value written by the parent process.
Incorrect.

Checkpoint 13.3.16. Page Tables.

Where is the page table stored?

in memory (RAM)
Correct!
on disk
Incorrect.
in the page table base register (PTBR)
Incorrect.
none of these
Incorrect.

Checkpoint 13.3.17. Virtual Address Translation.

Why is the offset part of a virtual address unchanged when translated to a physical address?

The offset is used for indexing within a page, which is the same offset for indexing into the frame.
Correct!
The offset bits determine the page number, which is the same as the frame number in the physical address.
Incorrect.
The OS ensures that virtual pages and physical frames are different sizes.
Incorrect.
The Memory Management Unit (MMU) modifies only the offset during address translation.
Incorrect.

Hint.

The size of a page is the same as the size of a frame.

Checkpoint 13.3.18. Virtual and Physical Memory.

Which of the following statements are true?

The virtual pages of a process must be laid out in consecutive virtual addresses.
Correct!
Physical frames should be larger than virtual pages so that pages can fit within them.
Incorrect.
Physical frames allocated to a process must be laid out in consecutive physical addresses.
Incorrect.

Checkpoint 13.3.19. Page Table Entries.

For each term associated with Page Table Entry (PTE) match it with its definition.

Frame Number
Specifies the physical frame number where a virtual page is stored in physical memory
Valid Bit
Indicates if the PTE contains a correct mapping
Dirty Bit
Indicates if the in-RAM copy of the page has been modified

Checkpoint 13.3.20. Address Translation.

Order the following steps for translating a virtual address to a physical address when there is not a page fault.

Divide the address bits into the page offset (lower order bits) and page number bits (higher order bits).
---
Use the value in the PTBR combined with the page number value to compute the address of the PTE.
---
Verify the PTE valid bit is 1.
---
Construct the physical address using the page offset bits as the lower order bits and the frame number bits from the PTE as the higher order bits of the physical address.
---
Notice that the PTE valid bit is 0. #distractor
---
Trigger a page fault exception. #distractor
---
Update the PTE with the frame number and set the valid bit to 0. #distractor

Checkpoint 13.3.21. Handling a Page Fault.

Order the following steps for translating a virtual address to a physical address when there is a page fault.

Divide the address bits into the page offset (low order bits) and page number (high order bits).
---
Use the value in the PTBR combined with the page number to compute the address of the PTE.
---
Notice that the PTE valid bit is 0.
---
Trigger a page fault exception.
---
Find a free frame of RAM.
---
Read the page from disk into the free frame of RAM.
---
Update the PTE with the frame number and set the valid bit to 1.
---
Construct the physical address using the page offset bits as the lower order bits and the frame number bits from the PTE as the higher order bits of the physical address.
---
Restart the process at the instruction that caused the page fault.
---
Notice that the PTE valid bit is 1. #distractor
---
Update the PTE with the frame number and set the valid bit to 0. #distractor

Checkpoint 13.3.22. Paging During Context Switches.

Which of the following actions must the OS take during a context switch to support paging?

Save the current process’s PTBR value, restore the new process’s PTBR value
Correct!
Set the valid bit of each page in the current process’s page table to 0.
Incorrect.
Set the valid bit of each page in the new process’s page table to 1.
Incorrect.

Prev Top Next