Paging concept and page tables in Operating Systems - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the size of a page in bytes.

Operating Systems

page_size = 2[1]12  # Page size is 2 to the power of 12 bytes

Drag options to blanks, or click blank then click option'

A*

B**

C//

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '*' instead of '**' for exponentiation.

Using '//' which is integer division.

Using '+' which adds numbers instead of exponentiation.

The page size is calculated as 2 raised to the power of 12, which equals 4096 bytes. The '**' operator is used for exponentiation in Python.

2fill in blank

medium

Complete the code to calculate the number of pages given the total memory size.

Operating Systems

num_pages = total_memory_size [1] page_size

Drag options to blanks, or click blank then click option'

A//

B*

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '*' which multiplies instead of dividing.

Using '+' or '-' which do not calculate pages correctly.

Integer division '//' is used to find how many whole pages fit into the total memory size.

3fill in blank

hard

Fix the error in the code to correctly access the frame number from a page table entry.

Operating Systems

frame_number = page_table[page_number][1] 12

Drag options to blanks, or click blank then click option'

A>>

B-

C+

D<<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<<' which shifts bits left, increasing the number.

Using '+' or '-' which do not manipulate bits.

The '>>' operator shifts bits to the right, which is used here to get the frame number by removing the offset bits (12 bits for page size 4096).

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps page numbers to frame numbers for pages with even frame numbers.

Operating Systems

page_frame_map = {page: page_table[page] [1] 12 for page in range(num_pages) if (page_table[page] [2] 12) % 2 == 0}

Drag options to blanks, or click blank then click option'

A>>

B<<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<<' which shifts bits left incorrectly.

Using '+' or '-' which do not perform bit shifts.

Both blanks use the right shift operator '>>' to extract the frame number by shifting 12 bits to the right.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps page numbers to frame numbers only if the frame number is greater than 1000.

Operating Systems

filtered_map = { [1] : page_table[[2]] [3] 12 for [2] in range(num_pages) if (page_table[[2]] [3] 12) > 1000 }

Drag options to blanks, or click blank then click option'

Apage

Bpage_number

C>>

D<<

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Using '<<' instead of '>>' for bit shifting.

The dictionary maps each page to its frame number by right shifting 12 bits. The variable 'page' is used consistently for clarity.

Practice

1. What is the main purpose of paging in an operating system?

easy

A. To encrypt data in memory

B. To increase the CPU clock speed

C. To manage input/output devices

D. To divide memory into fixed-size blocks called pages

Solution

1. Step 1: Understand the concept of paging

   Paging divides the computer's memory into fixed-size blocks called pages to simplify memory management.

2. Step 2: Identify the main purpose

   The main goal is to break memory into pages, not to affect CPU speed, encryption, or device management.

3. Final Answer:

   To divide memory into fixed-size blocks called pages -> Option D

4. Quick Check:

   Paging = dividing memory into pages [OK]

Hint: Paging means splitting memory into equal blocks [OK]

Common Mistakes:

• Confusing paging with CPU speed improvement
• Thinking paging encrypts memory data
• Mixing paging with device management

2. Which of the following correctly describes a page table?

easy

A. A table that stores the mapping from virtual pages to physical frames

B. A list of all running processes in the system

C. A table that stores CPU instructions

D. A list of input/output devices connected to the system

Solution

1. Step 1: Define the role of a page table

   A page table maps virtual memory pages to physical memory frames, helping the OS locate data.

2. Step 2: Eliminate incorrect options

   Page tables do not list processes, CPU instructions, or devices.

3. Final Answer:

   A table that stores the mapping from virtual pages to physical frames -> Option A

4. Quick Check:

   Page table = virtual to physical mapping [OK]

Hint: Page tables map virtual pages to physical memory [OK]

Common Mistakes:

• Confusing page tables with process lists
• Thinking page tables store CPU instructions
• Mixing page tables with device lists

3. Consider a system with a page size of 4 KB and a virtual address 0x1A3F. Which part of the address identifies the page number?

medium

A. The lower 12 bits

B. The middle 8 bits

C. The higher bits except the lower 12 bits

D. The entire 16 bits

Solution

1. Step 1: Calculate bits for page offset

   Page size is 4 KB = 2^12 bytes, so lower 12 bits represent the offset within a page.

2. Step 2: Identify page number bits

   The page number is given by the higher bits of the address excluding the lower 12 bits (offset).

3. Final Answer:

   The higher bits except the lower 12 bits -> Option C

4. Quick Check:

   Page number = address bits above offset bits [OK]

Hint: Page offset = log2(page size) bits; rest is page number [OK]

Common Mistakes:

• Confusing offset bits with page number bits
• Using entire address as page number
• Selecting wrong bit segments

4. A page table entry is missing for a virtual page, causing a page fault. What is the most likely cause?

medium

A. The CPU clock speed is too low

B. The page is not currently loaded in physical memory

C. The virtual address is invalid

D. The page size is too large

Solution

1. Step 1: Understand what causes a page fault

   A page fault occurs when the page table does not have an entry for the requested virtual page, meaning it is not in physical memory.

2. Step 2: Eliminate unrelated causes

   CPU speed, invalid address, or page size do not directly cause missing page table entries.

3. Final Answer:

   The page is not currently loaded in physical memory -> Option B

4. Quick Check:

   Page fault = missing page in physical memory [OK]

Hint: Page fault means page not in physical memory [OK]

Common Mistakes:

• Blaming CPU speed for page faults
• Assuming invalid address always causes page fault
• Confusing page size with page fault cause

5. In a system with 32-bit virtual addresses and 4 KB page size, how many entries does a single-level page table have?

hard

A. 1,048,576 entries

B. 4,096 entries

C. 65,536 entries

D. 16,384 entries

Solution

1. Step 1: Calculate number of pages

   Page size = 4 KB = 2^12 bytes, virtual address size = 32 bits.

2. Step 2: Calculate entries in page table

   Number of pages = 2^(32 - 12) = 2^20 = 1,048,576 entries.

3. Final Answer:

   1,048,576 entries -> Option A

4. Quick Check:

   Entries = 2^(address bits - offset bits) [OK]

Hint: Entries = 2^(virtual bits - page offset bits) [OK]

Common Mistakes:

• Using page size directly as number of entries
• Confusing total address bits with offset bits
• Calculating entries incorrectly as 2^16 or 2^14