Demand paging in Operating Systems - Mini Project: Build & Apply

Understanding Demand Paging

📖 Scenario: You are learning how an operating system manages memory using demand paging. Demand paging loads pages into memory only when they are needed, which helps save memory space.

🎯 Goal: Build a simple step-by-step explanation of demand paging using a dictionary to represent pages and their status, then simulate checking and loading pages on demand.

📋 What You'll Learn

Create a dictionary called pages with page numbers as keys and their status ('not loaded') as values

Create a variable called page_to_load and set it to the page number 3

Write a loop using for page, status in pages.items() to check if page_to_load is in the dictionary and update its status to 'loaded' if it is 'not loaded'

Add a final statement that sets a variable page_status to the status of page_to_load in the pages dictionary

💡 Why This Matters

🌍 Real World

Operating systems use demand paging to efficiently manage memory by loading only the pages a program needs at a given time.

💼 Career

Understanding demand paging is important for roles in system administration, software development, and IT support where memory management concepts are essential.

1

Create the pages dictionary

Create a dictionary called pages with these exact entries: 1: 'not loaded', 2: 'not loaded', 3: 'not loaded', 4: 'not loaded', 5: 'not loaded'.

Operating Systems

# Create the pages dictionary with page numbers 1 to 5 all set to 'not loaded'
pages = {
    1: 'not loaded',
    2: 'not loaded',
    3: 'not loaded',
    4: 'not loaded',
    5: 'not loaded'
}
# Your code here

Hint

Use curly braces to create a dictionary with keys 1 to 5 and all values set to 'not loaded'.

2

Set the page to load

Create a variable called page_to_load and set it to the integer 3.

Operating Systems

pages = {
    1: 'not loaded',
    2: 'not loaded',
    3: 'not loaded',
    4: 'not loaded',
    5: 'not loaded'
}

# Create a variable page_to_load and set it to 3
page_to_load = 3
# Your code here

Hint

Just assign the number 3 to the variable page_to_load.

3

Load the page on demand

Use a for loop with variables page and status to iterate over pages.items(). Inside the loop, check if page equals page_to_load and if its status is 'not loaded'. If both are true, update pages[page] to 'loaded'.

Operating Systems

pages = {
    1: 'not loaded',
    2: 'not loaded',
    3: 'not loaded',
    4: 'not loaded',
    5: 'not loaded'
}

page_to_load = 3

# Use a for loop to check and load the page on demand
for page, status in pages.items():
    if page == page_to_load and status == 'not loaded':
        pages[page] = 'loaded'
# Your code here

Hint

Use a for loop to go through each page and check if it matches page_to_load and is 'not loaded'. Then change its status to 'loaded'.

4

Set the page status variable

Create a variable called page_status and set it to the value of pages[page_to_load].

Operating Systems

pages = {
    1: 'not loaded',
    2: 'not loaded',
    3: 'not loaded',
    4: 'not loaded',
    5: 'not loaded'
}

page_to_load = 3

for page, status in pages.items():
    if page == page_to_load and status == 'not loaded':
        pages[page] = 'loaded'

# Set page_status to the status of page_to_load
page_status = pages[page_to_load]
# Your code here

Hint

Assign the status of the page number stored in page_to_load to the variable page_status.

Practice

1. What is the main purpose of demand paging in operating systems?

easy

A. To load pages into memory only when they are needed

B. To load all program pages into memory before execution

C. To increase the size of physical memory

D. To prevent any page faults during program execution

Solution

1. Step 1: Understand demand paging concept

   Demand paging means loading pages only when the program tries to use them, not before.

2. Step 2: Compare options with concept

   To load pages into memory only when they are needed matches this idea, while others describe different or incorrect behaviors.

3. Final Answer:

   To load pages into memory only when they are needed -> Option A

4. Quick Check:

   Demand paging = load pages on demand [OK]

Hint: Demand paging loads pages only when accessed [OK]

Common Mistakes:

• Thinking all pages load before execution
• Confusing demand paging with increasing memory size
• Assuming demand paging prevents all page faults

2. Which of the following is the correct description of a page fault in demand paging?

easy

A. When all pages are loaded into memory at once

B. When the CPU finishes executing all instructions

C. When the operating system shuts down

D. When a program tries to access a page not currently in memory

Solution

1. Step 1: Define page fault in demand paging

   A page fault occurs when a program accesses a page that is not in physical memory.

2. Step 2: Match definition with options

   When a program tries to access a page not currently in memory correctly describes this event; others describe unrelated events.

3. Final Answer:

   When a program tries to access a page not currently in memory -> Option D

4. Quick Check:

   Page fault = access missing page [OK]

Hint: Page fault happens on missing page access [OK]

Common Mistakes:

• Confusing page fault with program completion
• Thinking page fault happens during shutdown
• Believing page fault means all pages loaded

3. Consider a system using demand paging. If a program accesses pages in this order: 2, 4, 2, 5, 4, which pages cause page faults assuming none are initially in memory?

medium

A. No page faults occur because pages are preloaded

B. Pages 2, 4, and 5 cause page faults

C. Pages 2 and 5 cause page faults, but not 4

D. Only page 2 causes a page fault

Solution

1. Step 1: Identify initial memory state and page accesses

   No pages are in memory initially. Access order is 2, 4, 2, 5, 4.

2. Step 2: Determine which accesses cause page faults

   First access to page 2 causes a fault (not in memory). Then page 4 causes a fault. The second access to page 2 is in memory, no fault. Access to page 5 causes a fault. Last access to page 4 is in memory, no fault.

3. Final Answer:

   Pages 2, 4, and 5 cause page faults -> Option B

4. Quick Check:

   First access to new pages = page fault [OK]

Hint: First access to each new page causes fault [OK]

Common Mistakes:

• Assuming repeated pages cause faults
• Thinking pages preload automatically
• Ignoring initial empty memory state

4. A system using demand paging is experiencing frequent page faults. Which of the following could be a likely cause?

medium

A. The program accesses many pages not yet loaded

B. The CPU clock speed is too high

C. The hard disk is disconnected

D. The program is very small and fits in memory

Solution

1. Step 1: Understand causes of frequent page faults

   Frequent page faults happen when a program accesses many pages not currently in memory.

2. Step 2: Evaluate options for likely cause

   The program accesses many pages not yet loaded matches this cause. CPU speed or disk disconnection do not directly cause page faults. A small program fitting in memory would cause fewer faults.

3. Final Answer:

   The program accesses many pages not yet loaded -> Option A

4. Quick Check:

   Many new page accesses = frequent faults [OK]

Hint: Many new pages accessed cause faults [OK]

Common Mistakes:

• Blaming CPU speed for page faults
• Confusing hardware issues with paging faults
• Assuming small programs cause many faults

5. In a demand paging system, if the page replacement algorithm always removes the page that was least recently used, what is the main benefit of this approach?

hard

A. It increases the total number of pages in memory

B. It prevents any page faults from occurring

C. It reduces the number of page faults by keeping frequently used pages in memory

D. It loads all pages into memory before execution

Solution

1. Step 1: Understand least recently used (LRU) page replacement

   LRU removes the page not used for the longest time, keeping frequently used pages in memory.

2. Step 2: Identify benefit of LRU in demand paging

   This reduces page faults by avoiding removal of pages likely to be used soon.

3. Final Answer:

   It reduces the number of page faults by keeping frequently used pages in memory -> Option C

4. Quick Check:

   LRU keeps frequent pages, reduces faults [OK]

Hint: LRU keeps frequent pages, lowers faults [OK]

Common Mistakes:

• Thinking LRU increases memory size
• Believing LRU prevents all faults
• Confusing LRU with preloading all pages