Demand paging in Operating Systems - Time & Space Complexity

Time Complexity: Demand paging

O(n)

Understanding Time Complexity

When using demand paging, the system loads pages only when needed. We want to understand how the time to access memory changes as the number of pages grows.

How does the number of page accesses affect the total time spent loading pages?

Scenario Under Consideration

Analyze the time complexity of the following simplified demand paging process.

for each memory_access in program:
    if page not in memory:
        load page from disk
    access data in page

This code checks if a page is in memory for each access and loads it if missing, then accesses the data.

Identify Repeating Operations

Look for repeated actions that affect time.

• Primary operation: Checking and loading pages for each memory access.
• How many times: Once per memory access in the program.

How Execution Grows With Input

As the number of memory accesses increases, the system may need to load more pages, which takes extra time.

Input Size (memory accesses)	Approx. Operations (page checks and loads)
10	Up to 10 page checks and some page loads
100	Up to 100 page checks and more page loads
1000	Up to 1000 page checks and many page loads

Pattern observation: The time grows roughly in direct proportion to the number of memory accesses, since each access may cause a page load.

Final Time Complexity

Time Complexity: O(n)

This means the time to handle memory accesses grows linearly with the number of accesses, because each access may trigger a page load.

Common Mistake

[X] Wrong: "Demand paging always takes constant time per access because pages are loaded only once."

[OK] Correct: The first time a page is accessed, loading it from disk takes extra time, so the total time depends on how many unique pages are accessed, not just the number of accesses.

Interview Connect

Understanding how demand paging affects time helps you explain how operating systems manage memory efficiently. This skill shows you can think about how system design impacts performance.

Self-Check

What if the program accesses pages in a repeating pattern that fits entirely in memory? How would the time complexity change?

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