10 Paging (OS) MCQ

10 multiple-choice numerical questions on paging in operating systems for UGC NET, SET, ISRO and Other exam.

Question 1: Page Size Calculation

Q1: If a system uses a 32-bit address space and a page size of 4 KB, how many pages are there in the address space?

• A) 1024

• B) 4096

• C) 1048576

• D) 262144

Answer: C) 1048576

Explanation:
The number of pages is calculated as:

$$\frac{\text{Total address space}}{\text{Page size}} = \frac{2^{32} \text{ bytes}}{2^{12} \text{ bytes}} = 2^{20} = 1048576 \text{ pages}$$

Question 2: Number of Frames

Q2: If the physical memory is 1 GB and the page size is 4 KB, how many frames are available in the physical memory?

• A) 256

• B) 1024

• C) 256000

• D) 262144

Answer: D) 262144

Explanation:
The number of frames is given by:

$$\frac{\text{Physical memory}}{\text{Page size}} = \frac{1 \text{ GB}}{4 \text{ KB}} = \frac{2^{30} \text{ bytes}}{2^{12} \text{ bytes}} = 2^{18} = 262144 \text{ frames}$$

Question 3: Page Number and Offset

Q3: In a system with a 16-bit address space and a page size of 256 bytes, what are the page number and offset for the address 0x1234?

• A) Page number: 0x12, Offset: 0x34

• B) Page number: 0x123, Offset: 0x4

• C) Page number: 0x1, Offset: 0x234

• D) Page number: 0x1234, Offset: 0x0

Answer: A) Page number: 0x12, Offset: 0x34

Explanation:
Page number is obtained by dividing the address by the page size (256 bytes = 2⁸).

$$0x1234 / 256 = 0x12$$

Offset is the remainder:

$$0x1234 \mod 256 = 0x34$$

Question 4: Address Translation

Q4: If the logical address is 0x1A3 and the page size is 64 bytes, what is the physical address assuming the page table entry for the page is 0x02?

• A) 0x02A3

• B) 0x00A3

• C) 0x0223

• D) 0x0243

Answer: C) 0x0223

Explanation:
Page number: $0x1A3 / 64 = 0x6$
Offset: $0x1A3 \mod 64 = 0x23$
Physical address: The page table entry provides frame number 0x02, so the physical address is:

$$\text{Physical address} = (\text{Frame number} \times \text{Page size}) + \text{Offset} = (0x02 \times 64) + 0x23 = 0x0223$$

Question 5: Effective Access Time (EAT)

Q5: If the memory access time is 100 ns and the page fault service time is 20 ms, what is the effective access time if the page fault rate is 0.001?

• A) 20.1 ms

• B) 200 ns

• C) 20.0001 ms

• D) 200.1 ns

Answer: B) 200 ns

Explanation:

$$EAT = (1 - p) \times \text{Memory access time} + p \times \text{Page fault time}$$ $$= (1 - 0.001) \times 100 + 0.001 \times 20,000,000$$ $$= 99.9 + 20000 = 200 ns$$

Question 6: Page Table Size

Q6: For a system with a 64-bit address space and a page size of 4 KB, what is the size of the page table assuming each page table entry is 8 bytes?

• A) 64 GB

• B) 32 MB

• C) 64 MB

• D) 128 MB

Answer: D) 128 MB

Explanation:
Number of pages:

$$\frac{2^{64}}{2^{12}} = 2^{52}$$

Page table size:

$$2^{52} \times 8 \text{ bytes} = 2^{55} \text{ bytes} = 128 \text{ MB}$$

Question 7: Number of Pages

Q7: In a system with a 48-bit virtual address space and a page size of 8 KB, how many pages are there?

• A) $2^{35}$

• B) $2^{36}$

• C) $2^{34}$

• D) $2^{33}$

Answer: C) $2^{34}$

Explanation:

$$\text{Number of pages} = \frac{2^{48}}{2^{13}} = 2^{34}$$

Question 8: TLB Hit Ratio

Q8: If the TLB hit time is 20 ns, the memory access time is 100 ns, and the TLB hit ratio is 90%, what is the effective memory access time?

• A) 102 ns

• B) 108 ns

• C) 110 ns

• D) 120 ns

Answer: B) 108 ns

Explanation:

$$EAT = (\text{Hit ratio} \times (\text{TLB time} + \text{Memory time})) + (\text{Miss ratio} \times (\text{TLB time} + 2 \times \text{Memory time}))$$ $$= (0.9 \times 120) + (0.1 \times 220) = 108$$

Question 9: Page Table Levels

Q9: In a system with a 32-bit address space and 4 KB pages, what is the number of page table levels if each page table entry is 4 bytes and each level can map a maximum of 1024 entries?

• A) 2

• B) 3

• C) 4

• D) 5

Answer: B) 3

Question 10: Effective Access Time with Page Faults

Q10: If the memory access time is 200 ns, the page fault service time is 10 ms, and the page fault rate is 0.0005, what is the effective access time?

• A) 200 ns

• B) 205 ns

• C) 202 ns

• D) 502 ns

Answer: D) 502 ns