Question

1.
2.
3.
4.
Answer (Detailed Solution Below)

Option 4 : 1,048,576 / 1,048,576

Advertisement

Detailed Solution

सही उत्तर 1,048,576 / 1,048,576 है।

Key Points
Helpful?
PrepSnapprepsnap.in • Part -A

Question:

A system uses 32-bit logical addresses with a 4 KB page size. How many entries does each process's page table contain? / एक सिस्टम 4 KB पेज आकार के साथ 32-बिट लॉजिकल पते उपयोग करता है। प्रत्येक प्रक्रिया की पेज तालिका में कितनी प्रविष्टियाँ होंगी?

1.
1,048,575 / 1,048,575
2.
524,288 / 524,288
3.
2,097,152 / 2,097,152
4.
1,048,576 / 1,048,576

Correct Answer: 1,048,576 / 1,048,576

Number of pages = 2^32 / 2^12 = 2^20 = 1,048,576, so the page table needs exactly that many entries. / पेजों की संख्या = 2^32 / 2^12 = 2^20 = 1,048,576, अतः पेज तालिका में ठीक इतनी ही प्रविष्टियाँ चाहिए।

Downloaded from PrepSnap (prepsnap.in)