VIRAM SHAH
help mesolve this question
doubt 1 can PA be more than VA
doubt 2 in page table all PTE MUST be of same size? as it store frame no which is fixed
what is the answer
as per answer key option D
there is no reply for mentor side
@viramshah We will check on our ends once and let you know.
@saikatmohanta2003 will be wating for your reply
wrong answer
KG AI MentorKG AI Mentor
Quick AnswerOption B (24, 24, 24) is correct. Yes — physical-address bits can exceed virtual-address bits. PTEs are treated fixed-size here (4 bytes) so each must hold the same frame-number field. Key StepsPage size = 4 KB → offset = 12 bits (bits 0–11). Physical address = 36 bits → frame-number bits = 36 − 12 = 24. That’s the number of bits needed to identify any physical frame. Each PTE (level‑1 and level‑2 point to the next-level page‑table frame; level‑3 points to a data page frame) must contain that frame-number field → 24, 24, 24. Given PTE size is 4 bytes = 32 bits, the remaining 32 − 24 = 8 bits are for flags/unused; this does not change the required 24 frame bits. Note: GATE keys occasionally have typos; the correct reasoning gives option B. — KG AI Mentor 🤖
KG AI MentorKG AI Mentor
Quick AnswerYes, physical-address bits can exceed virtual-address bits (PA > VA). Page-table entries (PTEs) are normally a fixed size (here 4 bytes) at every level. The required bits are 24, 24 and 24 (option b). Key StepsPage size = 4 KB → offset = 12 bits. Physical address = 36 bits → frame number bits = 36 − 12 = 24. Every PTE (first and second levels point to the next-level page-table frame; third-level points to the data page frame) must contain that frame-number field → 24 bits each. Given PTE size is 4 bytes (32 bits), the remaining bits (32 − 24 = 8) are for flags/unused; that doesn't change the 24-bit frame-number requirement. Code-like summary: offset_bits = 12 frame_number_bits = physical_addr_bits - offset_bits = 36 - 12 = 24 answer = 24, 24, 24— KG AI Mentor 🤖
@kg_ai_mentor but gate provide answer key is option D
