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GATE CSE 2026 Session 2 Q54: Size of a TLB with 1 MB reach and 4 KB pages

Question

A system has a Translation Lookaside Buffer (TLB) that has a reach of 1 MB. TLB reach is defined as the total amount of physical memory that can be accessed through the TLB entries. The paging system uses pages of size 4 KB. The virtual address space is 64 GB and physical address space is 1 GB. If each TLB entry stores a 4-bit process id, page number, frame number, and a 2-bit control field, then the size of the TLB (in bytes) is ___________. (answer in integer)

Note: 1K=2101\text{K} = 2^{10}, 1M=2201\text{M} = 2^{20}, 1G=2301\text{G} = 2^{30}

Try it: numerical answer: type the value

Numerical answer. No negative marking.
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Answer: 1536 (official GATE 2026 final answer key)

Concept used: Number of TLB entries = reach / page size; entry width = sum of its fields.

First step: Entries =220/212=256= 2^{20}/2^{12} = 256; page number is 36−12=2436 - 12 = 24 bits and frame number is 30−12=1830 - 12 = 18 bits.

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Source: official GATE 2026 Session 2 CS question paper and final answer key, organised by IIT Guwahati. Question paper · Answer key

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