GATE 2026 CS (CS2) – Question 54
A system has a Translation Lookaside Buffer (TLB) with reach 1 MB. The paging system uses pages of size 4 KB. The virtual address space is 64 GB and the 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: $1\text{ K}=2^{10}$, $1\text{ M}=2^{20}$, and $1\text{ G}=2^{30}$.
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Correct answer: 1536
Explanation
A TLB reach of 1 MB with page size 4 KB means the number of TLB entries is $$\frac{2^{20}}{2^{12}}=256.$$ The virtual address space is 64 GB, so virtual addresses are 36 bits and the page number field has $36-12=24$ bits. The physical address space is 1 GB, so physical addresses are 30 bits and the frame number field has $30-12=18$ bits. Each TLB entry stores 4 bits of process id, 24 bits of page number, 18 bits of frame number, and 2 control bits, for a total of $48$ bits or 6 bytes. Thus the TLB size is $$256\times6=1536\text{ bytes}.$$ Hence the answer is 1536.