Computer Sciences > GATE 2026 SET-2 > Computer Organization
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=210, 1M=220, 1G=230
Correct : 1536
The correct answer is 1536 bytes
Number of TLB entries: TLB reach = entries × page size → 220 = N × 212 → N = 28 = 256 entries.
Page number bits: Virtual address space = 64 GB = 236 bytes. Virtual pages = 236 / 212 = 224. Page number = 24 bits.
Frame number bits: Physical address space = 1 GB = 230 bytes. Physical frames = 230 / 212 = 218. Frame number = 18 bits.
Bits per TLB entry: 4 (process ID) + 24 (page number) + 18 (frame number) + 2 (control) = 48 bits = 6 bytes.
Total TLB size: 256 entries × 6 bytes = 1536 bytes.
Similar Questions
Consider a system with 2KB direct mapped data cache with a block size of 64bytes. The system has a physical address space of 64KB and a word length of 16bits. D...
Consider the given C-code and its corresponding assembly code, with a few operands U1-U4 being unknown. Some useful information as well as the semantics of each...
A 4 kilobyte (KB) byte-addressable memory is realized using four 1 KB memory blocks. Two input address lines (IA4 and IA3) are connected to the chip select (CS)...
Total Unique Visitors
Loading......