Practice Question - 2
Duration: 2 min
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This educational video segment presents a solved problem from the GATE 2017 examination concerning computer architecture, specifically focusing on direct-mapped cache memory calculations. The instructor analyzes a byte-addressable machine with 2^32 bytes of main memory and a block size of 32 bytes. The core task is to determine the bit size of the tag field given a cache with 512 lines. The visual content displays structured diagrams partitioning the physical address into Tag, Cache Line (Index), and Block Offset components. The solution involves calculating the offset bits from the block size, deriving the index bits from the number of cache lines, and subtracting these from the total address width to find the tag size.
Chapters
0:00 – 1:51 00:00-01:51
The instructor introduces a GATE 2017 problem on direct-mapped cache memory, displaying the question text: 'Consider a machine with byte addressable memory of 2^32 bytes divided into blocks of size 32 bytes.' The visual diagrams illustrate the physical address structure, explicitly labeling 'Main Memory (Physical Address) 32' bits. The instructor breaks down the address fields, showing that a block size of 32 bytes requires 5 bits for the Block Offset (since 2^5 = 32). The diagram further indicates that with 512 cache lines, the Cache Line index requires 9 bits (since 2^9 = 512). The final calculation partitions the 32-bit address into Block Number (27 bits) and Block Offset (5 bits), leading to the conclusion that the Tag field size is 18 bits, derived by subtracting the index and offset bits from the total address width.
The video provides a concise walkthrough of cache address mapping, emphasizing the relationship between memory parameters and bit allocation. Key evidence includes the on-screen text specifying '2^32 bytes' memory size and '512 cache lines,' alongside diagrams showing the 32-bit address split. The instructor demonstrates that Block Offset is determined by log2(block size) and Cache Line Index by log2(number of lines). The Tag field is the remainder. This problem reinforces the standard formula: Total Address Bits = Tag + Index + Offset. The solution confirms that for a 32-bit address, with 5 offset bits and 9 index bits, the Tag must be 18 bits. This aligns with option (C) in the multiple-choice question presented.