When direct mapping is used, a certain block of the cache can only establish a mapping relationship with a fixed number of main memory blocks, and a certain block of the main memory can only correspond to one cache block.
Example:
The cache and main memory use a fully associative address mapping method. The cache capacity is 4MB, divided into 4 blocks, each block is 1MB, and the main memory capacity is 256MB. If the main memory read and write time is 30ns, the cache read and write time is 3ns, and the average read and write time is 3.27ns, then the cache hit rate is ___(3)___%. If the address conversion table is as follows, when the main memory address is 8888888H, the cache address is ___(4)___H.
Address conversion table
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(3)A. 90 B. 95 C. 97 D. 99
(4)A. 488888 B. 388888 C. 288888 D.188888
The main memory capacity is 256MB, which is represented by 28 binary addresses and corresponds to 7 hexadecimal numbers 8888888H. Each IMB block indicates that the address within the block is a 20-bit binary number, which is a 5-bit hexadecimal number 88888. The remaining 2 hexadecimal digits 88H represent the block address (fully associative mapping).
1. The Cache with a capacity of 64 blocks is mapped in a set associative manner. The word block size is 128 words, and each group consists of 4 blocks. If the main memory capacity is 4096 blocks and is addressed by word, then the main memory address should be __(7)__ bits, and the main memory area number should be __(8)__ bits. (7) A.16 B.17 C.18 D.19
(8) A.5 B.6 C.7 D.
Analysis: This question involves the knowledge of the working storage principle of Cache. Cache is a copy of a local area of main memory and is used to store currently active programs and data. Copying the contents of a local range from main memory to cache allows the CPU to read data from cache at high speed, which is much faster than accessing main memory. Cache has three mapping methods.
Here, since the main memory capacity is 4096 blocks (or "pages"), and each block is 128 words, the main memory address is 4096*128=2^n; n=19.
The main memory can be divided into 4096/64=64 groups, so the main memory area number is 2^n=64; n=6.
Answer choice (7) D (8) B
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