5.5.1 Each cache block consists of four 8-byte words. Th e total off set is 5 bits. Th ree of those 5 bits is the word off set (the off set into an 8-byte word). Th e remaining two bits are the block off set. Two bits allows us to enumerate 2^2 = 4 words. 5.5.2 Th ere are fi ve index bits. Th is tells us there are 2^5 = 32 lines in the cache. 5.5.3 Th e ratio is 1.21. Th e cache stores a total of 32 lines * 4 words/block * 8 bytes/ word = 1024 bytes = 8192 bits. In addition to the data, each line contains 54 tag bits and 1 valid bit. Th us, the total bits required = 8192 + 54*32 + 1*32 = 9952 bits. S-6 Chapter 5 Solutions 5.5.4 5.5.5 4/12 = 33%. 5.5.6 <index, tag, data> <0, 3, Mem[0xC00]-Mem[0xC1F]> <4, 2, Mem[0x880]-Mem[0x89f]> <5, 0, Mem[0x0A0]-Mem[0x0Bf]> <7, 0, Mem[0x0e0]-Mem[0x0ff]>

这是一些与计算机组成相关的问题和答案，具体如下：

5.5.1：每个缓存块由四个8字节的单词组成。总偏移量为5位。其中三位是单词偏移量（对8字节单词的偏移量）。剩下的两位是块偏移量。两位可以枚举2^2 = 4个单词。

5.5.2：有五个索引位。这告诉我们缓存中有2^5 = 32个行。

5.5.3：比率为1.21。缓存存储了总共32行 * 4字/块 * 8字节/字 = 1024字节 = 8192位。除了数据之外，每行还包含54个标记位和1个有效位。因此，所需的总位数=8192 + 54*32 + 1*32 = 9952位。

5.5.4：4/12 = 33%。

5.5.5：这里没有给出问题，因此无法回答。

5.5.6：<索引，标记，数据> <0, 3, Mem[0xC00]-Mem[0xC1F]> <4, 2, Mem[0x880]-Mem[0x89f]> <5, 0, Mem[0x0A0]-Mem[0x0Bf]> <7, 0, Mem[0x0e0]-Mem[0x0ff]>。