Question

A computer has an 8 GB memory with 64 bit word sizes.(1 word = 4 bytes).Each block of memory stores 16 words. The computer has a direct-mapped cache of 128 blocks. The computer uses word level addressing. What is the address format?

Answer 1

Answer:

8GB = 1GW (64 bit word = 8 bytes/word), which requires 30 bits (log 2 1G=30). The memory address is divided into line number and word number. With 16 words per line, we need 4 bits for the line number leaving 26 bits of the line number: 26 –4.