Assume a system has a Main Memory Access Time of 90 ns that is supported by a Ca

Question

Assume a system has a Main Memory Access Time of 90 ns that is supported by a Cache Memory having an Access Time of 30 and a Hit Rate of 96%. Calculate the Effective Access Time. Suppose a computer using Set-Associative cache has 2^24 words of main memory and a cache of 64 blocks where each cache block contains 16 words. If the cache is 4-way set-associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and word fields? If the cache is 8-way set-associative, what is the format of a memory address as seen by the cache?

Explanation / Answer

39 . Effective Access Time = Hit Rate* Cache Access Time + Miss Rate * main MemoryAcess Time

=> 0.96*30 + 0.04*90
=>28.8 + 3.6
=> 32.4 ns


40. a) 224 words of main memory implies we have 24 bits in an address. Cache contains 26 [64] blocks, but each set must have 4 blocks[4 way set associative ], so we have 26/22=24 sets. Therefore our 24-bit address is divided into 16 bits for the tag field, 4 bits for the set field, and 4 bits for the word field.

b) The 64 blocks in cache must now be divided into sets with 8 blocks each [8 way set associative ], implying we have 26/23=23 sets. The tag field would now have 17 bits, the set field 3 bits, and the word field 4 bits.

Note : 16 words [Given in question ] will require 4 bits

Thanks, let me know if there is any concern.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.