Question

Question 3 The access time of a cache is 80 ns and the access time of main memory is 1200 ns. We have 85% of instructions are directed to read while 15% is for writes. Hit ratio is 92%. A write through procedure is used. A. Give the average access time considering only the read requests B. Give average access time for both read and write requests. C. Give the overall hit ratio given the write cycle as well. Question 4 A digital computer has a memory of 32K X 16 and a cache memory of 2K (2048) words. The cache uses a direct mapping with block size of eight worc. How many bits are there in tag, index, blocks, and word fields of address format? How many bits are there in each word of the cache? Show division of functions and include valid bit. A. B. C. The cache has how many blocks?

Answer 1

From the question:

Access time of a cache is 80ns.

Access time of a main memory is 1200ns.

Instructions directed to read is 85% = 0.85

Instructions directed to write is 15% = 0.15

Hit ratio = 92% = 0.92

Miss ratio = 8% = 0.08

a.

What is the average access time of the system considering only memory read requests.

average access time to read = hit ratio x cache access time + (1 - hit ratio) x main memory access time

average access time to read = 0.92 x 80 + (1 - 0.92) x 1200 =169.6 ns

b.

Average access time of the system for both read and write requests

Read average would take those 85% of overall requests.

average read access time is 169.6 ns

Read average would take those 85% of overall requests = 0.85 x 190 = 161.5 ns

Write average would take those 15% of overall requests and the main memory access time of 1000ns.

= 0.15 x 1000ns.

=150ns

Average access time of the system for both read and write requests = 161.5ns + 150 ns = 311.5 ns

c)

What is the hit ratio taking into consideration the write cycles

Consider the write cycles means that we should discard write requests from the given overall hit ratio.

So we have hit ratio read = read_requests_percentage x hit_ratio = 0.85 x 0.92 = 0.782

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