Question
Consider a cache with 64 blocks and a block size of 16 bytes. The byte address of 1200 maps to ______ block number.(A) 10(B) 11(C) 64(D) 16
Step 1
The memory block number is calculated by dividing the byte address by the block size. In this case, the byte address is 1200 and the block size is 16 bytes. So, the memory block number is $\frac{1200}{16} = 75$. Show more…
Show all steps
Your feedback will help us improve your experience
Vysakh M and 65 other AP CS educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the number of bits in the cache index and tag for a direct mapped cache of size $32 \mathrm{~KB}$ with block size of 32 bytes. The CPU generates 48 -bit addresses. (A) 33,15 (B) 15,10 (C) 10,33 (D) 15,33
Computer Organization and Architecture
Cache and Main Memory, Secondary Storage
Consider a 32 -bit microprocessor that has an on-chip $16 \mathrm{~KB}$ four-way set associative cache. Assume that the cache has a line size of four 32 -bit words. Then the word in the memory location ABCDE8F8 will be mapped to (A) 143 rd set (B) 815 th set (C) 255 th set (D) 0 th set
Given the following specifications for an external cache memory: Four-way set associative, Line size of two 16 -bit words; Able to accommodate a total of $4 \mathrm{~K} 32$ -bit words from main memory. Used with a 16-bit processor that issues 24-bit address. Then the number of bits used to represent set field is (A) 2 -bits (B) 10 -bits (C) 12 -bits (D) 14 -bits
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD