Suppose we have a byte-addressable computer using direct mapping with 16-bit main memory addresses and 32 blocks of cache. If each block contains 8 bytes, determine the size of the block field.
Added by Rhonda Y.
Close
Step 1
Since we have 32 blocks of cache, we need to find the smallest number of bits that can represent 32. To do this, we can use the formula: log2(N), where N is the number of blocks. In this case, N = 32. Using the formula, we have log2(32) = 5. Show more…
Show all steps
Your feedback will help us improve your experience
Anecia Mcmurrin-Bala and 54 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
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
Computer Organization and Architecture
Cache and Main Memory, Secondary Storage
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
Given a 4-way set associative cache with 64 sets (S). Each cache block contains 32 bytes of data (B). Provide answers to each of the following questions. How many bits (b) will be needed for block offset (B)? b = bits. How many bits (s) will be needed for sets (S)? s = bits. What is the size of this cache? size = bytes.
Akash M.
Recommended Textbooks
Computer Science and Information Technology
Introduction to Programming Using Python
Computer Science - An Overview
Watch the video solution with this free unlock.
EMAIL
PASSWORD