Assignment 3 Answers - CGS 3269 Computer Architecture...

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CGS 3269 Computer Architecture Concepts Summer 2011 Solution for Assignment #3 1. How many bits would you need to address a 2M x 32 memory if a. The memory is byte-addressable? There are 2M × 4 bytes which equals 2 × 2^20 × 2^2 = 2^23 total bytes, so 23 bits are needed for an address. b. The memory is word-addressable? There are 2M words which equals 2 × 2^20= 2^21, so 21 bits are required for an address 2. How many bits are required to address a 4M x 16 main memory if a. Main memory is byte-addressable? There are 4M × 2 bytes which equals 2^2 × 2^20 × 2 = 2^23 total bytes, so 23 bits are needed for an address b. Main memory is word-addressable? There are 4M words which equals 2^2 × 2^20 = 2^22, so 22 bits are required for an address 3. Suppose we have 4 memory modules instead of 8 (refer to the slides in topic 4.6 in the lecture. Draw the memory modules with the addresses they contain using: a. High-order interleaving and b. Low-order interleaving.
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4. Consider the MARIE program below. a. List the hexadecimal code for each instruction.
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Assignment 3 Answers - CGS 3269 Computer Architecture...

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