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A Digital computer has a memory unit with 32 bits per word (memory element). The instruction set consists of 260 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. bits. How many bits are needed for the opcode? bits How many bits are left for the address part of the instruction? MBytes What is the maximum allowable size for memory (in MegaBytes) that can be accessed directly?

A Digital computer has a memory unit with 32 bits per word (memory element). The instruction set consists of 260 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. bits. How many bits are needed for the opcode? bits How many bits are left for the address part of the instruction? MBytes What is the maximum allowable size for memory (in MegaBytes) that can be accessed directly?

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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A Digital computer has a memory unit with 32 bits per word (memory element). The instruction set consists of 260 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. bits. How many bits are needed for the opcode? bits How many bits are left for the address part of the instruction? MBytes What is the maximum allowable size for memory (in MegaBytes) that can be accessed directly?
Transcribed Image Text:A Digital computer has a memory unit with 32 bits per word (memory element). The instruction set consists of 260 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. bits. How many bits are needed for the opcode? bits How many bits are left for the address part of the instruction? MBytes What is the maximum allowable size for memory (in MegaBytes) that can be accessed directly?
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