Consider a word-addressable computer with 32 bits per word. The instruction set consists of 30 different All instructions have an operation code field, a mode field to specify one of 7 possible addressing modes, a register address field to specify one of 60 available registers, and a memory address field. Each instruction is stored in one word of memory. What is the maximum allowable size for memory? Use KB, MB, or GB as a unit.

Consider a word-addressable computer with 32 bits per word. The instruction set consists of 30 different All instructions have an operation code field, a mode field to specify one of 7 possible addressing modes, a register address field to specify one of 60 available registers, and a memory address field. Each instruction is stored in one word of memory. What is the maximum allowable size for memory? Use KB, MB, or GB as a unit.

Name: Consider a word-addressable computer with 32 bits per word. The instru
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Description: Consider a word-addressable computer with 32 bits per word. The instruction set consists of 30 different All instructions have an operation code field, a mode field to specify one of 7 possible addressing modes, a register address field to specify on

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter4: Processor Technology And Architecture

Section: Chapter Questions

Problem 2PE: If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the...

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Consider a word-addressable computer with 32 bits per word. The instruction set consists of 30 different instructions. All instructions have an operation code field, a mode field to specify one of 7 possible addressing modes, a register address field to specify one of 60 available registers, and a memory address field. Each instruction is stored in one word of memory. What is the maximum allowable size for memory? Use KB, MB, or GB as a unit.Instruction size = 32bit , allow instruction = 30 add bit = 5 Registere = 60 add bit = 8Modes = 7 add bit = 3Then: 5+8+3+address field = 32 so address filed = 16bitsNow maximum allowable size for memory = 2 ^16 * 4 bytes = 256bKB
Suppose that you have a computer with a memory unit of 24 bits per word. In this computer, the assembly program's instruction set consists of 198 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. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. How many additional instructions could be added to this instruction set without exceeding the assigned number of bits? Discuss and show your calculations.
The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify one of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:a. How large must the mode field be?b. How large must the register field be?c. How large must the address field be?d. How large is the opcode field?
The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with four fields: an opcode field; a mode field to specify one of seven addressing modes; a register address field to specify one of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:Q.) How large is the opcode field?
A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address, similar to Marie). Each instruction is stored in one word of memory. a) How many bits are needed for the opcode? b) How many bits are left for the address part of the instruction? c) What is the maximum allowable size for memory?
A certain computer has a memory of 1M words, and each word is 32 bits long. Each instruction is 32 bits long and is consisted of an opcode field, a register address field to specify one of 32 registers, and a memory address field. How large must the register field be? How large must the address field be? How many different opcodes can be supported by this format? Show your reasoning
The memory unit of a computer has 2M words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 4 addressing modes; a register address field to specify one of 9 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following: b) How large must the register field be?
A computer has a 256K word addressable memory Module with 16 bits per word. The instruction set consists of 166 different instructions. 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. Based on the above, answer the following questions: a. How many bits are there in the main memory? (Represent it in power of 2) b. How many bits are needed for the opcode? c. How many bits are left for the address part of the instruction? d. How many additional instructions can be added to the existing 166 without affecting the assigned size of the opcode part? Justify.
A computer employs RAM chips of 128 x 8 and ROM chips of 512 x 8. The computer system needs 1K bytes of RAM, 2K bytes of ROM, and two interface units, each with two registers. A memory-mapped 1/0 configuration is used. The two highest-order bits of the 16-bit address bus are assigned 11 for RAM, 10 for ROM, and 01 for interface registers.a. How many RAM and ROM chips are needed?b. Draw a memory-address map for the system.c. Give the address range in hexadecimal for RAM, ROM, and interface