A digital computer has a memory unit with 30 bits per word. The instruction set

Question

A digital computer has a memory unit with 30 bits per word. The instruction set consists of 58 different operations. All instructions have an operation code part (opcode) and two address fields: one for a memory address and one for a register address. This particular system includes eight general-purpose, user-addressable registers. Registers may be loaded directly from memory, and memory may be updated directly from the registers. Direct memory-to-memory data movement operations are not supported. Each instruction stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are needed to specify the register? c. How many bits are left for the memory address part of the instruction? d. What is the maximum allowable size for the memory? e. What is the largest unsigned binary number that can be accommodated in one word of the memory?

Explanation / Answer

since it has 58 operations 2^6 can be accomodated in this.

So 6 bits are required for opcode representation.

Since it has 8 registers you will be 2^3 = 3 bits to specify the register number.

so in total we have 30 bits.

30-6-3-1 = 20 bits (1 bit to specify whether register or memory)

Maximum allowable size will be 2^20 = 1 MB

It should be 0xFFFFF in hexadecimal as 20 bits are there.

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