Question

A digital computer has a memory unit with 32 bits per word. The instruction set consists of 147 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 memory?

e) What is the largest unsigned binary number that can be accommodated in one word of memory?

Answer 1

a) How many bits are needed for the opcode?

8 bits

Explanation:

147 opcode implies 2⁸ = 256 bits

256 > 147

b) How many bits are needed to specify the register?

3 bits

Explanation:

Number of registers is 8

2³= 8

c) How many bits are left for the memory address part of the instruction?

21 bits

Explanation:

32 - 8 - 3 = 21

d) What is the maximum allowable size for memory?

16MB

Explanation:

Maximum memory = 2²⁴ = 2⁴ x 2²⁰ = 16MB

e) What is the largest unsigned binary number that can be accommodated in one word of memory?

2³² - 1

Explanation:

Word length is 32 bits

Then, Largest unsigned binary number is 2³² - 1

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