A digital computer has a memory unit with 48 bits per word. The instruction set

Question

A digital computer has a memory unit with 48 bits per word. The instruction set consists of 240 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 needed to specify the register? How many bits are left for the address part of the instruction?

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

Explanation / Answer

Q: a. How many bits are needed for the opcode?

A: 2^8 = 256 which is larger than 240.

Therefore the operation code needs 8 bits.

Q: b. How many bits are left for the address part of the instruction?

A: 48-8=40. There sare 48 bits available for the address part of the instruction.

Q: c. What is the largest unsigned binary number that can be accommodated in one word of memory?

A: 2^48-l i.e = 281,474,976,710,656 -1 = 281,474,976,710,655

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