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

(a)
we need to complete 240 operations so 2^8=256 and 2^7=128 .
So we need minimum 8 bits for opcode because 128 < 240.

(b)

(total bits - opcode )/4=48-8/4=10 bits
10 bits are required to specify the register .

Left bits = (48)-(8+10) = 30 bits

(c) now largest unsigned binary number = (2^48) - 1 is the largest unsigned binary number that can be accommodated in one word of memory .

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