Question

A digital computer has a memory unit with 26 bits per word. The instruction set consists of756 different operations. All instructions have an operation code part (opcode) and anaddress part (allowing for only one address). Each instruction is stored in one word ofmemory.

Answer 1

Answer:

hello your question is incomplete attached below is missing part of the  question

Determine how many bits are needed for the opcode

answer : 7 bits

Explanation:

Given data:

memory unit = 26 bits per word

Instruction set consists = 756 different operations

Determine how many bits are needed for the opcode

First we determine The number of bits required to represent 756 operations

= $Log^{756} _{2}$ = 9.56 ≈  10 bits

finally the number of bits needed for the opcode can be determined by

$2^9 = 512 ,$   and $2^{10} = 1024$

since 2^10 > 756 hence the number of bits will be 10 bits