[8 points] A digital computer has a memory unit with 24 bits per word. The instr

Question

[8 points] A digital computer has a memory unit with 24 bits per word. The instruction set consists of 135 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.

How many bits are needed for the opcode?

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

What is the maximum allowable size for memory?

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

Explanation / Answer

Answer:---------------

How many bits are needed for the opcode?=>>>>>>>>>>>>

As there are 135 instructions, So there will be 8 bits for representing one operation, Thus, 8 bits are needed for the opcode.

How many bits are left for the address part of the instruction?=>>>>>>>>

An instruction is stored ia a word with 24 bits. So, there will be (24 - 8)=16 bits bits are left for the address part of the instruction.

What is the maximum allowable size for memory?=>>>>>>>>>>>>>>>

Maximum allowable size for memory will be 216 .

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

the largest unsigned binary number that can be accommodated in one word of memory will be, (111111111111111111111111)2 means 224 - 1

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