Instruction Format

Question

Consider a computer has 64 registers and support 15 different instructions. Each instruction has 4 fields i.e. opcode, source register, destination register and immediate value of 6 bits. If each instruction in byte aligned and 50 instructions are present in memory, then the size of memory needed to store these instructions are ________ (in byte).

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I know the answer

My question is, in this question what is difference between computer has 15 instruction and memory has 50 instructions?

Comments

will u provide the detail sol plz? i don't know how to solve this problem

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Number of opcodes (ADD, LOAD, MUL etc.) are 15.
These opcodes are used for writing the memory instructions when we design the main memory.
The combination of these 15 opcodes can be used to write as many as 50 instructions.

These instructions can be one-address , two-address, can have one or more registers etc. That is a separate thing.

Say, while designing our memory we are writing instructions like this:

ADD R1,R2 (......say this instruction comes 10 times)
LOAD X (......say this instruction comes 14 times)
STORE (......say this instruction comes 16 times)
PUSH A (......say this instruction comes 5 times)
MUL X,Y (......say this instruction comes​​​​​​​ 5 times)

So total 50 instructions we have used to design our main memory.

 

Answer 1

My question is, in this question what is difference between computer has 15 instruction and memory has 50 instructions?

CPU/Processor has 15 Instructions mean "15 Instruction Types..i.e. Instruction set has 15 types of instructions or you can say 15 Opcodes... like ADD, MUL, DIV, MOV, LOAD...etc"

 Memory has 50 instructions mean that the Program stored has 50 Instructions....like there could be ADD 10 times, JMP 20 times..etc..

Comments

So what is the answer here? Is it 150?

I mean 50 x 3 bytes per instruction??

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yes 150

Answer 2

Opcode

Rs

Rd

Imm.

instruction size = (opcode)+R_s + R_d + immediate value = 4+6+6+6=22bits

memory is byte a=Addressable  so we will need 24 bit or 3 Bytes

there are 50 instructions in program so memory needed =  3 * 50 bytes = 150 Bytes

Answer 3

Computer supports 15 different instruction, it means our Opcode could be 15, that is we have 15 different operations like DIV, MUL, ADD, SUB etc.

And there are 50 instructions in memory, that is Program has 50 such instruction in which Opcode can have 15 different operations.