# GATE | GATE-CS-2016 (Set 2) | Question 41

• Difficulty Level : Easy
• Last Updated : 21 Nov, 2019

Consider a processor with 64 registers and an instruction set of size twelve. Each instruction has five distinct fields, namely, opcode, two source register identifiers, one destination register identifier, and a twelve-bit immediate value. Each instruction must be stored in memory in a byte-aligned fashion. If a program has 100 instructions, the amount of memory (in bytes) consumed by the program text is ____________

[Note that this was originally a Fill-in-the-Blanks question]
(A) 100
(B) 200
(C) 400
(D) 500

Explanation: One instruction is divided into five parts,

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• (1): The opcode- As we have instruction set of size 12, an instruction opcode can be identified by 4 bits, as 2^4=16 and we cannot go any less.
• (2) & (3): Two source register identifiers- As there are total 64 registers, they can be identified by 6 bits. As they are two i.e. 6 bit + 6 bit.
• (4): One destination register identifier- Again it will be 6 bits.
• (5): A twelve bit immediate value- 12 bit.

```= 4 + 6 + 6 + 6 + 12
= 34 bit
= 34/8 byte
= 4.25 byte ```

As given Each instruction must be stored in memory in a byte-aligned fashion,4.25 is not byte alignment, memory address should be 0,1,2,3,4,5,6,7…….. so it should be 5 bytes.

As there are 100 instructions, we have a size of 5*100= 500 bytes.

Hence (D) 500 is the answer.

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