GATE CSE 2014 | Set 1 | Question: 9

Question

A machine has a $32\text{-bit}$ architecture, with $1\text{-word}$ long instructions. It has $64$ registers, each of which is $32$ bits long. It needs to support $45$ instructions, which have an immediate operand in addition to two register operands. Assuming that the immediate operand is an unsigned integer, the maximum value of the immediate operand is ____________

Comments

1.What is meaning of 1 word long instruction? and  2. What would be the length of instruction if it was 2 word long instruction ? @GO Classes

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$1\space word$ long instruction means that the it is $32$ bits (since we have $32$ bit architecture).

If the instruction was of $2\space words$ long then it would be $64$ bits long.

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Instruction size is 32 bits as in question they mentioned 32-bit archeicture

size of register is given to be 32 bits but this information is of no use to solve the question !

Answer 1

$64$ registers means $6$ bits $(\lceil \log_2 64 \rceil = 6)$ for a register operand. So, $2$ register operands require $12$ bits. Now, $45$ instructions require another $6$ bits for opcode $(\lceil \log_2 45 \rceil = 6)$. So, totally $18$ bits. Now we have $32 - 18 = 14$ bits left for the immediate operand. So, the max value will be $2^{14} - 1 = 16383$ (as the operand is unsigned we do not need a sign bit and with $14$ bits we can represent from $0$ to $2^{14} -1$)

Comments

same doubt @srestha

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But ain't instruction length 16 bits (1 word, given in question)?

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@GATE 5 It is given 32 bit architecture which is also the word length. 

Answer 2

Instruction 32 bit = (6+6+6+14) bit 

⌈log45⌉=6 bit opcode

R1 ⌈log64⌉=6 bit

R2 ⌈log64⌉=6 bit

remaining 14 bit oprand

It needs to support 45 instructions,   means types of instruction supported is of45 

Comments

how to know operand are 6 bit means register are 64 bit so what are relation between register and operand size .?

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Instruction size if of 32 bits.

according to given information here we have 2 different types oprands and one opcode.

so first find register mode oprand bits, how -> given total 64 rigesters, take log(64)=6

now find opcode field bits, how -> given 45 types of instructions, take ceil{log(45)}=6

now find  immediate operand field bits, how -> total bits- other field bits

                                                                  = 32 - (6+6+6)    (2 times 6 bcz two register oprands given)

                                                                   = 14

*note: base of log is 2.

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@mint why here we left other remaining bits for 45 instrustions means from 2^6=64

45 will b in use so some will left ...why we not use them ?

Answer 3

1 Word = 32 bits

Each instruction has 32 bits

To support 45 instructions, opcode must contain 6-bits

Register operand1 requires 6 bits, since the total registers
are 64.

Register operand 2 also requires 6 bits.

14-bits are left over for immediate Operand Using 14-bits,
we can give maximum 16383,

Since 2^14=16384(from 0 to 16383) 

Comments

Had it been 2-word long instruction ,the instruction length would be 64-bit long ?

@MiNiPanda

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@HeadShot Yes. Then the no. of bits for operand would increase.

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@Anu007  
\[
\text{ 2's Complement Range} = \left[ -2^{n-1}, 2^{n-1} - 1 \right]
\]

Answer 4

Answer : 16383

no. of bits for 45 instructions / opcode = ⌈$\log _{2}45$⌉ = 6

no. of bits for each register operand = ⌈$\log _{2}64$⌉ =6

6 (opcode)

6 (r1)

6 (r2)

p  (immediate)

no. of immediate operand bits = no. of bits for machine - (6 + 6 + 6 )

no. of immediate operand bits = 32 - 18 = 14 bits 

for max value of immediate operand you have to put all 1's in all 14 immediate bits 

so, max value of immediate operand  = $2^{14}$-1 = 16383

Comments

Why  -1???

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for n bits, maximum unsigned number that can be represented is ( 2^n - 1 ),
for example : consider 4 bits 
now, maximum unsigned number possible is ( 1111 ), which is equivalent to 15, which is equivalent to 2^4 -1 

Answer 5

16,383.

32 bit architecture means word size of CPU is 32 bit, which means instruction size = 32 bit.

It is given that register memory size is 64 register, to access 1 register 6 bits are required(log 64 base2)

The size of ISA(instruction set architecture) = 45 which require 6 bit for opcode(as 5 won’t be sufficient)

So, now the feilds of instruction will have bifurcation like this,

opcode+register1+register2+Immediate value = 32  bit

Immediate value = 32 – (6+6+6) = 14

Now, Immediate value is given to be unsigned and the range of unsigned is 0 to 2^(n) – 1

So maximum unsigned value that can be represented in this instruction = 2^(14) – 1 = 16,383.

Answer 6

1. Bit Allocation:

• Opcode: To support 45 instructions, we need 6 bits for the opcode ($2^6 = 64,$ which is greater than $45$).
• Register Operands: To address 64 registers, we need 6 bits for each register operand ($2^6 = 64$). So, two register operands require 12 bits.
• Immediate Operand: The remaining bits in the 32-bit instruction word are available for the immediate operand. This leaves $32 - 6 - 12 = 14$ bits for the immediate operand.

2. Maximum Value:

• With 14 bits, the maximum unsigned integer value that can be represented is 2^14 - 1 =  16383

Therefore, the maximum value of the immediate operand in this machine architecture is $16383$.