Respuesta :
Answer:
Explanation:
1) Effect on R-type instructions:
log2(128) = 7 [since 128=27]
opcode = 6
rs = 7
rt = 7
rd = 7
shamt = 5
funct = 6
Here all the bits ends up with 2 additional bits. As there are 6 bits for opcode in R-type instruction the size of word is 44.
.2) Effect on I-type Instruction:
log2(128) = 7 [since 128=27]
opcode = 6
rs = 7
rt = 7
immediate = 16
The word size of the I-type instruction is also 44.
3)
The programs will take long time to do the operations if the instruction encoding is long.
It means the format must be larger as the increases are not aligned.
The code will decrease if there are more registers.
Because more registers can hold more information without going to memory.
As this leads to fewer calls to load or store the code will decrease.
Number of bits needed to address a register is 7 bits
MIPS register file :
Given information in question:
Number of MIPS registers = 128
Number of bits needed = log₂128
Number of bits needed = 7 bits
Increasing number bits for opcode = (6 + 2)
Increasing number bits for opcode = 8
Computation:
1. R-type instruction
Op-code = 6 bits
rs = 5 bits
rt = 5 bits
rd = 5 bits
shamt = 5 bits
funct = 6 bits
The size of the opcode field will be increased by two bits, to eight bits.
The size of the rs,rt, and rd fields is also increased to 7 bits.
2.I-type instruction
Op-code = 6 bits
rs = 5 bits
rt = 5 bits
Immediate = 16 bits
The size of the opcode field will be increased by two bits, to eight bits.
The size of the rs,rt fields is also increased to 7 bits.
3. More complicated operations will be implemented in one instructions rather than numerous instructions as a result of the reduced register leaking problem. The program's size will be reduced as a result of this.
The size of the instruction word will be raised when the needed bits are added to the opcode and register fields, which will increase the size of the programmed.
Find out more information about 'Opcode'.
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