If the set of MIPS instructions were changed to accommodate 128 records theory question

Question

If the set of MIPS instructions were changed to accommodate 128 records and 4 times more instructions type-I, which would be the largest hexadecimal immediate value that could be supported, keeping the 32-bit instructions?

Answer 1

Assuming you mean registers not records:

Most of what you're asking about is a matter of encodings which are done with bits.

4x more means 2 more bits are needed, because log₂(4) = 2, or in other words, 4 = 2².

The current Type-I instructions:

I instructions are converted into machine code words in the following format:

opcode  rs         rt   IMM
6 bits  5 bits  5 bits  16 bits

The current register fields are 5 bits which can accommodate designation of 2⁵ = 32 different registers. To take that to 128 registers, you'd need log₂(128) = 7 (also that 2⁷ = 128) bits, meaning 2 more bits than now.

So in total:

4x more opcodes:   2 more bits are needed.
4x more registers: 2 more bits are needed per register field for a total of 4 more bits

That all adds up to 6 more bits required, which would leave 10 bits left for the immediate field instead of the current 16.

10 bits can be used to represent an unsigned range from 0 to 2¹⁰-1 or from 0 to 1023. Or 10 bits can be used to represent a signed range from -2⁹ to +2⁹-1, or -512 to 511. There are other encodings possible (one's complement, or, say ranging, from 1 to 1024), but there are only 2¹⁰ = 1024 possible values that can be differentiated in 10 bits.

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