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?

  • 1
    What are records? Do you perhaps mean registers?
    – Erik Eidt
    Oct 29, 2016 at 17:02

1 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.

  • Thank you, I'm still new to all of "this" and sometimes questions like these can feel overwhelming. Oct 29, 2016 at 17:56

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