Recent content by johnnyrocket

I meant to say 2x as fast on N processors.

I think the correct formula is as follows, but I'm not sure: Speedup = 1/((non parallel) +(parallel/# processors)). So given that they tell me the run time is 2x as fast on 6 processors, I'm guessing that is speedup. So I would say: 2 = 1/((.40)+(.60/N)) Which makes N = 6?

I'm not exactly sure what he's suggesting. The 2nd part of the question asks for mips rating for the single and N processor systems, so that's what I was trying to solve for. Can you offer me any guidance? Am I on the right track?

Ok. I think the MIPS equation is: MIPS = number of Instructions / (time to execute * 1 million) So I would say that I can assume that if each instruction takes 1 cycle, with a 4GHz processor, I can execute 4 *10^9 instructions per second. So: MIPS = 16million / (.0004 seconds to execute...

I need some help with this problem I am trying to solve if anyone can help. Each instruction on a certain computer system requires 1 cycle for any type of instruction. This system includes N processors and is used to execute a program contain 16 million instructions and 60% of the code...

if the $a0 register has a 32 bit value in it, how will shifting it right 31 bits produce anything other than a 1 or 0 (the 1st bit)? If I have 10000000000000000000000000000000 and I shift right 31 bits I am left with a 1(with 31 0's in front of it to make it 32 bits) right?

thanks for the reply. If I use 1 instruction to put the immediate value into the register, and the 2nd value to shift, that should do it right? li $a, value srl $v,$a,31

I need to write a MIPS program (using only 2 true-ops) to take in 1 signed integer argument into register $a0, and return (into $v0) a 1 if the input was negative, and a 0 (into $v0) otherwise. Can anyone offer help. I'm not sure since I can only use 2 instructions.