a. If there is no forwarding or hazard detection, we need to insert nops to ensure correct execution. In this case, we have a data hazard between the first and second instructions, since the second instruction uses the value of $s5 that is being modified by the first instruction.
To avoid this hazard, we can insert a nop after the first instruction:
Add $s5,$s2,$s1
Nop
Lw $s3,4($s5)
Lw $s2,0($s2)
Or $s3,$s5,$s3
Sw $s3,0($s5)
We also have a data hazard between the second and third instructions, since the third instruction uses the value of $s2 that is being modified by the second instruction. To avoid this hazard, we can insert a nop after the second instruction:
Add $s5,$s2,$s1
Nop
Lw $s3,4($s5)
Nop
Lw $s2,0($s2)
Or $s3,$s5,$s3
Sw $s3,0($s5)
b. Now, we can use nops only when a hazard cannot be avoided by changing or rearranging these instructions. In this case, we can rearrange the second and third instructions to avoid the hazard between them:
Add $s5,$s2,$s1
Lw $s2,0($s2)
Lw $s3,4($s5)
Or $s3,$s5,$s3
Sw $s3,0($s5)
We still have a data hazard between the first and second instructions, but it cannot be avoided by rearranging the instructions. Therefore, we need to insert a nop to ensure correct execution:
Add $s5,$s2,$s1
Nop
Lw $s2,0($s2)
Lw $s3,4($s5)
Or $s3,$s5,$s3
Sw $s3,0($s5)
This is the final solution that uses nops only when necessary to avoid hazards.
To know more about data hazard, visit: https://brainly.com/question/17184351
#SPJ4
