Question

(10pts) (A) Identify hazards (including type of the hazard) in the following code. Write hazards next to each instruction. Write none if there is no hazard. Assume that each instruction could have more than one hazard and I5 does not create a control hazard, Type of Hazards Instructions I: LABEL:lw Ss2, 0(Ss0) none 12: 13: 14 15: 16: add Ss1, Ss6, $sl add Stl, Ss0, $s2 and St1, St, $s3 sw St1 0(Ss0) beq St1, St7, LABEL (B) How many clock cycles does it take to resolve hazards assuming no forwarding is available and EX is only allowed after WB cycle?

Answer 1

actually a data harard arises if one instruction needs data that is not ready yet.

LABEL : lw $s2, 0($s0) it is not hazardous because it is ready to perform operation.

add $t1,$s0,$s2 it is data hazardous because it not ready to perform due to insufficient data

and $t1, $t1, $s3 it is also data hazardous because it is not ready to perform due to insufficient data

sw $t1, 0($s0) it is also data hazardous

beq $t1,$t7, Label it is also data hazardous

add $s1, $s6, $s1 it is of none it is not hazardous

forwarding is performed only via the register file. Branch outcomes and targets are not known until the end of the end of the execute stage. all instructions are introduced to the pipeline prior to the point are flushed.