1) Note: If a data hazard is NOT prevented by data forwarding, then the pipeline stalls. Assume data forwarding is working, does the following code stall? "stall" means it needs an extra nop instruction inserted, If the code stalls then list the labels of the instruction(s) that cause a stall. (Assume you have completed part2a and part2b of the project.) This question is asking if stall signal is set to '1', four cases. See lecture 20 for four cases, and sample pipeline stages. L1: lw $3, 50($3) L2: add $2, $3, $4 L3: sw $2, 20($2) L4: lw $5, 30($2) L5: lw $6, 40($5) L6: or $5, $6, $5 2) Draw the pipeline diagram, you may use the style similar to Lecture 19 and 20, or similar to HW7 simplified to use one clock at one ns for every stage. There are no nop instructions to be fetched, yet show the effect of hazard prevention by repeating the stage name when an instruction is stalled. We have full data forwarding and hazard prevention that automatically inserts nop as needed into the pipeline. (Assume you have completed part2a and part2b of the project.) add $5, $6, $7 lw $6, 60($5) sub $7, $6, $5 or $8, $4, $6 3) How many total clock cycles, start to completion, are required by your expanded code in part 2) 4) Assuming all instructions are in the cache: If the nop instruction was actually in the code, rather than being inserted into the pipeline, would the total clock cycles be the same as in part 3)