Problem 1: Loop Unrolling [18 points] In this problem, we will use the ...

Instruction Reservation Station IS EX WR Comments (if appropriate)
L.D F2, 0(R1) Integer 1 1 2 3
L.D F4, 8(R1) Integer 2 2 3 4
DIV.D F6,F2,F4 FP Mul/Div 1 3 5-19 20 RAW on F4
MUL.D F8,F6,F6 FP Mul/Div 2 4 26-40 41 RAW on F6, Structural Hazard on FU
ADD.D F6,F2,F4 FP Add 1 5 6-9 10
MUL.D F10,F6,F6 FP Mul/Div 3 6 11-25 26 RAW on F6
S.D F8, 0(R1) Integer 1 7 42 RAW on F8
S.D F10, 8(R1) Integer 2 8 27 RAW on F10
DADDI R1,R1,16 Integer 3 9 10 11
BNEZ R1, LOOP Integer 4 10 12 RAW on R1
L.D F2, 0(R1) Integer 3 11 13 14 Structural Hazard on FU
L.D F4, 8(R1) Integer 5 12 14 15 Structural Hazard on FU
DIV.D F6,F2,F4 FP Mul/Div 4 13 20-34 35 RAW on F2 and F4, Str. Haz. On FU
MUL.D F8,F6,F6 FP Mul/Div 1 20 41-55 56 Str. Haz. on Reservation Station + FU
ADD.D F6,F2,F4 FP Add 1 21 22-25 27 CDB Conflict
MUL.D F10,F6,F6 FP Mul/Div 3 26 35-49 50 RAW on F6, Str. Haz. on Res. Sta. + FU
S.D F8, 0(R1) Integer 3 27 57 RAW on F8
S.D F10, 8(R1) Integer 2 28 51 RAW on F10
DADDI R1,R1,16 Integer 4 29 30 31
BNEZ R1, LOOP Integer 5 30 32 RAW on R1
Table 2. Execution profile using Tomasulo’s algorithm.
Problem 3 [8 points]
Part A [3 points]
What technique does Tomasulo’s employ to eliminate WAR and WAW hazards Why does it
work Why doesn’t it also eliminate RAW
Tomasulo’s uses register renaming to avoid these hazards. It works because by changing the
5
destination registers, the 2 instructions which cause the hazard no longer need to access the
same
register thus eliminating the hazards. It doesn’t work on RAW dependencies because no
matter
what we rename the register to, the ’reading’ instruction still needs the result from the
’writing’
instruction in order to execute.
Part B [2 points]