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Lecture 6: Static ILP

Lecture 6: Static ILP. Topics: loop analysis, SW pipelining, predication, speculation (Section 2.2, Appendix G) Please hand in Assignment 1 now Assignment 2 posted; due in a week. Smart Schedule. Loop: L.D F0, 0(R1) stall

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Lecture 6: Static ILP

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  1. Lecture 6: Static ILP • Topics: loop analysis, SW pipelining, predication, speculation • (Section 2.2, Appendix G) • Please hand in Assignment 1 now • Assignment 2 posted; due in a week

  2. Smart Schedule Loop: L.D F0, 0(R1) stall ADD.D F4, F0, F2 stall stall S.D F4, 0(R1) DADDUI R1, R1,# -8 stall BNE R1, R2, Loop stall Loop: L.D F0, 0(R1) DADDUI R1, R1,# -8 ADD.D F4, F0, F2 stall BNE R1, R2, Loop S.D F4, 8(R1) • By re-ordering instructions, it takes 6 cycles per iteration instead of 10 • We were able to violate an anti-dependence easily because an • immediate was involved • Loop overhead (instrs that do book-keeping for the loop): 2 • Actual work (the ld, add.d, and s.d): 3 instrs • Can we somehow get execution time to be 3 cycles per iteration?

  3. Loop Unrolling Loop: L.D F0, 0(R1) ADD.D F4, F0, F2 S.D F4, 0(R1) L.D F6, -8(R1) ADD.D F8, F6, F2 S.D F8, -8(R1) L.D F10,-16(R1) ADD.D F12, F10, F2 S.D F12, -16(R1) L.D F14, -24(R1) ADD.D F16, F14, F2 S.D F16, -24(R1) DADDUI R1, R1, #-32 BNE R1,R2, Loop • Loop overhead: 2 instrs; Work: 12 instrs • How long will the above schedule take to complete?

  4. Scheduled and Unrolled Loop Loop: L.D F0, 0(R1) L.D F6, -8(R1) L.D F10,-16(R1) L.D F14, -24(R1) ADD.D F4, F0, F2 ADD.D F8, F6, F2 ADD.D F12, F10, F2 ADD.D F16, F14, F2 S.D F4, 0(R1) S.D F8, -8(R1) DADDUI R1, R1, # -32 S.D F12, 16(R1) BNE R1,R2, Loop S.D F16, 8(R1) • Execution time: 14 cycles or 3.5 cycles per original iteration

  5. Loop Unrolling • Increases program size • Requires more registers • To unroll an n-iteration loop by degree k, we will need (n/k) • iterations of the larger loop, followed by (n mod k) iterations • of the original loop

  6. Automating Loop Unrolling • Determine the dependences across iterations: in the • example, we knew that loads and stores in different iterations • did not conflict and could be re-ordered • Determine if unrolling will help – possible only if iterations • are independent • Determine address offsets for different loads/stores • Dependency analysis to schedule code without introducing • hazards; eliminate name dependences by using additional • registers

  7. Superscalar Pipelines Integer pipeline FP pipeline Handles L.D, S.D, ADDUI, BNE Handles ADD.D • What is the schedule with an unroll degree of 4?

  8. Superscalar Pipelines Integer pipeline FP pipeline Loop: L.D F0,0(R1) L.D F6,-8(R1) L.D F10,-16(R1) ADD.D F4,F0,F2 L.D F14,-24(R1) ADD.D F8,F6,F2 L.D F18,-32(R1) ADD.D F12,F10,F2 S.D F4,0(R1) ADD.D F16,F14,F2 S.D F8,-8(R1) ADD.D F20,F18,F2 S.D F12,-16(R1) DADDUI R1,R1,# -40 S.D F16,16(R1) BNE R1,R2,Loop S.D F20,8(R1) • Need unroll by degree 5 to eliminate stalls • The compiler may specify instructions that can be issued as one packet • The compiler may specify a fixed number of instructions in each packet: • Very Large Instruction Word (VLIW)

  9. Loop Dependences • If a loop only has dependences within an iteration, the loop • is considered parallel  multiple iterations can be executed • together so long as order within an iteration is preserved • If a loop has dependeces across iterations, it is not parallel • and these dependeces are referred to as “loop-carried” • Not all loop-carried dependences imply lack of parallelism

  10. Examples For (i=1000; i>0; i=i-1) x[i] = x[i] + s; For (i=1; i<=100; i=i+1) { A[i+1] = A[i] + C[i]; S1 B[i+1] = B[i] + A[i+1]; S2 } For (i=1; i<=100; i=i+1) { A[i] = A[i] + B[i]; S1 B[i+1] = C[i] + D[i]; S2 } For (i=1000; i>0; i=i-1) x[i] = x[i-3] + s; S1

  11. Examples For (i=1000; i>0; i=i-1) x[i] = x[i] + s; No dependences For (i=1; i<=100; i=i+1) { A[i+1] = A[i] + C[i]; S1 B[i+1] = B[i] + A[i+1]; S2 } S2 depends on S1 in the same iteration S1 depends on S1 from prev iteration S2 depends on S2 from prev iteration For (i=1; i<=100; i=i+1) { A[i] = A[i] + B[i]; S1 B[i+1] = C[i] + D[i]; S2 } S1 depends on S2 from prev iteration S1 depends on S1 from 3 prev iterations Referred to as a recursion Dependence distance 3; limited parallelism For (i=1000; i>0; i=i-1) x[i] = x[i-3] + s; S1

  12. Constructing Parallel Loops If loop-carried dependences are not cyclic (S1 depending on S1 is cyclic), loops can be restructured to be parallel For (i=1; i<=100; i=i+1) { A[i] = A[i] + B[i]; S1 B[i+1] = C[i] + D[i]; S2 } A[1] = A[1] + B[1]; For (i=1; i<=99; i=i+1) { B[i+1] = C[i] + D[i]; S3 A[i+1] = A[i+1] + B[i+1]; S4 } B[101] = C[100] + D[100]; S1 depends on S2 from prev iteration S4 depends on S3 of same iteration

  13. Finding Dependences – the GCD Test • Do A[ai + b] and A[ci + d] refer to the same element? • Restrict ourselves to affine array indices (expressible as • ai + b, where i is the loop index, a and b are constants) – • example of non-affine index: x[y[i]] • For a dependence to exist, must have two indices j and k • that are within the loop bounds, such that • aj + b = ck + d; • aj – ck = d – b; • G = GCD(a,c); • (aj/G - ck/G) = (d-b)/G; • If (d-b)/G is not an integer, the initial equality can not be true

  14. Software Pipeline?! L.D ADD.D S.D DADDUI BNE L.D ADD.D S.D DADDUI BNE L.D ADD.D S.D DADDUI BNE L.D ADD.D S.D DADDUI BNE … L.D ADD.D Loop: L.D F0, 0(R1) ADD.D F4, F0, F2 S.D F4, 0(R1) DADDUI R1, R1,# -8 BNE R1, R2, Loop DADDUI BNE … L.D ADD.D DADDUI BNE

  15. Software Pipeline Original iter 1 L.D ADD.D S.D L.D ADD.D S.D Original iter 2 L.D ADD.D S.D Original iter 3 L.D ADD.D S.D Original iter 4 L.D ADD.D S.D New iter 1 L.D ADD.D S.D New iter 2 L.D ADD.D New iter 3 L.D New iter 4

  16. Software Pipelining Loop: L.D F0, 0(R1) ADD.D F4, F0, F2 S.D F4, 0(R1) DADDUI R1, R1,# -8 BNE R1, R2, Loop Loop: S.D F4, 16(R1) ADD.D F4, F0, F2 L.D F0, 0(R1) DADDUI R1, R1,# -8 BNE R1, R2, Loop • Advantages: achieves nearly the same effect as loop unrolling, but • without the code expansion – an unrolled loop may have inefficiencies • at the start and end of each iteration, while a sw-pipelined loop is • almost always in steady state – a sw-pipelined loop can also be unrolled • to reduce loop overhead • Disadvantages: does not reduce loop overhead, may require more • registers

  17. Title • Bullet

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