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Published byElias Critchlow Modified about 1 year ago

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1 ILP (Recap)

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2 Basic Block (BB) ILP is quite small –BB: a straight-line code sequence with no branches in except to the entry and no branches out except at the exit –average dynamic branch frequency 15% to 25% => 4 to 7 instructions execute between a pair of branches –Plus instructions in BB likely to depend on each other To obtain substantial performance enhancements, we must exploit ILP across multiple basic blocks Simplest: loop-level parallelism to exploit parallelism among iterations of a loop Instruction Level Parallelism

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3 Loop Unrolling Example: Key to increasing ILP For the loop: for (i=1; i<=1000; i++) x(i) = x(i) + s; ; The straightforward MIPS assembly code is given by: Loop: L.DF0, 0 (R1) ADD.DF4, F0, F2 S.D F4, 0(R1) SUBI R1, R1, # 8 BNE R1,Loop InstructionInstruction Latency in producing resultusing result clock cycles FP ALU opAnother FP ALU op3 FP ALU opStore double2 Load doubleFP ALU op1 Load doubleStore double0 Integer opInteger op0

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4 Loop Showing Stalls and Code Re-arrangement 1 Loop:LD F0,0(R1) 2stall 3ADDDF4,F0,F2 4stall 5stall 6 SD0(R1),F4 7 SUBIR1,R1,8 8 BNEZR1,Loop 9 stall 9 clock cycles per loop iteration. 1Loop:LDF0,0(R1) 2Stall 3ADDDF4,F0,F2 4SUBIR1,R1,8 5BNEZR1,Loop 6SD8(R1),F4 Code now takes 6 clock cycles per loop iteration Speedup = 9/6 = 1.5 The number of cycles cannot be reduced further because: The body of the loop is small The loop overhead (SUBI R1, R1, 8 and BNEZ R1, Loop)

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5 Basic Loop Unrolling Concept: 4n iterations n iterations 4 iterations

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6 Unroll Loop Four Times to expose more ILP and reduce loop overhead 1 Loop:LDF0,0(R1) 2ADDDF4,F0,F2 3SD0(R1),F4 ;drop SUBI & BNEZ 4LDF6,-8(R1) 5ADDDF8,F6,F2 6SD-8(R1),F8 ;drop SUBI & BNEZ 7LDF10,-16(R1) 8ADDDF12,F10,F2 9SD-16(R1),F12 ;drop SUBI & BNEZ 10LDF14,-24(R1) 11ADDDF16,F14,F2 12SD-24(R1),F16 13SUBIR1,R1,#32 14BNEZR1,LOOP 15stall x (2 + 1)= 27 clock cycles, or 6.8 cycles per iteration (2 stalls after each ADDD and 1 stall after each LD) 1 Loop:LDF0,0(R1) 2LDF6,-8(R1) 3LDF10,-16(R1) 4LDF14,-24(R1) 5ADDDF4,F0,F2 6ADDDF8,F6,F2 7ADDDF12,F10,F2 8ADDDF16,F14,F2 9SD0(R1),F4 10SD-8(R1),F8 11SD-16(R1),F8 12SUBIR1,R1,#32 13BNEZR1,LOOP 14SD8(R1),F16 14 clock cycles or 3.5 clock cycles per iteration The compiler (or Hardware) must be able to: Determine data dependency Do code re-arrangement Register renaming

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7 Loop-Level Parallelism (LLP) Analysis Loop-Level Parallelism (LLP) analysis focuses on whether data accesses in later iterations of a loop are data dependent on data values produced in earlier iterations. e.g. in for (i=1; i<=1000; i++) x[i] = x[i] + s; the computation in each iteration is independent of the previous iterations and the loop is thus parallel. The use of X[i] twice is within a single iteration. Thus loop iterations are parallel (or independent from each other). Loop-carried Dependence: A data dependence between different loop iterations (data produced in earlier iteration used in a later one) – limits parallelism. Instruction level parallelism (ILP) analysis, on the other hand, is usually done when instructions are generated by the compiler.

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8 LLP Analysis Example 1 In the loop: 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 */ } (Where A, B, C are distinct non-overlapping arrays) –S2 uses the value A[i+1], computed by S1 in the same iteration. This data dependence is within the same iteration (not a loop-carried dependence). does not prevent loop iteration parallelism. –S1 uses a value computed by S1 in an earlier iteration, since iteration i computes A[i+1] read in iteration i+1 (loop-carried dependence, prevents parallelism). The same applies for S2 for B[i] and B[i+1] These two dependences are loop-carried spanning more than one iteration preventing loop parallelism.

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9 LLP Analysis Example 2 In the loop: for (i=1; i<=100; i=i+1) { A[i] = A[i] + B[i]; /* S1 */ B[i+1] = C[i] + D[i]; /* S2 */ } –S1 uses the value B[i] computed by S2 in the previous iteration (loop-carried dependence) –This dependence is not circular: S1 depends on S2 but S2 does not depend on S1. –Can be made parallel by replacing the code with the following: A[1] = A[1] + B[1]; for (i=1; i<=99; i=i+1) { B[i+1] = C[i] + D[i]; A[i+1] = A[i+1] + B[i+1]; } B[101] = C[100] + D[100]; Loop Start-up code Loop Completion code

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10 LLP Analysis Example 2 Original Loop: A[100] = A[100] + B[100]; B[101] = C[100] + D[100]; A[1] = A[1] + B[1]; B[2] = C[1] + D[1]; A[2] = A[2] + B[2]; B[3] = C[2] + D[2]; A[99] = A[99] + B[99]; B[100] = C[99] + D[99]; A[100] = A[100] + B[100]; B[101] = C[100] + D[100]; A[1] = A[1] + B[1]; B[2] = C[1] + D[1]; A[2] = A[2] + B[2]; B[3] = C[2] + D[2]; A[99] = A[99] + B[99]; B[100] = C[99] + D[99]; 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]; A[i+1] = A[i+1] + B[i+1]; } B[101] = C[100] + D[100]; Modified Parallel Loop: Iteration 1 Iteration 2 Iteration 100Iteration 99 Loop-carried Dependence Loop Start-up code Loop Completion code Iteration 1 Iteration 98 Iteration 99 Not Loop Carried Dependence.....

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