EECS 583 Class 18: Iterative Modulo Scheduling – University of Michigan - Prof. Scott Mahl, Study notes of Electrical and Electronics Engineering

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EECS 583 – Class 18 Iterative Modulo Scheduling Part II^ University of Michigan^ March 21, 2005^ Note: The Msched example slides from the last lecturewere wrong, use the ones from these notes

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Schedule for the Rest of the Semester^ Y^ 4 more lectures^ »^ 3/21 (today) – Finish modulo scheduling^ »^ 3/23 – Partitioning for multicluster processors (Kevin Fan)^ »^ 3/28 – Compiler-controlled data prefetching (Manjunath Kudlur)^ »^ 3/30 – Register allocation^ Y^ Exam – Monday, April 4^ Y^ Paper presentations by you guys^ »^ 4/6 – Memory group^ »^ 4/11 – Code generation group^ »^ 4/13 – Opti group (part I)^ »^ 4/18 – Opti group (part II)^ Y^ Final project demos^ »^ 4/20 or 4/25-4/27^ Y^ SIG meetings this week – same time as last week

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Modulo Scheduling - Driver^ Y^ compute MII^ Y^ II = MII^ Y^ budget = BUDGET_RATIO * number of ops^ Y^ while (schedule is not found) do^ »^ iterative_schedule(II, budget)^ »^ II++^ Y^ Budget_ratio is a measure of the amount of backtracking that can beperformed before giving up and trying a higher II

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Modulo Scheduling – Iterative Scheduler^ Y^ iterative_schedule(II, budget)^ »^ compute op priorities^ »^ while (there are unscheduled ops and budget > 0) do^ y

op = unscheduled op with the highest priority y min = early time for op (E(Y)) y max = min + II – 1 y t = find_slot(op, min, max) y schedule op at time t^ X^ /* Backtracking phase – undo previous scheduling decisions */^ X^ Unschedule all previously scheduled ops that conflict with op y budget--

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Modulo Scheduling Example^ resources: 4 issue, 2 alu, 1 mem, 1 br^ latencies: add=1, mpy=3, ld = 2, st = 1, br = 1

Step1: Compute to loop into form that uses LC

for (j=0; j<100; j++)^ b[j] = a[j] * 26

LC = 99 1: r3 = load(r1) Loop:2: r4 = r3 * 263: store (r2, r4)4: r1 = r1 + 45: r2 = r2 + 47: brlc Loop

Loop:^ 1: r3 = load(r1)2: r4 = r3 * 263: store (r2, r4)4: r1 = r1 + 45: r2 = r2 + 46: p1 = cmpp (r1 < r9)7: brct p1 Loop

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Example – Step 2^ resources: 4 issue, 2 alu, 1 mem, 1 br^ latencies: add=1, mpy=3, ld = 2, st = 1, br = 1

Step 2: DSA convert LC = 99^

LC = 99

1: r3 = load(r1)2: r4 = r3 * 263: store (r2, r4)4: r1 = r1 + 45: r2 = r2 + 47: brlc Loop

1: r3[-1] = load(r1[0])2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

Loop:^

Loop:

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Example – Step 4

Step 4 – Calculate priorities (MAX height to pseudo stop node) 1,1 0,0 1 2,00,0 2 3,0 0,0 (^3) 0,01,11,1 0,0 (^4) 0,0 1,1 (^5) 0,0 1,1 7

Iter^

Iter 1: H = 5 2: H = 3 3: H = 0 4: H = 0 5: H = 0 7: H = 0

1: H = 5 2: H = 3 3: H = 0 4: H = 4 5: H = 0 7: H = 0

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Example – Step 5

Schedule brlc at time II - 1

resources: 4 issue, 2 alu, 1 mem, 1 br latencies: add=1, mpy=3, ld = 2, st = 1, br = 1

Unrolled Schedule Rolled Schedule

LC = 99^

1: r3[-1] = load(r1[0]) Loop:2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

4 5 6 brmemalu1alu 0

MRT

X

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Example – Step 7^ Step7: Schedule the highest priority op^ Op4: E = 0, L = 1^ Place at time 0 (0 % 2)

Unrolled Schedule Rolled Schedule

LC = 99^

1: r3[-1] = load(r1[0]) Loop:2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

4 5 6 brmemalu1alu0 X X 0

MRT

X

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Example – Step 8^ Step8: Schedule the highest priority op^ Op2: E = 2, L = 3^ Place at time 2 (2 % 2)

Unrolled Schedule Rolled Schedule

LC = 99^

1: r3[-1] = load(r1[0]) Loop:2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

4 5 6 brmemalu1alu0 X XX 0

MRT

X

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Example – Step 10^ Step10: Schedule the highest priority op^ Op5: E = 0, L = 1^ Place at time 1 (1 % 2)

Unrolled Schedule Rolled Schedule

LC = 99^

1: r3[-1] = load(r1[0]) Loop:2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

4 5 6 brmemalu1alu0 XX X XX 0

MRT

X

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Example – Step 11^ Step11: calculate ESC, SC = max unrolled sched length / ii^ unrolled sched time of branch = rolled sched time of br + (iiesc)^ SC = 6 / 2 = 3, ESC = SC – 1^ time of br = 1 + 22 = 5

Unrolled Schedule Rolled Schedule

LC = 99^

1: r3[-1] = load(r1[0]) Loop:2: r4[-1] = r3[-1] * 263: store (r2[0], r4[-1])4: r1[-1] = r1[0] + 45: r2[-1] = r2[0] + 4remap r1, r2, r3, r47: brlc Loop

4 5 6 brmemalu1alu0 XX X XX 0

MRT

X

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Example – Dynamic Execution of the Code

time: ops executed

LC = 99 ESC = 2 p1[0] = 1

Loop:^ 1: r3[-1] = load(r1[0]) if p1[0]^ 2: r4[-1] = r3[-1] * 26 if p1[1]^ 4: r1[-1] = r1[0] + 4 if p1[0]^ 3: store (r2[0], r4[-1]) if p1[2]^ 5: r2[-1] = r2[0] + 4 if p1[0]^ 7: brlc Loop if p1[2]

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Class Problem^ latencies: add=1, mpy=3, ld = 2, st = 1, br = 1

How many resources of each type are required to achieve an II=1 schedule? If the resources are non-pipelined, how many resources of each type are required to achieve II=1 Assuming pipelined resources, generate the II=1 modulo schedule. for (j=0; j<100; j++)^ b[j] = a[j] * 26^ LC = 99^ 1: r3 = load(r1) Loop:2: r4 = r3 * 263: store (r2, r4)4: r1 = r1 + 45: r2 = r2 + 47: brlc Loop