Instr Sched 1

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Instruction Scheduling - Part 1

Y.N. Srikant

Department of Computer ScienceIndian Institute of Science

Bangalore 560 012

NPTEL Course on Compiler Design

Y.N. Srikant Instruction Scheduling
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Outline

Instruction Scheduling

Simple Basic Block Scheduling

Automaton-based SchedulingInteger programming based schedulingOptimal Delayed-load Scheduling (DLS) for treesTrace, Superblock and Hyperblock scheduling

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Instruction Scheduling

Reordering of instructions so as to keep the pipelines of

functional units full with no stalls

NP-Complete and needs heuristcs

Applied on basic blocks (local)

Global scheduling requires elongation of basic blocks

(super-blocks)

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Instruction Scheduling - Motivating Example

time: load - 2 cycles, op - 1 cycle

This code has 2 stalls, at i3 and at i5,

due to the loads

i1 i2 i4

i3

i5

i7

i6

load load load

add

sub

st

mult

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Scheduled Code - no stalls

There are no stalls, but dependences are indeed satisfied

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Definitions - Dependences

Consider the following code:

i1 : r1 load(r2)i2 : r3 r1 + 4i3 : r1 r4 + r5

The dependences are

i1 i2 (flow dependence) i2 i3 (anti-dependence)

i1 o i3 (output dependence)

anti- and ouput dependences can be eliminated by register

renaming


G
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Dependence DAG

full line: flowdependence, dash line: anti-dependence

dash-dot line: outputdependence

some anti- and output dependences are because memorydisambiguation could not be done

st

add

mult st

add

ldld

add sub

i1

i3 i4

i7

i8

i6

i2

i5

i9


B i Bl k S h d li
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Basic Block Scheduling

Basic block consists of micro-operation sequences (MOS),

which are indivisible

Each MOS has several steps, each requiring resources

Each step of an MOS requires one cycle for executionPrecedence constraints and resource constraints must besatisfied by the scheduled program

PCs relate to data dependences and execution delaysRCs relate to limited availability of shared resources


Th B i Bl k S h d li P bl
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The Basic Block Scheduling Problem

Basic block is modelled as a digraph, G = (V, E)

R: number of resourcesNodes (V): MOS; Edges (E): PrecedenceLabel on node v

resource usage functions, v(i) for each step of the MOSassociated with v

length l(v) of node vLabel on edge e: Execution delay of the MOS, d(e)

Problem: Find the shortest schedule : V N such thate= (u, v) E, (v) (u) d(e) and

i,

vVv(i (v)) R, where

length of the schedule is maxvV

{(v) + l(v)}


I t ti S h d li P d d R
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Instruction Scheduling - Precedence and Resource

Constraints


A Si l Li t S h d li Al ith
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A Simple List Scheduling Algorithm

Find the shortest schedule : V N, such that precedenceand resource constraints are satisfied. Holes are filled with

NOPs.FUNCTION ListSchedule (V,E)

BEGIN

Ready = root nodes of V; Schedule= ;

WHILE Ready= DO

BEGINv = highest priority node in Ready;

Lb = SatisfyPrecedenceConstraints (v, Schedule, );(v) = SatisfyResourceConstraints (v, Schedule, , Lb);

Schedule = Schedule + {v};Ready = Ready {v} + {u | NOT (u Schedule)AND (w, u) E, w Schedule};

END

RETURN ;

ENDY.N. Srikant Instruction Scheduling

List Scheduling Ready Queue Update
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List Scheduling - Ready Queue Update


Constraint Satisfaction Functions
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Constraint Satisfaction Functions

FUNCTION SatisfyPrecedenceConstraint(v, Sched, )

BEGINRETURN ( max

uSched(u) + d(u, v))

END

FUNCTION SatisfyResourceConstraint(v, Sched, , Lb)BEGIN

FOR i := Lb TO DO

IF 0 j < l(v), v(j) +uSched

u(i + j (u)) R THEN

RETURN (i);END


Precedence Constraint Satisfaction
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Precedence Constraint Satisfaction

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List Scheduling - Priority Ordering for Nodes


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List Scheduling - Priority Ordering for Nodes

1 Height of the node in the DAG (i.e., longest path from the

node to a terminal node2 Estart, and Lstart, the earliest and latest start times

Violating Estartand Lstartmay result in pipeline stallsEstart(v) = max

i=1, ,k(Estart(ui) + d(ui, v))

where u1, u2, , uk are predecessors of v. Estart value ofthe source node is 0.Lstart(u) = min

i=1, ,k(Lstart(vi) d(u, vi))

where v1, v2, , vk are successors of u. Lstart value of thesink node is set as its Estart value.

Estart and Lstart values can be computed using atop-down and a bottom-up pass, respectively, eitherstatically (before scheduling begins), or dynamically duringscheduling


Estart and Lstart Computation
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Estart and Lstart Computation


List Scheduling - Slack
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List Scheduling Slack

1 A node with a lower Estart (or Lstart) value has a higher

priority2 Slack

=Lstart

Estart

Nodes with lower slack are given higher priorityInstructions on the critical path may have a slack value ofzero and hence get priority

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Simple List Scheduling - Example - 2


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Simple List Scheduling Example 2

latenciesadd,sub,store: 1 cycle; load: 2 cycles; mult: 3 cycles

path lengthand slackare shown on the left side and right

side of the pair of numbers in parentheses

ld

sub

mult st

add

st

add

ld

add

5(3, 3)0

6(2, 2)0

8(0, 0)0

3(2, 5)3 1(2, 7)5

7(0, 1)1

0(8, 8)0

2(6, 6)0

1(7, 7)0

i1

i3 i4

i7

i8

i6

i2

i5

i9

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Instr Sched 1

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