1. “Rate-Optimal” Resource-Constrained Software Pipelining
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2. Objectives Establish good bounds
Help compiler writers
Help architects
Study pragmatic issues when implemented as an option of compilers
Study its payoffs
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3. A Motivating Example for i = 0 to n do 0: a [i] = X + d [i - 2];
1: b [i] = a [i] * F + f [i - 2] + e [i - 2];
2: c [i] = Y - b [i];
3: d [i] = 2 * c [i];
4: e [i] = X - b [i];
5: f [i] = Y + b [i]
end
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4. Assume all statements have a delay = 12 L:
for ( i = 0; i < n; i ++) { 3 S :
a [i] = X + d [i - 2]; S :
b [i] = a [i] * F + f [i - 2] + e [i - 2]; 1 S :
c [i] = Y - b [i]; 1 2 2 S :
d [i] = 2 * c [i]; 3 2 2 S :
e [i] = X - b [i]; 4 4 5 S :
f [i] = Y + b [i] 5 }
Data Dependence Graph
Program Representation
Assume all statements have a delay = 1
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5. Question
How fast can L run (I.e.optimal computation rate) without resource constraints?
How many FUs it needs minimally to achieve the optimal rate?
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6. . Schedule A : # of FUs = 4 Schedule A: t (i, Sj) = 2i + tsj where:
ts2 = ts4 = ts5 = 2
ts3 = 3
iter #1 #2 #3
. . . S 1 S 1 2 S
, S
, S S 2 4 5 3 S S 3 1 4 S
, S
, S S 2 4 5 5 S S 3 1 6 S
, S
, S 2 4 5 7 S . 3
Schedule A : # of FUs = 4
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7. Can we do better?
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8. . Schedule B : # of FUs = 3 Schedule B : t (i, Sj) = 2i + tsj where:
ts2 = ts4 = 2
ts3 = ts5 = 3
iter #1 #2 #3
. . . S 1 S 1 2 S
, S S 2 4 3 S S S 3, 5 1 4 S
, S S 2 4 5 S S S 3, 5 1 6 S
, S 2 4 7 S S . 3, 5
Schedule B : # of FUs = 3
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9. Problem Statements Assumptions: homogeneous function units
Problem 0: Given a loop L, determine a “rate-optimal” schedule for L which uses minimum # of FUs.
Problem I (FIxed Rate SofTware Pipelining with minimum Resource - FIRST):
Given a loop L and a fixed initiation rate determine a schedule for L which uses minimum # of FUs.
Problem II (REsource Constrained SofTware Pipelining - REST)
Given a loop L and the # of Fus, determine an optimal schedule which can run under the given resource constraints.
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10. Problem Formulation of FIRST
How to formulate resource constraints into a linear form?
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11. all dependence constraints R is the maximum width
R = Max (width)
Min R
Subject to:
all dependence constraints
R is the maximum width
for a fixed rate (or period)
The SWP kernel:
--- The “frustum”
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12. Min (max ari) . R Frustum A Now R = max ari r [ 0, II - 1]
time r . 1 2
. . .
N-1 3
II-1
Min (max ari) R
Frustum A
Contribution of node i to step
r’s resource requirement
Now R = max ari
r [ 0, II - 1]
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13. FIRST: A Linear Programming Formulation Example
a00 a01 a02 a03 a04 a05
a10 a11 a12 a13 a14 a15
So for node I:
ti = ki II + (aoi , a1i ) * 1
Where ari = 1 means node i
start at step r
= 0 otherwise
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14. FIRST: A Linear Programming Formulation
Minimize R
Subject to
R: Max T T
= T
Periodic
Dependence
Constraints
and are integers
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15. Example Revisited . . . . . Min R R - ( a + a + a + a + a + a ) ³ R -
Subjected to: R - ( a + a + a + a + a + a ) 00 01 02 03 04 05 R - ( a + a + a + a + a + a ) 10 11 12 13 14 15 2 × k + × a + 1 × a = t 00 10 2 × k 1 + × a + 1 × a = t 1 01 11 . 2 × k + × a + 1 × a = t 5 05 15 5 a + a = 1 00 10 a + a = 1 01 11 . a + a = 1 05 15 t - t . 1 1 . t - t 1 4 - 3 t - t 1 5 - 3 . t , k , a 1 i ri
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16. Solution so, # of FUs = 3! Schedule C t(i, s) = 2i + ts where t0 = 0
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17. Linear Programming
A Linear Program is a problem that can be expressed in the following form:
minimize cx
subject to
Ax = b
x >= 0
where x: vector of variables to be solved
A: matrix of known coefficients
c, b: vectors of known coefficients
cx: it is called the objective function
Ax=b is called the constraints
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18. Linear Programming “programming” actually means “planning” here.
Importance of LP
many applications
the existence of good general-purpose techniques for finding optimal solutions
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19. Integer Linear Programming (ILP)
Integer programming have proved valuable for modeling many and diverse types of problems in :
planning,
routing,
scheduling,
assignment, and
design.
Industry applications:
transportation, energy, telecommunications, and manufacturing
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20. Integer Linear Programming
Classification:
mixed integer (part of variables)
pure integer (all variables)
zero-one (only takes 0 or 1)
Much harder to solve
Many existing tools and product
solve the LP/ILP problem and get optimal solution
help to model the problem, and then solve the problem
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21. How to handle cases where di 1 for node i
Quiz
How to handle cases where di for node i ?
>=
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22. a((r - 1)%II)i The “Trick”
At time step r, the # of FUs required
Contributed by node i
{Started at steps in [0, (t + di-1)%II]}
a((r - 1)%II)i
Note: if actor i requires a FU at time r or 0 otherwise.
So objective function becomes:
Max
Min
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23. How to extend the FIRST formulation to heterogeneous function units?
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24. Heterogeneous Function Units
Solution Hints: “weighted Sum”!
For FU of type k:
Mk = max
" r Î [ , II - 1 ]
and so, the objective should be
min Ck Mk
Where h is the total # of FU types
Mk - 0
FU(i) = k
" k Î [ , h - 1 ],
" r Î [ , II - 1 ]
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25. Resource - constraint rate-optimal software pipelining problem
Solution of the REST
Resource - constraint rate-optimal software pipelining problem
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26. Hints Based on scheme for “FIRST” and play “trial-and improve”
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27. Solution Space for REST
MII = max { RecMII, ResMII}
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28. MII = max {RecMII, ResMII }
Note: The bounds of initiation interval
1. Loop-carried dependence constraints:
2. Resource constraints
As a result:
RecMII =
Max
cycles C
d(C)
m(C)
ResMII =
# of nodes
# of FUs
MII = max {RecMII, ResMII }
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29. Other Work
Extend the technique to “more benchmarks” and establish “bounds”
Extend the framework to include register constraints
Extend the model for pipelined architectures (with structural “hazards”)
Extend the model to consider superscalar and multithreaded with dynamic scheduling support.
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