Slides - ETH TIK

Embedded Systems
Exercise 8: Integer Linear Programming
and Iterative Algorithms
01. June 2016
Georgia Giannopoulou (ETZ G77)
[email protected]
Swiss Federal
Institute of Technology
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Computer Engineering
and Networks Laboratory
Synthesis Problem Specification (1)
Sequence Graph GS(VS,ES)
nop
T2
T1
T3
nop
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Synthesis Problem Specification (2)
Resource graph, GR(VR,ER)
Operations, VS
Resources, VT
4
T1
r1
(r1)=1
r2
(r2)=1
2
T2
3
T3
Execution
times, wi
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Integer Linear Programming
Principle
Synthesis Problem
transformation into ILP
Integer Linear Program (ILP)
optimization of ILP
Solution of ILP
back interpretation
Solution of Synthesis Problem
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Question 1 – Integer Linear Program
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nop
Earliest starting time li
T2
T1
T3
nop
T1
nop
nop
T2
T3
0
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t
nop
Earliest starting time li
T2
T1
ASAP
T3
nop
T1
nop
nop
T2
T3
0
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t
nop
Earliest starting time li
T2
T1
ASAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
0
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t
nop
Latest starting time hi
T2
T1
ALAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
0
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t
nop
Latest starting time hi
T2
T1
ALAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
Lmax
0
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t
nop
Latest starting time hi
T2
T1
ALAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
h3 =6
Lmax
0
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t
nop
Latest starting time hi
T2
T1
ALAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
h3 =6
Lmax
0
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t
nop
Latest starting time hi
T2
T1
ALAP
T3
nop
T1
nop
nop
T2
T3
l3 = 2
h3 =6
Lmax
0
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t
nop
Binary variables xi,t
T2
T1
T3
nop
T1
nop
nop
T2
T3
x3,t
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0
0
0
0
0
0
1
0
0
0
0
1
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t
nop
Binary variables xi,t
T2
T1
T3
nop
T1
nop
nop
T2
T3
x3,t
0
0
0
0
0
0
1
0
0
0
8
9
x3,6 = 1
0
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t
nop
Binary variables xi,t
T2
T1
T3
nop
T1
nop
nop
T2
T3
x3,t
0
0
0
0
1
0
0
0
0
0
6
7
8
9
x3,4 = 1
0
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t
nop
Binary variables xi,t
T2
T1
T3
nop
T1
nop
An operation cannot
have more than one
starting point!
nop
T2
T3
x3,t
0
0
0
0
1
0
0
0
0
0
6
7
8
9
x3,4 = 1
0
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t
nop
Starting times  (Ti)
T2
T1
T3
nop
T1
nop
nop
T2
T3
Starting time of Ti
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x3,t
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0
0
0
0
0
1
0
0
0
0
0
1
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t
nop
Starting times  (Ti)
T2
T1
T3
nop
T1
nop
nop
T2
T3
Starting time of Ti
x3,t
0
0
0
0
0
1
0
0
0
0
x3,5 = 1   (T3) = 5
0
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t
nop
Starting times  (Ti)
T2
T1
T3
nop
T1
nop
nop
T2
T3
Starting time of Ti
x3,t
0
0
0
1
0
0
0
0
0
0
7
8
9
x3,3 = 1   (T3) = 3
0
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t
nop
Precedence (timing) constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
0
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t
nop
Precedence (timing) constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
 (T2)
0
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t
nop
Precedence (timing) constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
 (T2)
 (T3)
0
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t
nop
Precedence (timing) constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
 (T2)
 (T3)
 w(T2)
0
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
0
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
x2,t
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0
1
0
0
0
0
0
0
0
0
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
x2,t
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0
1
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
x2,t
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0
0
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
x2,t
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0
0
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
1
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t
nop
Resource constraints
T2
T1
T3
for T2:
t = 0: x2,0
t = 1: x2,0 + x2,1
t = 2: x2,1 + x2,2
t = 3: x2,2 + x2,3
T1
nop
nop
T2
T3
…
x2,t
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0
0
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T3
0
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T1 :
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T3
0
0
0
0
0
1
1
1
1
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
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T3
T1 :
0
0
0
0
0
1
1
1
1
0
T2 :
0
1
1
0
0
0
0
0
0
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T1 :
T2 :
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0
0
0
0
0
T3
1
1
1
1
0
0
0
0
0
0
+
0
1
1
0
0
=
0
1
1
0
0
1
1
1
1
0
0
1
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t
nop
Resource constraints
T2
T1
T3
nop
T1
nop
nop
T2
T1 :
T2 :
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0
0
1
1
1
T3
1
0
0
0
0
0
0
0
0
0
+
0
1
1
0
0
=
0
1
2
1
1
1
0
0
0
0
0
1
2
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t
Question 1
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Exec. Times
Allocation
D+ = 1
Dx = 2
α+ = 2
αx = 2
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Question 1 – Solution Steps
(1) Perform ASAP & ALAP scheduling to find li, hi


Question: How to select Lmax for ALAP?
Hint: “Lmax is chosen such that a feasible solution to the problem
(scheduling under resource constraints) exists.”
(2) Define binary variables xi,t
(3) Define the starting times of the operations τ(vi)
(4) Write down the set of constraints (precedence, resource)
(5) Specify the objective function
(6) Modify your ILP formulation for a different objective (1c)
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Question 1a – Solution
Lmax = 7
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Question 1a – Solution
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Question 1a – Solution
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Question 1a – Solution
...
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Question 1a – Solution
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Question 1c – Solution
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Question 2 – Iterative Algorithms
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Iterative Algorithms
displacement
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Iteration interval
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Iterative Algorithms
Extended sequence graph
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Iterative Algorithms
P=2
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Question 2
(1) Draw extended sequence graph
(2) Formulate dependency constraints
(3) Compute minimum iteration interval P and latency L
without and with functional pipelining
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Question 2 – Solution
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Question 2 – Solution
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Question 2 – Solution
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