Transcript Lecture 8: Linear Programming

ELEC 5270/6270 Spring 2009
Low-Power Design of Electronic Circuits
Linear Programming – A Mathematical
Optimization Technique
Vishwani D. Agrawal
James J. Danaher Professor
Dept. of Electrical and Computer Engineering
Auburn University, Auburn, AL 36849
[email protected]
http://www.eng.auburn.edu/~vagrawal/COURSE/E6270_Spr09/course.html
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
1
What is Linear Programming
Linear programming (LP) is a
mathematical method for selecting the
best solution from the available solutions
of a problem.
 Method:

 State
the problem and define variables whose
values will be determined.
 Develop a linear programming model:


Write the problem as an optimization formula (a linear
expression to be minimized or maximized)
Write a set of linear constraints
 An
available LP solver (computer program) gives
the values of variables.
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ELEC6270 Spring 09, Lecture 8
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Types of LPs
LP – all variables are real.
 ILP – all variables are integers.
 MILP – some variables are integers,
others are real.
 A reference:


S. I. Gass, An Illustrated Guide to Linear
Programming, New York: Dover, 1990.
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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A Single-Variable Problem
Consider variable x
 Problem: find the maximum value of x
subject to constraint, 0 ≤ x ≤ 15.
 Solution: x = 15.

Constraint satisfied
x
0
15
Solution
x = 15
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Single Variable Problem (Cont.)


Consider more complex constraints:
Maximize x, subject to following constraints:




x≥0
5x ≤ 75
6x ≤ 30
x ≤ 10
0
(1)
(2)
(3)
(4)
5
10
15
(2)
(3)
x
(1)
(4)
All constraints
satisfied
Solution, x = 5
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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A Two-Variable Problem

Manufacture of chairs and tables:
 Resources available:
 Material: 400 boards of wood
 Labor: 450 man-hours
 Profit:
 Chair: $45
 Table: $80
 Resources needed:
 Chair



Table



5 boards of wood
10 man-hours
20 boards of wood
15 man-hours
Problem: How many chairs and how many tables should be
manufactured to maximize the total profit?
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ELEC6270 Spring 09, Lecture 8
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Formulating Two-Variable Problem
Manufacture x1 chairs and x2 tables to
maximize profit:
P = 45x1 + 80x2 dollars
 Subject to given resource constraints:

boards of wood,
5x1 + 20x2 ≤ 400
 450 man-hours of labor, 10x1 + 15x2 ≤ 450
 x1 ≥ 0
 x2 ≥ 0
 400
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
(1)
(2)
(3)
(4)
7
Solution: Two-Variable Problem
40
Tables, x2
30
Best solution: 24 chairs, 14 tables
Profit = 45×24 + 80×14 = 2200 dollars
(1) 20
(24, 14)
10
(3)
(4)
0
0
10
20
30
40
50
Chairs, x1
60
(2)
70
80
90
increasing
decresing
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Change Profit of Chair to $64/Unit
Manufacture x1 chairs and x2 tables to
maximize profit:
P = 64x1 + 80x2 dollars
 Subject to given resource constraints:

boards of wood,
5x1 + 20x2 ≤ 400
 450 man-hours of labor, 10x1 + 15x2 ≤ 450
 x1 ≥ 0
 x2 ≥ 0
 400
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
(1)
(2)
(3)
(4)
9
Solution: $64 Profit/Chair
40
Tables, x2
30
Best solution: 45 chairs, 0 tables
Profit = 64×45 + 80×0 = 2880 dollars
(1) 20
(24, 14)
10
(3)
(4)
0
0
Copyright Agrawal, 2009
10
20
30
40
50
Chairs, x1
ELEC6270 Spring 09, Lecture 8
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(2)
70
80
90
increasing
decresing
10
A Dual Problem
Explore an alternative.
 Questions:

Should we make tables and chairs?
 Or, auction off the available resources?


To answer this question we need to know:
What is the minimum price for the resources that
will provide us with same amount of revenue
from sale as the profits from tables and chairs?
 This is the dual of the original problem.

Copyright Agrawal, 2009
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Formulating the Dual Problem

Revenue received by selling off resources:
For each board, w1
 For each man-hour, w2

Minimize 400w1 + 450w2
 Subject to constraints:

 5w1
+ 10w2
 20w1 + 15w2
 w1
≥0
 w2
≥0
Copyright Agrawal, 2009
≥ 45
≥ 80
Resources:
Material: 400 boards
Labor: 450 man-hrs
Profit:
Chair: $45
Table: $80
Resources needed:
Chair
5 boards of wood
10 man-hours
Table
20 boards of wood
15 man-hours
ELEC6270 Spring 09, Lecture 8
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The Duality Theorem

If the primal has a finite optimum solution,
so does the dual, and the optimum values
of the objective functions are equal.
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Primal-Dual Problems

Primal problem



Fixed resources
Maximize profit








5x1 + 20x2
10x1 + 15x2
x1
x2

Variables:









5w1 + 10w2
20w1 + 15w2
w1
w2
≥ 45
≥ 80
≥0
≥0
Solution:
x1 = 24 chairs, x2 = 14 tables
Profit = $2200
Copyright Agrawal, 2009
w1 ($ value/board of wood)
w2 ($ value/man-hour)
Minimize value 400w1+450w2
Subject to:
≤ 400
≤ 450
≥0
≥0
Solution:

Fixed profit
Minimize value
x1 (number of chairs)
x2 (number of tables)
Maximize profit 45x1+80x2
Subject to:

Dual Problem

Variables:



ELEC6270 Spring 09, Lecture 8


w1 = $1, w2 = $4
value = $2200
14
LP for n Variables
n
minimize
Σ
cj xj
Objective function
j =1
n
subject to
Σ aij xj
≤ bi, i = 1, 2, . . ., m
j =1
n
Σ cij xj
= di, i = 1, 2, . . ., p
j =1
Variables: xj
Constants: cj, aij, bi, cij, di
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Algorithms for Solving LP

Simplex method


Ellipsoid method


L. G. Khachiyan, “A Polynomial Algorithm for Linear Programming,”
Soviet Math. Dokl., vol. 20, pp. 191-194, 1984.
Interior-point method


G. B. Dantzig, Linear Programming and Extension, Princeton, New
Jersey, Princeton University Press, 1963.
N. K. Karmarkar, “A New Polynomial-Time Algorithm for Linear
Programming,” Combinatorica, vol. 4, pp. 373-395, 1984.
Course website of Prof. Lieven Vandenberghe (UCLA),
http://www.ee.ucla.edu/ee236a/ee236a.html
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Basic Ideas of Solution methods
Extreme points
Constraints
Extreme points
Objective
function
Objective
function
Simplex: search on extreme points.
Complexity: polynomial in n, number of
variables
Copyright Agrawal, 2009
Constraints
Interior-point methods: Successively
iterate with interior spaces of
analytic convex boundaries.
Complexity: O(n3.5L), L = no. of int. values
ELEC6270 Spring 09, Lecture 8
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Integer Linear Programming (ILP)
Variables are integers.
 Complexity is exponential – higher than LP.
 LP relaxation

Convert all variables to real, preserve ranges.
 LP solution provides guidance.
 Rounding LP solution can provide a nonoptimal solution.

Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
18
Traveling Salesperson Problem (TSP)
6
4
12
5
27
1
18
12
15
20
19
10
2
3
Copyright Agrawal, 2009
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ELEC6270 Spring 09, Lecture 8
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Solving TSP: Five Cities
Distances (dij) in miles (symmetric TSP, general TSP is asymmetric)
City
j=1
j=2
j=3
j=4
j=5
i=1
0
18
10
12
27
i=2
18
0
5
12
20
i=3
10
5
0
15
19
i=4
12
12
15
0
6
i=5
27
20
19
6
0
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Search Space: No. of Tours

Asymmetric TSP tours
Five-city problem: 4 × 3 × 2 × 1 = 24 tours
 Ten-city problem: 362,880 tours
 15-city problem: 87,178,291,200 tours
 50-city problem: 49! = 6.08×1062 tours
Time for enumerative search assuming 1 μs
per tour evaluation
=
1.93×1055 years

Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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A Greedy Heuristic Solution
Tour length = 10 + 5 + 12 + 6 + 27 = 60 miles (non-optimal)
City
j=1
j=2
j=3
j=4
j=5
i=1
(start)
0
18
10
12
27
i=2
18
0
5
12
20
i=3
10
5
0
15
19
i=4
12
12
15
0
6
i=5
27
20
19
6
0
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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ILP Variables, Constants and Constraints
x14 ε [0,1]
d14 = 12
4
5
x15 ε [0,1]
d15 = 27
x12 ε [0,1]
d12 = 18
1
Integer variables:
xij = 1, travel i to j
xij = 0, do not travel i to j
x13 ε [0,1]
d13 = 10
Real constants:
dij = distance from i to j
2
3
x12 + x13 + x14 + x15 = 1
four other similar equations
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Objective Function and ILP Solution
5
i-1
Minimize ∑
∑ xij × dij
i=1 j=1
∑ xij = 1
j≠i
Copyright Agrawal, 2009
for all i
xij
j=1
2
3
4
5
i=1
0
0
1
0
0
2
1
0
0
0
0
3
0
1
0
0
0
4
0
0
0
0
1
5
0
0
0
1
0
ELEC6270 Spring 09, Lecture 8
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ILP Solution
d54 = 6
4
5
d45 = 6
1
d21 = 18
d13 = 10
2
3
d32 = 5
Total length = 45
but not a single tour
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ELEC6270 Spring 09, Lecture 8
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Additional Constraints for Single Tour

Following constraints prevent split tours.
For any subset S of cities, the tour must
enter and exit that subset:
∑ xij ≥ 2 for all S, |S| < 5
i ε S
j ε S
Remaining
set
At least two
arrows must cross
this boundary.
Any subset
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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ILP Solution
d41 = 12
4
d54 = 6
5
1
d25 = 20
d13 = 10
2
3
d32 = 5
Total length = 53
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Characteristics of ILP



Worst-case complexity is exponential in number of
variables.
Linear programming (LP) relaxation, where integer
variables are treated as real, gives a lower bound on the
objective function.
Recursive rounding of relaxed LP solution to nearest
integers gives an approximate solution to the ILP
problem.

K. R. Kantipudi and V. D. Agrawal, “A Reduced Complexity
Algorithm for Minimizing N-Detect Tests,” Proc. 20th International
Conf. VLSI Design, January 2007, pp. 492-497.
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Why ILP Solution is
Exponential?
LP solution
Must try all
2n roundoff
points
Second variable
found in
polynomial time
(bound on ILP
solution)
Constraints
First variable
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
Objective
(maximize)
29
ILP Example: Test Minimization


A combinational circuit has n test vectors that detect m
faults. Each test detects a subset of faults. Find the
smallest subset of test vectors that detects all m faults.
ILP model:


Assign an integer variable ti ε [0,1] to ith test vector such that ti =
1, if we select ti, otherwise ti= 0.
Define an integer constant fij ε [0,1] such that fij = 1, if ith vector
detects jth fault, otherwise fij = 0. Values of constants fij are
determined by fault simulation.
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ELEC6270 Spring 09, Lecture 8
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Test Minimization by ILP
n
Σ ti
minimize
Objective function
i=1
n
subject to
Σ fij ti
≥ 1,
j = 1, 2, . . ., m
i=1
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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3V3F: A 3-Vector 3-Fault Example
Test vector i
fij
i=1
i=2
i=3
Variables: t1, t2, t3 ε [0,1]
Minimize t1 + t2 + t3
Fault j
j=1
1
1
0
Subject to:
t1 + t2 ≥ 1
j=2
0
1
1
j=3
1
0
1
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
t2 + t3 ≥ 1
t1 + t3 ≥ 1
32
3V3F: Solution Space
t3
ILP solutions
(optimum)
1
Non-optimum
solution
1st LP solution
(0.5, 0.5, 0.5)
Rounding and
2nd ILP solution
(1.0, 0.5, 0.5)
t2
1
Rounding and
3rd LP solution
(1.0, 1.0, 0.0)
1
t1
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
33
3V3F: LP Relaxation and Rounding
ILP – Variables: t1, t2, t3 ε [0,1]
LP relaxation: t1, t2, t3 ε (0.0, 1.0)
Minimize t1 + t2 + t3
Solution: t1 = t2 = t3 = 0.5
Subject to:
Recursive rounding:
t1 + t2 ≥ 1
t2 + t3 ≥ 1
t1 + t3 ≥ 1
(1) round one variable, t1 = 1.0
Two-variable LP problem:
Minimize t2 + t3
subject to t2 + t3 ≥ 1.0
LP solution t2 = t3 = 0.5
(2) round a variable, t2 = 1.0
ILP constraints are satisfied
solution is t1 = 1, t2 = 1, t3 = 0
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
34
Recursive Rounding Algorithm
1. Obtain a relaxed LP solution. Stop if
each variable in the solution is an
integer.
2. Round the variable closest to an integer.
3. Remove any constraints that are now
unconditionally satisfied.
4. Go to step 1.
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
35
Recursive Rounding


ILP has exponential complexity.
Recursive rounding:




ILP is transformed into k LPs with progressively reducing
number of variables.
A solution that satisfies all constraints is guaranteed; this
solution is often close to optimal.
Number of LPs, k, is the size of the final solution, i.e., the
number of non-zero variables in the test minimization
problem.
Recursive rounding complexity is k × O(np), where k ≤ n,
n is number of variables.
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
36
Four-Bit ALU Circuit
ILP
Recursive rounding
Initial
vectors
Vectors
CPU s
Vectors
CPU s
285
14
0.65
14
0.42
400
13
1.07
13
1.00
500
12
4.38
13
3.00
1,000
12
4.17
12
3.00
5,000
12
12.95
12
9.00
10,000
12
34.61
12
17.0
16,384
12
87.47
12
37.0
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
37
ILP vs. Recursive Rounding
100
ILP
CPU s
75
Recursive
Rounding
50
25
0
0
Copyright Agrawal, 2009
5,000
10,000
ELEC6270 Spring 09, Lecture 8
15,000 Vectors
38
N-Detect Tests (N = 5)
Relaxed LP/Recur. rounding
ILP (exact)
Circuit
Unoptimized
vectors
c432
608
196.38
197
1.0
197
1.0
c499
379
260.00
260
1.2
260
2.3
c880
1,023
125.97
128
14.0
127
881.8
c1355
755
420.00
420
3.2
420
4.4
c1908
1,055
543.00
543
4.6
543
6.9
c2670
959
477.00
477
4.7
477
7.2
c3540
1,971
467.25
477
72.0
471
20008.5
c5315
1,079
374.33
377
18.0
376
40.7
c6288
243
52.52
57
39.0
57
34740.0
c7552
2,165
841.00
841
52.0
841
114.3
Copyright Agrawal, 2009
Lower
bound
Min.
Min.
CPU s
vectors
vectors
ELEC6270 Spring 09, Lecture 8
CPU s
39
Finding LP/ILP Solvers



R. Fourer, D. M. Gay and B. W. Kernighan, AMPL: A Modeling
Language for Mathematical Programming, South San Francisco,
California: Scientific Press, 1993. Several of programs described in
this book are available to Auburn users.
B. R. Hunt, R. L. Lipsman, J. M. Rosenberg, K. R. Coombes, J. E.
Osborn and G. J. Stuck, A Guide to MATLAB for Beginners and
Experienced Users, Cambridge University Press, 2006.
Search the web. Many programs with small number of variables can
be downloaded free.
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
40
A Circuit Optimization Problem

Given:
 Circuit
netlist
 Cell library with multiple versions for each cell

Select cell versions to optimize a specified
characteristic of the circuit. Typical
characteristics are:
Area
 Power
 Delay

Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Example: NAND(X), X = 0 or 1

X=0
Delay = nominal delay (d)
 Power = 3 × nominal power (p)


X=1
Delay = 2 × nominal delay (d)
 Power = 0.5 × nominal power (p)

Gate delay = (1 – X) d + 2 X d
 Power = 3(1 – X) p + 0.5 X p

Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Constraint: Gate k in the Circuit
Ti = signal arrival time at ith input
 Tk = signal arrival time at gate output
 Tk ≥ Ti + (1 – Xk) dk + 2 Xk dk, for all i
 Where, dk = nominal delay of gate
Xk = 0 or 1, specifies version of cell

Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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Clock Constraints
Tk = 0, for all primary inputs
 Tk ≤ clock period, for all primary outputs
Combinational Logic
Register
Register

Clock
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
44
Optimization Function

Minimize
∑ 3(1 – Xk) pk + 0.5 Xk pk
all k
where pk = nominal power consumption of
gate k
Copyright Agrawal, 2009
ELEC6270 Spring 09, Lecture 8
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