Spring 2010, Feb 22...ELEC 7770: Advanced VLSI Design (Agrawal)1 ELEC 7770 Advanced VLSI Design...

Spring 2010, Feb 22 . . .Spring 2010, Feb 22 . . . ELEC 7770: Advanced VLSI Design (Agrawal)ELEC 7770: Advanced VLSI Design (Agrawal) 11

ELEC 7770ELEC 7770Advanced VLSI DesignAdvanced VLSI Design

Spring 2010Spring 2010 Linear Programming – A Mathematical Linear Programming – A Mathematical

Optimization TechniqueOptimization TechniqueVishwani D. AgrawalVishwani D. Agrawal

James J. Danaher ProfessorJames J. Danaher ProfessorECE Department, Auburn UniversityECE Department, Auburn University

Auburn, AL 36849Auburn, AL [email protected]

http://www.eng.auburn.edu/~vagrawal/COURSE/E7770_Spr10/course.html

mailto:[email protected]

http://www.eng.auburn.edu/~vagrawal/COURSE/E7770_Spr08/course.html


What is Linear ProgrammingWhat 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.


Types of LPsTypes 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.


A Single-Variable ProblemA Single-Variable Problem

Consider variable x Problem: find the maximum value of x subject to

constraint, 0 ≤ x ≤ 15. Solution: x = 15.

0 15

Constraint satisfied

x

Solution x = 15


Single Variable Problem (Cont.)Single Variable Problem (Cont.) Consider more complex constraints: Maximize x, subject to following constraints:

x ≥ 0 (1) 5x ≤ 75 (2) 6x ≤ 30 (3) x ≤ 10 (4)

0 5 10 15x (1)(2)

(3)(4)

All constraints satisfied

Solution, x = 5


A Two-Variable ProblemA 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

5 boards of wood 10 man-hours

Table 20 boards of wood 15 man-hours

Problem: How many chairs and how many tables should be manufactured to maximize the total profit?


Formulating Two-Variable ProblemFormulating Two-Variable Problem

Manufacture x1 chairs and x2 tables to maximize profit:

P = 45x1 + 80x2 dollars Subject to given resource constraints:

400 boards of wood, 5x1 + 20x2 ≤ 400 (1) 450 man-hours of labor, 10x1 + 15x2 ≤ 450 (2) x1 ≥ 0 (3) x2 ≥ 0 (4)


Solution: Two-Variable ProblemSolution: Two-Variable Problem

Chairs, x1

Tabl

es, x

2

(1)

(2)0 10 20 30 40 50 60 70 80 90

40

30

20

10

0

(24, 14)

Profi

t increasing

decresing

P = 2200

P = 0

Best solution: 24 chairs, 14 tablesProfit = 45×24 + 80×14 = 2200 dollars

(3)(4)

Material constraint

Man-power constraint


Change Profit of Chair to $64/UnitChange Profit of Chair to $64/Unit

Manufacture x1 chairs and x2 tables to maximize profit:

P = 64x1 + 80x2 dollars Subject to given resource constraints:

400 boards of wood, 5x1 + 20x2 ≤ 400 (1) 450 man-hours of labor, 10x1 + 15x2 ≤ 450 (2) x1 ≥ 0 (3) x2 ≥ 0 (4)


Solution: $64 Profit/ChairSolution: $64 Profit/Chair

Chairs, x1

Tabl

es, x

2

(1)

(2)

Profit increasing

decresing

P = 2880

P = 0

Best solution: 45 chairs, 0 tablesProfit = 64×45 + 80×0 = 2880 dollars

0 10 20 30 40 50 60 70 80 90

(24, 14)

40

30

20

10

0(3)

(4)

Material constraint

Man-power constraint


A Dual ProblemA 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.


Formulating the Dual ProblemFormulating 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 ≥ 45 20w1 + 15w2 ≥ 80 w1 ≥ 0 w2 ≥ 0

Resources:Material: 400 boards Labor: 450 man-hrs

Profit:Chair: $45Table: $80

Resources needed:Chair

5 boards of wood10 man-hours

Table20 boards of wood15 man-hours


The Duality TheoremThe Duality Theorem

If the primal has a finite optimum solution, so If the primal has a finite optimum solution, so does the dual, and the optimum values of the does the dual, and the optimum values of the objective functions are equal.objective functions are equal.

Reference:Reference:G. Strang, G. Strang, Linear Algebra and Its ApplicationsLinear Algebra and Its Applications. . Fort Worth: Harcourt Brace Javanovich College Fort Worth: Harcourt Brace Javanovich College Publishers, third edition, 1988.Publishers, third edition, 1988.


Primal-Dual ProblemsPrimal-Dual Problems Primal problem

Fixed resources Maximize profit

Variables: x1 (number of chairs) x2 (number of tables)

Maximize profit 45x1+80x2 Subject to:

5x1 + 20x2 ≤ 400 10x1 + 15x2 ≤ 450 x1 ≥ 0 x2 ≥ 0

Solution: x1 = 24 chairs, x2 = 14 tables Profit = $2200

Dual Problem Fixed profit Minimize value

Variables: w1 ($ value/board of wood) w2 ($ value/man-hour)

Minimize value 400w1+450w2 Subject to:

5w1 + 10w2 ≥ 45 20w1 + 15w2 ≥ 80 w1 ≥ 0 w2 ≥ 0

Solution: w1 = $1, w2 = $4 value = $2200


LP for LP for nn Variables Variables n

minimize Σ cj xj Objective functionj =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: xjConstants: cj, aij, bi, cij, di


Algorithms for Solving LPAlgorithms for Solving LP

Simplex methodSimplex method G. B. Dantzig, G. B. Dantzig, Linear Programming and ExtensionLinear Programming and Extension, Princeton, New , Princeton, New

Jersey, Princeton University Press, 1963.Jersey, Princeton University Press, 1963. Ellipsoid methodEllipsoid method

L. G. Khachiyan, “A Polynomial Algorithm for Linear Programming,” L. G. Khachiyan, “A Polynomial Algorithm for Linear Programming,” Soviet Math. DoklSoviet Math. Dokl., vol. 20, pp. 191-194, 1984.., vol. 20, pp. 191-194, 1984.

Interior-point methodInterior-point method N. K. Karmarkar, “A New Polynomial-Time Algorithm for Linear N. K. Karmarkar, “A New Polynomial-Time Algorithm for Linear

Programming,” Programming,” CombinatoricaCombinatorica, vol. 4, pp. 373-395, 1984., vol. 4, pp. 373-395, 1984. Course website of Prof. Lieven Vandenberghe (UCLA), Course website of Prof. Lieven Vandenberghe (UCLA),

http://www.ee.ucla.edu/ee236a/ee236a.html

http://www.ee.ucla.edu/ee236a/ee236a.html


Basic Ideas of Solution methodsBasic Ideas of Solution methods

Constraints

Extreme points

Objective function Constraints

Extreme points

Objective function

Simplex: search on extreme points.Complexity: polynomial in n, number of variables

Interior-point methods: Successivelyiterate with interior spaces of analytic convex boundaries.Complexity: O(n3.5L), L = no. of int. values


Integer Linear Programming (ILP)Integer Linear Programming (ILP)

Variables are integers.Variables are integers. Complexity is exponential – higher than LP.Complexity is exponential – higher than LP. LP relaxationLP relaxation

Convert all variables to real, preserve ranges.Convert all variables to real, preserve ranges. LP solution provides guidance.LP solution provides guidance. Rounding LP solution can provide a non-optimal Rounding LP solution can provide a non-optimal

solution.solution.


Traveling Salesperson Problem (TSP)Traveling Salesperson Problem (TSP)

1

32

5

412

27

18

10

5

20

1215

19

6


Solving TSP: Five CitiesSolving TSP: Five CitiesDistances (dij) in miles (symmetric TSP, general TSP is asymmetric)

CityCity j=1j=1 j=2j=2 j=3j=3 j=4j=4 j=5j=5

i=1i=1 00 1818 1010 1212 2727

i=2i=2 1818 00 55 1212 2020

i=3i=3 1010 55 00 1515 1919

i=4i=4 1212 1212 1515 00 66

i=5i=5 2727 2020 1919 66 00


Search Space: No. of ToursSearch Space: No. of Tours

Asymmetric TSP toursAsymmetric TSP tours Five-city problem: 4 Five-city problem: 4 × 3 × 2 × 1 = 24 tours× 3 × 2 × 1 = 24 tours Ten-city problem: 362,880 toursTen-city problem: 362,880 tours 15-city problem: 87,178,291,200 tours15-city problem: 87,178,291,200 tours 50-city problem: 49! = 6.08×1050-city problem: 49! = 6.08×1062 tours tours

Time for enumerative search assuming 1 Time for enumerative search assuming 1 μμs per tour s per tour evaluationevaluation == 1.93×101.93×1055 years years


A Greedy Heuristic SolutionA Greedy Heuristic Solution

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

Tour length = 10 + 5 + 12 + 6 + 27 = 60 miles (non-optimal)


ILP Variables, Constants and ConstraintsILP Variables, Constants and Constraints

1

32

5

4 d14 = 12

d15 = 27

d12 = 18

d13 = 10

x14 ε [0,1]

x15 ε [0,1]

x12 ε [0,1]

x13 ε [0,1]

x12 + x13 + x14 + x15 = 1 four other similar equations

Integer variables:xij = 1, travel i to jxij = 0, do not travel i to j

Real constants:dij = distance from i to j


Objective Function and ILP SolutionObjective Function and ILP Solution 5 i - 1

Minimize ∑ ∑ xij × dij i = 1 j = 1

xijxij j=1j=1 22 33 44 55 i=1i=1 00 00 11 00 0022 11 00 00 00 0033 00 11 00 00 0044 00 00 00 00 1155 00 00 00 11 00

∑ xij = 1 for all i j ≠ i


ILP SolutionILP Solution

1

32

5

4

d13 = 10

d45 = 6

Total length = 45 but not a single tour

d54 = 6

d21 = 18

d32 = 5


Additional Constraints for Single TourAdditional Constraints for Single Tour

Following constraints prevent split tours. For any Following constraints prevent split tours. For any subset S of cities, the tour must enter and exit subset S of cities, the tour must enter and exit that subset:that subset:

∑ xij ≥ 2 for all S, |S| < 5i ε S j ε S

Any subset

Remaining set At least two

arrows must crossthis boundary.


ILP SolutionILP Solution

1

32

5

4

d13 = 10

d41 = 12

Total length = 53

d54 = 6

d25 = 20

d32 = 5


ILP Example: Test MinimizationILP Example: Test Minimization A combinational circuit has A combinational circuit has nn test vectors that detect test vectors that detect mm faults. faults.

Each vector detects a subset of faults. Find the smallest Each vector detects a subset of faults. Find the smallest subset of test vectors that each fault is detected by at least subset of test vectors that each fault is detected by at least NN vectors.vectors.

Simulate vectors without dropping faults.Simulate vectors without dropping faults.

T1 T2 . . Tj . . . TnF1 1 0 0 1 1 0 1 0 0

F2 0 0 1 1 0 0 0 1 0

. . . . . . . . . .

. . . . . . . . . .

Fj 1 0 0 1 1 0 0 1 1

. . . . . . . . . .

Fm 0 1 1 1 0 0 0 0 1

Faul

ts

Test vectors

fij = 1, if test Tidetects fault Fj


Test Minimization by ILPTest Minimization by ILP

n minimize Σ ti Objective function

i=1 n

subject to Σ fij ti ≥ N, j = 1, 2, . . ., mi=1

Construct an ILP model:1.Assign an integer variable ti ε [0,1] to ith test vector such that ti = 1, if we select ti, otherwise ti= 0.2.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.


N-Detect Tests (N-Detect Tests (NN = 5) = 5)

CircuitCircuit Unoptimized vectorsUnoptimized vectorsILP (exact)ILP (exact)

Minimum vectorsMinimum vectors CPU sCPU s

c432c432 608608 197197 1.01.0

c499c499 379379 260260 2.32.3

c880c880 1,0231,023 127127 881.8881.8

c1355c1355 755755 420420 4.44.4

c1908c1908 1,0551,055 543543 6.96.9

c2670c2670 959959 477477 7.27.2

c3540c3540 1,9711,971 471471 20008.520008.5

c5315c5315 1,0791,079 376376 40.740.7

c6288c6288 243243 5757 34740.034740.0

c7552c7552 2,1652,165 841841 114.3114.3


Why ILP Solution is Exponential?Why ILP Solution is Exponential?LP solutionfound inpolynomial time(bound on ILPsolution)

Must try all2n roundoffpoints

First variable

Sec

ond

varia

ble

Constraints

Objective(maximize)


Characteristics of ILPCharacteristics of ILP Worst-case complexity is exponential in number of Worst-case complexity is exponential in number of

variables.variables. Linear programming (LP) relaxation, where integer Linear programming (LP) relaxation, where integer

variables are treated as real, gives a lower bound on the variables are treated as real, gives a lower bound on the objective function.objective function.

Recursive roundingRecursive rounding of relaxed LP solution to nearest of relaxed LP solution to nearest integers gives an approximate solution to the ILP integers gives an approximate solution to the ILP problem.problem. K. R. Kantipudi and V. D. Agrawal, “A Reduced Complexity K. R. Kantipudi and V. D. Agrawal, “A Reduced Complexity

Algorithm for Minimizing Algorithm for Minimizing NN-Detect Tests,” -Detect Tests,” Proc. 20Proc. 20thth International International Conf. VLSI DesignConf. VLSI Design, January 2007, pp. 492-497., January 2007, pp. 492-497.


Recursive Rounding AlgorithmRecursive Rounding Algorithm

1.1. Obtain a relaxed LP solution. Obtain a relaxed LP solution. Stop if each if each variable in the solution is an integer.variable in the solution is an integer.

2.2. Round the variable closest to an integer.Round the variable closest to an integer.3.3. Remove any constraints that are now Remove any constraints that are now

unconditionally satisfied.unconditionally satisfied.4.4. Go to step 1.Go to step 1.


Complexity of ApproximationComplexity of Approximation Recursive rounding:Recursive rounding:

ILP is transformed into k LPs with progressively reducing ILP is transformed into k LPs with progressively reducing number of variables, where k is the size of the solution.number of variables, where k is the size of the solution.

A solution that satisfies all constraints is guaranteed; this A solution that satisfies all constraints is guaranteed; this solution is often close to optimal.solution is often close to optimal.

Number of LPs, k, is the size of the final solution, i.e., the Number of LPs, k, is the size of the final solution, i.e., the number of non-zero variables in the test minimization number of non-zero variables in the test minimization problem.problem.

Recursive rounding complexity is k × O(nRecursive rounding complexity is k × O(npp), where k ≤ n, ), where k ≤ n, n is number of variables.n is number of variables.


Four-Bit ALU CircuitFour-Bit ALU Circuit

Initial Initial vectorsvectors

ILPILP Recursive roundingRecursive rounding

VectorsVectors CPU sCPU s VectorsVectors CPU sCPU s

285285 1414 0.650.65 1414 0.420.42

400400 1313 1.071.07 1313 1.001.00

500500 1212 4.384.38 1313 3.003.00

1,0001,000 1212 4.174.17 1212 3.003.00

5,0005,000 1212 12.9512.95 1212 9.009.00

10,00010,000 1212 34.6134.61 1212 17.017.0

16,38416,384 1212 87.4787.47 1212 37.037.0


ILP vs. Recursive RoundingILP vs. Recursive Rounding

0 5,000 10,000 15,000 Vectors

100

75

50

25

0

ILP

Recursive Rounding

CPU

s


N-Detect Tests (N-Detect Tests (N N = 5)= 5)

CircuitCircuit Unoptimized Unoptimized vectorsvectors

Relaxed LP/Recur. Relaxed LP/Recur. roundingrounding ILP (exact)ILP (exact)

Lower Lower boundbound

Min. Min. vectorsvectors CPU sCPU s Min. Min.

vectorsvectors CPU sCPU s

c432c432 608608 196.38196.38 197197 1.01.0 197197 1.01.0

c499c499 379379 260.00260.00 260260 1.21.2 260260 2.32.3

c880c880 1,0231,023 125.97125.97 128128 14.014.0 127127 881.8881.8

c1355c1355 755755 420.00420.00 420420 3.23.2 420420 4.44.4

c1908c1908 1,0551,055 543.00543.00 543543 4.64.6 543543 6.96.9

c2670c2670 959959 477.00477.00 477477 4.74.7 477477 7.27.2

c3540c3540 1,9711,971 467.25467.25 477477 72.072.0 471471 20008.520008.5

c5315c5315 1,0791,079 374.33374.33 377377 18.018.0 376376 40.740.7

c6288c6288 243243 52.5252.52 5757 39.039.0 5757 34740.034740.0

c7552c7552 2,1652,165 841.00841.00 841841 52.052.0 841841 114.3114.3

A Primal-Dual Solution (A Primal-Dual Solution (NN = 1) = 1)


Circuit Name

Lower bound

on vectors

Recursive LP minimization Primal-dual minimization

Unopt. vectors

LPCPU s

Minimized vectors

Unopt. vectors

TotalCPU s

Minimized vectors

c432 27 608 2.00 36 983 5.52 31c499 52 379 1.00 52 221 1.35 52c880 13 1023 31.00 28 1008 227.21 25

c1355 84 755 5.00 84 507 1.95 84c1908 106 1055 8.00 107 728 2.50 107c2670 44 959 9.00 84 1039 17.41 79c3540 78 1971 197.00 105 2042 276.91 95c5315 37 1079 464.00 72 1117 524.53 67c6288 6 243 78.00 18 258 218.9 17c7552 65 2165 151.00 145 2016 71.21 139

M. A. Shukoor and V. D. Agrawal, “A Primal-Dual Solution to Minimal Test Generation Problem,” Proc. 12th IEEE VLSI Design & Test Symp. (VDAT08), 2008, pp. 269-279.


Finding LP/ILP SolversFinding LP/ILP Solvers

R. Fourer, D. M. Gay and B. W. Kernighan, R. Fourer, D. M. Gay and B. W. Kernighan, AMPL: A Modeling AMPL: A Modeling Language for Mathematical ProgrammingLanguage for Mathematical Programming, South San Francisco, , South San Francisco, California: Scientific Press, 1993. Several of programs described in California: Scientific Press, 1993. Several of programs described in this book are available to Auburn users.this book are available to Auburn users.

B. R. Hunt, R. L. Lipsman, J. M. Rosenberg, K. R. Coombes, J. E. B. R. Hunt, R. L. Lipsman, J. M. Rosenberg, K. R. Coombes, J. E. Osborn and G. J. Stuck, Osborn and G. J. Stuck, A Guide to MATLAB for Beginners and A Guide to MATLAB for Beginners and Experienced UsersExperienced Users, Cambridge University Press, 2006., Cambridge University Press, 2006.

Search the web. Many programs with small number of variables can Search the web. Many programs with small number of variables can be downloaded free.be downloaded free.

A Circuit Optimization ProblemA Circuit Optimization Problem

Given:Given: Circuit netlistCircuit netlist Cell library with multiple versions for each cellCell library with multiple versions for each cell

Select cell versions to optimize a specified Select cell versions to optimize a specified characteristic of the circuit. Typical characteristic of the circuit. Typical characteristics are:characteristics are: AreaArea PowerPower DelayDelay


Example: Cell(X), X = 0 or 1Example: Cell(X), X = 0 or 1

X = 0X = 0 Delay = dDelay = d Power = 3 Power = 3 × × pp

X = 1X = 1 Delay = 2 Delay = 2 × × dd Power = 0.5 Power = 0.5 × × pp

Cell delay = (1 – X) d + 2 X dCell delay = (1 – X) d + 2 X d Power = 3(1 – X) p + 0.5 X pPower = 3(1 – X) p + 0.5 X p


ILP Model: Minimum Power & DelayILP Model: Minimum Power & Delay

Ti = signal arrival time at ith input; Ti = 0 for all PIsTi = signal arrival time at ith input; Ti = 0 for all PIs Tk = signal arrival time at cell outputTk = signal arrival time at cell output Tk Tk ≥ Ti + ≥ Ti + (1 – Xk) dk + 2 Xk dk, for all i(1 – Xk) dk + 2 Xk dk, for all i

Where, Where, dk = nominal delay of gatedk = nominal delay of gateXk = 0 or 1, specifies version of cellXk = 0 or 1, specifies version of cell

Minimize Minimize αα TPO + ∑ [3(1 – Xk) pk + 0.5 Xk pk] TPO + ∑ [3(1 – Xk) pk + 0.5 Xk pk] αα is constant is constant all kall k


kth Cell Arrival time = Tk

Arrival time = T1

Ti

Given Clock SpecificationGiven Clock Specification Tj = 0, for all primary inputs jTj = 0, for all primary inputs j Tk Tk ≤ clock period, for all primary outputs k≤ clock period, for all primary outputs k Tk Tk ≥ Ti + ≥ Ti + (1 – Xk) dk + 2 Xk dk, for all gates k with input i(1 – Xk) dk + 2 Xk dk, for all gates k with input i


Combinational Logic

Reg

iste

r

Reg

iste

r

Clock

Minimum Power DesignMinimum Power Design

MinimizeMinimize ∑∑ 3(1 – Xk) pk + 0.5 Xk pk 3(1 – Xk) pk + 0.5 Xk pkall kall k

wherewhere pk = nominal power consumption of pk = nominal power consumption of kth cellkth cell


Logic MinimizationLogic Minimization Consider a four-variable function, {2,4,6,8,9,10,12,13,15}Consider a four-variable function, {2,4,6,8,9,10,12,13,15} Karnaugh map shows prime implicants (PI) found by Quine-Karnaugh map shows prime implicants (PI) found by Quine-

McCluskey procedure.McCluskey procedure. Find the minimum number of Pis to cover all minterms.Find the minimum number of Pis to cover all minterms.

Fall 2008, Oct 13 . . .Fall 2008, Oct 13 . . . ELEC2200-002 Lecture 5ELEC2200-002 Lecture 5 4545

1 1 1

1 1

1

1 1 1

A

B

C

DEPI’s

Non-EPI’s Non-EPI’s

Select a Minimal Set of PI’sSelect a Minimal Set of PI’sCovered by EPI → x x x x x

Minterm → 2 4 6 8 9 10 12 13 15

PI1 x x x x

PI2 x x

PI3 x x

PI4 x x

PI5 x x

PI6 x x

PI7 x x


1. First select essential prime implicants (EPIs).2. Cover remaining minterms with smallest number of prime

implicants (Pis).

Cover Remaining MintermsCover Remaining MintermsRemaining minterms → 2 4 6 10

PI2 x x

PI3 x x

PI4 x x

PI5 x

PI6 x


Integer linear program (ILP): Define integer {0,1} variables, xk = 1, select PIk; xk = 0, do not select PIk.Minimize k xk, subject to following constraints:

x2 + x3 ≥ 1 (cover minterm 2)x4 + x5 ≥ 1 (cover minterm 4)x2 + x4 ≥ 1 (cover minterm 6)x3 + x6 ≥ 1 (cover minterm 10)

A solution is x3 = x4 = 1, x2 = x5 = x6 = 0

Minimized FunctionMinimized Function

F(A,B,C,D) =F(A,B,C,D) = PI1PI1 + PI3 + PI4 + + PI3 + PI4 + PI7PI7 == AACC + +B CB CD D ++A BA BD D + + A B DA B D


1 1 1

1 1

1

1 1 1

A

B

C

DEPI’s in MSOP

Selected PIs Pis not selected

Copyright Agrawal, 2007Copyright Agrawal, 2007 ELEC6270 Spring 09, Lecture 10ELEC6270 Spring 09, Lecture 10 4949

Comb. Circuit Power OptimizationComb. Circuit Power Optimization

Given a set of test vectorsGiven a set of test vectors Reorder vectors to minimize the number of Reorder vectors to minimize the number of

transitions at primary inputstransitions at primary inputs

Combinational circuit(tested by exhaustive

vectors)

010101010011001100001111

01111000Rearranged vector set 00110011 produces 7 transitions

00011110

11 transitions


Reducing Comb. Test Power Reducing Comb. Test Power

1 1 0 0 01 0 1 0 01 0 1 0 11 0 1 1 1

V1 V2 V3

V4 V5

3 4

1

3 223

2

1

1

Original tests:V1 V2 V3 V4 V5

10 input transitions

Traveling salesperson problem (TSP) finds the shortest distance closed path (or cycle) to visit all nodes exactly once.But, we need an open loop solution.

Reordered tests:V1 V3 V5 V4 V21 0 0 0 11 1 0 0 01 1 1 0 01 1 1 1 0

5 input transitions


Open-Loop TSP Open-Loop TSP

Add a node V0 at distance 0 from all other nodes.Add a node V0 at distance 0 from all other nodes. Solve TSP for the new graph.Solve TSP for the new graph. Delete V0 from the solution.Delete V0 from the solution.

V1 V2 V3

V4 V5

3 4

1

322

3

2

11V0

0

00

0

0

Combinational Vector OrderingCombinational Vector Ordering See: P. Wray, “See: P. Wray, “Minimize test power for benchmark circuit c6288 by Minimize test power for benchmark circuit c6288 by

optimal ordering of vectors,” ELEC 6270 Class Project Report, Spring optimal ordering of vectors,” ELEC 6270 Class Project Report, Spring 2009, 2009, www.eng.auburn.edu/~vagrawal/COURSE/E6270_Spr09/PROJECT/WRAY/

TSP has exponential complexity; good heuristics are available.TSP has exponential complexity; good heuristics are available. For other extensions:For other extensions:

V. Dabholkar, S. Chakravarty, I Pomeranz and S. Reddy, V. Dabholkar, S. Chakravarty, I Pomeranz and S. Reddy, “Techniques for Minimizing Power Dissipation in Scan and “Techniques for Minimizing Power Dissipation in Scan and Combinational Circuits During Test Application,” Combinational Circuits During Test Application,” IEEE Trans. CADIEEE Trans. CAD, , vol. 17, no. 12, pp. 1325-1333, Dec. 1998.vol. 17, no. 12, pp. 1325-1333, Dec. 1998.

Typical average power saving:Typical average power saving: 30-50%30-50% 50-60% with vector repetition (to satisfy peak power)50-60% with vector repetition (to satisfy peak power) ? ? ? With inserted vectors (to satisfy peak power)? ? ? With inserted vectors (to satisfy peak power)


http://www.eng.auburn.edu/~vagrawal/COURSE/E6270_Spr09/PROJECT/WRAY/


Traveling Salesperson ProblemTraveling Salesperson Problem A. V. Aho, J. E. Hopcroft anf J. D. Ullman, A. V. Aho, J. E. Hopcroft anf J. D. Ullman, Data Data

Structures and AlgorithmsStructures and Algorithms, Reading, , Reading, Massachusetts: Addison-Wesley, 1983.Massachusetts: Addison-Wesley, 1983.

E. Horowitz and S. Sahni, E. Horowitz and S. Sahni, Fundamentals of Fundamentals of Computer AlgorithmsComputer Algorithms, Computer Science Press, , Computer Science Press, 1984.1984.

B. R. Hunt, R. L. Lipsman, J. M. Rosenberg, K. B. R. Hunt, R. L. Lipsman, J. M. Rosenberg, K. R. Coombes, J. E. Osborn and G. J. Stuck, R. Coombes, J. E. Osborn and G. J. Stuck, A A Guide to MATLAB for Beginners and Guide to MATLAB for Beginners and Experienced UsersExperienced Users, Cambridge University , Cambridge University Press, 2006.Press, 2006.

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