1 computer science and decision making fred roberts, rutgers university

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Computer Science and Decision Making

Fred Roberts, Rutgers University

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Computer Science and Decision Making

•Many recent applications in CS involve issues/problems of long interest in decision theory:

social choice or consensuspreference, utilityconflict and cooperationallocationincentivesmeasurement

•Methods developed in decision theory, often in the social sciences, beginning to be used in CS

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CS and DM •CS applications place great strain on SS-DM methods

Sheer size of problems addressedComputational power of agents an issueLimitations on information possessed by playersSequential nature of repeated applications

•Thus: Need for new generation of SS-DM methods

•Also: These new methods will provide powerful tools to social scientists/decision makers.

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CS and DM: Outline 1. Consensus Rankings2. Meta-search and Collaborative Filtering3. Computational Approaches to Information

Management in Group Decision Making4. Large Databases and Inference5. Consensus Computing, Image Processing6. Computational Intractability of Consensus

Functions7. Electronic Voting8. Software and Hardware Measurement9. Power of a Voter10.Sequential DM: Port of Entry Inspections

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Consensus Rankings: Social Choice • Relevant social science problems: voting, group

decision making• Goal: based on everyone’s opinions, reach a “consensus” • Typical opinions expressed as:

“first choice”ranking of all alternativesscores classifications

• Long history of research on such problems.

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Consensus Rankings Background: Arrow’s Impossibility Theorem: • There is no “consensus method” that satisfies

certain reasonable axioms about how societies should reach decisions.

• Input to Arrow’s Theorem: rankings of alternatives (ties allowed).• Output: consensus ranking.

Kenneth ArrowNobel prize winner

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Consensus Rankings

• There are widely studied and widely used consensus methods (that violate one or

more of Arrow’s conditions).

• One well-known consensus method: “Kemeny-Snell medians”: Given set of rankings, find ranking minimizing sum of distances to other rankings.

• Kemeny-Snell medians are having surprising new applications in CS.

John Kemeny,pioneer in time sharingin CS

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Consensus Rankings• Kemeny-Snell distance between rankings: twice the

number of pairs of candidates i and j for which i is ranked above j in one ranking and below j in the other + the number of pairs that are ranked in one ranking and tied in another.

a bx y-zy xzOn {x,y}: +2On {x,z}: +2On {y,z}: +1d(a,b) = 5.

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Consensus Rankings

• Kemeny-Snell median: Given rankings a1, a2, …, ap, find a ranking x so that

d(a1,x) + d(a2,x) + … + d(ap,x) is minimized.

• x can be a ranking other than a1, a2, …, ap.

• Sometimes just called Kemeny median.

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Consensus Rankings a1 a2 a3

Fish Fish ChickenChicken Chicken FishBeef Beef Beef

• Median = a1.

• If x = a1:d(a1,x) + d(a2,x) + d(a3,x) = 0 + 0 + 2 = 2

is minimized.• If x = a3, the sum is 4.• For any other x, the sum is at least 1 + 1 + 1 = 3.

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Fish Chicken BeefChicken Beef FishBeef Fish Chicken

• Three medians = a1, a2, a3.

• This is the “voter’s paradox” situation.

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• Note that sometimes we wish to minimize

d(a1,x)2 + d(a2,x)2 + … + d(ap,x)2

• A ranking x that minimizes this is called a Kemeny-Snell mean.

• In this example, there is one mean: the ranking declaring all three alternatives tied.

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• If x is the ranking declaring Fish, Chicken and Beef tied, then

d(a1,x)2 + d(a2,x)2 + … + d(ap,x)2 = 32 + 32 + 32 = 27.

• Not hard to show this is minimum.

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Consensus Rankings

Theorem (Bartholdi, Tovey, and Trick, 1989; Wakabayashi, 1986): Computing the Kemeny median of a set of rankings is an NP-complete problem.

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Consensus Rankings

Okay, so what does this have to do with practical computer science questions?

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Consensus Rankings

I mean really practical computer science questions.

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Meta-search and Collaborative Filtering Meta-search

• A consensus problem• Combine page rankings from several search

engines• Dwork, Kumar, Naor, Sivakumar (2000):

Kemeny-Snell medians good in spam resistance in meta-search (spam by a page if it causes meta-search to rank it too highly)

• Approximation methods make this computationally tractable

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Meta-search and Collaborative Filtering

Collaborative Filtering

• Recommending books or movies• Combine book or movie ratings• Produce ordered list of books or movies to

recommend• Freund, Iyer, Schapire, Singer (2003):

“Boosting” algorithm for combining rankings.• Related topic: Recommender Systems

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Meta-search and Collaborative Filtering

A major difference from SS-DM applications:

• In SS-DM applications, number of voters is large, number of candidates is small.

• In CS applications, number of voters (search engines) is small, number of candidates (pages) is large.

• This makes for major new complications and research challenges.

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Computational Approaches to Information Management in Group Decision Making

Representation and Elicitation

• Successful group decision making requires efficient elicitation of information and efficient representation of the information elicited.

• Old problems in the social sciences.• Computational aspects becoming a focal point

because of need to deal with massive and complex information.

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• Example I: Preferences are key components in decision making applications.

• “I prefer beef to fish”

• Extracting and representing preferences is key in group decision making applications.

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• “Brute force” approach: For every pair of alternatives, ask which is preferred to the other.

• Often computationally infeasible.

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• In many applications (e.g., collaborative filtering), important to elicit preferences automatically.

• CP-nets introduced as tool to represent preferences succinctly and provide ways to make inferences about preferences (Boutilier, Brafman, Doomshlak, Hoos, Poole 2004).

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• Example II: combinatorial auctions.• Auctions increasingly used in business and

government.• Information technology allows complex auctions with huge number of bidders.• There are key decision makingproblems arising in auctions.

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Representation and Elicitation• Bidding functions maximizing expected profit

can be exceedingly difficult to compute.• Determining the winner of an auction can be

extremely hard. (Rothkopf, Pekec, Harstad 1998)

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Combinatorial Auctions

• Multiple goods auctioned off.• Submit bids for combinations of goods.• This leads to NP-complete allocation problems.• Might not even be able to feasibly express all

possible preferences for all subsets of goods.• Rothkopf, Pekec, Harstad (1998): determining

winner is computationally tractable for many economically interesting kinds of combinations.

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Combinatorial Auctions

• Decision maker needs to elicit preferences from all agents for all plausible combinations of items in the auction.

• Similar problem arises in optimal bundling of goods and services.

• Elicitation requires exponentially many queries in general.

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Representation and Elicitation• Challenge: Recognize situations in which

efficient elicitation and representation is possible.

• One result: Fishburn, Pekec, Reeds (2002)• Even more complicated: When objects in

auction have complex structure. • Problem arises in:

Legal reasoning, sequential decision making, automatic decision devices, collaborative filtering.

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Large Databases and Inference • Decision makers consult massive data sets.• Real data often in form of sequences.• Example: Bioinformatics.• GenBank has over 7 million sequences

comprising 8.6 billion bases. • The search for similarity or patterns has

extended from pairs of sequences to finding patterns that appear in common in a large number of sequences or throughout the database: “consensus sequences”

• Emerging field of “Bioconsensus”: applies SS-DM consensus methods to biological databases.

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Large Databases and Inference

Why look for such patterns?

Similarities between sequences or parts of sequences lead to the discovery of shared phenomena.

For example, it was discovered that the sequence for platelet derived factor, which causes growth in the body, is 87% identical to the sequence for v-sis, a cancer-causing gene. This led to the discovery that v-sis works by stimulating growth.

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Large Databases and Inference Example

Bacterial Promoter Sequences studied by Waterman (1989):

RRNABP1: ACTCCCTATAATGCGCCATNAA: GAGTGTAATAATGTAGCCUVRBP2: TTATCCAGTATAATTTGTSFC: AAGCGGTGTTATAATGCC

Notice that if we are looking for patterns of length 4, each sequence has the pattern TAAT.

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Bacterial Promoter Sequences studied by Waterman (1989):

RRNABP1: ACTCCCTATAATGCGCCATNAA: GAGTGTAATAATGTAGCCUVRBP2: TTATCCAGTATAATTTGTSFC: AAGCGGTGTTATAATGCC

Notice that if we are looking for patterns of length 4, each sequence has the pattern TAAT.

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However, suppose that we add another sequence:

M1 RNA: AACCCTCTATACTGCGCG

The pattern TAAT does not appear here.However, it almost appears, since the pattern

TACT appears, and this has only one mismatch from the pattern TAAT.

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However, suppose that we add another sequence:

M1 RNA: AACCCTCTATACTGCGCG

The pattern TAAT does not appear here.However, it almost appears, since the pattern

TACT appears, and this has only one mismatch from the pattern TAAT.

So, in some sense, the pattern TAAT is a good consensus pattern.

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We make this precise using best mismatch distance.

Consider two sequences a and b with b longer than a.

Then d(a,b) is the smallest number of mismatches in all possible alignments of a as a consecutive subsequence of b.

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a = 0011, b = 111010

Possible Alignments:111010111010 1110100011 0011 0011

The best-mismatch distance is 2, which is achieved in the third alignment.

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Large Databases and Inference Smith-Waterman

•Let be a finite alphabet of size at least 2 and be a finite collection of words of length L on . •Let F() be the set of words of length k 2 that are our consensus patterns. (Assume L k.)•Let = {a1, a2, …, an}. •One way to define F() is as follows. •Let d(a,b) be the best-mismatch distance. •Consider nonnegative parameters sd that are monotone decreasing with d and let F(a1,a2, …, an) be all those words w of length k that maximize

S(w) = isd(w,ai)

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Large Databases and Inference•We call such an F a Smith-Waterman consensus.•In particular, Waterman and others use the parameters

sd = (k-d)/k.

Example:

•An alphabet used frequently is the purine/pyrimidine alphabet {R,Y}, where R = A (adenine) or G (guanine) and Y = C (cytosine) or T (thymine). •For simplicity, it is easier to use the digits 0,1 rather than the letters R,Y.

•Thus, let = {0,1}, let k = 2. Then the possible pattern words are 00, 01, 10, 11.

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Large Databases and Inference•Suppose a1 = 111010, a2 = 111111. How do we find F(a1,a2)?•We have:

d(00,a1) = 1, d(00,a2) = 2d(01,a1) = 0, d(01,a2) = 1d(10,a1) = 0, d(10,a2) = 1d(11,a1) = 0, d(11,a2) = 0

S(00) = sd(00,ai) = s1 + s2,

S(01) = sd(01,ai) = s0 + s1

S(10) = sd(10,ai) = s0 + s1

S(11) = sd(11,ai) = s0 + s0

•As long as s0 > s1 > s2, it follows that 11 is the consensus pattern, according to Smith-Waterman’s consensus.

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Example:•Let ={0,1}, k = 3, and consider F(a1,a2,a3) where a1 = 000000, a2 = 100000, a3 = 111110. The possible pattern words are: 000, 001, 010, 011, 100, 101, 110, 111.

d(000,a1) = 0, d(000,a2) = 0, d(000,a3) = 2,d(001,a1) = 1, d(001,a2) = 1, d(001,a3) = 2,

d(100,a1) = 1, d(100,a2) = 0, d(100,a3) = 1, etc.

S(000) = s2 + 2s0, S(001) = s2 + 2s1, S(100) = 2s1 + s0, etc.

•Now, s0 > s1 > s2 implies that S(000) > S(001). •Similarly, one shows that the score is maximized by S(000) or S(100). • Monotonicity doesn’t say which of these is highest.

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The Special Case sd = (k-d)/k

•Then it is easy to show that the words w that maximize s(w) are exactly the words w that minimize

id(w,ai).

•In other words: In this case, the Smith-Waterman consensus is exactly the median.

Algorithms for computing consensus sequences such as Smith-Waterman are important in modern molecular biology.

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Large Databases and Inference Other Topics in “Bioconsensus”

• Alternative phylogenies (evolutionary trees) are produced using different methods and we need to choose a consensus tree.

• Alternative taxonomies (classifications) are produced using different models and we need to choose a consensus taxonomy.

• Alternative molecular sequences are produced using different criteria or different algorithms and we need to choose a consensus sequence.

• Alternative sequence alignments are produced and we need to choose a consensus alignment.

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Other Topics in “Bioconsensus”

• Several recent books on bioconsensus.• Day and McMorris [2003]• Janowitz, et al. [2003]

• Bibliography compiled by Bill Day: In molecular biology alone, hundreds of papers using consensus methods in biology.

• Large database problems in CS are being approached using methods of “bioconsensus” having their origin in SS-DM.

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Consensus Computing, Image Processing • Old SS-DM problem: Dynamic modeling of

how individuals change opinions over time, eventually reaching consensus.

• Often use dynamic models on graphs• Related to neural nets.

• CS applications: distributed computing.• Values of processors in a network are updated

until all have same value.

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Consensus Computing, Image Processing • CS application: Noise removal in digital images• Does a pixel level represent noise?• Compare neighboring pixels.• If values beyond threshold, replace pixel value

with mean or median of values of neighbors.• Related application in distributed computing.• Values of faulty processors are replaced by those

of neighboring non-faulty ones.• Berman and Garay (1993) use “parliamentary

procedure” called cloture

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Consensus Computing, Image Processing • Side comment: same models are being applied in

“computational and mathematical epidemiology”.

• Modeling the spread of disease through large social networks.

SARSMeasles

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Computational Intractability of Consensus Functions

• How hard is it to compute the winner of an election?

• We know counting votes can be difficult and time consuming. • However:

• Bartholdi, Tovey and Trick (1989): There are voting schemes where it can be computationally intractable to determine who won an election.

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• So, is computational intractability necessarily bad?

• Computational intractability can be a good thing in an election: Designing voting systems where it is computationally intractable to “manipulate” the outcome of an election by “insincere voting”:Adding votersDeclaring voters ineligibleAdding candidatesDeclaring candidates ineligible

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• Given a set A of all possible candidates and a set I of all possible voters.

• Suppose we know voter i’s ranking of all candidates in A, for every voter i.

• Given a subset of I of eligible voters, a particular candidate a in A, and a number k, is there a set of at most k ineligible voters who can be declared eligible so that candidate a is the winner?

• Bartholdi, Tovey, Trick (1989): For some consensus functions (voting rules), this is an NP-complete problem.

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• Given a set A of all possible candidates and a set I of all possible voters.

• Suppose we know voter i’s ranking of all candidates in A, for every voter i.

• Given a subset of I of eligible voters, a particular candidate a in A, and a number k, is there a set of at most k eligible voters who can be declared ineligible so that candidate a is the winner?

• Bartholdi, Tovey, Trick (1989): For some consensus functions (voting rules), this is an NP-complete problem.

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• Software agents may be more likely to manipulate than individuals (Conitzer and Sandholm 2002):Humans don’t think about manipulatingComputation can be tedious.Software agents are good at running

algorithmsSoftware agents only need to have code for

manipulation written once.All the more reason to develop

computational barriers to manipulation.

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• Stopping those software agents:

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• Conitzer and Sandholm (2002):Earlier results of difficulty of manipulation depend

on large number of candidatesNew results: manipulation possible with some

voting methods if smaller number (bounded number) of candidates)

In weighted voting, voters may have different numbers of votes (as in US presidential elections, where different states (= voters) have different numbers of votes). Here, manipulation is harder.Manipulation difficult when uncertainty about

others’ votes.

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• Conitzer and Sandholm (2006):Try to find voting rules for which

manipulation is usually hard.Why is this difficult to do?One explanation: under one reasonable

assumption, it is impossible to find such rules.

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Aside: Electronic Voting

• Issues:CorrectnessAnonymityAvailabilitySecurityPrivacy

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Electronic VotingSecurity Risks in Electronic Voting

• Threat of “denial of service attacks”• Threat of penetration attacks involving a

delivery mechanism to transport a malicious payload to target host (thru Trojan horse or remote control program)

• Private and correct counting of votes• Cryptographic challenges to keep votes private• Relevance of work on secure multiparty

computation

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Electronic Voting

Other CS Challenges:

• Resistance to “vote buying”• Development of user-friendly interfaces• Vulnerabilities of communication path between

the voting client (where you vote) and the server (where votes are counted)

• Reliability issues: random hardware and software failures

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Software & Hardware Measurement • Theory of measurement developed by

mathematical social scientists.• Measurement theory studies ways to combine

scores obtained on different criteria.• A statement involving scales of measurement is considered meaningful if its

truth or falsity is unchanged under acceptable transformations of all scales involved.

• Example: It is meaningful to say that I weigh more than my daughter.

• That is because if it is true in kilograms, then it is also true in pounds, in grams, etc.

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Software & Hardware Measurement • Measurement theory has studied what statements you

can make after averaging scores.• Think of averaging as a consensus (DM) method.• One general principle: To say that the average score of

one set of tests is greater than the average score of another set of tests is not meaningful (it is meaningless) under certain conditions.

• This is often the case if the averaging procedure is to take the arithmetic mean: If s(xi) is score of xi, i = 1, 2, …, n, then arithmetic mean is

is(xi)/n.• Long literature on what averaging methods lead to

meaningful conclusions.

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Software & Hardware Measurement A widely used method in hardware measurement:

Score a computer system on different benchmarks.

Normalize score relative to performance of one base system

Average normalized scoresPick system with highest average.Fleming and Wallace (1986): Outcome can

depend on choice of base system. Meaningless in sense of measurement theoryLeads to theory of merging normalized scores

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Software & Hardware Measurement Hardware Measurement

417 83 66 39,449 772

244 70 153 33,527 368

134 70 135 66,000 369

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

Data from Heath, Comput. Archit. News (1984)

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Software & Hardware Measurement Normalize Relative to Processor R

417

1.00

83

1.00

66

1.00

39,449

1.00

772

1.00

244

.59

70

.84

153

2.32

33,527

.85

368

.48

134

.32

70

.85

135

2.05

66,000

1.67

369

.45

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

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Software & Hardware Measurement Take Arithmetic Mean of Normalized Scores

417

1.00

83

1.00

66

1.00

39,449

1.00

772

1.00

244

.59

70

.84

153

2.32

33,527

.85

368

.48

134

.32

70

.85

135

2.05

66,000

1.67

369

.45

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

ArithmeticMean

1.00

1.01

1.07

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417

1.00

83

1.00

66

1.00

39,449

1.00

772

1.00

244

.59

70

.84

153

2.32

33,527

.85

368

.48

134

.32

70

.85

135

2.05

66,000

1.67

369

.45

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

ArithmeticMean

1.00

1.01

1.07

Conclude that machine Z is best

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Software & Hardware Measurement Now Normalize Relative to Processor M

417

1.71

83

1.19

66

.43

39,449

1.18

772

2.10

244

1.00

70

1.00

153

1.00

33,527

1.00

368

1.00

134

.55

70

1.00

135

.88

66,000

1.97

369

1.00

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

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417

1.71

83

1.19

66

.43

39,449

1.18

772

2.10

244

1.00

70

1.00

153

1.00

33,527

1.00

368

1.00

134

.55

70

1.00

135

.88

66,000

1.97

369

1.00

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

ArithmeticMean

1.32

1.00

1.08

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417

1.71

83

1.19

66

.43

39,449

1.18

772

2.10

244

1.00

70

1.00

153

1.00

33,527

1.00

368

1.00

134

.55

70

1.00

135

.88

66,000

1.97

369

1.00

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

ArithmeticMean

1.32

1.00

1.08

Conclude that machine R is best

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Software and Hardware Measurement • So, the conclusion that a given machine is best

by taking arithmetic mean of normalized scores is meaningless in this case.

• Above example from Fleming and Wallace (1986), data from Heath (1984)

• Sometimes, geometric mean is helpful.• Geometric mean is

is(xi)n

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Software & Hardware Measurement Normalize Relative to Processor R

417

1.00

83

1.00

66

1.00

39,449

1.00

772

1.00

244

.59

70

.84

153

2.32

33,527

.85

368

.48

134

.32

70

.85

135

2.05

66,000

1.67

369

.45

BENCHMARK

R

M

Z

PROCESSOR

E F G H I

GeometricMean

1.00

.86

.84

Conclude that machine R is best

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Software & Hardware Measurement Now Normalize Relative to Processor M

417

1.71

83

1.19

66

.43

39,449

1.18

772

2.10

244

1.00

70

1.00

153

1.00

33,527

1.00

368

1.00

134

.55

70

1.00

135

.88

66,000

1.97

369

1.00

BENCHMARK

R

M

Z

PROCESSOR

E F G H IGeometricMean

1.17

1.00

.99

Still conclude that machine R is best

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Software and Hardware Measurement• In this situation, it is easy to show that the conclusion

that a given machine has highest geometric mean normalized score is a meaningful conclusion.

• Even meaningful: A given machine has geometric mean normalized score 20% higher than another machine.

• Fleming and Wallace give general conditions under which comparing geometric means of normalized scores is meaningful.

• Research area: what averaging procedures make sense in what situations? Large literature.

• Note: There are situations where comparing arithmetic means is meaningful but comparing geometric means is not.

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Unknown function u = F(a1,a2,…,an)

The values a1, a2, …, an are scores and F is some averaging function

Luce's idea (“Principle of Theory Construction”): If you know the scale types of the ai and the scale type of u and you assume that an admissible transformation of each of the ai leads to an admissible transformation of u, you can derive the form of F.

Admissible transformation: pounds into grams, Fahrenheit into Centigrade. Scale type determined by class of admissible transformations.

Ratio scales, interval scales, …

Technical Aside: How Should We Average Scores?

R. Duncan Luce

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Example: a1, a2, …, an are independent ratio scales, u is a ratio scale.

F: (+)n +

F(a1,a2,…,an) = u F(1a1,2a2,…,nan) = u,

1 > 0, 2 > 0, n > 0, > 0, depends on a1, a2, …, an.

•Thus we get the functional equation:

(*) F(1a1,2a2,…,nan) = A(1,2,…,n)F(a1,a2,…,an),

A(1,2,…,n) > 0

How Should We Average Scores?

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(*) F(1a1,2a2,…,nan) = A(1,2,…,n)F(a1,a2,…,an),

A(1,2,…,n) > 0

Theorem (Luce 1964): If F: (+)n + is continuous and satisfies (*), then there are > 0, c1, c2, …, cn so that


F (a1;a2;:::;an) = ¸ac11 ac22 :::a

cnn :

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(Aczél, Roberts, Rosenbaum 1986): It is easy to see that the assumption of continuity can be weakened to continuity at a point, monotonicity, or boundedness on an (arbitrarily small, open) n-dimensional interval or on a set of positive measure. Call any of these assumptions regularity.


Janos Aczél“Mr Functional Equations”

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Theorem (Aczél and Roberts 1989): If in addition F satisfies reflexivity and symmetry, then = 1 and c1 = c2 = … = cn = 1/n , so F is the geometric mean.

Reflexivity: F(a,a,...,a) = a

Symmetry: F(a1,a2,…,an) = F(a(1),a(2),…,a(n)) for all permutations of {1,2,…,n}


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Sometimes You Get the Arithmetic Mean

Example: a1, a2, …, an interval scales with the same unit and independent zero points; u an interval scale.

Functional Equation:

(**) F(a1+1,a2+2,…,an+n) = A(,1,2,…,n)F(a1,a2,…,an) + B(,1,2,…,n)

A(,1,2,…,n) > 0


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Functional Equation:

(**) F(a1+1,a2+2,…,an+n) = A(,1,2,…,n)F(a1,a2,…,an) + B(,1,2,…,n)

A(,1,2,…,n) > 0

Solutions to (**) (Aczél, Roberts, and Rosenbaum 1986):

1, 2, …, n, b arbitrary constants

(no continuity assumptions needed)


F (a1;a2;:::;an) =nX

i=1

¸ iai +b

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Theorem (Aczél and Roberts 1989):

(1). If in addition F satisfies reflexivity, then

(2). If in addition F satisfies reflexivity and symmetry, then i= 1/n for all i, and b = 0, i.e., F is the arithmetic mean.


P ni=1

¸ i =1, b=0:

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Software and Hardware Measurement

• Message from measurement theory to computer science (and DM):

Do not perform arithmetic operations on data without paying attention to whether the conclusions you get are meaningful.

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Power of a VoterShapley-Shubik Power Index

• Think of a “voting game”• Shapley-Shubik index measures the power of each player in a multi-player game.• Consider a game in which some coalitions of players win and some lose, with no subset of a losing coalition winning.

Lloyd Shapley

MartinShubik

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Power of a Voter

Shapley-Shubik Power Index• Consider a coalition forming at random, one

player at a time.• A player i is pivotal if addition of i throws

coalition from losing to winning. • Shapley-Shubik index of i = probability i is

pivotal if an order of players is chosen at random.

• Power measure applying to more general games than voting games is called Shapley Value.

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Power of a Voter

Example: Shareholders of CompanyShareholder 1 holds 3 shares.Shareholders 2, 3, 4, 5, 6, 7 hold 1 share each.A majority of shares are needed to make a decision.Coalition {1,4,6} is winning.Coalition {2,3,4,5,6} is winning.

Shareholder 1 is pivotal if he is 3rd, 4th, or 5th.So shareholder 1’s Shapley value is 3/7.Sum of Shapley values is 1 (since they are probabilities)Thus, each other shareholder has Shapley value

(4/7)/6 = 2/21

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Power of a Voter

Example: United Nations Security Council

•15 member nations•5 permanent members

China, France, Russia, UK, US

•10 non-permanent•Permanent members have veto power•Coalition wins iff it has all 5 permanent membersand at least 4 of the 10 non-permanent members.

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Power of a Voter

Example: United Nations Security Council

•What is the power of eachMember of the SecurityCouncil? •Fix non-permanent member i.•i is pivotal in permutations in which all permanent members precede i and exactly 3 non- permanent members do.•How many such permutations are there?

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Power of a VoterExample: United Nations Security Council•Choose 3 non-permanent members preceding i. •Order all 8 members preceding i.•Order remaining 6 non-permanent members.•Thus the number of such permutations is:

C(9,3) x 8! x 6! = 9!/3!6! x 8! x 6! = 9!8!/3!•The probability that i is pivotal = power of non-permanent member =

9!8!/3!15! = .001865•The power of a permanent member =

[1 – 10 x .001865]/5 = .1963.•Permanent members have 100 times power of non-permanent members.

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Power of a Voter

•There are a variety of other power indices used in game theory and political science (Banzhaf index, Coleman index, …)

•Need calculate them for huge games•Mostly computationally intractable

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Power of a Voter: Allocation/Sharing Costs and Revenues

• Shapley-Shubik power index and more general Shapley value have been used to allocate costs to different users in shared projects.Allocating runway fees in airportsAllocating highway fees to trucks of

different sizesUniversities sharing library facilitiesFair allocation of telephone calling

charges among users sharing complex phone systems (Cornell’s experiment)

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Power of a Voter: Allocating/Sharing Costs and Revenues

Allocating Runway Fees at Airports• Some planes require longer runways.

Charge them more for landings. How much more?

• Divide runways into meter-long segments.

• Each month, we know how many landings a plane has made.

• Given a runway of length y meters, consider a game in which the players are landings and a coalition “wins” if the runway is not long enough for planes in the coalition.

98


Allocating Runway Fees at Airports• A landing is pivotal if it is the first

landing added that makes a coalition require a longer runway.

• The Shapley value gives the cost of the yth meter of runway allocated to a given landing.

• We then add up these costs over all runway lengths a plane requires and all landings it makes.

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Multicasting

• Applications in multicasting.• Unicast routing: Each packet sent from a

source is delivered to a single receiver.• Sending it to multiple sites: Send multiple

copies and waste bandwidth.• In multicast routing: Use a directed tree connecting source to all receivers.• At branch points, a packet is duplicated as necessary.

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Multicasting

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Multicasting

• Multicast routing: Use a directed tree connecting source to all receivers.

• At branch points, a packet is duplicated as necessary.

• Bandwidth is not directly attributable to a single receiver.

• How to distribute costs among receivers?• One idea: Use Shapley value.

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Allocating/Sharing Costs and Revenues• Feigenbaum, Papadimitriou, Shenker (2001):

no feasible implementation for Shapley value in multicasting.

• Note: Shapley value is uniquely characterized by four simple axioms.

• Sometimes we state axioms as general principles we want a solution concept to have.

• Jain and Vazirani (1998): polynomial time computable cost-sharing algorithmSatisfying some important axiomsCalculating cost of optimum multicast tree within

factor of two of optimal.

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104

Algorithms for Port of Entry Inspection for WMDs

Joint work with Saket Anand, David Madigan, Richard Mammone, Saumitr Pathak, Philip Stroud

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Port of Entry Inspection Algorithms

•Goal: Find ways to intercept illicit nuclear materials and weapons

destined for the U.S. via the maritime transportation system

•Currently inspecting only small % of containers arriving at ports

•Even inspecting 8% of containers in Port of NY/NJ might bring international trade to a halt (Larrabbee 2002)

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Port of Entry Inspection Algorithms•Aim: Develop decision support algorithms that will help us to “optimally” intercept illicit materials and weapons subject to limits on delays, manpower, and equipment

•Find inspection schemes that minimize total “cost” including “cost” of false positives and false negatives

Mobile Vacis: truck-mounted gamma ray imaging system

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Sequential Decision Making Problem•Stream of containers arrives at a port•The Decision Maker’s Problem:

•Which to inspect?•Which inspections next based on previous results?

•Approach: –“decision logics”–combinatorial optimization methods–Builds on ideas of Stroud and Saeger at Los AlamosNational Laboratory–Need for new models and methods

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Sequential Diagnosis Problem•Such sequential diagnosis problems arise in many areas:

–Communication networks (testing connectivity, paging cellular customers, sequencing tasks, …)–Manufacturing (testing machines, fault diagnosis, routing customer service calls, …)–Artificial intelligence/CS (optimal derivation strategies in knowledge bases, best-value satisficing search, coding decision trees, …)–Medicine (diagnosing patients, sequencing treatments, …)

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Sequential Decision Making Problem•Containers arriving to be classified into categories.•Simple case: 0 = “ok”, 1 = “suspicious”

•Inspection scheme: specifies which inspections are to be made based on previous observations

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Sequential Decision Making Problem

•Containers have attributes, each in a number of states

•Sample attributes:–Levels of certain kinds of chemicals or biological materials–Whether or not there are items of a certain kind in the cargo list–Whether cargo was picked up in a certain port

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•Currently used attributes:–Does ship’s manifest set off an “alarm”?–What is the neutron or Gamma emission count? Is it above threshold?–Does a radiograph image come up positive?–Does an induced fission test come up positive?

Gamma ray detector

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•We can imagine many other attributes•Concern with general algorithmic approaches.•Seek a methodology not tied to today’s technology.•Detectors are evolving quickly.

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•Simplest Case: Attributes are in state 0 or 1

•Then: Container is a binary string like 011001

•So: Classification is a decision function F that assigns each binary string to a category.

011001 F(011001)

If attributes 2, 3, and 6 are present and others are not, assign container to category F(011001).

114


•If there are two categories, 0 and 1, decision function F is a boolean function.

Example: F(000) = F(111) = 1, F(abc) = 0 otherwise

This classifies a container as positive iff it has none of the attributes or all of them.

1 =

115


•Given a container, test its attributes until know enough to calculate the value of F.

•An inspection scheme tells us in which order to test the attributes to minimize cost.

•Even this simplified problem is hard computationally.

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Sequential Decision Making Problem•This assumes F is known.•Simplifying assumption: Attributes are independent.•At any point we can stop inspecting and output the value of F based on outcomes of inspections so far.•Complications: May be precedence relations in the components (e.g., can’t test attribute a4 before testing a6. •Or: cost may depend on attributes tested before.•F may depend on variables that cannot be directly tested or for which tests are too costly.

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•Such problems are hard computationally.•There are many possible boolean functions F.•Even if F is fixed, problem of finding a good classification scheme (to be defined precisely below) is NP-complete. •Several classes of functions F allow for efficient inspection schemes:

–k-out-of-n systems–Certain series-parallel systems–Read-once systems–“regular” systems–Horn systems

118

Sensors and Inspection Lanes•n types of sensors measure presence or absence of the n attributes. •Many copies of each sensor.•Complication: different characteristics of sensors.•Entities come for inspection.•Which sensor of a given type to use?•Think of inspection lanes and queues.•Besides efficient inspection schemes, could decrease costs by:

–Buying more sensors–Change allocation of containers to sensor lanes.

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Binary Decision Tree Approach•Sensors measure presence/absence of attributes.

•Binary Decision Tree (BDT): –Nodes are sensors or categories (0 or 1)–Two arcs exit from each sensor node, labeled left and right.–Take the right arc when sensor says the attribute is present, left arc otherwise

120

Binary Decision Tree Approach

•Reach category 1 from the root only through the path a0 to a1 to 1.

•Container is classified in category 1 iff it has both attributes a0 and a1 .

•Corresponding boolean function F(11) = 1, F(10) = F(01) = F(00) = 0.

Figure 1

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Binary Decision Tree Approach•Reach category 1 from the root by:a0 L to a1 R a2 R 1 ora0 R a2 R1

•Container classified in category 1 iff it hasa1 and a2 and not a0 or a0 and a2 and possibly a1 .

•Corresponding boolean function F(111) = F(101) = F(011) = 1, F(abc) = 0 otherwise.

Figure 2

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Binary Decision Tree Approach•This binary decision tree corresponds to the same boolean function

F(111) = F(101) = F(011) = 1, F(abc) = 0 otherwise.

However, it has one less observation node ai. So, it is more efficient if all observations are equally costly and equally likely.

Figure 3

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Binary Decision Tree Approach•Even if the boolean function F is fixed, the problem of finding the “optimal” binary decision tree for it is very hard (NP-complete).

•For small n = number of attributes, can try to solve it by brute force enumeration.

•Even for n = 4, not practical. •(n = 4 at Port of Long Beach-Los Angeles)

Port of Long Beach

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Binary Decision Tree ApproachPromising Approaches:

•Heuristic algorithms, approximations to optimal.•Special assumptions about the boolean function F. •For “monotone” boolean functions, integer programming formulations give promising heuristics.•Stroud and Saeger enumerate all “complete,” monotone boolean functions and calculate the least expensive corresponding binary decision trees.•Their method practical for n up to 4, not n = 5.

125

Binary Decision Tree ApproachMonotone Boolean Functions:

•Given two strings x1x2…xn, y1y2…yn

•Suppose that xi yi for all i implies that F(x1x2…xn) F(y1,y2…yn).•Then we say that F is monotone. •Then 11…1 has highest probability of being in category 1.

126


Incomplete Boolean Functions:

•Boolean function F is incomplete if F can be calculated by finding at most n-1 attributes and knowing the value of the input string on those attributes•Example: F(111) = F(110) = F(101) = F(100) = 1, F(000) = F(001) = F(010) = F(011) = 0. •F(abc) is determined without knowing b (or c).•F is incomplete.

127

Binary Decision Tree ApproachComplete, Monotone Boolean Functions:

•Stroud and Saeger: “brute force” algorithm for enumerating binary decision trees implementing complete, monotone boolean functions and choosing least cost BDT. •Feasible to implement up to n = 4.•n = 2:

–There are 6 monotone boolean functions.–Only 2 of them are complete, monotone–There are 4 binary decision trees for calculating these 2 complete, monotone boolean functions.

128


Complete, Monotone Boolean Functions:

•n = 3:–9 complete, monotone boolean functions.–60 distinct binary trees for calculating them

•n = 4:–114 complete, monotone boolean functions.–11,808 distinct binary decision trees for calculating them.–(Compare 1,079,779,602 BDTs for all boolean functions)

129


Complete, Monotone Boolean Functions:

•n = 5:–6894 complete, monotone boolean functions–263,515,920 corresponding binary decision trees.

•Combinatorial explosion! •Need alternative approaches; enumeration not feasible!•(Even worse: compare 5 x 1018 BDTs corresponding to all boolean functions)

130

Cost Functions

•Stroud-Saeger method applies to more sophisticated cost models, not just cost = number of sensors in the BDT.•Using a sensor has a cost:

–Unit cost of inspecting one item with it–Fixed cost of purchasing and deploying it–Delay cost from queuing up at the sensor station

•Preliminary problem: disregard fixed and delay costs. Minimize unit costs.

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Cost Functions: Delay Costs•Tradeoff between fixed costs and delay costs: Add more sensors cuts down on delays.•Stochastic process of containers arriving•Distribution of delay times for inspections•Use queuing theory to find average delay times under different models

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Cost Functions:Unit Costs

Tree Utilization•Complication: Assume cost depends on how many nodes of BDT are actually visited during an “average” container’s inspection. (This is sum of unit costs.)•Depends on characteristics of population of entities being inspected. •I.e., depends on “distribution” of containers. •In our early models, we assume we are given probability of sensor errors and probability of bomb in a container.•This allows us to calculate “expected” cost of utilization of the tree Cutil.

133

Cost Functions

•Cost of false positive: Cost of additional tests.

–If it means opening the container, it’s very expensive.

•Cost of false negative: –Complex issue.–What is cost of a bomb going off in Manhattan?

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Cost Functions: Sensor Errors•One Approach to False Positives/Negatives:Assume there can be Sensor Errors•Simplest model: assume that all sensors checking for attribute ai have same fixed probability of saying ai is 0 if in fact it is 1, and similarly saying it is 1 if in fact it is 0.•More sophisticated analysis later describes a model for determining probabilities of sensor errors. •Notation: X = state of nature (bomb or no bomb)

Y = outcome (of sensor or entire inspection process).

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Cost Function used for Evaluating the Decision Trees.

CTot = CFalsePositive *PFalsePositive + CFalseNegative *PFalseNegative + Cutil

CFalsePositive is the cost of false positive (Type I error)CFalseNegative is the cost of false negative (Type II error)PFalsePositive is the probability of a false positive occurringPFalseNegative is the probability of a false negative occurringCutil is the expected cost of utilization of the tree.

137

Stroud Saeger Experiments• Stroud-Saeger ranked all trees formedfrom 3 or 4 sensors A, B, C and D according to increasing tree costs. • Used cost function defined above. • Values used in their experiments:

– CA = .25; P(YA=1|X=1) = .90; P(YA=1|X=0) = .10;– CB = 10; P(YC=1|X=1) = .99; P(YB=1|X=0) = .01;– CC = 30; P(YD=1|X=1) = .999; P(YC=1|X=0) = .001;– CD = 1; P(YD=1|X=1) = .95; P(YD=1|X=0) = .05;

– Here, Ci = unit cost of utilization of sensor i. • Also fixed were: CFalseNegative, CFalsePositive, P(X=1)

138

Stroud Saeger Experiments: Our Sensitivity Analysis

• We have explored sensitivity of the Stroud-Saeger conclusions to variations in values of these three parameters.

• We estimated high and low values for these parameters.n = 3 (use sensors A, B, C)

• We chose one of the values from the interval of values and then explored the highest ranked tree as the other two were chosen at random in the interval of values. 10,000 experiments for each pair of fixed values.

• We looked for the variation in the top-ranked tree and how the top-rank related to choice of parameter values.

• Very surprising results.

139

Conclusions from Sensitivity Analysis

• Considerable lack of sensitivity to modification in parameters for trees using 3 sensors.

• Very few optimal trees.

• Very few boolean functions arise among optimal and near-optimal trees.

• Similar results for trees using 4 sensors.

140

Stroud Saeger Experiments: Our Sensitivity Analysis

– CFalseNegative was varied between 25 million and 10 billion dollars• Low and high estimates of direct and indirect costs

incurred due to a false negative.

– CFalsePositive was varied between $180 and $720

• Cost incurred due to false positive

(4 men * (3 -6 hrs) * (15 – 30 $/hr)– P(X=1) was varied between 1/10,000,000 and

1/100,000

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Frequency of Top-ranked Trees when CFalseNegative and CFalsePositive are Varied

• 10,000 randomized experiments (randomly selected values of CFalseNegative and CFalsePositive from the specified range of values) for the median value of P(X=1).

• The above graph has frequency counts of the number of experiments when a particular tree was ranked first or second, or third and so on.

• Only three trees (7, 55 and 1) ever came first. 6 trees came second, 10 came third, 13 came fourth.

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• 10,000 randomized experiments for the median value of CFalsePositive.

• Only 2 trees (7 and 55) ever came first. 4 trees came second. 7 trees came third. 10 and 13 trees came 4th and 5th respectively.

Frequency of Top-ranked Trees when CFalseNegative and P(X=1) are Varied

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• 10,000 randomized experiments for the median value of CFalseNegative. • Only 3 trees (7, 55 and 1) ever came first. 6 trees came second. 10

trees came third. 13 and 16 trees came 4th and 5th respectively.

Frequency of Top-ranked Trees when P(X=1) and CFalsePositive are Varied

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Values of CFalseNegative and CFalsePositive when Tree 7 was Ranked First

• This is a graph of CFalsePositive against CFalseNegative values obtained from the randomized experiments. The black dots represent points at which tree 7 scored first rank.

145


• Tree 55 fills up the lower area in the range of CFalseNegative and CFalsePositive values.

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• Tree 1 fills up the upper area in the range of CFalseNegative and CFalsePositive.

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Values of CFalseNegative and CFalsePositive for the Three First Ranked Trees

• Trees 7, 55 and 1 fill up the entire area in the range of CFalseNegative and CFalsePositive among themselves.

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Values of CFalseNegative and P(X=1) when Tree 7 was Ranked First

• Tree 7 again fills up the major area in the range of CFalseNegative and P(X=1).

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Values of CFalseNegative and P(X=1) when Tree 55 was Ranked First

• Tree 55 fills up the rest of the area in the range of CFalseNegative and P(X=1).

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Values of CFalseNegative and P(X=1) for First Ranked Trees

• Together trees 7 and 55 fill up the entire region of CFalseNegative and P(X=1).

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Values of CFalsePositive and P(X=1) When Tree 7 was Ranked First

• Tree 7 fills up the major area in the range of CFalsePositive and P(X=1).

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Values of CFalsePositive and P(X=1) when Tree 55 was Ranked First

• Tree 55 fills up the upper area in the range of CFalsePositive and P(X=1).

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Values of CFalsePositive and P(X=1) when Tree 1 was Ranked First

• Tree 1 fills up the lower area in the range of CFalsePositive and P(X=1).

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Values of CFalsePositive and P(X=1) for First Ranked Trees

• Trees 7, 55 and 1 fill up the entire area in the range of CFalsePositive and P(X=1) among themselves.

155

Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors

• Second set of computer experiments: n = 4

(use sensors, A, B, C, D).

• Same values as before.

• Experiment 1: Fix values of two of CFalseNegative,

CFalsePositive, P(X=1) and vary the third through their interval of possible values.

• Experiment 2: Fix a value of one of CFalseNegative,

CFalsePositive, P(X=1) and vary the other two.

• Do 10,000 experiments each time.

• Look for the variation in the highest ranked tree.

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• Experiment 1: Fix values of two of CFalseNegative, CFalsePositive, P(X=1) and vary the third.

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CTot vs CFalseNegative for Ranked 1 Trees (Trees 11485(9651) and 10129(349))

Only two trees ever were ranked first, and one, tree 11485, was ranked first in 9651 out of 10,000 runs.

158

CTot vs CFalsePositive for Ranked 1 Trees (Tree no. 11485 (10000))

One tree, number 11485, was ranked first every time.

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CTot vs P(X=1) for Ranked 1 Trees (Tree no. 11485(8372), 10129(488), 11521(1056))

Three trees dominated first place. Trees 10201(60), 10225(17) and 10153(7) also achieved first rank but with relatively low frequency.

160


• Experiment 2: Fix the values of one of CFalseNegative, CFalsePositive, P(X=1) and vary the others.

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Frequency of First Ranked Trees when Two Parameters (CFalseNegative and CFalsePositive) were Varied Keeping P(X=1)

Constant at Randomly Selected Values.

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Frequency of First Ranked Trees when Two Parameters (CFalseNegative and P(X=1)) were Varied Keeping CFalsePositive


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The experiments were repeated for 20 different randomly selected values of CFalsePositive

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Frequency of First Ranked Trees when Two Parameters (P(X=1) and CFalsePositive) were Varied Keeping CFalseNegative


0.95 1 1.05 1.1 1.15 1.2

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Trees coming first -9541 10129 10153 10201 10225 11485 11521

10,000 randomized experiments with randomly selected values of CFalseNegative

The experiments were repeated for 20 different randomly selected values of CFalseNegative

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Modeling Sensor Errors•One Approach to Sensor Errors: Modeling Sensor Operation

•Threshold Model:–Sensors have different discriminating power–Many use counts (e.g., Gamma radiation counts)–See if count exceeds threshold–If so, say attribute is present.

165

Modeling Sensor ErrorsThreshold Model:

•Sensor i has discriminating power Ki, threshold Ti

•Attribute present if counts exceed Ti

•Calculate fraction of objects in each category whose readings exceed T•Seek threshold values that minimize all costs: inspection, false positive/negative•Assume readings of category 0 containers follow a Gaussian distribution and similarly category 1 containers•Simulation approach

166

Probability of Error for Individual Sensors

• For ith sensor, the type 1 (P(Yi=1|X=0)) and type 2 (P(Yi=0|X=1)) errors are modeled using Gaussian distributions. – State of nature X=0 represents absence of a bomb.– State of nature X=1 represents presence of a bomb. i represents the outcome (count) of sensor i. – Σi is variance of the distributions– PD = prob. of detection, PF = prob. of false pos.

Ki

P(i|X=1)P(i|X=0)

Ti

Characteristics of a typical sensori

167

Modeling Sensor ErrorsThe probability of false positive for the ith sensor is computed as:

P(Yi=1|X=0) = 0.5 erfc[Ti/√2]The probability of detection for the ith sensor is computed as:

P(Yi=1|X=1) = 0.5 erfc[(Ti-Ki)/(Σ√2)]

erfc = complementary error function erfc(x) = (1/2,x2)/sqrt()

The following experiments have been done using sensors A, B, C and using:

KA = 4.37; ΣA = 1KB = 2.9; ΣB = 1KC = 4.6; ΣC = 1

We then varied the individual sensor thresholds TA, TB and TC from -4.0 to +4.0 in steps of 0.4. These values were chosen since they gave us an “ROC curve” for the individual sensors over a complete range P(Yi=1|X=0) and P(Yi=1|X=1)

168

Frequency of First Ranked Trees for Variations in Sensor Thresholds

• 68,921 experiments were conducted, as each Ti was varied through its entire range. • The above graph has frequency counts of the number of experiments when a particular

tree was ranked first. There are 15 such trees. Tree 37 had the highest frequency of attaining rank one.

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Conclusions from Sensitivity Analysis: Recapitulation

• Considerable lack of sensitivity to modification in parameters for trees using 3 or 4 sensors.

• Very few optimal trees.

• Very few boolean functions arise among optimal and near-optimal trees.

170

Some Complications•More complicated cost models; bringing in costs of delays•More than two values of an attribute

(present, absent, present with probability > 75%, absent with probability at least 75%) (ok, not ok, ok with probability > 99%, ok with probability between 95% and 99%)

•Inferring the boolean function from observations (partially defined boolean functions)

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Some Research Challenges•Explain why conclusions are so insensitive to variation in parameter values.•Explore the structure of the optimal trees and compare the different optimal trees.•Develop less brute force methods for finding optimal trees that might work if there are more than 4 attributes.•Develop methods for approximating the optimal tree.

172

Port of Entry Inspection: Closing Remark

•Recall that the “cost” of inspection includes the cost of failure, including failure to foil a terrorist plot.•There are many ways to lower the total “cost” of inspection:

Use more efficient orders of inspection.Find ways to inspect more containers.Find ways to cut down on delays at inspection lanes.

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Concluding Comment• In recent years, interplay between CS and biology has transformed major parts of Bio into an information science.• Led to major scientific breakthroughs in

biology such as sequencing of human genome.

• Led to significant new developments in CS, such as database search.

• The interplay between CS and SS-DM not nearly as far along.

• Moreover: problems are spread over many disciplines.

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Concluding Comment

• However, CS/SS-DM interplay has already developed a unique momentum of its own.

• One can expect many more exciting outcomes as partnerships between computer scientists and social scientists/decision theorists expand and mature.

1 computer science and decision making fred roberts, rutgers university

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