Algorithms for Port of Entry Inspection for WMDs

1
Algorithms for Port of Entry Inspection for WMDs
Fred S. Roberts DyDAn Center Rutgers University
2
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)

3
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 alarms (false
  positives) and failed alarms (false negatives)

Mobile Vacis truck-mounted gamma ray imaging
system
4
Port of Entry Inspection Algorithms

• My work on port of entry inspection has gotten me
  and my students to some remarkable places.

Me on a Coast Guard boat in a tour of the harbor
in Philadelphia
5
(No Transcript)
6
Sequential Decision Making Problem

• Stream of containers arrives at a port
• The Decision Makers 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 Alamos
• National Laboratory
• Need for new models
• and methods

7
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, )
• Medicine (diagnosing patients, sequencing
  treatments, )

8
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

9
Sequential Decision Making Problem

• 0s and 1s suggest binary digits (bits)
• Bit String A sequence of bits
• 0001, 1101,
• Boolean Function A function that assigns to each
  bit string a 0 or a 1.
• Bit String x B(x)
• 00 1
• 01 0 B(00) 1, B(10) 0
• 10 0
• 11 1

10
Sequential Decision Making Problem

• Bit strings are of course crucial carriers of
  information in modern computers
• They are used to encode detailed instructions and
  are translated to a sequence of on-off
  instructions for switches in the computer.
• We will use them differently.

11
Counting Bit Strings and Boolean Functions

• To understand the difficulty of the container
  inspection problem, we will need to count bit
  strings and Boolean functions.
• Product Rule of Counting If something can happen
  in n1 ways, and no matter how the first thing
  happens, a second thing can happen in n2 ways,
  then the two things together can happen in n1 x
  n2 ways.

12
Counting Bit Strings and Boolean Functions

• Product Rule of Counting (More Generally) If
  something can happen in n1 ways, and no matter
  how the first thing happens, a second thing can
  happen in n2 ways, and no matter how the first
  two things happen, a third thing can happen in n3
  ways, and , then all the things together can
  happen in n1 x n2 x n3 x ways.

13
Counting Bit Strings and Boolean Functions

• How many bit strings are there of 3 bits?
• There are 2 choices for the first bit.
• No matter how it is chosen, there are 2 choices
  for the second bit.
• No matter how the first 2 bits are chosen, there
  are 2 choices for the third bit.
• Get 2 x 2 x 2 23 8

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Counting Bit Strings and Boolean Functions

• How many bit strings are there of 3 bits?
• 000, 001, 010, 011, 100, 101, 110, 111
• How many bit strings are there of 5 bits?
• 25
• How many bit strings are there of n bits?
• 2n

15
Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of 3
  variables?

16
Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of 3
  variables?
• There are 23 8 bit strings of 3 variables
• For each such bit string, the Boolean function
  assigns it one of 2 different values (0 or 1).

17
Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of 3
  variables?
• There are 23 8 bit strings of 3 variables
• For each such bit string, the Boolean function
  assigns it one of 2 different values (0 or 1).
• There are 28 such Boolean functions.

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Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of n
  variables?

19
Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of n
  variables?
• There are 2n bit strings of n variables
• For each such bit string, the Boolean function
  assigns it one of 2 different values (0 or 1).

20
Counting Bit Strings and Boolean Functions

• How many Boolean functions are there of n
  variables?
• There are 2n bit strings of n variables
• For each such bit string, the Boolean function
  assigns it one of 2 different values (0 or 1).
• There are such Boolean functions.

21
Counting Bit Strings and Boolean Functions
How big is ? When n 4, this is 65,536
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Counting Bit Strings and Boolean Functions
The problem of making a detailed design of a
digital computer usually involves finding a
practical circuit implementation of certain
functional behavior. A computer device
implements a Boolean function. In the early days
of computing, the goal was to create a catalogue
listing a good circuit implementation of each
Boolean function. This would require 65,536
entries!
23
Counting Bit Strings and Boolean Functions

• Of course, luckily, by symmetry, some of these
  examples are equivalent
• For example, if we interchange two variables.
• In the early days of computing (1951), a group
  from the Harvard Computation Laboratory
  painstakingly listed all 65,536 Boolean functions
  of 4 variables and determined which were
  equivalent.
• (There were 222 types.)

24
Sequential Decision Making ProblemFor Container
Inspection

• 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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Sequential Decision Making Problem

• Currently used attributes
• Does ships 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
26
Sequential Decision Making Problem

• We can imagine many other attributes
• Research at DyDAn is concerned with general
  algorithmic approaches.
• We seek a methodology not tied to todays
  technology.
• Detectors are evolving quickly.

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Sequential Decision Making Problem

• Simplest Case Attributes are in state 0 or 1
• Then Container is a bit 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, assign
container to category F(011001).
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Sequential Decision Making Problem

• 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
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Sequential Decision Making Problem

• What if there are three categories, 0 ½, and 1?.
• Example
• F(000) 0, F(111) 1, F(abc) 1/2 otherwise
• This classifies a container as positive if it has
  all of the attributes, negative if it has none of
  the attributes, and uncertain if it has some but
  not all of the attributes.
• How many such decision functions are there if
  there are three attributes?

30
Sequential Decision Making Problem

• What if there are three categories, 0 ½, and 1?
• How many such decision functions are there if
  there are three attributes?

31
Sequential Decision Making Problem

• What if there are three categories, 0 ½, and 1?.
• How many such decision functions are there if
  there are three attributes?
• There are 23 bit strings of length 3.

32
Sequential Decision Making Problem

• What if there are three categories, 0 ½, and 1?.
• How many such decision functions are there if
  there are three attributes?
• There are 23 bit strings of length 3.
• For each of them, there are 3 choices for the
  value of F.

33
Sequential Decision Making Problem

• What if there are three categories, 0 ½, and 1?.
• How many such decision functions are there if
  there are three attributes?
• There are 23 bit strings of length 3.
• For each of them, there are 3 choices for the
  value of F.
• Thus, there are such decision functions.

34
Sequential Decision Making Problem

• 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 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., cant 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.

36
Sequential Decision Making Problem

• 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 it is hard in a precise
  computer science sense.

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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
• waiting on line for inspection
• Besides efficient inspection
• schemes, could decrease costs by
• Buying more sensors
• Change allocation of containers to sensor lanes.

38
Trees

• A tree for us is a directed graph.
• It has nodes (vertices).
• Directed edges or arcs head
• from a vertex to another.
• There are no cycles
• (you cant double back on
• yourself)
• We will deal with rooted trees One node is a
  root.
• All arcs point downwards in our diagrams,
  starting from the root.
• If each node has two or zero outgoing (downwards)
  arcs, we have a binary tree.
• Nodes with no outgoing arcs are called leaves.

39
Binary Decision Tree Approach

• Sensors measure presence/absence of attributes
  so 0 or 1
• Use two categories 0, 1
• Binary Decision Tree
• Nodes are sensors or categories
• 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

40
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 only through the path a1
• to a0 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.
• Note Different tree, same function

Figure 1
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Binary Decision Tree Approach

• Reach category 1 from the
• root only through the path a0
• to 1 or a0 to a1 to 1.
• Container is classified in category 1 iff it has
  attribute
• a0 or attribute a1 .
• Corresponding Boolean function
• F(11) 1, F(10) F(01) 1, F(00) 0.

Figure 1
43
Binary Decision Tree Approach

• Can you find another binary
• tree that calculates the same
• Boolean function?
• Sure, just interchange
• nodes a0 and a1

Figure 1
44
Binary Decision Tree Approach

• Reach category 1 from
• the root by
• a0 L to a1 R a2 R 1 or
• a0 R a2 R1
• Container classified in category 1 iff it has
• a1 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
45
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
46
Binary Decision Tree Approach

• So we have seen that a given Boolean function may
  correspond to different binary decision trees.
• How do we find a binary decision tree
  corresponding to a Boolean function?
• How do we find a least cost one?

Port of Long Beach
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Binary Decision Tree Approach

• So we have seen that a given Boolean function may
  correspond to different binary decision trees.
• How do we find a binary decision tree
  corresponding to a Boolean function F?
• Brute force method
• Start with root a0, put both left and right arcs
  to an a1.
• From each a1 put both left and right arcs to a2,
  etc.
• At the end, each path through the tree
  corresponds to a bit string x and let that
• path end at F(x).

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Binary Decision Tree Approach
49
Binary Decision Tree Approach
x F(x) x F(x)
111 1 011 1
110 1 010 0
101 0 001 0
100 1 000 0
1
1
1
0
1
0
0
0
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Binary Decision Tree Approach

• Even if the Boolean function F is fixed, the
  problem of finding the least cost binary
  decision tree for it is very hard (NP-complete).
• For small n number of attributes, can try to
  solve it by trying all possible binary decision
  trees corresponding to the Boolean function F.
• Even for n 4, not practical. (n 4 at Port of
  Long Beach-Los Angeles)

Port of Long Beach
51
Binary Decision Tree Approach

• Promising 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 (Los Alamos
• National Lab) 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.

52
Binary Decision Tree Approach

• Monotone Boolean Functions
• Given two bit strings x1x2xn, y1y2yn
• Suppose that xi ? yi for all i implies that
  F(x1x2xn) ? F(y1,y2yn).
• Then we say that F is monotone.
• Then 111 has highest probability of being in
  category 1.

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Binary Decision Tree Approach

• Monotone Boolean Functions
• Given two bit strings x1x2xn, y1y2yn
• Suppose that xi ? yi for all i implies that
  F(x1x2xn) ? F(y1,y2yn).
• Then we say that F is monotone.
• Example
• n 4, F(x) 1 iff x has at least two 1s.
• F(1100) F(0101) F(1011) 1, F(1000) 0,
  etc.
• Is this monotone?
• Yes

54
Binary Decision Tree Approach

• 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.

55
Binary Decision Tree Approach

• Complete, Monotone Boolean Functions
• Stroud and Saeger algorithm for enumerating
  binary decision trees implementing complete,
  monotone Boolean functions.
• Feasible to implement up to n 4.
• Then you can find least cost tree by enumerating
  all binary decision trees corresponding to a
  given Boolean function and repeating this for
  all complete, monotone Boolean functions.

56
Binary Decision Tree Approach

• Complete, Monotone Boolean Functions
• Stroud and Saeger algorithm for enumerating
  binary decision trees implementing complete,
  monotone Boolean functions.
• 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.

57
Binary Decision Tree Approach

• Complete, Monotone Boolean Functions
• n 3
• 9 complete, monotone Boolean functions.
• 60 distinct binary trees for calculating them
• Counting methods more complicated than simple
  ones we have described before
• All counts here are from Stroud and Saeger

58
Binary Decision Tree Approach

• Complete, Monotone Boolean Functions
• 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)

59
Binary Decision Tree Approach

• 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)

60
Cost Functions

• So far, we have figured one binary decision tree
  is cheaper than another if it has fewer nodes.
• This is oversimplified.
• There are more complex costs involved than number
  of sensors in a tree.

61
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.

62
Cost Functions

• Simplification so far Disregard characteristics
  of population of entities being inspected.
• Only count number of observation (attribute)
  nodes in the tree.
• Unit Cost Complication How many nodes of the
  decision tree are actually visited during average
  containers inspection? Depends on distribution
  of containers.
• Answer can depend on probability of sensor errors
  and probability of bomb in a container.

63
Cost Functions Delay Costs

• Tradeoff between fixed costs and delay costs Add
  more sensors cuts down on delays.
• More sophisticated models describe the process of
  containers arriving
• There are differing delay times for inspections
• Use queuing theory to find average delay times
  under different models

64
Cost Functions Delay Costs

• There are differing delay times for inspections
• Use queuing theory to find average delay times
  under different models
• Explorations for Students Explore the
  fundamental principles of queueing theory

65
Cost FunctionsUnit CostsTree Utilization

• Complication Assume cost depends on how many
  nodes of BDT are actually visited during an
  average containers 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. (Details omitted
  here.)

66
Cost Functions

• Cost of false positive Cost of additional tests.
  
• If it means opening the container, its very
  expensive.
• Cost of false negative
• Complex issue.
• What is cost of a bomb going off in Manhattan?

67
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).

68
Probability of Error for The Entire Tree

• State of nature is zero (X 0), absence of a bomb

State of nature is one (X 1), presence of a bomb
Probability of false positive (P(Y1X0)) for
this tree is given by
Probability of false negative (P(Y0X1)) for
this tree is given by
P(Y1X0) P(YA1X0) P(YB1X0)
P(YA1X0) P(YB0X0) P(YC1X0)
Pfalsepositive
P(Y0X1) P(YA0X1) P(YA1X1)
P(YB0X1)P(YC0X1) Pfalsenegative
69
Probability of Error for The Entire Tree

• State of nature is zero (X 0), absence of a bomb

Note the use of the product rule probability is
ratio between number of ways something we are
interested in can happen and total number ways
things can happen.
Probability of false positive (P(Y1X0)) for
this tree is given by
P(Y1X0) P(YA1X0) P(YB1X0)
P(YA1X0) P(YB0X0) P(YC1X0)
Pfalsepositive
70
Probability of Error for The Entire Tree

• State of nature is zero (X 0), absence of a bomb

Also uses the sum rule of counting If one event
can happen in n1 ways and a second event in n2
different ways, then there are n1n2 ways in
which either the first event or the second event
can occur (but not both).
Probability of false positive (P(Y1X0)) for
this tree is given by
P(Y1X0) P(YA1X0) P(YB1X0)
P(YA1X0) P(YB0X0) P(YC1X0)
Pfalsepositive
71
Aside Illustrating the Sum Rule
For example, how many bit strings are there of 4
bits so that there are either exactly three 1s
or exactly three 0s? For bit strings of three
1s, we have four choices for the place to put
the 0, so there are 4 such strings. For bit
strings with three 0s, there are similarly 4
such strings. By the sum rule, there are 8 bit
strings with exactly three 1s or exactly three
0s, since it is impossible to have both. 1110,
1101, 1011, 0111, 0001, 0010, 0100, 1000
72
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
occurring PFalseNegative is the probability of a
false negative occurring Cutil is the expected
cost of utilization of the tree.
73
Cost Function used for Evaluating the Decision
Trees.
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
occurring PFalseNegative is the probability of a
false negative occurring Cutil is the expected
cost of utilization of the tree. PFalsePositive
and PFalseNegative are calculated from the tree.
Cutil is calculated from tree and probabilities
of bomb in container and probability of sensor
errors. CFalsePositive, CFalseNegative are input
given information.
74
Stroud Saeger Experiments

• Stroud-Saeger ranked all trees formed
• from 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(YA1X1) .90 P(YA1X0) .10
• CB 10 P(YC1X1) .99 P(YB1X0) .01
• CC 30 P(YD1X1) .999 P(YC1X0) .001
• CD 1 P(YD1X1) .95 P(YD1X0) .05
• Here, Ci unit cost of utilization of sensor i.
  
• Also fixed were CFalseNegative, CFalsePositive,
  P(X1)

75
Sensitivity Analysis

• When parameters in a model are not known exactly,
  the results of a mathematical analysis can change
  depending on the values of the parameters.
• It is important to do a sensitivity analysis let
  the parameter values vary and see if the results
  change.
• So, do the least cost trees change if we change
  values like probability of a bomb, cost of a
  false positive, etc?

76
Stroud Saeger Experiments Our Sensitivity
Analysis

• We have explored sensitivity of the Stroud-Saeger
  conclusions to variations in values of the three
  parameters
• CFalseNegative, CFalsePositive, P(X1)
• (Work of Anand, Madigan, Mammone, Pathak,
  Roberts)
• We estimated high and low values for the
  parameters.

77
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(X1) was varied between 1/10,000,000 and
  1/100,000

78
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(X1) was varied between 1/10,000,000 and
  1/100,000
• Extra assignment think about whether these
  ranges are reasonable and how one would determine
  them?

79
Stroud Saeger Experiments Our Sensitivity
Analysis

• n 3 (use sensors A, B, C)
• Varied the parameters
• CFalseNegative, CFalsePositive, P(X1)
• We chose the value of one of these parameters
  from the interval of values
• Then explored the highest ranked tree as the
  other two were chosen at random in the interval
  of values.
• 10,000 experiments for each fixed value.
• We looked for the variation in the top-ranked
  tree and how the top-rank related to choice of
  parameter values.
• Very surprising results.

80
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(X1).
• 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.

81
Frequency of Top-ranked Trees when CFalseNegative
and P(X1) are Varied

• 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.

82
Frequency of Top-ranked Trees when P(X1) and
CFalsePositive are Varied

• 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.

83
Most Frequent Tree Groups Attaining the Top Three
Ranks.

• Trees 7, 9 and 10

All the three decision trees have been generated
from the same Boolean function. Both Tree 9 and
Tree 10 are ranked second and third more than 99
of the times when Tree 7 is ranked first.
84
Most Frequent Tree Groups Attaining the Top Three
Ranks

• Trees 55, 57 and 58

All three trees correspond to the same Boolean
function. Tree ranked 57 is second 96 of the
times and tree 58 is third 79 of the times when
tree 55 is ranked first.
85
Most Frequent Tree Groups Attaining the Top Three
Ranks

• Trees 1, 3, and 2

All three trees correspond to the same Boolean
function. Tree 3 is ranked second 98 of times
and tree 2 is ranked third 80 of the times when
tree 1 is ranked first.
86
Stroud Saeger Experiments 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(X1) and vary
  the third through their interval of possible
  values.
• Experiment 2 Fix a value of one of
  CFalseNegative, CFalsePositive, P(X1) and vary
  the other two.
• Do 10,000 experiments each time.
• Look for the variation in the highest ranked
  tree.

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Tree Structure For Top Trees
Tree number 11485
Tree number 10129
88
Tree Structure For Top Trees
Tree number 11485
Tree number 10129
Explorations for Students Compare the Boolean
functions corresponding to these two trees. See
where they differ. How close are they? Can you
make closeness precise?
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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.

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Conclusions from Sensitivity Analysis

• 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.
• Surprising results.

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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.

Pallet vacis
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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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Research Team

• Saket Anand, Rutgers, ECE graduate student
• Endre Boros, Rutgers, Operations Research
• Elsayed Elsayed, Rutgers, Ind. Systems
  Engineering
• Liliya Fedzhora, Rutgers, Operations Res. grad.
  student
• Paul Kantor, Rutgers, Schl. of Infor. Library
  Studies
• Abdullah Karaman, Rutgers Ind. Syst. Eng. grad.
  student
• Alex Kogan, Rutgers, Business School
• Paul Lioy, Rutgers/UMDNJ, Environmental and
  Occupational Health and Sciences Institute
• David Madigan, Rutgers, Statistics
• Richard Mammone, Rutgers, Center for Advanced
  Information Processing
• S. Muthukrishnan, Rutgers, Computer Science
• Saumitr Pathek, Rutgers ECE graduate student
• Richard Picard, Los Alamos, Statistical Sciences
  Group
• Fred Roberts, Rutgers, DIMACS Center
• Kevin Saeger, Los Alamos, Homeland Security
• Phillip Stroud, Los Alamos, Systems Engineering
  and Integration Group
• Hao Zhang, Rutgers Ind. Systems Eng., graduate
  student

94

• Collaborators on Sensitivity Analysis
• Saket Anand
• David Madigan
• Richard Mammone
• Saumitr Pathak
• Research Support
• Office of Naval Research
• National Science Foundation
• Los Alamos National Laboratory
• Rick Picard
• Kevin Saeger
• Phil Stroud

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Mathematics can help with homeland security!