Algorithms for Port of Entry Inspection for WMDs Fred S. Roberts DIMACS Center, Rutgers University Philip D. Stroud Los Alamos National Laboratory 1 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) 2 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: truckmounted gamma ray imaging system 3 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 Alamos National Laboratory –Need for new models – and methods 4 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, …) 5 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 6 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 7 Sequential Decision Making Problem •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 8 Sequential Decision Making Problem •We can imagine many other attributes •This project is concerned with general algorithmic approaches. •We seek a methodology not tied to today’s technology. •Detectors are evolving quickly. 9 Sequential Decision Making Problem •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). 10 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= 11 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. 12 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. 13 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. •Several classes of functions F allow for efficient inspection schemes: –k-out-of-n systems –Certain series-parallel systems –Read-once systems –“regular” systems 14 –Horn systems 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. 15 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 16 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 17 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 18 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 19 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. Port of Long Beach •Even for n = 4, not practical. •(n = 4 at Port of Long Beach-Los Angeles) 20 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 enumerate all “complete,” monotone boolean functions and calculate the least expensive corresponding binary decision trees. •Their method practical for n up to 4, 21 not n = 5. Binary Decision Tree Approach Monotone 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. 22 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. 23 Binary Decision Tree Approach Complete, 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 24 functions. Binary Decision Tree Approach 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) 25 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) 26 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 27 costs. Minimize unit costs. 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 28 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 29 utilization of the tree Cutil. 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? 30 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 31 process). 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 A A C 0 B 0 B 0 C 1 1 Probability of false positive (P(Y=1|X=0)) for this tree is given by 0 1 1 Probability of false negative (P(Y=0|X=1)) for this tree is given by P(Y=1|X=0) = P(YA=1|X=0) * P(YB=1|X=0) + P(YA=1|X=0) *P(YB=0|X=0)* P(YC=1|X=0) P(Y=0|X=1) = P(YA=0|X=1) + P(YA=1|X=1) *P(YB=0|X=1)*P(YC=0|X=1) Pfalsepositive Pfalsenegative 32 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. 33 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(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) 34 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. 35 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 36 Frequency of Top-ranked Trees when CFalseNegative and CFalsePositive are Varied 7000 1st 2nd 3rd 4th 5th 6000 Frequency 5000 4000 3000 2000 1000 0 0 10 20 30 40 50 60 Tree no. • • 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. 37 Frequency of Top-ranked Trees when CFalseNegative and P(X=1) are Varied 8000 1st 2nd 3rd 4th 5th 7000 6000 Frequency 5000 4000 3000 2000 1000 0 0 10 20 30 40 50 60 Tree no. • 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. 38 Frequency of Top-ranked Trees when P(X=1) and CFalsePositive are Varied 7000 1st 2nd 3rd 4th 5th 6000 Frequency 5000 4000 3000 2000 1000 0 0 10 20 30 40 50 60 Tree no. • 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 39 trees came third. 13 and 16 trees came 4th and 5th respectively. 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. 40 Values of CFalseNegative and CFalsePositive when Tree 55 was Ranked First • Tree 55 fills up the lower area in the range of CFalseNegative 41 and CFalsePositive values. Values of CFalseNegative and CFalsePositive when Tree 1 was Ranked First 900 800 700 CFalsePositive 600 500 400 300 200 100 0 0 1 2 3 4 C 5 6 7 8 9 10 9 FalseNegative x 10 • Tree 1 fills up the upper area in the range of CFalseNegative and 42 CFalsePositive. 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. 43 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). 44 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). 45 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). 46 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). 47 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). 48 Values of CFalsePositive and P(X=1) when Tree 1 was Ranked First -5 x 10 1 0.9 0.8 0.7 P(X=1) 0.6 0.5 0.4 0.3 0.2 0.1 0 0 100 200 300 400 C 500 600 700 800 900 FalsePositive • Tree 1 fills up the lower area in the range of CFalsePositive and P(X=1). 49 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. 50 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. 51 • Look for the variation in the highest ranked tree. Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors • Experiment 1: Fix values of two of CFalseNegative, CFalsePositive, P(X=1) and vary the third. 52 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. 53 CTot vs CFalsePositive for Ranked 1 Trees (Tree no. 11485 (10000)) One tree, number 11485, was ranked first every time. 54 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) and55 10153(7) also achieved first rank but with relatively low frequency. Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors • Experiment 2: Fix the values of one of CFalseNegative, CFalsePositive, P(X=1) and vary the others. 56 Frequency of First Ranked Trees when Two Parameters (CFalseNegative and CFalsePositive) were Varied Keeping P(X=1) Constant at Randomly Selected Values. 5 2 Trees coming first -9541 10129 10153 10201 11485 11521 x 10 1.8 1.6 1.4 Frequency 1.2 1 0.8 0.6 0.4 0.2 0 0 2000 4000 6000 Tree number 8000 10000 12000 10,000 randomized experiments with randomly selected values of P(X=1) The experiments were repeated for 20 different randomly selected values of P(X=1) 57 Frequency of First Ranked Trees when Two Parameters (CFalseNegative and P(X=1)) were Varied Keeping CFalsePositive Constant at Randomly Selected Values. 14 4 xTrees 10 coming first -505 4695 5105 5129 7353 9541 10129 10153 10201 10225 11485 11521 12 Frequency 10 8 6 4 2 0 0 2000 4000 6000 Tree number 8000 10000 12000 10,000 randomized experiments with randomly selected values of CFalsePositive The experiments were repeated for 20 different randomly selected values of CFalsePositive 58 Frequency of First Ranked Trees when Two Parameters (P(X=1) and CFalsePositive) were Varied Keeping CFalseNegative Constant at Randomly Selected Values. 4 15 x 10 Trees coming first -9541 10129 10153 10201 10225 11485 11521 Frequency 10 5 0 0.95 1 1.05 1.1 Tree number 1.15 1.2 4 x 10 10,000 randomized experiments with randomly selected values of CFalseNegative The experiments were repeated for 20 different randomly selected values of CFalseNegative 59 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. 60 Modeling Sensor Errors Threshold 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 61 •Simulation approach 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=0) Ti P(i|X=1) i Characteristics of a typical sensor 62 Modeling Sensor Errors The 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 = 1 KB = 2.9; ΣB = 1 KC = 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) 63 Frequency of First Ranked Trees for Variations in Sensor Thresholds 18000 16000 14000 Frequency 12000 10000 8000 6000 4000 2000 0 0 10 20 30 40 50 60 Tree no. • 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. 64 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. 65 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) 66 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 67 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. 68 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 69 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 70