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In order to check the performance in terms of CPU utilization, we designed a benchmark application that represents service logic that is bounded by computation resources. This application is sent and executed in a DINA node. Without any load the application requires about 2 seconds to complete. Then, we added further load by executing a number of load application representing the logic of other (different) services running on the same DINA node. As one can see from Figure 8.6, when the load increases the time it takes the application to complete increases as well. The Total time in this gure is the time taken to load and execute the application, while the Execution time is the amount of

Currently, MN detection represents a widely used parameter, easily performed, which also allows molecular approaches in studying the effects of many clastogenic or aneugenic agents (Venier et al, 1997) Various studies have shown that the peripheral erythrocytes of sh have a high incidence of micronuclei after exposure to different pollutants under eld and laboratory conditions Al-Sabti (1994) observed that selenium, mercury, methylmercury and their mixtures induce micronuclei under laboratory conditions in the binucleated erythrocytes of Prussian carp (Carassius auratus gibelio) in a dose-dependent manner The exposure by injection of ve carcinogenic-mutagenic chemicals (a atoxin B1, aroclor 1254, benzidine, benzo[a]pyrene and 20-methylcholanthrene) of three species of cyprinids (common carp, C carpio; tench, T tinca; and grass carp, C idella) enhanced the frequency of micronuclei in their erythrocytes (Al-Sabti, 1986a) Pacheco and Santos (1996) observed a signi cant increase in micronuclei in erythrocytes from eels (Anguilla anguilla) exposed under laboratory conditions.

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1 For each pair of attributes A and B, search for a set SAB U {A, B}, so that A B | SAB holds, that is, so that A and B are conditionally inde pendent given SAB If there is no such set SAB , connect the two attributes by an undirected edge If a directed acyclic perfect map G = (U, E) is to be found, the following two steps have to be carried out in order to direct the edges: 2 For each pair of nonadjacent attributes A and B with a common neighbor C SAB , direct the edges towards C, that is, A C B / 3 Recursively direct all undirected edges according to the rules: If for two adjacent attributes A and B there is a strictly directed path from A to B not including A B, then direct the edge towards B.

A key rule, regularly adopted in adversarial trial systems (McEwan, 1998) is that the lawyers who call a witness may not ask leading questions. A question is leading if the terms of the question suggest the answer. Did you see anything is not a leading

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If there are three attributes A, B, and C with A and B not adjacent, A C, and C B, then direct the edge C B If neither rule is applicable, arbitrarily direct an undirected edge Obviously, the first step of this algorithm implements the idea stated above: only those attributes are connected by an edge for which there is no set of attributes that renders them conditionally independent Note that for this step it is crucial that a perfect map actually exists Otherwise edges may be omitted that are needed in an independence map (see also below) The reasons underlying the second step of the algorithm are as follows: if a set SAB of attributes renders two attributes A and B conditionally independent, it must block all paths from A to B in the graph (otherwise this conditional independence statement would not be represented by the graph).

33. Campbell, Focus on Cyber-Fraud, Internal Auditor, 30. 34. See note 32, 16 19. 35. The Threat to U.S. National Security from Loss of U.S. Intellectual Property to Foreign

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It follows that it must also block a path that runs via a common neighbor C of A and B However, if this common neighbor C is not in SAB , the only way in which this path can be blocked (given SAB ) is by the fact that it has converging edges at C (cf the definition of d-separation32 ) The first rule of the third step simply exploits the fact that the resulting graph must be acyclic If there already is a directed path from A to B not including the edge connecting A and B, directing the edge from B towards A would introduce a directed cycle Therefore it must be directed the other way.

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