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So we have y x0 < 0 On the other hand, it is evident that y x will become and remain positive after x reaches a suf ciently large value Thus, there must exist one and only one zero z > x0 of y x for x P 0; I such that y x 0 for x z and y x > 0 otherwise The result to be proved then follows j Theorem 1336 shows that if an LRD uid is modeled by a Markov process, then the Markov model may indeed underestimate the loss probability of the underlying LRD traf c A similar result holds for the conditional expected loss ratio Theorem 1337 Suppose that S obeys the Pareto distribution (136) for L t , and E S is the same for L t and M t .



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For given B < I, C < I, r > C, and w < B,.

static long bufferTest(int repeat){ StringBuffer buffer = new StringBuffer(1024); String[] strings = new String[repeat]; Runtime runtime = Runtime.getRuntime(); long freeMemory; long timeStart = System.currentTimeMillis(); long initialMemory = runtime.freeMemory(); for(int loop = repeat; --loop >= 0; ){ buffer.insert(0, "" + loop); strings[loop] = buffer.toString(); freeMemory = runtime.freeMemory(); Test.test.println("Used: " + (initialMemory freeMemory) + ", total: " + runtime.totalMemory() + " " + strings[loop]); initialMemory = freeMemory; } return System.currentTimeMillis() - timeStart; }

we have E ljM n r, w 0 w ! E ljL n r, w 0 w if y E ljM n r, w 0 w < E ljL n r, w 0 w otherwise, where y u0 t w ; u0 x





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communication costs, as there is in the general problem On the other hand, problems that are similar to the scheduling problem and seem to be easy at rst sight, such as the KNAPSACK or the BINPACKING problems (Cormen et al [42], Garey and Johnson [73]), are NP-complete, too Theorem 43 (NP-Completeness Cost-Free Communication) Let G = (V, E, w) be a task graph and Pc0 a parallel system The decision problem SCHED-C0(G, Pc0 ) associated with the scheduling problem is as follows Is there a schedule S for G on Pc0 with length sl(S) T , T Q+ SCHED-C0(G, Pc0 ) is NP-complete Proof The same proof as for Theorem 41 is used with one small modi cation for the construction of an instance of SCHED-C0(G, Pc0 ):

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x P 0; u0 , t w is given by Eq. (13.2), u0 is given by Eq. (13.4), and z > 0 is the solution of ebx x 1 for x P 0; I and b 1 a 1 . Proof. We rst recall a well-known result (the generalized mean value theorem) in elementary analysis. Suppose that F u and G u are continuous on a; b and differentiable on a; b , and GH u T 0 for a < u < b, where H indicates derivation with respect to u. Then there exists at least one x P a; b such that F b F a F H x : G b G a GH x Now let a 0; b u0 and denote u0 t w du P S> u0 u by F u if S obeys the Pareto distribution (13.6) and by G u if S is exponentially distributed. Both F u and G u are continuous on 0; u0 and differentiable on 0; u0 . According to the above result and Eq. (13.3), E ljL n r; w 0 w F u0 F 0 F H x ; E ljM n r; w 0 w G u0 G 0 GH x where x P 0; u0 , a u0 t w 1 ; F x g u0 x

we have F H x gy 1 a and GH x e a 1 gy . Replacing t w in Theorem 13.3.6 by y, then using the same arguments as that used in the proof of Theorem 13.3.6, we see that the result to be proved follows directly. j COMMENT 13.3.8. The assumption of Pareto distribution is not restrictive. Similar results hold for any other heavy-tailed distributions. One numerical example is given in Section 13.3.2.

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Tesseract is a very good OCR engine: https://github.com/tesseract-ocr/tesseract. The project has been launched by HP Labs and is now ...

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