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because each iteration Apl is random After making N iterations, the first random value G0 is obtained and stored in the vector storage_for_G The outer for-loop is applied EXPERIMENTS number of times collecting G values Finally, all values of G from the vector can be added to an object of the class Distribution (see 8) The last can estimate the average value of G for given b, A0 , M, and N Under other equal conditions, the average value of G is a function of b If we change b, then we get a new average value of G Our task is to determine the value of b* that maximizing average value of G This is a task for a class Optimizer, which goes beyond the scope of this book Useful algorithms and code suitable for optimizers and solvers are described in Press et al (1992).



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Instead of using a more efficient optimizer, we can arrive at the same solution by creating values of b from a systematic incrementation using small steps within the interval [0, 1] The step 0001 would mean a 01 percent allocation Clearly, in this way, one or more b values maximizing the average value G can be found in the interval This looks simple However, where do we get the random values Apl Random Apl Building a random numbers generator with a distribution the same as the empirical distribution found in the example above will be a significant diversion from the main focus of this book Because of that, I will not develop the framework for this process; however, I will describe one way in which this can be achieved.





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First, we need a uniform random number generator that returns fractional numbers within the interval [0, 1] (any other interval with known boundaries can be created with a simple transformation) The algorithms are described in Knuth (1998) The C code for four suitable functions ran0, ran1, ran2, and ran3 generating uniformly distributed numbers can be found in Press et al (1992) Particularly, the function ran2 is supplied with the following comment: ran2 provides perfect random numbers; a practical definition of perfect is that we will pay $1000 to the first reader who convinces us otherwise (by finding a statistical test that ran2 fails in a nontrivial way, excluding the ordinary limitations of a machine s floating-point representation) Second, one needs to transform uniformly distributed numbers into random numbers with a given distribution.

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0 2.2 4.4 6.6 8.8 11 13.2 15.4 17.6 19.8 22 1.000 1.000 1.000 0.999 0.997 0.991 0.959 0.819 0.533 0.286 0.161 1.000 1.000 1.000 1.000 0.999 0.992 0.952 0.801 0.531 0.299 0.173 0.993 0.989 0.983 0.971 0.935 0.820 0.578 0.321 0.161 0.084 0.048 U 0.7 0.8 Un = 4 Un = 2 Un = 3 Un = 4 Un = 2 Un = 3 0.9 Un = 3 0.998 0.997 0.995 0.989 0.967 0.885 0.683 0.421 0.226 0.121 0.071 1.000 1.000 0.999 0.998 0.995 0.983 0.920 0.719 0.424 0.219 0.121 0.999 0.998 0.997 0.994 0.988 0.969 0.885 0.648 0.351 0.172 0.093 1.000 1.000 0.999 0.998 0.992 0.967 0.863 0.626 0.361 0.192 0.109 0.999 0.997 0.995 0.991 0.981 0.939 0.795 0.519 0.270 0.135 0.075 0.994 0.992 0.988 0.980 0.963 0.907 0.731 0.440 0.215 0.104 0.057

In terms of the example building the Apl distribution and described a few sections earlier, this means that (1) the numbers are generated randomly and (2) the numbers, for instance, from the interval (50, 100] are generated three times more often than the numbers from the interval ( 100, 50] The ratios of occurrences for other intervals must also correspond to the empirical estimation Donald Knuth (1998) describes several algorithms and discusses the efficiency of different implementations The following is an illustration suitable for our case: Write the intervals and corresponding probabilities as a row of pairs ( 100, 50], 1/10; ( 50, 0], 3/10; (0, 50], 1/10; (50, 100], 3/10; (100, 150], 1/10; (150, 200], 1/10 Note that the sum of the probabilities is equal to 1.

Following the example, we take the average values of two corresponding boundaries of an interval and associate them with increasing sums of probabilities (cumulative probabilities) This gives a new row 75, 1/10; 25, 4/10; 25, 5/10; 75, 8/10; 125, 9/10; 175, 1 The cumulative probabilities divide the interval [0, 1] into subintervals.

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