-->l=3,n=4,N=2048,show=0,[DIST2,STD, MEAN, FREQ,STD1, MEAN1, FREQ1,FX, POPX]=frequCheck(l,n,N,show) l = 3. n = 4. N = 2048. show = 0. 0. 944. 0.1152344 92.1875 1. 1051. 0.1282959 102.63672 2. 1057. 0.1290283 103.22266 3. 1087. 0.1326904 106.15234 4. 1028. 0.1254883 100.39063 5. 1052. 0.1284180 102.73438 6. 980. 0.1196289 95.703125 7. 993. 0.1212158 96.972656 Number of Events n * N = 8192 POPX = 1. 1. 1. 7. 0. 1. 0. 2. 1. 0. 0. 4. 0. 0. 0. 0. FX = 0. 944. 0.1152344 92.1875 1. 1051. 0.1282959 102.63672 2. 1057. 0.1290283 103.22266 3. 1087. 0.1326904 106.15234 4. 1028. 0.1254883 100.39063 5. 1052. 0.1284180 102.73438 6. 980. 0.1196289 95.703125 7. 993. 0.1212158 96.972656 FREQ1 = 0.1326904 1. 0.1290283 1. 0.1284180 1. 0.1282959 1. 0.1254883 1. 0.1212158 1. 0.1196289 1. 0.1152344 1. MEAN1 = 0.125 STD1 = 0.0058149 FREQ = 106.15234 1. 103.22266 1. 102.73438 1. 102.63672 1. 100.39063 1. 96.972656 1. 95.703125 1. 92.1875 1. MEAN = 100. STD = 4.6519063 DIST2 = 6.9824219
If one includes mutation in gassimple0() then the data of empirical experiments show a different outcome depending from the frequency of the application of the mutation operator (cf. diagram 3.6.1). If one uses the mutation operator every 512 cycles (MT=512) then the results are more worse than without the mutation operator. If one selects MT=1024 then the results are better compared with MT=512, but not better than without the application of the mutation operator.
Increasing N to with and and comparing it (row 2) with frequCheck() with and with the original values of gassimple0() without mutation (row 1) with .
Gerd Doeben-Henisch 2012-03-31