Figuring the first "year's" population of 3-I
5.10204082*10e14 2-C will be produced per saturation cycle. One in 1.96*10e14 is the number of 3-I generated by 2-C organisms
countit = 0 original = 5.10204082*10^14 I3 = 0 population = 0 [start] population = population + original population = population * 1.01 I3 = I3 + (population/(1.96*10^14)) I3 = I3 * 1.02 countit = countit + 1 IF countit = 1000 then [end] GOTO [start] [end] print I3 end
answer: 5.44635665*10e12 or about 5.45*10e12 3-I at the end of "year one"
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