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I have a problem and I really don't know what kind of mathematical method should I apply to solve or model my problem. I would be thankful If anyone can give me some answer or help.

Suppose we have more than one option to select a set of numbers. like A={a1,a2,a3,..an} and B={b1,b2,b3,..,bn} (number of members in each set is equal to n) each ai or bi is an integer value. We also have a threshold value lets say "TH". If we choose Option A, the new value of set A becomes A={a1+1,a2+1,....an+1} and also same case for B or any other options. Each round we have options, my question is that which option every time we must select to increase the number of rounds until we have a number (ai or bi ..) reaches to the Threshold? (Assume we have a global set which contains sets A,B,.. I mean the numbers are finite) Im trying to find a criteria by which every time I choose a set which help me to increase the number of rounds(because once in a set we have a number which is greater than or equal the threshold, we should stop). Also considering that we have a periodic time "t" by which we decrease values of each sets(the set that is not currently in use) by "1".

-Example : A={7,4,5,5}, B={5,6,2,4} and TH = 8 If we choose A then after choosing we have A={8,5,6,6} and if we choose B we get={6,7,3,5}. So is better to select B, because after selecting A we have a value equal to our threshold.

-Another Example: A={5,5,3,4}, B={5,4,3,3} In this case we get A={6,6,4,5} and B={6,5,4,4} so maybe is better choose B again, because we have two 6 in A but one in B and most probably A goes to threshold sooner, so we choose B ?

-Third Example A={6,4,3,3}, B={5,4,3,3} : considering that we have a periodic time "t" by which we decrease values of each sets by 1 .meaning that each t time, t time, we decrease the values by 1. So If in third example we choose B then we are giving a time to A to have that "6" value becomes to 5.

Thanks in advance.

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2 Answers 2

It looks simple to me unless I am missing something. If any number in the Sequence A is one below or at the threshold reject it for B if B has all its values 2 or more below. Keep doing this until you hit the threshold. Now if both A and B are acceptable (meaning largest value in A and largest in B are 2 or more below threshold) pick the one whose maximum value is the lowest. That one will take youu to the threshold at least one round later than the other one. If the largest value in A equals the largest in B it doesn't matter which one you pick. Either one will get you to the threshold in the same round.

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If you have to choose only one set, pick the one with the lowest maximum value. If it is A, you get TH-max(A)-1 picks before you stop. Since you are subtracting the maximum, you want it as small as possible. If you can change sets between picks, the order doesn't matter, just always pick one that won't hit the threshold. If there are n sets in total, you get n*TH-max(A)-max(B)-...-1 picks before you have to hit the threshold.

If you change the values with a period, you have to say how often you pick a set. If you pick sets with the same period the values decrease, you will never hit the threshold.

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