Maximizing the time we reach to a threshold in a series of numbers

[moji]

Hello everyone,

 

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 second example we choose B then we are giving a time to A to have that "6" value becomes to 5.

 

Thanks in advance.

[Daedalus]

This sounds more like a math question than it does an engineering question.

 

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).

In this case, I would simplify the rules because it really doesn't matter which set we choose to increment. Basically, you are solving for the maximum number of rounds you can go before a set terminates (a member of the set equals the value of T). It would be easier to take the largest member in each set:

 

[math]S_1=\{a_i\}_{i=1}^{n}, \ \ M_1=\text{max} \ S_1[/math]

[math]S_2=\{b_i\}_{i=1}^{n}, \ \ M_2=\text{max} \ S_2[/math]

[math]S_3=\{c_i\}_{i=1}^{n}, \ \ M_3=\text{max} \ S_3[/math]

etc...

 

Calculate the number of rounds [math]R_1, \ R_2, \ R_3, \ \text{...}\, , \ [/math] each set can make before terminating:

 

[math]R_1 = T - M_1 - 1[/math]

[math]R_2 = T - M_2 - 1[/math]

[math]R_3 = T - M_3 - 1[/math]

etc...

 

Add up the number of rounds for each set to get the total number of rounds:

 

[math]R_{\, \text{total}} = \sum_{k=1}^{m} R_k[/math]

 

If you want to include the round that terminates the process, just add one:

 

[math]R_{\, \text{total}} = 1 + \sum_{k=1}^{m} R_k[/math]

 

 

Now for the next part of the problem:

 

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".

...

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 second example we choose B then we are giving a time to A to have that "6" value becomes to 5.

It sounds like the process waits for time [math]t[/math] to elapse after making a choice each round (increment one set and decrement the others). In that case, the process would not terminate because we can decrement the set we incremented the previous round, and increment the set we just decremented.

 

[math]R_0: \ A=\{6,4,3,3\}, \ B=\{5,4,3,3\}[/math]

[math]R_1: \ A=\{5,3,2,2\}, \ B=\{6,5,4,4\}[/math]

[math]R_2: \ A=\{6,4,3,3\}, \ B=\{5,4,3,3\}[/math]

etc..

Edited May 18, 2012 by Daedalus