Pawel Wembley 1 Posted November 8, 2016 Share Posted November 8, 2016 (edited) I would appreciate very much your help or direction to solve the problem. Looks like equation Q=SUM(ABS(B-Ai)) has got min and max values with the assumptions 0<B<1 0<=Ai<=1 SUM(A1:AK)=1 and i is a natural number from 1 to K Looks like the Qmax and Qmin depend on B and K only. I failed to find general solution for Qmax and Qmin. I attach the pdf with the problem written in Word Equation instead of Excel manner Q.pdf Edited November 8, 2016 by Pawel Wembley Link to post Share on other sites

renerpho 9 Posted November 20, 2016 Share Posted November 20, 2016 (edited) Hello Pawel. The mimimum [math]Q_{min}=\max(KB-1,0)[/math] is calculated as follows: [math]\sum_{i=1}^{K}|B-A{_i}|\geq 0[/math] is trivial, and reached if all [math]A{_i}[/math] are equal to [math]B[/math]. If [math]KB-1>0[/math] then the minimum is not reached at 0, but at [math]\sum_{i=1}^{K}|B-A{_i}|\stackrel{|x|\geq x}{\geq}\sum_{i=1}^{K}(B-A{_i})\[/math] [math]=KB-\sum_{i=1}^{K}A{_i}\ \stackrel{\sum_{i=1}^{K}A{_i}=1}{=}KB-1[/math]. This minimum value is reached if and only if [math]A{_i}<B \forall i[/math]. In the case [math]KB-1<0[/math], you have [math]B<\frac{1}{K}[/math], so at least one of the [math]A{_i}[/math] is larger than [math]B[/math] (by the Pigeon Principle). Which means that [math]KB-1[/math] can't be reached. In that case, the mimimum is 0. For the maximum [math]Q_{max}=1+B(K-2)[/math]: [math]\sum_{i=1}^{K}|B-A{_i}| \stackrel{extend}{=}\sum_{i=1}^{K}(B+A{_i})-\sum_{i=1}^{K}(B+A_{i}-|B-A{_i}|)[/math] [math]=KB+1-\sum_{i=1,A{_i}>B}^{K}(B+A{_i}+B-A{_i})-\sum_{i=1,A{_i}\leq B}^{K}(B+A{_i}-B+A{_i})[/math] [math]=KB+1-\sum_{i=1,A{_i}>B}^{K}(2B)-\sum_{i=1,A{_i}\leq B}^{K}(2A{_i})[/math]. This will get maximal if you have [math]A{_i}=0 \forall A{_i}\leq B[/math], when it will be equal to [math]KB+1-2B \cdot 1_{A{_i}>B}[/math]. If one of the [math]A{_i}[/math] is equal to [math]1[/math] (and all other [math]A_{i}[/math] are [math]0[/math]), this will take its maximum value [math]KB+1-2B \cdot 1=1+B(K-2)[/math]. Edited November 20, 2016 by renerpho 2 Link to post Share on other sites

Pawel Wembley 1 Posted November 21, 2016 Author Share Posted November 21, 2016 Hi Quark, it is great. Big thanks and regards. 1 Link to post Share on other sites

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