Re: optimization routine using conjugate gradient (or other) method?

*To*: mathgroup at smc.vnet.net*Subject*: [mg83769] Re: optimization routine using conjugate gradient (or other) method?*From*: dantimatter <google at dantimatter.com>*Date*: Fri, 30 Nov 2007 05:17:03 -0500 (EST)*References*: <9889841.1196176137975.JavaMail.root@m35> <fim9d1$s0c$1@smc.vnet.net>

thanks to all for responding. this is exactly what i needed. may i ask one further question: how does one search the solution landscape to determine if there are other local minima? many thanks, dan On Nov 29, 5:56 am, DrMajorBob <drmajor... at bigfoot.com> wrote: > My result seems close to Daniel Lichtblau's, although I've rewritten the > code: > > Clear[a, h, ahmodT, sqSum, func] > h[t_][x_List] := h[t] /@ x > h[t_][x_] = (1/33 UnitStep[a, 64.5 - a] > Piecewise[{{UnitStep[32.25 - a], > t < 127.75}, {0.5 (UnitStep[32.25 - a] + > UnitStep[32.25 - (t - 127.75) - a] + > UnitStep[a - 64.5 + (t - 127.75)]), 127.75 <= t < 160}}, > 0.5 (UnitStep[32.25 - a] + UnitStep[a - 32.25])] // > Rationalize // PiecewiseExpand // FullSimplify) /. a -> x; > ahmodT[a_List][t_] := a.h[t]@Mod[t - Range@160, 160] > sqSum[a_List] := #.# &[m - ahmodT[a] /@ time] > func[a_] := > Module[{an = Normalize[a, Total]}, 50 an.Log[an] + 1/2 sqSum[a] > ] > > time = {0, 12, 24, 36, 48, 60, 84, 96, 108, 120, 132, 144, 156}; > m = {0.149, 0.223, 0.721, 2.366, 2.580, 5.904, 7.613, 11.936, 11.943, > 8.477, 6.683, 4.641, 4.238}; > a = Array[f, 160]; > > The objective function is VERY complicated: > > func[a] // LeafCount > > 107403 > > But NMinimize finishes in under 81 seconds: > > constraints = Thread[a > .0001]; > Timing[{obj1, rule1} = NMinimize[{func[a], constraints}, a];] > obj1 > a1 = a /. rule1; > {lo, hi} = Through[{Min, Max}@a1] > > {80.89, Null} > > -237.938 > > {0.811749, 15.0187} > > A random search doesn't find anything better (or even nearly as good): > > randomStart := RandomReal[{.001, 16}, Length@a] > Timing[Table[func[randomStart], {5000}] // Min] > > {142.891, -158.832} > > But that's a tiny sample considering the dimensions!! > > I also tried FindMinimum, but it seemed to take FOREVER, so I aborted. > > Bobby > > On Tue, 27 Nov 2007 05:08:26 -0600, dantimatter <dantimat... at gmail.com> = > = > > wrote: > > > > > > > > > Hello all, > > > I have a system that I'd like to optimize in 160 dimensions. I have > > all the equations I need to set up the problem, but I have no idea how= > > to use Mathematica to actually do the optimization. You've all been > > so helpful in the past, I'm hoping maybe some of you have suggestions > > as to how I should proceed. My problem is as follows: > > > I have an array of 160 numbers a = Array[f, 160]. I don't know any = > of > > the a[[j]] but I'd like to determine them through the optimization > > process. The sum of all elements in the array is atot = Sum[a[[j]],= > > {j, 160}]. > > > I also have two fixed arrays, time and m, where time = {0, 12, 24, 3= > 6, > > 48, 60, 84, 96, 108, 120, 132, 144, 156} and m = {0.149, 0.223, 0.72= > 1, > > 2.366, 2.580, 5.904, 7.613, 11.936, 11.943, 8.477, 6.683, 4.641, > > 4.238}. > > > Lastly, I have a somewhat-complicated "time shift function" h[a_, > > t_] := > > 1/33 UnitStep[a] UnitStep[ > > 64.5 - a] Piecewise[{{UnitStep[32.25 - a] , > > t < 127.75}, {0.5 (UnitStep[32.25 - a] + > > UnitStep[32.25 - (t - 127.75) - a] + > > UnitStep[a - 64.5 + (t - 127.75)]), 127.75 <= t < 160}}, > > 0.5 (UnitStep[32.25 - a] + UnitStep[a - 32.25])] > > > Now with all this stuff set up, the function I'd like to minimize is > > > func = 50 Sum[(a[[j]]/atot) Log[a[[j]]/atot], {j, 160}] + 1/2 > > Sum[(m[[i]] - > > Sum[a[[j]] h[Mod[time[[i]] - j, 160], time[[i]]], {j, 160}])^2, > > {i, Length[time]}] > > > The problem is complicated by the fact that the conjugate gradient > > optimization routine may go through a point where the a[[j]] have > > negative values, which makes it impossible to compute Log[a[[j]]]. > > So, an alternative formula is used instead of the a[[j]]/atot) > > Log[a[[j]]/atot]: > > > 1/atotp (eps Log[eps/atotp] - (eps - a[[j]]) (1 + Log[eps/atotp]) + > > (eps - a[[j]])^2/(2 eps)) > > > in this case, 'eps' is a predefined small positive number and 'atotp' > > is the sum of a[[j]] such that a[[j]] is equal to or greater than > > 'eps'. > > > So, does anyone have any suggestions as to how I could code this in > > Mathematica? Is it something that's even possible? Please let me > > know if you have any ideas. > > > Many thanks, > > Dan > > -- = > > DrMajor... at bigfoot.com

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