
Math Software Math Software  Mathematica, Matlab, Calculators, Graphing Software 
 LinkBack  Thread Tools  Display Modes 
January 10th, 2012, 08:53 AM  #1 
Newbie Joined: Dec 2011 Posts: 8 Thanks: 0  mathematica
hi all I am new in using mathematica , I need to excute this algorithem but I have no idea how to start hope you could help me. 1set n(max)=(cbcop)/(cmea+cit). 2set n=1 3set m(max)=(cbcit*n)/(cop+ cmea*n) 4set m=1 5find f such that (cop*f*m+cmea*n*m+cit*n) <=cb 6calculate G(f,l,n). 7set m=m+1 and repeat steps and 6 until m=m(max). 8set n=n+1,and repeat steps 3 to 7until n=n(max). 9 the optimal solution G"(f",l",n") can be obtained as min G(f,l,n). 
January 12th, 2012, 11:28 AM  #2 
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: mathematica Code: nMax=(cbcop)/(cmea+cit); gfln=Reap[Do[mMax=(cbcit*n)/(cop+ cmea*n); Do[f=cb(cmea*n*m+cit*n)/(cop*m);(*ASSUMING cop*m never equal to zero*) Sow[G(f,l,n)][[2,1]]; ,{m,mMax}];(*ASSUMING you meant to write "repeat steps 5 and 6"*) ,{n,nMax}]]; optimalsolution=Min[gfln] 
January 12th, 2012, 12:59 PM  #3 
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: mathematica
Note: Correcting my errors in translating 6calculate G(f,l,n). I Apologize. Code: nMax=(cbcop)/(cmea+cit); gfln=Reap[Do[mMax=(cbcit*n)/(cop+ cmea*n); Do[f=cb(cmea*n*m+cit*n)/(cop*m);(*ASSUMING cop*m never equal to zero*) Sow[G[f,l,n][[2,1]]]; ,{m,mMax}];(*ASSUMING you meant to write "repeat steps 5 and 6"*) ,{n,nMax}]]; optimalsolution=Min[gfln] 
January 12th, 2012, 01:47 PM  #4 
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: mathematica
Another error because at the moment I can't test this before posting. I apologize. Code: nMax=(cbcop)/(cmea+cit); gfln=Reap[Do[mMax=(cbcit*n)/(cop+ cmea*n); Do[f=cb(cmea*n*m+cit*n)/(cop*m);(*ASSUMING cop*m never equal to zero*) Sow[G[f,l,n]]; ,{m,mMax}];(*ASSUMING you meant to write "repeat steps 5 and 6"*) ,{n,nMax}]][[2,1]]; optimalsolution=Min[gfln] 
January 12th, 2012, 02:27 PM  #5 
Newbie Joined: Dec 2011 Posts: 8 Thanks: 0  Re: mathematica
Thank you very much dear Bill...you saved my life .But I need your help again ,by this code i get the minimum of G(f,l,n),i need to obtain the values of f ,l and n that correspond this minimum..how could i do it ? Code: u = 1.0; d = 1000.0; m = 2.0; p = 0.1; \[Sigma] = 2.5; cb = 7200.0; cop = 20.05; cmea = 4.65; cit = 130.0; nMax = (cb  cop)/(cmea + cit); gfkn = Reap[Do[kMax = (cb  cit*n)/(cop + cmea*n); Do[f = cb  (cmea*n*k + cit*n)/(cop*k);(*ASSUMING cop* k never equal to zero*) Sow[E^(2 (m + Log[d] + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1)))^(1/2) Log[p/(1  p)])) (1  2 E^((3 (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))) (1 + Log[p/(1  p)]^2/( 1 + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))))))/(2 n)) + E^(( 6 (u^2 + 3 (6 \[Sigma]^2 E^(2 m))/(f^2 k (k + 1) (2 k + 1))) (1 + Log[p/(1  p)]^2/( 1 + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))))))/n))];, {k, kMax}];, {n, nMax}]][[2, 1]]; optimalsolution = Min[gfkn] 
January 12th, 2012, 08:57 PM  #6 
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: mathematica
There are two parts to the answer. First part. Instead of Sow[G[f,l,n]]; use Sow[{G[f,l,n],f,l,n}]; That will save the value AND the parameters for that value as a list. Second part. Now Min[gfln] is no longer sufficient to extract the data you want, but Sort can be a substitute. try First[Sort[gfln,(First[#1]<First[#2])&]] What that does is sort your gfln list on the value of the function only and then discard all but the smallest entry. As always with Mathematica, there are at least a dozen other ways of doing anything, at least several of which are incomprehensible. 
February 13th, 2012, 10:43 AM  #7 
Newbie Joined: Dec 2011 Posts: 8 Thanks: 0  Re: mathematica
thank you very much, I still have a problem with this code ,the parameter k should change when I change the cb,cop,cmea, or cit ,but this code keep giving me k=1 for all cases what should i do d = 100.0; m = 2.0; u = 1.0; p = 0.1; \[Sigma] = 2.5; cb = 1700.0; cop = 29.05; cmea = 60.65; cit = 18.0; nMax = (cb  cop)/(cmea + cit); gfkn = Reap[Do[kMax = (cb  cit*n)/(cop + cmea*n); Do[f <= (cb  (cmea*n*k + cit*n))/(cop*k); Sow[{E^(2 (m + Log[d] + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1)))^(1/2) Log[p/(1  p)])) (1  2 E^((3 (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))) (1 + Log[p/(1  p)]^2/( 1 + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))))))/(2 n)) + E^(( 6 (u^2 + 3 (6 \[Sigma]^2 E^(2 m))/(f^2 k (k + 1) (2 k + 1))) (1 + Log[p/(1  p)]^2/( 1 + (u^2 + 3/\[Pi]^2 (6 \[Sigma]^2 E^(2 m))/( f^2 k (k + 1) (2 k + 1))))))/n)), f, k, n}];, {k, 1, kMax}];, {n, 1, nMax}]][[2, 1]]; First[ Sort[gfkn, (First[#1] < First[#2]) &]] 
February 14th, 2012, 10:27 AM  #8 
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: mathematica
I am concerned by your line: ... Do[f <= (cb  (cmea*n*k + cit*n))/(cop*k); ... Did you perhaps mean Do[f=(cb  (cmea*n*k + cit*n))/(cop*k); because otherwise you are just comparing an uninitialized f against an expression and then throwing away the result. But that still doesn't seem to change your k when I fiddle with cb,cop,cmea,cit. If I change your last line of code to Sort[gfkn, OrderedQ[{First[#1], First[#2]}] &] // MatrixForm then it appears it is correctly sorting on the first column and for small values of the first column k just seems to usually be 1. Can you provide some specific values for cb,cop,cmea,cit where k should not be 1? 

Tags 
mathematica 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integral for Mathematica  The_Fool  Calculus  5  July 10th, 2012 01:24 PM 
Mathematica  phoebe  Math Software  3  February 2nd, 2012 09:17 AM 
Mathematica  sivela  Math Software  2  April 15th, 2010 04:30 PM 
Mathematica Help  carXunderwater  Math Software  1  March 4th, 2010 09:12 PM 
Mathematica  billymac00  Number Theory  5  September 6th, 2009 10:59 PM 