you created R to hold 600 things, things indexable as R[1]..R[600] mnths=600 R=zeros(Float64,mnths)
later you bring Julia's attention to R[600+1], just one step 'out of bounds' when i is 600, i+1 is 601 In Julia, the top part of the range is included That loop should be for i in 3:(600-1) R[i+1] = .. end On Friday, August 21, 2015 at 4:37:48 PM UTC-4, Tj Midkiff wrote: > > I am trying to write a small script to help my company out. I came across > Julia on a web search naturally when looking for more speed. I am just > getting my feet wet with programming so please be patient with me. > > This is one piece of the code that I know is very inefficient, so any help > is greatly appreciated. FYI, the final version will include pulling the > initial variables from a SQL database that could include 100k records. So > basically this code could be ran 100k times in the final version at any > time (along with many more calculations). Please comment anywhere I could > add more efficiency, because speed is absolutely critical. > > I have a spreadsheet I can provide that shows the methodology a little > clearer if it would help. > > My main question though is why am I getting the BoundsError()... > > Thanks for all of the help. > > > qi=3454.0 > di=0.6 > b=0.9 > mnths=600 > > AI=(1/b)*((1-di)^-b-1) > ai=AI/12 > q(t)=qi/(1+b*ai*t)^(1/b) > Q=[q(t-1) for t=1:3] > a=((Q[2]/Q[3])^b-1)/b > mOil=zeros(Float64,mnths) > > #Is it worth creating a function here? > mOil[1]=(Q[1]^b/((1-b)*ai))*(Q[1]^(1-b)-Q[2]^(1-b)) > mOil[2]=(Q[2]^b/((1-b)*a))*(Q[2]^(1-b)-Q[3]^(1-b)) > R=zeros(Float64,mnths) > R[1]=mOil[2]/(mOil[2]-mOil[1])+b > for i=1:2 > R[i+1]=R[1]-i*b > end > > for i=3:600 #Getting a BoundsError() here > #mOil[i]=mOil[i-1]*(R[i]/(R[i]-1)) > R[i+1]=R[1]-i*b > end > > >
