Index exceeds matrix dimensions. - In running lsqnonlin optimization.

Question

Pietro Lindia 2015-2-28

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/180850-index-exceeds-matrix-dimensions-in-running-lsqnonlin-optimization

评论： Matt J 2015-2-28

Hi everybody, I am trying to running the following optimization problem under lsqnonlin function. But I got stuck at line ' [Params, Fval, Residuals, Exitflag] = lsqnonlin(@(Params) NSobjP(Params, BondsCF, Bonds.DirtyPrices), Params0(i,:), lb, ub, options); ' where the optimization is supposed to start. Here is the code:

Nbonds = length(Bonds.Prices); 
BondsCF = cell(Nbonds, 1);
ObsYTM = zeros(Nbonds, 1);
Time2Mat = zeros(Nbonds, 1);
for i = 1:Nbonds
       % Bonds future cashflows
      BondsCF{i} = czbondfuturecf(Bonds.Coupon(i), Bonds.Issue(i), Bonds.Maturity(i), Bonds.Settle);
      BondsFig = czbondkeyfigures(Bonds.Coupon(i), Bonds.Issue(i), Bonds.Maturity(i), Bonds.Settle,  Bonds.Prices(i));
      ObsYTM(i) = BondsFig.YTM; 
      Time2Mat(i) = BondsFig.TimeToMaturity; 
      Bonds.DirtyPrices(i, 1) = Bonds.Prices(i);   
  end
  MaxTime2Mat = max(Time2Mat);
  MinTime2Mat = min(Time2Mat);
% Set initial parameters
LongYTM = ObsYTM(Time2Mat == max(Time2Mat)); % YTM for the longest maturity
ShortYTM = ObsYTM(Time2Mat == min(Time2Mat)); % YTM for the shortest maturity
LongYTM =  log(1+LongYTM); % Convert the LongYTM to the continuous compounding
beta0 = LongYTM; % beta0 = YTM with the maximal time to maturity
beta1 = ShortYTM - beta0; %YTM with the minimal time 2 maturity - beta0
% some bounds for lambda and gamma
MaxMat4lambda = 0.5*MaxTime2Mat; 
MaxMat4lambda = min(MaxMat4lambda, 10);
options = optimset('LargeScale', 'off', 'MaxIter', 3e5, 'MaxFunEvals', 3e5, 'TolFun', 1e-5, 'TolX', 1e-5, 'Display', 'off');   
lambda0 = 0.6;
OptMinLambda = fminsearch(@(lambda) LambdaLoading(lambda, MaxMat4lambda), lambda0, options);
% ==============================================
lambdaMin = OptMinLambda;
lambdaMax = 30;
lb = [-inf, -inf, -inf, lambdaMin]; %lowerbound
ub = [ inf,  inf,  inf, lambdaMax]; %upperbound  
    fprintf('\n\n======================== RUNNING ESTIMATION ============================\n');
  beta2 = 0;
  lambda = 0.62; 
  if lambda < lambdaMin
    lambda = lambdaMin+0.05;
  end
  Params0 = [beta0 beta1 beta2 lambda];           
%'lsqnonlin'         
options = optimset('MaxIter', 3e5, 'MaxFunEvals', 3e5, 'TolFun', 1e-12, 'TolX', 1e-12, 'Display', 'off', 'Algorithm', 'trust-region-reflective');   
% Fitting Price
[Params, Fval, Residuals, Exitflag] = lsqnonlin(@(Params) NSobjP(Params, BondsCF, Bonds.DirtyPrices), Params0(i,:), lb, ub, options);

It seems that ' Index exceeds matrix dimensions ' is a bullshit error but still I can't figure it out. At the beginning I thought it was a matter of Params dimension, but since I specified in NSobjP function that is a vector and since lsqnonlin gives vectors as output, I cannot see the solution.

Million thanks in advance :)

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Answer 1

Luuk van Oosten 2015-2-28

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在 MATLAB Online 中打开

What you are trying to do is taking a value (which you try to index) but it does not exist, because it exceeds the dimensions of the matrix. Maybe sound a bit vague (or like bullshit as you call it...) but here an easy example of what is happening:

Imagine you have the 3x3 matrix called 'a'.

a=[1 2 3;6 1 8; 9 1 5]

find a(3,3) (or any other value; a(1,2))

a(3,3) = 5

now try to find a(4,3).

Index exceeds matrix dimensions.

for the obvious reason that you try to to look in the 4th row (while the matrix 'a' only has 3 rows...)... ergo: your index (4,2) is exceeding the matrix dimensions.

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Matt J 2015-2-28

Pietro Lindia commented:

Thanks a lot Luuk. You have been very helpful. You gave me the hint to solve it. Indeed I have only one vector for initial parameter Params0 , while in the input of lsqnonlin there was written Params0(i,:).

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