super efficiency dea error: Index exceeds matrix dimensions

2019 12 2
1 个回答
更新时间：2019 12 2
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0 个投票

X=[];%input matrix
Y=[];%output matrix
n=size(X',1);
m=size(X,1);
s=size(Y,1);
epsilon=10^-10;
f=[zeros(1,n) -epsilon*ones(1,m+s) 1];
A=zeros(1,n+m+s+1);
b=0;
LB=zeros(n+m+s+1,1);
UB=[];
LB(n+m+s+1)=-Inf;
for i=1:n
Aeq=[[X(:,1:i-1),zeros(m,1),X(:,i+1:n)] eye(m) zeros(m,s) -X(:,i) [Y(:,1:i-1),zeros(s,1),Y(:,i+1:n)] zeros(s,m) -eye(s) zeros(s,1)]
beq=[zeros(m,1) Y(:,i)];
w(:,i)=linprog(f,A,b,Aeq,beq,LB,UB);
end
w
lambda=w(1:n,:);
s_minus=w(n+1:n+m,:);
s_plus=w(n+m+1:n+m+s,:);
theta=w(n+m+s+1,:);
I'm using the code above to solve a se-dea model with 38 DMUs, 11 input and 1 output. The code was found in a paper but it reports "Index exceeds matrix dimensions". I'm wondering why this happens and how to solve it. Thanks very much.

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回答（1 个）

Hiro Yoshino
Hiro Yoshino 2019-12-2

0 个投票

X=[];%input matrix
Y=[];%output matrix
Obviously the input dimensions for X, Y are zero, and thus
n=size(X',1);
m=size(X,1);
s=size(Y,1);
become zero. Am I taking it correctly?
why not using "break points" so you can check the line to know to what extent your code is OK?
This way you can check the code literally line by line.

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Kerwin Liu
Kerwin Liu 2019-12-2
Sorry but I was using
X=[];%input matrix
Y=[];%output matrix
simply because the X matrix is 11×38 and Y is 1×38 so I omit them to control the text length. I think the error occurs in the following line
for i=1:n
Aeq=[[X(:,1:i-1),zeros(m,1),X(:,i+1:n)] eye(m) zeros(m,s) -X(:,i) [Y(:,1:i-1),zeros(s,1),Y(:,i+1:n)] zeros(s,m) -eye(s) zeros(s,1)]
beq=[zeros(m,1) Y(:,i)];
w(:,i)=linprog(f,A,b,Aeq,beq,LB,UB);
end
as I checked the code by line

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