Closed Syk-yr closed 2 years ago
Syk-yr
commented
2 years ago Matrix P=[ 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 26 -50 0 0 0 0 0 0 0 0 0 0 0 51 -50 0 0 0 0 0 0 0 0 0 0 0 51 -50 0 0 0 0 0 0 0 0 0 0 0 26 ]
S-Dafarra
commented
2 years ago Hy @Syk-yr, thanks for opening the issue. Just to make sure we are aligned, is the matrix P the cost hessian?
S-Dafarra
commented
2 years ago Hy @Syk-yr, thanks for opening the issue. Just to make sure we are aligned, is the matrix P the cost hessian?
To answer myself, I remembered later that OSQP calls the hessian matrix P. Then, to answer your initial question
In the QP problem solving, why the lower triangular matrix using the same data can be solved successfully, but the solution of the upper triangular matrix fails
In general, the matrix P is supposed to be symmetric matrix. You can check why this is necessary at https://math.stackexchange.com/questions/307381/why-do-we-assume-that-a-matrix-in-quadratic-form-is-symmetric.
For this reason, OSPQ requires the user to only insert the upper triangular part of the matrix, because it assumes it to be symmetric. If you pass only the upper triangular part of the matrix, it basically assumes you have the following matrix P
P =
2 0 0 0 0 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0 0 0 0 0
0 0 2 0 0 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 26 -50 0 0
0 0 0 0 0 0 0 0 -50 51 -50 0
0 0 0 0 0 0 0 0 0 -50 51 -50
0 0 0 0 0 0 0 0 0 0 -50 26This matrix is not positive definite (you can simply check the determinant), that instead is a requirement https://osqp.org/docs/index.html. This si why the QP fails
S-Dafarra
commented
2 years ago Note that you can obtain the symmetric version of the matrix you have in https://github.com/robotology/osqp-eigen/issues/114#issuecomment-1054208677 by doing (P + P')/2, obtaining
P =
2 0 0 0 0 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0 0 0 0 0
0 0 2 0 0 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 26 -25 0 0
0 0 0 0 0 0 0 0 -25 51 -25 0
0 0 0 0 0 0 0 0 0 -25 51 -25
0 0 0 0 0 0 0 0 0 0 -25 26That is positive definite.
Syk-yr
commented
2 years ago Thank you very much for your answer, I see where my mistake is.
------------------ 原始邮件 ------------------ 发件人: "Stefano @.>; 发送时间: 2022年3月4日(星期五) 凌晨0:27 收件人: @.>; 抄送: @.>; @.>; 主题: Re: [robotology/osqp-eigen] In the QP problem solving, why the lower triangular matrix using the same data can be solved successfully, but the solution of the upper triangular matrix fails (Issue #114)
请注意，您可以获得矩阵的对称版本(评论)通过做( P + P ')/ 2 ，获得 P = 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 26 -25 0 0 0 0 0 0 0 0 0 0 -25 51 -25 0 0 0 0 0 0 0 0 0 0 -25 51 -25 0 0 0 0 0 0 0 0 0 0 -25 26 这是肯定的。
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In the QP problem solving, why the lower triangular matrix using the same data can be solved successfully, but the solution of the upper triangular matrix fails