Additional usage info for btran and ftran?

Hi Leigh,

Let's consider an LP in standard form: min cT x, A x = b, x >= 0. After using problem.solve(), A is partitioned into a basic part A_B and a nonbasic part A_N.

FTRAN is just a transposed version of BTRAN in the following sense. FTRAN(z) solves (finds y) such that A_B y = z. If z = e_i, the i'th unit vector, the solution y corresponds to the i'th column of A_B-1, the inverse of the basis matrix A_B. In contrast, BTRAN(z) solves A_BT y = z, such that for z = e_i, you obtain the i'th row vector of A_B-1.

Therefore, your code above exits with matrix tableau containing A_B-1. You could also use problem.ftran(y), in which case you would have to transpose the last line: tableau[ i , : ] becomes tableau[ :, i] in that case.

Note that both methods require an array of length problem.attributes.rows as input. There is no FTRAN operation on the variables (columns) of the problem.

This being said, it would be good to understand better what you are trying to achieve. The optimal simplex tableau is a table that holds information on the matrix, the right hand side, and the (reduced) cost vector, but not the variables.

All this information, you can query more easily than by computing the linear algebra by hand:

• use problem.getRCost() to obtain the reduced costs for each column.
• use problem.getbasis() to get more detailed information about the current basis
• use problem.getlpsolval() to get all information regarding solution values, slacks, dual multipliers, and reduced costs from one function call