Simplex2P
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package maxima
Return the dual solution vector y (cost row (M) under the slack columns).
Find the best variable x_l to enter the basis.
Find the best variable x_l to enter the basis. Determine the index of entering variable corresponding to column l (e.g., using Dantiz's Rule or Bland's Rule). Return -1 to indicate no such column. t(M).firstPos (jj) // use Bland's rule (lowest index, +ve) t(M).argmaxPos (jj) // use Dantiz's rule (min index, +ve, cycling possible)
Show the flaw by printing the error message.
Show the flaw by printing the error message.
the method where the error occurred
the error message
Initialize the basis to the slack and artificial variables.
Initialize the basis to the slack and artificial variables. Perform row operations to make cost row (t(M)) zero in artificial var columns. If b(i) is negative have a surplus and artificial variable, otherwise, just a slack variable.
Find the best variable x_k to leave the basis given that x_l is entering.
Find the best variable x_k to leave the basis given that x_l is entering. Determine the index of the leaving variable corresponding to row k using the Min-Ratio Rule. Return -1 to indicate no such row.
the entering variable (column)
Return the value of the objective function f(x) = c x.
Pivot on entry (k, l) using Gauss-Jordan elimination to replace variable x_k with x_l in the basis.
Pivot on entry (k, l) using Gauss-Jordan elimination to replace variable x_k with x_l in the basis.
the leaving variable (row)
the entering variable (column)
Return the primal solution vector x (only the basic variables are non-zero).
Remove the artifical variables and reset the cost row (M) in the tableau.
Run the simplex algorithm starting from an initial BFS and iteratively find a non-basic variable to replace a variable in the current basis so long as the objective function improves.
Solve the LP maximization problem using two phases if neceassry.
Solve the LP maximization problem using two phases if neceassry. Note: phase I is always minimization. Two phases are necessary if the number of artificial variables R > 0.
Convert the current tableau and basis to a string suitable for display.
Convert the current tableau and basis to a string suitable for display.
This class solves Linear Programming (LP) problems using a tableau based Simplex Algorithm. Given a constraint matrix 'a', limit/RHS vector 'b' and cost vector 'c', find values for the solution/decision vector 'x' that maximize the objective function f(x), while satisfying all of the constraints, i.e.,
maximize f(x) = c x subject to a x <= b, x >= 0
In case of "a_i x >= b_i", use -b_i as an indicator of a ">=" constraint. The program will flip such negative b_i back to positive as well as use a surplus and artificial variable instead of the usual slack variable, ie., a_i x <= b_i => a_i x + s_i = b_i // use slack variable s_i with coeff 1 a_i x >= b_i => a_i x + s_i = b_i // use surplus variable s_i with coeff -1 For each ">=" constraint, an artificial variable is introduced and put into the initial basis. These artificial variables must be removed from the basis during Phase I of the Two-Phase Simplex Algorithm. After this, or if there are no artificial variables, Phase II is used to find an optimal value for x and the optimum value for f.
Creates an MM-by-nn simplex tableau with -- [0..M-1, 0..N-1] = a (constraint matrix) -- [0..M-1, N..M+N-1] = s (slack/surplus variable matrix) -- [0..M-1, M+N..nn-2] = r (artificial variable matrix) -- [0..M-1, nn-1] = b (limit/RHS vector) -- [M, 0..nn-2] = c (cost vector)