advanced19 min studyLesson 282 in Mathematics

Standard forms and basic feasible solutions

Translate constraints without changing the optimization problem and connect algebraic bases to vertices.

On this page

What you’ll understand

  • Translate inequalities and unrestricted variables into equality standard form.
  • Compute a basic solution and test its feasibility.
  • Explain why a degenerate vertex can have several bases.

A solver needs a precise representation, not just a sketch of a feasible region. Equality standard form packages linear constraints into a matrix equation and nonnegative variables, making rank and linear-system tools available. The translation must preserve feasible choices and objective values even if it adds coordinates. Once that bookkeeping is secure, a basis selects a small invertible subsystem whose solution produces a candidate vertex. Feasibility remains a separate test: solving equations correctly does not guarantee nonnegative coordinates.

Translate with explicit reconstruction

Our equality standard form minimizes cTxc^Tx subject to Ax=bAx=b and x≥0x\ge0. A constraint aTx≤ba^Tx\le b becomes aTx+s=ba^Tx+s=b with slack s≥0s\ge0; a greater-than inequality uses a nonnegative surplus subtracted instead. An unrestricted coordinate becomes u−vu-v with u,v≥0u,v\ge0, and a lower bound can be removed by a shift. A maximization objective can be negated. Keep an explicit reconstruction map back to original variables and retain any constant introduced into the objective by shifts.

Splitting a free variable is not a one-to-one correspondence: adding the same positive number to both parts leaves the original coordinate unchanged. Nevertheless every original feasible point has an extended feasible representative, and every extended feasible point reconstructs a valid original point with the same objective. This is the right equivalence for optimization. Added redundancy can affect rank and algorithm behavior, so a translation should be economical. Equalities that contradict one another must be detected rather than silently deleted when preparing a full-row-rank system.

min⁡{cTx:Ax=b, x≥0},aTx≤b  ⟺  ∃s≥0: aTx+s=b\min\{c^Tx:Ax=b,\ x\ge0\},\qquad a^Tx\le b\iff \exists s\ge0:\ a^Tx+s=b
The slack records unused capacity exactly; it is not an approximation to the original inequality.↔ Scroll wide equations horizontally.

Choose independent columns

Assume AA has mm independent rows and nn columns. A basis is a set of mm column indices whose square submatrix BB is invertible. Set every nonbasic coordinate to zero and solve BxB=bBx_B=b. This is a basic solution. It is a basic feasible solution only if every basic coordinate is nonnegative. No sign condition on B−1B^{-1} alone is required: the sign of the particular product with bb is what matters. Factorizations normally solve this system without explicitly storing an inverse.

For equality standard form with full row rank, the basic feasible solutions are exactly the extreme points. One direction follows because a feasible perturbation supported only on independent basic columns must vanish. For the other direction, the positive-coordinate columns of an extreme point must be independent; a dependence would permit small positive and negative perturbations while keeping nonnegativity. Extend those columns to a basis using full row rank. This argument also explains why a vertex has at most mm positive coordinates, even when many variables are available.

xN=0,xB=B−1b,x is basic feasible  ⟺  B−1b≥0x_N=0,\qquad x_B=B^{-1}b,\qquad x\text{ is basic feasible}\iff B^{-1}b\ge0
The basis defines a candidate algebraically; the final coordinate test decides primal feasibility.↔ Scroll wide equations horizontally.

Zero basic variables and initialization

A basic feasible solution is degenerate when at least one basic coordinate equals zero. Several different bases may then describe the same vector, because a zero coordinate can exchange its role with another zero coordinate without moving the point. Do not identify the number of bases with the number of vertices. In a nondegenerate basic feasible solution all mm basic coordinates are positive, making its positive support determine the basis uniquely. Degeneracy is common when a model contains redundant-looking active restrictions or balances that force zero activity.

A feasible starting basis is not always visible. Phase I introduces artificial variables and minimizes their sum after arranging a convenient nonnegative right-hand side. A zero Phase I optimum establishes feasibility of the original equalities with nonnegative variables; a positive optimum certifies infeasibility, provided the auxiliary optimization was itself solved correctly. Artificial variables must then be removed or handled through redundant-row cleanup before optimizing the original objective. Numerical tolerances complicate the distinction between zero and a small positive residual, so report residuals and scaling rather than asserting exact feasibility from a rounded display.

Two bases, one vertex

For x1+x2+s1=2x_1+x_2+s_1=2 and x1+s2=2x_1+s_2=2 with all four variables nonnegative, compare the bases (x1,s1)(x_1,s_1) and (x1,s2)(x_1,s_2).

  1. For basis (x1,s1)(x_1,s_1), set x2=s2=0x_2=s_2=0. The second equation forces x1=2x_1=2, and the first then gives s1=0s_1=0. The basis columns are independent.

  2. For basis (x1,s2)(x_1,s_2), set x2=s1=0x_2=s_1=0. The first equation gives x1=2x_1=2, and the second gives s2=0s_2=0. These basis columns are also independent.

  3. Both basic solutions equal (x1,x2,s1,s2)=(2,0,0,0)(x_1,x_2,s_1,s_2)=(2,0,0,0). Each basis contains one zero basic coordinate, so both are degenerate representations of the same feasible vertex.

The result

Distinct invertible bases need not produce distinct basic feasible vectors. Both computed bases are feasible and degenerate.

Explore the model

Interactive lab

Interactive lab · matrix

Basis invertibility versus conditioning

Pause and Reset; set scale to 1 and shear to 0. Predict determinant one. For a fresh comparison Reset and set scale to 0.25 and shear to 2, then Play. The determinant stays positive, but the singular-value ratio grows: solving in a nearly aligned basis can amplify perturbations.

Basis invertibility versus conditioningThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn invertible basis permits a unique algebraic solve; good conditioning is an additional numerical question. This triangular two-dimensional family does not choose LP bases or test right-hand-side feasibility.

Check your understanding

A full-row-rank standard-form system has an invertible basis B and B⁻¹b = (2, −1). What follows?

Try it yourself

Write max⁡3z\max 3z subject to z≤2z\le2 with unrestricted zz in equality minimization standard form, and recover an optimal original solution.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Split the free variable into two nonnegative parts and introduce one slack; negate the objective.

Compare with the solution

Use z=u−vz=u-v, u,v≥0u,v\ge0, and u−v+s=2u-v+s=2, s≥0s\ge0. Minimize −3u+3v-3u+3v. The basis containing uu alone gives u=2,v=s=0u=2,v=s=0, hence z=2z=2 and original objective six. The inequality z≤2z\le2 bounds that objective above by six, proving optimality despite the nonunique split representation.

Take it with you

  • Standard-form equivalence needs a reconstruction map and objective bookkeeping.
  • A basis solve becomes feasible only after nonnegativity is checked.
  • Degenerate vertices can have multiple bases because some basic variables are zero.

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