# Multiple solutions for both primal and dual

If matrix $A$ in an LP (or $A^T$ in its dual) has full row (column- in dual) rank, is it possible that both primal and dual have multiple solutions?

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According to the table you have mentioned, there is no record for multiple implies multiple, every multiple solution will be expressed as degenerate in dual. Therefore, this is impossible that both have multiple solution.

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Thank you for your reply. The question now is if it is impossible to have multiple solutions for both primal and dual, why the first line of the table is not: --Multiple implies "Unique" and degenerate'-- (like the third line) – Quick Dec 15 '11 at 1:41
com, are you sure? If that's the case, then is the table confusing in omitting "unique" in the first line? – BCLC Dec 7 '14 at 22:25

In page 147 of (Sierksma, 2001), an example (Example 2) is given for this case. That is, "In case of (a), a dual optimal solution can either be multiple or non-multiple ...".

Primal Optimal Solution                     Dual Optimal Solution
(a) Multiple                     implies    Degenerate
(b) Unique and nondegenerate     implies    Unique and nondegenerate
(c) Multiple and nondegenerate   implies    Unique and degenerate
(d) Unique and degenerate        implies    Multiple


The example is as follows:

\begin{align} (P)&&&\min{}& &\quad{} 5x_2 \\ &&& \text{s.t.}& & - 2x_2 \geq{} 0 \\ && &&x_1& + 4x_2 \geq{} 3 \\ &&&& x_1&,\phantom{+4} x_2 \geq{} 0 \\ \end{align}

\begin{align} (D)&&&\max{}& &\quad{} 3y_2 \\ &&& \text{s.t.}& &\phantom{4}\quad{} y_2 \leq{} 0 \\ && &&-2y_1& + 4y_2 \leq{} 5 \\ &&&& y_1&,\phantom{+4} y_2 \geq{} 0 \\ \end{align}

The solution for $(P)$ is $(x_1, x_2)=(a, 0)$ with $a\geq{}3$ where the solution of $(D)$ is $(y_1, y_2) = (b, 0)$ with $b\geq{}0$. Therefore, yes, it is possible for both primal and dual to have multiple optimal solutions.

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