 Solution

In the unnormalized embedding, takeThen the first equation is $Ae-b+\bar{b}=0$, the second is $c-e-\bar{c}=0$, and the third is $-c^{T}e-1+\bar{z}=0$. The fourth becomesThe multiplier elimination preserves this point because $q=c-e-\bar{c}=0$ and hence $Hq=0$. Every coordinate of $y$ is one. Dividing by $n^{\ast }=2n+4$ therefore giveswhich satisfies $A^{\ast }{y}^{(0)}=0$, $1^T{y}^{(0)}=1$, and strict positivity. Its existence is independent of feasibility of the original input problem.