A two-resource allocation LP and its central path

Substituting $x_2=1-x_1$, stationarity of the barrier objective gives $-t-1\mathrm{/}{x}_1+1\mathrm{/}{x}_2=0$, or $1\mathrm{/}{x}_2-1\mathrm{/}{x}_1=t$. Multiplying by $x_1{x}_2>0$ and using $x_1+x_2=1$ yields $t{x}_1^2+(2-t)x_1-1=0$. Its roots are $(t-2\pm \sqrt{t^2+4})\mathrm{/}(2t)$; because $\sqrt{t^2+4}>\mathrm{\mid }t-2\mathrm{\mid }$, only the plus root is positive. Thus $x_1=(t-2+\sqrt{t^2+4})\mathrm{/}(2t)$, and rationalizing $1-x_1$ gives $x_2=2\mathrm{/}(t+2+\sqrt{t^2+4})>0$. These coordinates sum to one and satisfy stationarity. The restricted second derivative is $1\mathrm{/}{x}_1^2+1\mathrm{/}{x}_2^2>0$, so the stationary point is the unique exact central point. At $t_0=1$ these become the values in Given, whose reciprocal difference is $1$; hence the supplied start is exact and its decrement is zero.

The central inequality multipliers are ${\lambda }_i=1\mathrm{/}(t{x}_i)>0$. Their difference is ${\lambda }_2-{\lambda }_1=(1\mathrm{/}t)(1\mathrm{/}{x}_2-1\mathrm{/}{x}_1)=1$, so the equality multiplier $y=1-{\lambda }_1=2-{\lambda }_2$ satisfies both dual stationarity equations. Every primal feasible vector has objective $x_1+2x_2=1+x_2\ge 1$, while $(1,0)$ attains $1$, so $p^{\ast }=1$ and the objective excess is $x_2$. The primal-dual gap is $x_1{\lambda }_1+x_2{\lambda }_2=2\mathrm{/}t$; hence $x_2\le 2\mathrm{/}t$ by [F1]. By [F3], $\nu =2$. For $0<\epsilon <3$, put $K=\lceil \log (3\mathrm{/}\epsilon )\mathrm{/}\log (1+1\mathrm{/}(8\sqrt{2}))\rceil$, supply exact solves of the scalar restricted Newton equations, and choose $N_{out}\ge K$ and $N_{in}\ge 2$. Their scalar Hessian is positive by step 1.1. With the exact $t_0=1$ start from that step and $p^{\ast }=1$, all hypotheses of [F2] hold, so at most $K$ path updates suffice, with at most two corrections each.