Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / T / T2 / b / ii / Solution

The first two active constraints give $x_1+2x_2=4$, $3x_1+2x_2=6$, hence $x=(1,3/2{)}^T$. The remaining left side is $4+3/2=11/2<6$, so it is feasible, with objective $5/2$. The associated dual has $y\ge 0$ and equalities $y_1+3y_2+4y_3=1$, $2y_1+2y_2+y_3=1$. The vector $y=(1/4,1/4,0{)}^T$ is feasible and has value $4/4+6/4=5/2$. ThusEqual objectives certify both optima even with unrestricted primal variables. The positive first two dual entries force both active constraints at any primal optimum, so that optimum is unique. The positive slack in the third primal constraint forces $y_3=0$ at any dual optimum, and the two dual equalities then determine its unique value.