Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / T / T2 / b / Solution

The dual prices $u,v\ge 0$ solveThe optimal tableau gives $(u,v)=c_B^T{B}^{-1}=(6,3)B^{-1}=(3/4,3/2)$. This vector is dual feasible and has objective $21$, equal to the primal value, proving optimality.

Thus an extra unit of R is worth $3/4$ while the current basis stays feasible. Increasing its supply by $\delta$ changes the basic values toFor increases, feasibility is retained exactly up to $\delta =4/3$. Therefore the firm can buy up to $4/3$ extra units at a marginal value of $3/4$ per unit. The reduced profits do not change during this right-hand-side variation.

For the proposed fourth good, the reduced profit is $8-(4u+4v)=8-9=-1$. Its profit is below the resource opportunity cost, so it is not worth producing at the current optimum. The old dual solution also satisfies the new good's inequality, certifying that introducing it cannot improve the optimum.