Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / T / 2 / c / iii

Changing only the second profit coefficient leaves the dual prices at $(1,1)$, because the basic products remain one and three. The nonbasic product has nonpositive reduced cost exactly when $c_2\le 3(1)+2(1)=5$. ThereforeThere is no mathematical lower bound; if profit coefficients are additionally restricted to be nonnegative, intersect this range with $[0,\infty )$. At $c_2=5$, alternate optimal solutions are possible.