x+2y 10 2x + 3y 18 f = 5x +4y Solution Let z1 and z2 be the dual values... hence initial table... x y z1 z2 RHS -1 -2 1 0 -10 -2 -3 0 1 -18 -5 -4 0 0 0 first x enters.... and z2 leaves... x y z1 z2 RHS 0 -1/2 1 -1/2 1 1 3/2 0 - 1/2 9 0 7/2 0 -5/2 45 This is the optimal solution as there is no candidate left which can enter and can further increase the value of the function... Hence the optimal value of f is.. 45 corresponding to x=9 and y=0......