1. Note that there was a misprint (now corrected) in the original version of this question: the 3 in the right-hand side of the third constraint should be a 2.
(a). We have , so the current basis is optimal as long as . Within this range, .
(b). Small changes in c1 don't affect the optimal , but . We have and . Note that y3 is URS because the third constraint is an equality, but we need and , so the range of values in which this is true is . If c1 was slightly less than 1, x4 would enter the basis. The x4 column . The only positive entry is in the x1 row, so x1 would leave the basis.
2.(a). The Dual Simplex Method is appropriate here because the
initial tableau is dual-feasible but not primal feasible.
(b). The initial tableau is
s2 leaves the basis. The ratios are 1/2 for x3 and 3/1 for x4. Having the minimum ratio, x3 enters the basis.
The basic solution of the primal is x1 = x2 = x4 = 0, x3 = 11/2, s1 = 17, s2 = 0, s3 = -21/2. The basic solution of the dual is y1 = 0, y2 = 1/2, y3 = 0, e1 = 5/2, e2 = 2, e3 = 0, e4 = 5/2, with (for both primal and dual) z = -11/2.
3. The value of the dual variable corresponding
to product III is
the reduced cost of the primal variable xIII.
It represents the increase
in the contribution to the profit of one unit of
product III that would be necessary in order to make
it profitable to produce product III; or, put another
way, the reduction in profit for every unit of product
III that is produced. It is the difference between
the sum of the prices of the resources consumed
(according to the shadow prices)
and the contribution to the profit for one unit of
this product.