[3] First, we can rewrite the primal problem as follows
minimize z=cTx
subject to Ax ≥b1(1)
Ax ≤b2(2)
x≥0 (3)
Since we have ”2” constraints and ”1” variables in the primal, we have ”2”
variables and ”1” constraints in the dual. Then, let’s introduce a new variable
y= (y1, y2)Tin the dual. Then, the dual to this problem is
maximize z=bT
1y1+bT
2y2
subject to ATy1+ATy2≤c(4)
y1≥0 (5)
y2≤0 (6)
•For (4): since (3) x≥0, (4) is consistent with the canonical form of
maximization problem (Ax ≤b).
•For (5): since (1) is consistent with the canonical form of minimization
problem (Ax ≥b), (5) y1is nonnegative.
•For (6): since (2) is reversed from the canonical form of minimization
problem, (6) y2is nonpositive.
[4] Dual to this problem is
min z=y1+ 2y2
−y1+ 2y2≥ −1
y1−y2≥ −1
y1≤0
y2≥0
And, the optimal solution of primal is x= (0,1)T, and the optimal solution of
dual is y= (−1,0)T. Thus, cTx=−1” = ” −1 = bTy(strong duality holds).
[5.a.] The optimal solution in the dual is y∗= (7
2,1
2)T.
[5.b.] An optimal solution in the primal is x∗= (0,1
3,1
3)T: use the complemen-
tary slackness xT
∗(ATy∗−c) = 0 and yT
∗(Ax∗−b) = 0. (we did this in class)
[Note] In the discussion section, we solved ]13 in the book (p. 159). For (b),
we derived three equations for x∗= (x1, x2, x3)Tusing the above two comple-
mentary slackness conditions and strong duality theorem (optimal values are
equal). But, among these three equations, we have only ”two linearly inde-
pendent” equations, which means that we have ”two” equations for ”three”
1
