SGT

1

Yes, because it is a linear programme.

2

${\max }_{x_1,x_2}{x}_1+x_2$ becomes $\min -x_1-x_2$ as we can only find dual for minimisation problem

$$\begin{array}{r}{x}_1+\frac{x_2}{2}-1\le 0\\ \frac{x_1}{3}+x_2-1\le 0\\ x_1\ge 0\\ x_2\ge 0\end{array}$$

[picture for formulas]

$L(x,v)=-x_1-x_2+u_1(x_1+\frac{x_2}{2}-1)+u_2(\frac{x_1}{3}+x_2-1)-u_3{x}_1-u_4{x}_2$

$$\begin{array}{rlrl}g(x_1,u)& =x_1+\frac{x_2}{2}-1& & (=0)\\ g(x_2,u)& =\frac{x_1}{3}+x_2-1& & (=0)\text{ set to zero to find optimal value}\end{array}$$

Dual function will be: $\max -u_1-u_2$ subject to

$$\begin{array}{rl}{x}_1+\frac{x_2}{2}-1& =0\\ \frac{x_1}{3}+x_2-1& =0\\ u_1,u_2,u_3,u_4& \ge 0\end{array}$$

3

strong optimality <= slater’s condition satisfied primal problem has a feasible point in interior of feasible set

4

min f(x) s.t. g(x) <= 0

kkt constraints: stationary contraints —> nabla_x L = nabla_f + lambda nabla g = 0 primal feasibility —> g(x) <= 0 dual feasibility —> lambda >= 0 complementary slackness —> lambda g(x) = 0

[pic; camera messed up tho]

5

graphically 1.4 is the maximum sum (this is correct)

u_3*=0 u_4*=0 u_1*=4/5 u_2*=2/5

x_1*=3/5 x_2*=4/5

x_1*+x_2*=1.4