# LPDuality

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### Recall the linear program from LinearProgram

maximize x1+x2 subject to x1 ≥ 0, x2 ≥ 0, x1+2x2 ≤ 1, 3x1+x2 <= 1. Any feasible solution (e.g. x=(1/4, 1/4)) to the linear program provides a lower bound on OPT.

How can we get upper bounds? For any feasible solution x,

x1+x2
≤ x1+2x2 since x2 ≥ 0
≤ 1 since x1+2x2 ≤ 1 is a constraint.

#### Idea: try combinations of the constraints.

goal: bound x1+x2 for any feasible x.
know: x1+2x2 ≤ 1 and 3x1+x2 ≤ 1.

Linear combinations of these constraints give (all) other valid constraints:

a (x1+2x2) + b(3x1+x2) ≤ a*1 + b*1 (provided a,b ≥ 0)

Rewriting:

(a+3b) x1 + (2a+b) x2 ≤ a+b . (#1)

Want

x1+x2 ≤ (a+3b) x1 + (2a+b) x2 (#2)

for then (#1) above implies x1+x2 ≤ a+b.

Desired inequality (#2) will be true as long as 1 ≤ a+3b and 1 ≤ (2a+b),
in which case we know any feasible x satisfies x1+x2 ≤ a+b.

minimize a+b subject to
a+3*b ≥ 1
2*a+b ≥ 1
a,b ≥ 0

Exercise: determine optimal solution using pen and paper. (Geometric interpretation of feasible dual solutions as valid half-spaces.)

#### General form:

primal: minimize b.y subject to Ay ≥ c, y ≥ 0.

is dual of

dual: maximize c.x subject to ATx ≤ b, x ≥ 0

weak duality: cost of any feasible solution to the primal ≥ cost of any feasible solution to the dual

proof: b.y ≥ (ATx)y = xAy = x(Ay) ≥ x.c

strong duality: cost of optimal solution to the primal = cost of optimal solution to the dual

Notes:

• constraint in primal <--> variable in dual
• variable in primal <--> constraint in dual
• dual of dual is primal

Examples:

References:

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