Agenda Duality (quickly) Piecewise linearity Start chapter 4

Duality y1:y1: y2:y2: min y y 1 ≥ 0 y 2 ≤ 0 7 y y 2 x 1 : 6 y y 2 ≥2 x 2 : -1 y 2 = 4 x 3 : 2 y 1 ≤ -1.5

Piecewise Linear Functions min x c 1 (x 1 ) + c 2 x 2 s.t.x 1 +x 2 >= d x >= 0 Idea: z = c 1 (x 1 ) min x,z z + c 2 x 2 s.t.x 1 +x 2 >= d x >= 0 z >= s 1 x 1 z >= s 2 x 1 + t c 1 (x 1 ) slope s 1 slope s 2 intercept t

Among the Constraints min x c T x s.t.Ax <= b f(x 1 )+3x 2 <= 7 x >= 0 f(x 1 ) piecewise linear introduce z replace f(x 1 )+3x 2 <= 7 by z+3x 2 <= 7 z >= s 1 x 1 z >= s 2 x 1 + t

Alternative Formulation Idea: x 1 =z 1 +z 2, c 1 (x 1 )=s 1 z 1 + s 2 z 2 min x,z s 1 z 1 + s 2 z 2 + c 2 x 2 s.t.x 1 +x 2 >= d x >= 0 x 1 =z 1 +z 2 z >= 0 z 1 <= 5 c 1 (x 1 ) 5 slope s 1 slope s 2

Nonconvex Case These approaches work if –min problem or ≤ constraint and –function convex: slopes increasing –opposite for max problem or ≥ constraint want function concave: slopes decreasing Otherwise: need binary variables…

f 1, f 2, f 3 binary 0/1 variables Idea: f i =1 if y in piece i f 1 +f 2 +f 3 =1 y=z 1 +z 2 +z 3 the contribution from the different pieces 0≤z 1 ≤4f 1, 0≤z 2 ≤3f 2, 0≤z 3 ≤3f 3 c(y) = z 1 *(slope1) +z 2 *(slope2)+2f 2 +z 3 *(slope3)+8f 3 Nonconvex Case (p204) c 1 (y)

Chapter 4: Assignment Problem