linear programming Flashcards
linear programming intro:
convex optimisation problem to minimise f(x) subject to g1<=0 … gm(x)<=0, x in Ω where Ω is convex and all functions are convex, linear programming means all functions are linear and Ω=Rn
convex hull:
convS of a set S, the intersection of all convex sets containing S, so s=convS if S is convex
convex combination:
linear combo (k)Σ(i=1)λixi such that λi>=0 and (k)Σ(i=1)λi=1
convex hulls and combos:
convex sets are closed under convex combos, the set of all convex combos of points in a set S is the convex hull of S
hyperplane:
H={x in Rn: ⟨a,x⟩=b=a^(T)x}, so a solution set, it’s convex
halfspace:
a hyperplane divides Rn into two sides, which are these
H-={x in Rn: ⟨a,x⟩<=b}, H+={x in Rn: ⟨a,x⟩>=b}, these are convex also
2d version of hyperplane separation theorem:
C is a nonempty closed convex set, x Not in C, then there exists a point y* in C such that ||x-y||=min||x-y||, equivalent to saying y=argmin||x-y||, minimising the distance between them
also for all z in C, ⟨z-y,x-y⟩<=0 (so z-y* and x-y* form an obtuse angle) - look in the notes for visual for this but essentially x-y* is perpendicular to the edge of C and z is in C so ofc the angle is obtuse it has to be at least 90 degrees
continuous function on a closed bounded set:
must attain a max and a min in that set
hyperplane separation theorem:
let C be a closed convex set, x Not in C, then there exists a hyperplane H such that C within intH- and x in intH+
essentially you can draw a line between the edge of C and x
linear programming:
maximise ⟨c,x⟩ subject to Ax<=b, A in R^(mxn), x,c in Rn, b in Rm, inequality means each inequality in each row
feasible set:
F={x in Rn: Ax<=b}, set of all possible candidates, always convex but can be empty
polyhedron:
the set of all feasible sets
how to check if a lin programming problem has a solution:
either find a vector from F or show that assuming Ax<=b has a solution makes a contradiction so F must be empty, ⟨λ,Ax⟩<=⟨λ,b⟩ find λ such that (m)Σ(i=1)λiai=0 and ⟨λ,b⟩<0, if this answer makes no sense rewrite this with more detail I just don’t know if we need it all, if it makes sense remove this disclaimer lmao