Linear Programming in 2D

Linear Programming in 2D

1 Linear Programminggeometric problem formulationproblem specification

2 Incremental 2D Linear Programmingadding constraints one by onealgorithm for 2D bounded LP

MCS 481 Lecture 10Computational Geometry

Jan Verschelde, 8 February 2019

Computational Geometry (MCS 481) Linear Programming in 2D L-10 8 February 2019 1 / 21

linear programming in d variables

In standard form, a linear programming problem is given by1 an objective function in d variables, and2 a finite number n of contraints.

Maximize c1x1 + c2x2 + . . .+ cdxdsubject to

a1,1x1 + a1,2x2 + · · ·+ a1,dxd ≤ b1a2,1x1 + a2,2x2 + · · ·+ a2,dxd ≤ b2

...an,1x1 + an,2x2 + · · ·+ an,dxd ≤ bn

If we want only one solution to a problem, then writing the problemas a linear programming problem can get that one solution faster.


geometric interpretation

Maximize 2x1 + x2 subject to −x2 ≤ 0, x2 ≤ 2,−x1 − x2 ≤ −1, −x1 + x2 ≤ 1, x1 − x2 ≤ 2, x1 + x2 ≤ 4.

(−1,−1)

(−1,+1)

(0,−1)

(0,+1)

(+1,+1)

(+1,−1)

1 x1

x2

1

(+2,+1)

P

Points on the same red line have the same value of 2x1 + x2.


geometric interpretation of half planes

Maximize 2x1 + x2 subject to −x2 ≤ 0, x2 ≤ 2,−x1 − x2 ≤ −1, −x1 + x2 ≤ 1, x1 − x2 ≤ 2, x1 + x2 ≤ 4.

(−1,−1)

(−1,+1)

(0,−1)

(0,+1)

(+1,+1)

(+1,−1)

1 x1

x2

1

(+2,+1)

P

For all half planes a1x1 + a2x2 ≤ b, with b = max(x1,x2)∈P

a1x1 + a2x2.


problem specification

On input are1 the coefficients (c1, c2, . . . , cd) of the objective function, and2 a tuple H of n half planes H = (h1,h2, . . . ,hn), given by

I outward pointing normal vectors (ai,1,ai,2, . . . ,ai,d ), andI corresponding bounds bi , for i = 1,2, . . . ,n.

Let P =n⋂

i=1

hi , the polyhedron defined by all half planes.

The output is the point (x1, x2, . . . , xn) ∈ P at whichthe objective function c1x1 + c2x2 + · · ·+ cdxd is maximized.

Questions:1 Does this problem always have a solution?2 Is the solution unique?


four cases

Answering the questions1 Does this problem always have a solution?2 Is the solution unique?

leads to the following case analysis:1 The solution is unique.2 The problem is infeasible if P = ∅.3 The problem is unbounded if the optimal value is∞.4 The solution is not unique if the coefficient vector of the objective

is parallel to one of the outward pointing normal vectors of P.

Exercise 1: If the coefficient vector of the objective is parallel to one ofthe outer normals of P, then the solution is not unique. Does theopposite direction hold as well? Can you prove the if and only if?Otherwise make a drawing why such a proof is not possible.


making problems bounded with unique solution

To make the problem bounded, we can impose restrictions as follows.

Let M be some large enough number.

For a given objective (c1, c2, . . . , cd), add the constraints:

mi =

{xi ≤ M, if ci > 0,−xi ≤ M, if ci ≤ 0,

for i = 1,2, . . . ,d .

Then m1 ∩m2 ∩ · · · ∩md is an orthogonal wedge.

In case the solution is not unique,then we choose the lexicographically smallest solution as the solution.

Any feasible program has thus a unique bounded solution,which is a vertex of P, called the optimal vertex.


adding constraints one by one

Let H = (h1,h2, . . . ,hn) be the tuple of half planes,with the added constraints m1 and m2.

Consider Pi as the region defined by1 m1 and m2, the added constraints, and2 the first i half planes of the tuple H.

Formally, for i = 0,1, . . . ,n:

Pi = m1 ∩m2 ∩ h1 ∩ h2 ∩ · · · ∩ hi .

The regions are getting smaller as we add more constraints:

P0 ⊇ P1 ⊇ P2 ⊇ · · · ⊇ Pn = P.


updating the optimal vertex

For i = 0,1, . . . ,n: the region Pi =i⋂

j=1

hi ∩m1 ∩m2,

has the optimal vertex vi with coordinates (xi,1, xi,2):c1xi,1 + c2xi,2 = max

(x1,x2)∈Pi

c1x1 + c2x2.

Lemma (update the optimal vertex)Consider adding hi to Pi−1 with current optimal vertex vi−1.

1 If vi−1 ∈ hi , then vi = vi−1.2 If vi−1 6∈ hi , then either Pi = ∅ or vi ∈ `, the line bounding hi .


two cases illustrated

We start with four half planes and the optimal vertex v4.1 Adding h5 does not change the optimal vertex, v5 = v4.2 Adding h6 changes the optimal vertex, v6 6= v5.

v4

(c1, c2)

P4

h5

v5

(c1, c2)

P5

h5

h6

v6

(c1, c2)

P6

Observe that v6 lies on the line of the half plane h6.


proof of the first case

1 If vi−1 ∈ hi , then vi = vi−1.

v4

(c1, c2)

P4

h5

v5

(c1, c2)

P5

If vi−1 ∈ hi , then vi−1 is the optimal vertex over Pi , as Pi ⊆ Pi−1,the maximum cannot increase over a smaller set, so vi = vi−1.


proof of the second case

2 If vi−1 6∈ hi , then either Pi = ∅ or vi ∈ `, the line bounding hi .

h5

v5

(c1, c2)

P5

h5

h6

v6

(c1, c2)

P6

If Pi = ∅, then we are done. Assume Pi 6= ∅.

Assume v6 6∈ `, we will derive a contradiction.


looking for a contradiction

Assume v6 6∈ `, we will derive a contradiction.Consider the line segment (vi−1, vi).

v5

h5

h6

v6

P6

(vi−1, vi) ∈ Pi−1, as Pi ⊆ Pi−1

As the objective is linear,moving from vi to vi−1 increasesthe objective monotonically.

Consider q ∈ (vi−1, vi) ∩ `.q exists because vi−1 6∈ hi

and vi ∈ Pi .As the objective is monotoneincreasing from vi to vi−1,its value at q is higher than at vi .


linear programming in one dimensionIf vi−1 6∈ hi and Pi 6= ∅, then vi ∈ `, the line bounding hi .How to compute vi?

The line ` has equation ai,1x1 + ai,2x2 = bi ,we eliminate x2: x2 = (bi − ai,1x1)/ai,2.

This gives rise to a one dimensional problem:

Maximize c1x1 + c2(bi − ai,1x1)/ai,2subject to

aj,1x1 + aj,2(bi − ai,1x1)/ai,2 ≤ bjj = 1,2, . . . , i − 1

Depending on the coefficient of x1, every inequality gives either anupper or a lower bound, so we left with A ≤ x1 ≤ B.

Lemma (cost of one dimensional linear programming)Solving an LP problem in 1 variable with i inequalities takes O(i) time.


algorithm for 2D bounded LPAlgorithm 2DBOUNDEDLP(H, c)

Input: H = (h1,h2, . . . ,hn), c = (c1, c2).Output: (x , y) optimal vertex or i if Pi = ∅.

1 set M constraints, v0 is corner of orthogonal wedge2 for i from 1 to n do3 if vi−1 ∈ hi then4 vi = vi−1

5 else if 1D LP is infeasible then6 return i

else7 vi = solution of 1D LP8 return vn

Exercise 2: Show the correctness of 2DBOUNDEDLP.


cost of incremental linear programmingTheorem (cost of incremental linear programming)

Algorithm 2DBOUNDEDLP takes O(n2) operations for n half planes.

The proof usesn∑

i=1

O(i) is O(n2).

h7 h6 h5 h4 h3

h2

h1v2

v3

v4v5

v6

v7

c


recommended assignments

We covered section 4.3 in the textbook.

Consider the following activities, listed below.

1 Write the solutions to exercises 1 and 2.2 Consider the exercises 10,11,12 in the textbook.


Linear Programming in 2D

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