EE236A Linear Programming Midterm

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Winter 2018 EE 236A Prof. Christina Fragouli TA M. Karmoose

EE236A Linear Programming Midterm Tuesday Feb. 13, 2018 Time: 12:00 - 1:45 NAME:

UID:

This exam has 3 questions, for a total of 24 points.

Closed book. No calculators. No electronic devices.

One page, letter-size, two-sided cheat-sheet allowed.

Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page.

Please, write your name and UID on the top of each loose sheet!

Good luck!

Problem

Mark

Total

P1

7

P2

8

P3

9

1

Problem 1 (7 points): Can you express the following problems as LPs? (If yes do so, if not, explain why). Note that we do not ask you to solve these problems, only to express them as LPs if possible.

1. (2 points) Let x ∈ Rn . min subject to

cT x ||x||1 ≤ 100

2. (2 points) For x1 , x2 ∈ R −2x1 − 3x2 min{x1 , x2 } ≤ 4 x1 , x2 ≥ 0

min subject to

3. (3 points) Assume that we want to design a game, where the output x takes one of n possible values ai , with probability prob(x = ai ) = pi , i = 1 . . . , n. We also want that the probability we get the output a1 is higher than the probability we get any other output. Moreover, we n P want to minimize the expected value of the output, E(x) = pi ai . Can this optimization i=1

be expressed as an LP?

2

Solution: 1. Yes, through the following LP min subject to

cT x 1T y ≤ 100 −y ≤ x ≤ y

2. No, because the constraint min{x1 , x2 } ≤ 4 does not specify a convex set.

3. Yes, through the following LP min

n P

p i ai

i=1

subject to

pi ≥ 0, i = 1, · · · , n n P pi = 1, i=1

p1 ≥ pi , i = 2, · · · , n

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Problem 2 (8 points): Assume you are given a set P that contains d distinct points pi in Rn , i = 1 . . . d.

We want to partition Rn into d regions, so that region Vi contains the point pi and all other points in Rn that are closer in Euclidean distance to pi than any other point in P. In other words, we wish to find the following regions (for i = 1, · · · , d): Vi = {x ∈ Rn | kx − pi k2 ≤ kx − pj k2 , for all j = 1, · · · , d, such that j 6= i}

(1)

This is called the Voronoi region of the point pi . The Voronoi diagram is the partition of Rn into Voronoi regions. In what follows, assume that d > n.

1. (2 points) Prove that the Voronoi regions are polyhedra. The vertices of these polyhedra are called Voronoi vertices. How many Voronoi vertices can these polyhedra have?

2. (3 points) Show that every Voronoi vertex is the center of a circle that goes through at least n + 1 of the points in P. 3. (3 points) Assume that one of the d points pi was lost (e.g. due to storage errors). Let the lost point be pd . This point pd belonged in a closed Voronoi region Vd that has q > n vertices (by closed we mean that the region does not contain infinity). The information that you have retained about the Voronoi regions are: 1) the other d − 1 points (pi , i = 1, · · · , d − 1), and 2) the set of the q Voronoi vertices for Vd . Let x̂1 , · · · , x̂q be the q Voronoi vertices of Vd .

Describe a way by which you can recover the missing point pd given the information that you have.

4

Solution: 1. Voronoi regions are indeed polyhedra. In fact, for a given Voronoi region i and a given point j 6= i, we can express the corresponding condition in the definition of Vi as kx− pi k22 ≤ kx− pj k22 (x− pi )T (x− pi ) ≤ (x− pj )2 (x− pj ) xT x + kpi k22 − 2pTi x ≤ xT x + kpj k22 − 2pTj x 2(pj − pi )T x ≤ kpj k22 − kpi k22 Therefore, Vi can be expressed as the intersection of d − 1 halfspaces, thus it is a polyhedron.

A vertex has to satisfy at least n inequalities with equalities. Since there are d − 1 distinct

Voronoi vertices. points, there can be at most d−1 n 2. A Voronoi vertex is a vertex in the polyhedron describing the Voronoi region. Therefore, it satisfies n inequalities of the polyhedron with equalities. Consider the Voronoi region V1 , and let x̂ be a Voronoi vertex of V1 . Then it satisfies kx̂ − p1 k2 = kx̂ − pj k2 ,

for j ∈ S

where S is a set of n indices, each of which are different than 1; without loss of generality, let S = {2, 3, · · · , n + 1}. Then x̂ is at equal distance from p1 , p2 , · · · , pn+1 , and therefore there is a circle centered at x̂ and passes by all these n + 1 points.

3. For every Voronoi vertex x̂j , j = 1, · · · , q, there are at least n+1 points from the set p1 , · · · , pd that are of equal distance to x̂j , and one of which is pd . Let this set of points be S(x̂k ). Then by definition, pd ∈ S(x̂j ). If all the sets S(x̂k ), k = 1, · · · , q are known, then one can find the missing point pd by solving the following set of equations kpd − x̂j k2 = kpk − x̂i k2 ,

for all j = 1, · · · , q, and for one k ∈ S(x̂j ).

These are q > n equations in n unknowns, and therefore can be used to solve for the point pd .

The missing part is to find the sets S(x̂j ), which can be accomplished as follows. For each x̂j , compute the distance between x̂j and all points p1 , · · · , pd−1 . Then, the points in S(x̂j ) will correspond to the set of points in p1 , · · · , pd−1 which has the smallest distance to x̂j . We prove that this is true by contradiction. That is, assume that there exists a point pk ∈ / S(x̂j ) where kpk − x̂j k < kpd − x̂j k. Then this means that x̂j does not lie in Vd and therefore cannot be a Voronoi vertex of Vd , which is a contradiction.

5

Problem 3 (9 points): Consider the following LP, with variables x ∈ Rn and z ∈ Rn (assume that c 6= 0), cT x −z ≤ x ≤ z 1T z = 2 z ≤ 1.

max subject to

(2)

(a) (2 points) Write the dual of this LP. (b) (4 points) Provide an optimal solution to the primal LP in (2), and argue why the solution you provided is optimal. Hint: you do not need to use the KKT conditions. (c) (1 point) Give an example of a vector c for which the optimal solution of (2) is not unique. (d) (2 points) Argue whether the optimal dual solution of the LP in (2) is unique or not.

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Solution: (a) The LP can equivalently be written in the following form (not that it is a minimization problem now): " # i x −cT 0T z     I −I " # 0   x   ≤ 0 , −I −I  z 0 I " # 1 h i x =2 0T 1T z

h

min

subject to

By using the standard dual form, we get that the dual problem is

max

subject to

  1 i λ   0T 0T −1T λ2  − 2 · y λ3   " # λ1 " # " # I −I 0  2  0 c y− = 0, λ  + −I −I I 1 0 3 λ 1 2 λ ≥ 0, λ ≥ 0, λ3 ≥ 0.

h

(b) We note first that the problem is only feasible if n ≥ 2 (otherwise the constraint 1T x = 2 can never be satisfied). The optimal solution can be constructed as follows. Assume that the elements of the vector c are ordered as follows: |c1 | ≥ |c2 | ≥ · · · ≥ |cn |. Then the optimal solution would be ( ( sign(c ) · 1 if i ≤ 2 1 if i ≤ 2 i x?i = , zi? = (3) 0 otherwise 0 otherwise To prove that the solution in (3) is optimal, we will use the following claim: any vertex in the polyhedron described in (2) will be of the following structure: ( sign(ci ) · 1 xi = 0

for i ∈ Z , zi = otherwise

( 1 0

for i ∈ Z otherwise

(4)

where Z ⊆ {1, · · · , n} is a subset of indices of size |Z| = 2. If this is true, then it is easy to see that any vertex (except for the one in (3)) will at most give you an objective value equal to the one corresponding to (3). Specifically, the objective function corresponding to a point as in (4) will be 2 X X cT x = |ci | ≤ |ci | = cT x? . i=1

i∈Z

7

Note that a point as in (4) has a number of active constraints equal to

2 |{z}

z variables equal to 1

+ |{z} 1 + 1T z=2

2 |{z}

2(n − 2) | {z }

+

x variables equal to ±1

= 2n + 1

n−2 variables of x equal to 0

therefore it is indeed a vertex (in fact, a degenerate one). What remains is to prove that no other vertex can exists (except for those of the form (4)). Consider a point for which there are v < 2 elements of z not equal to 1. In that case, there will be q ≥ v + 1 elements of z that are non-zero, out of which v elements are equal to 1. The sum of those q elements must add to 2.

For such a point, the total number of active constraints can at most be

v |{z}

z variables equal to 1

+ |{z} 1 + 1T z=2

q |{z}

x variables equal to ±z

+

2(n − q) | {z }

n−2 variables of x equal to 0

= 2n − q + v + 1.

There are two cases for v: 1) v = 0: This means that no elements of z are exactly equal to 1. Therefore there are at least q = 3 elements of z that are non-zero (if there are 2 only, then they have to be equal to 1, and therefore v would be equal to 2). Therefore, the maximum number of active constraints for such a point is 2n − q + v + 1 ≤ 2n − 3 + 0 + 1 = 2n − 2. Therefore, such a point cannot be a vertex.

2) v = 1: This means that there is exactly one element of z that is equal to 1. Therefore, there has to be at least two other elements of z that are non-zero (if there is only one, that it has to be equal to 1, and therefore v would be equal to 2). Therefore, q is at least equal to 3.

In this case, the maximum number of active constraints for such a point is 2n − q + v + 1 ≤ 2n − 3 + 1 + 1 = 2n − 1. Therefore, such a point cannot be a vertex. (c) If |c1 | > · · · > |cn | (i.e., all with strict inequality), then the optimal solution is unique. If, for example, |c` | = |c`+1 |, then the following points are all optimal solutions:     sign(c ) · 1 if i ≤ l − 1 1 if i ≤ ` − 1   i      sign(c ) · δ  if i = ` δ if i = ` i xi = , zi =   sign(ci ) · 1 − δ if i = ` + 1 1 − δ if i = ` + 1         0 otherwise 0 otherwise for any value of 0 ≤ δ ≤ 1. (d) Since the optimal solution in (4) is degenerate, then there exists multiple solutions to the dual program.

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