EE 236A Linear Programming Solutions homework 2

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Fall 2015 EE 236A Christina Fragouli

Linear Programming Solutions of Homework 2

Problem 1 (0.4 points): Linear programming models are used by many Wall Street firms to select a desirable bond portfolio. The following is a simplified version of such a model. Solodrex is considering investing in four bonds; $1,000,000 is available for investment. The expected annual return, the worst case annual return, and the “duration” of each bond are given in Table 1.

Table 1: Return and duration of bonds Bond 1 Bond 2 Bond 3 Bond 4

Expected return

Worst-case return

Duration

13% 8% 12% 14%

6% 8% 10% 9%

3 4 7 9

The duration of a bond is a measure of the bond’s sensitivity to interest rates. Solodrex wants to maximize the expected return from its bond investments, subject to three constraints. (a) The worst-case return of the bond portfolio must be at least 8%. (b) The average duration of the portfolio must be at least 6. For example, a portfolio that invested $600,000 in bond 1 and $400,000 in bond 4 would have an average duration of 600, 000(3) + 400, 000(9) = 5.4 1, 000, 000 (c) Because of diversification requirements, at most 40% of the total amount can be invested in a single bond.

Formulate an LP that will enable Solodrex to maximize the expected return on its investments.

Solution: Let xi be the amount of money invested in Bond i, i = 1, 2, 3, 4. Then the LP problem can be formulated as maximize subject to

0.13x1 + 0.08x2 + 0.12x3 + 0.14x4 0.06x1 + 0.08x2 + 0.1x3 + 0.09x4 ≥ 0.08(x1 + x2 + x3 + x4 ) 3x1 + 4x2 + 7x3 + 9x4 ≥ 6(x1 + x2 + x3 + x4 ) 0 ≤ xi ≤ 400, 000, i = 1, 2, 3, 4.

Problem 2 (0.4 points, Exer. 14 (a) in Linear Programming Exercises): An illumination problem. We consider an illumination system of m lamps, at positions l1 , · · · , lm ∈ R2 , illuminating n flat patches. The patches are line segments; the i-th patch is given by [vi , vi+1 ], where v1 , · · · , vn+1 ∈ R2 . The variables in the problem are the lamp powers p1 , · · · , pm , which can vary between 0 and 1.

The illumination at (the midpoint of) patch i is denoted by Ii . We will use a simple model for the illumination: m X −2 Ii = aij pj , aij = rij max{cos θij , 0}, j=1

where rij denotes the distance between lamp j and the midpoint of patch i, and θij denotes the angle between the upward normal of patch i and the vector from the midpoint of patch i to lamp j, as shown in the figure. This model takes into account “self-shading” (i.e., the fact that a patch is illuminated only by lamps in the halfspace it faces) but not shading of one patch caused by another.

Of course we could use a more complex illumination model, including shading and even reflections.

This just changes the matrix relating the lamp powers to the patch illumination levels.

The problem is to determine lamp powers that make the illumination levels close to a given desired illumination level Ides , subject to the power limits 0 ≤ pi ≤ 1.

Suppose we use the maximum deviation φ(p) = max |Ik − Ides | k=1,··· ,n

(1)

as a measure for the deviation from the desired illumination level. Formulate the illumination problem using this criterion as a linear programming problem.

Solution: We can formulate the problem as the following LP: minimize subject to

The variables are p ∈ Rm and t ∈ R.

t −t ≤ aTk p − Ides ≤ t, k = 1, · · · , n 0 ≤ p ≤ 1.

Problem 3 (0.4 points, Exer. 12 in Linear Programming Exercises): We are given p matrices Ai ∈ Rn×n , and we would like to find a single matrix X ∈ Rn×n that we can use as an approximate right-inverse for each matrix Ai , i.e., we would like to have Ai X ≈ I, i = 1, · · · , p We can do this by solving the following optimization problem with X as variable:

minimize max kI − Ai Xk∞ . i=1,··· ,p

(2)

Here kHk∞ is the infinity-norm or max-row-sum norm of a matrix H, defined as

kHk∞ = max

i=1,··· ,m

n X

|Hij |

j=1

if H ∈ Rm×n . Express problem (2) as an LP. You dont have to reduce the LP to a canonical form, as long as you are clear about what the variables are, what the meaning is of any auxiliary variables that you introduce, and why the LP is equivalent to the problem (2).

Solution: We can express the problem as the LP minimize subject to

t −Sk ≤ I − Ak X ≤ Sk , k = 1, · · · , p Sk 1 ≤ t1, k = 1, · · · , p.

The variables are X ∈ Rn×n , p matrices Sk ∈ Rn×n , and the scalar t. The inequalities in the first constraints are componentwise inequalities between matrices, i.e., they mean −(Sk )ij ≤ (I − Ak X)ij ≤ (Sk )ij . for i = 1, · · · , n and for j = 1, · · · , n. These pairs of inequalities are equivalent to |(I − Ak X)ij | ≤ (Sk )ij .

The second inequality is a componentwise inequality between vectors, i.e., it means

(Sk 1)i =

n X (Sk )ij ≤ t j=1

for i = 1, · · · , n.

Problem 4 (0.6 points, Exer. 31 in Linear Programming Exercises): Explain how you would solve the following problem using linear programming. You are given two sets of points in Rn : S1 = {x1 , · · · , xN },

S2 = {y1 , · · · , yM }.

You are asked to find a polyhedron P = {x|aTi x ≤ bi , i = 1, · · · , m} that contains the points in S1 in its interior, and does not contain any of the points in S2 : S1 ⊆ {x|aTi x < bi , i = 1, · · · , m},

S2 ⊆ {aTi x > bi , for at least one i} = Rn \P.

An example is shown in the figure, with the points in S1 shown as open circles and the points in S2 as filled circles.

•

•

•

0

\

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•

.__...

.

•

.

You can assume that the two sets are separable in the way described. Your solution method should return ai and bi , i = 1, · · · , m, given the sets S1 and S2 . The number of inequalities m is not specified, but it should not exceed M + N . You are allowed to solve one or more LPs or LP feasibility problems. The method should be efficient, i.e., the dimensions of the LPs you solve should not be exponential as a function of N and M .

Solution: Start with y = y1 and compute a1 , b1 such that aT1 y − b1 ≥ 1,

aT1 x − b1 ≤ −1,

x ∈ S1 .

Then take a point y ∈ S2 for which aT1 y − b1 ≤ 0, and find a2 , b2 such that aT2 y − b2 ≥ 1,

aT2 x − b2 ≤ −1,

x ∈ S1 .

Then take a point y ∈ S2 for which aT1 y − b1 ≤ 0 and aT2 y − b2 ≤ 0, and find a3 , b3 such that aT3 y − b3 ≥ 1,

aT3 x − b3 ≤ −1,

x ∈ S1 .

We continue until we have a polyhedron with all points of S2 in its exterior. At most M inequalities are needed.

Problem 5 (0.4 points, Exer. 33 (b)(e) in Linear Programming Exercises): Which of the following sets S are polyhedra? If possible, express S in inequality form, i.e., give matrices A and b such that S = {x|Ax ≤ b}.

Pn Pn 2 (a) S = {x ∈ Rn | x ≥ 0, 1T x = 1, i=1 xi ai = b2 }, where ai ∈ R, i = 1, · · · , n, i=1 xi ai = b1 , b1 ∈ R, and b2 ∈ R are given. (b) S = {x ∈ Rn | ||x − x0 || ≤ ||x − x1 ||}, where x0 , x1 ∈ Rn are given. S is the set of points that are closer to x0 than to x1 .

Solution: (a) S is a polyhedron. In fact the definition involves linear inequalities x > 0 and three equality constraints, so we only have to write the equality constraints as two inequalities. This yields a set of n + 6 inequalities: −xi 1T x −1T x P ai xi Pi − i ai xi P 2 a xi Pi i2 − i ai xi

≤ ≤ ≤ ≤ ≤ ≤ ≤

0, i = 1, · · · , n 1 −1 b1 −b1 b2 −b2 .

(b) This set is a polyhedron (in fact, a halfspace). By squaring both sides of the inequality ||x−x0 || ≤ ||x − x1 ||, we obtain the equivalent condition ||x − x0 ||2 ≤ ||x − x1 ||2 ⇔ −2xT0 x + xT0 x0 ≤ −2xT1 x + xT1 x1 ⇔ −2(x0 − x1 )T x ≤ ||x1 ||2 − ||x0 ||2 , which is a linear inequality.

Problem 6 (0.3 points, Exer. 35 (a) in Linear Programming Exercises): Is x̃ = (1, 1, 1, 1) an extreme point of the polyhedron P defined by the linear inequalities      −3 −1 −6 1 3      x1   5   −1 −2 7 1      x2        0 3 −10 −1     x  ≤  −8 ?   3     −7   −6 −11 −2 12  x4 1 6 −1 −3 4 If it is, find a vector c such that x̃ is the unique minimizer of cT x over P.

Solution: To show that x̃ is an extreme point, we apply the rank criterion of page 3-23. The set of active constraints at x̃ is J = {1, 2, 3, 4}, so we have to check the rank of the matrix   −1 −6 1 3    −1 −2  7 1 .

AJ =   0 3 −10 −1    −6 −11 −2 12 The rank of AJ is 4; therefore x̃ is an extreme point.

As vector c we can choose any (strictly) negative combination of the rows of AJ , for example,      8 −1 −1 0 −6 −1        16   −6 −2  3 −11  =   −1  . c = −ATJ 1 =       4   1 7 −10 −2   −1    −15 3 1 −1 12 −1 With this choice of c, we have cT x̃ = −1T AJ x̃ = −

X

bi = −13.

i∈J

To show that no other feasible x can have the same or a lower value of cT x, note that if Ax ≤ b, then X cT x = −1T AJ x ≥ − bi = −13, i∈J

with equality only if aTi x = bi for i = 1, 2, 3, 4. However, this is a set of four equations in four variables with a nonsingular coefficient matrix AJ . Therefore x = x̃.

Problem 7 (0.5 points, Exer. 16 in Linear Programming Exercises): Formulate the following problem as an LP:

maximize

n X

rj (xj )

j=1

subject to

n X

Aij xj ≤ cmax , i = 1, · · · , m j

(3)

j=1

xj ≥ 0, j = 1, · · · , n.

The functions rj are defined as ( rj (u) =

pj u 0 ≤ u ≤ qj disc pj qj + pj (u − qj ) u ≥ qj ,

(4)

where pj > 0, qj > 0 and 0 < pdisc < pj . The variables in the problem are xj , j = 1, · · · , n. The j parameters Aij , cmax , pj , qj and pdisc are given. i j

The variables xj in the problem represent activity levels (for example, production levels for different products manufactured by a company). These activities consume m resources, which are limited.

Activity j consumes Aij xj of resource i. (Ordinarily we have Aij 0, i.e., activity j consumes resource i. But we allow the possibility that Aij < 0, which means that activity j actually generates resource i as a by-product). The total resource consumption is additive, so the total of resource i consumed P is ci = nj=1 Aij xj . Each resource consumption is limited: we must have ci ≤ cmax , where cmax are i i given.

Activity j generates revenue rj (xj ), given by the expression (4). In this definition pj > 0 is the basic price, qj > 0 is the quantity discount level, and pdisc is the quantity discount price, for (the j disc product of) activity j. We have 0 < pj < pj . The total revenue is the sum of the revenues P associated with each activity, i.e., nj=1 rj (xj ). The goal in (3) is to choose activity levels that maximize the total revenue while respecting the resource limits.

Solution: We note that

rj (u) = min{pj u, pj qj + pdisc j (u − qj )}, so the problem is equivalent to the LP maximize

X

tj

j

subject to tj ≤ pj xj , j = 1, · · · , n tj ≤ pj xj + pdisc j (xj − qj ), j = 1, · · · , n n X

Aij xj ≤ cmax , i = 1, · · · , m i

j=1

xj ≥ 0, j = 1, · · · , n. with variables x ∈ Rn and t ∈ Rn .