Linear Programming
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
consider
...
Linear Programming
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
Expected return Worst-case return Duration
Bond 1 13% 6% 3
Bond 2 8% 8% 4
Bond 3 12% 10% 7
Bond 4 14% 9% 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)
1; 000; 000
= 5:4
(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 0:13x1 + 0:08x2 + 0:12x3 + 0:14x4
subject to 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 2
R2, illuminating n flat patches. The patches are line segments; the i-th patch is given by [vi; vi+1],
where v1; · · · ; vn+1 2 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:
Ii =
mXj
=1
aijpj; aij = rij −2 maxfcos θij; 0g;
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
k=1;··· ;n
jIk − Idesj (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 t
subject to −t ≤ aT k p − Ides ≤ t; k = 1; · · · ; n
0 ≤ p ≤ 1:
The variables are p 2 Rm and t 2 R.Problem 3 (0.4 points, Exer. 12 in Linear Programming Exercises):
We are given p matrices Ai 2 Rn×n, and we would like to find a single matrix X 2 Rn×n that we
can use as an approximate right-inverse for each matrix Ai, i.e., we would like to have
AiX ≈ I; i = 1; · · · ; p
We can do this by solving the following optimization problem with X as variable:
minimize max
i=1;··· ;p
kI − AiXk1: (2)
Here kHk1 is the infinity-norm or max-row-sum norm of a matrix H, defined as
kHk1 = max
i=1;··· ;m
nXj
=1
jHijj
if H 2 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 t
subject to −Sk ≤ I − AkX ≤ Sk; k = 1; · · · ; p
Sk1 ≤ t1; k = 1; · · · ; p:
The variables are X 2 Rn×n, p matrices Sk 2 Rn×n, and the scalar t. The inequalities in the first
constraints are componentwise inequalities between matrices, i.e., they mean
−(Sk)ij ≤ (I − AkX)ij ≤ (Sk)ij:
for i = 1; · · · ; n and for j = 1; · · · ; n. These pairs of inequalities are equivalent to
j(I − AkX)ijj ≤ (Sk)ij:
The second inequality is a componentwise inequality between vectors, i.e., it means
(Sk1)i =
nXj
=1
(Sk)ij ≤ t
for i = 1; · · · ; n
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