Download Math 171A Homework Assignment #2: Linear Programming and Linear Systems and more Assignments Mathematics in PDF only on Docsity!
Math 171A Homework Assignment # 2
Instructor: Jiawang Nie
Assigned Date: January 13, 2010 Due Date: January 20, 2010
- (8 points) Consider the feasible set F defined by the following constraints
x 1 + x 2 ≥ 4 , x 1 + 3x 2 ≥ 6 , 6 x 1 − x 2 ≤ 18 , 3 ≤ x 2 ≤ 6 , x 1 ≥ − 1.
(a) Express F in the standard form Ax ≥ b. Write down A and b explicitly.
Solution:
A =
b =
(b) Draw the set F graphically in the plane, and find all the corner points of F.
Solution: Like the previous homework, we draw all of the lines and find which side of
the lines the feasible region is on. Then by finding which lines intersect we can solve for
the corners. The corners that lie in the feasible region should be
(c) Solve the following LP
minimize 2 x 1 − 3 x 2
subject to Ax ≥ b
where A and b are from part (a).
Solution: Since we know the corner points of the feasible region from part (b), we need to
plug those points into the objective function and compare. The point which minimizes
2 x 1 − 3 x 2 is the point (− 1 , 6) with a value of −20.
(d) Compute the residual vector r(x) for all the constraints at the point ¯x = (2, 4), and find
the constraints whose residuals would decrease after a positive step α along the direction
p = (1, −2) emanating from the point ¯x.
Solution: The residual vector is defined as r(x) = Ax − b (despite what some of you
coming to office hours told me it was defined as). If ¯x = (2, 4) then
r(x) = Ax − b =
We now wish to consider what happens to the residual if we plug in x = ¯x + αp where
α > 0.
r(x) = r(¯x + αp) = A(¯x + αp) − b = r(¯x) + αAp
With
Ap =
and since α > 0 this means the original residual r(¯x) will decrease in the first four
components if we move along the direction p.
- (8 points) Consider the linear system Ax = b where
A =
(^) , b =
(a) Determine the rank of A. Is the system Ax = b compatible?
Solution: We can see that the { 2 , 3 , 4 } submatrix has a non-zero determinant, and since
it is a 3×3 submatrix, the rank of A is 3. Since A has full row rank, Ax = b is compatible
for any b.
(b) List all the submatrices of A that have the same rank as A.
Solution: The only logical way to do this is to write a loop which tests all of the 3 × 3
submatrices to see if their determinant is non-zero. The following list of basic sets will
indicate which 3 × 3 submatrices are invertible.
Furthermore, if we take any 3 × 4 submatrix containing any of these 3 × 3 submatrices
we get a 3 × 4 submatrix with rank 3. This gives us 5 more submatrices with rank 3,
which are { 1 , 2 , 3 , 4 }
{ 1 , 2 , 3 , 5 }
{ 1 , 2 , 4 , 5 }
{ 1 , 3 , 4 , 5 }
{ 2 , 3 , 4 , 5 }
and finally, A is trivially a submatrix of itself with rank 3.
basic sets refer to rows, not columns. So use the basic set { 1 , 2 , 4 } which specifies which
rows to use to form our 3 × 3 invertible submatrix. This submatrix C is given by
C =
to find a basic solution we will find a solution to Cz = f where f is the { 1 , 2 , 4 } entries
of b. Thus,
f =
and the solution to Cz = f is
z =
This z is then a basic solution to Ax = b.
If one chooses another basic set, they will find the same basic solution that we just
found. This is consistent with the fact that compatible overdetermined systems have
unique solutions.
