Ralph Vince - Portfolio Mathematics

Classical Portfolio Construction 255

Now suppose we decided to input a value of E=.18. Again, we begin
with the augmented matrix, which is exactly the same as in the last example
of E=.14, only the upper rightmost cell, that is the first cell in the right-
hand-side vector, is changed to reflect this new E of .18:

Starting Augmented Matrix

X 1 X 2 X 3 X 4 L 1 L 2 | Answer

0.095 0.13 0.21 0.085 0 0 | 0.18

1 11100| 1

0.1 −0.023 0.01 0 0.095 1 | 0

−0.023 0.25 0.079 0 0.13 1 | 0

0.01 0.079 0.4 0 0.21 1 | 0

0 0 0 0 0.085 1 | 0

Through the use of row operations...the identity matrix is obtained:

100000 | 0.21401=X 1

010000 | 0.22106=X 2

001000 | 0.66334=X 3

000100 |−.0981=X 4

000010 |−1.3197/.5=−2.639=L 1

000001 | 0.11217/.5=.22434=L 2

We then go about solving the matrix exactly as before, only this time we
get a negative answer in the fourth cell down of the right-hand-side vector.
Meaning, we should allocate a negative proportion, a disinvestment of 9.81%
in the savings account.
To account for this, whenever we get a negative answer for any of the
Xi’s—which means if any of the first N rows of the right-hand-side vector is
less than or equal to zero—we must pull that row+2 and that column out
of the starting augmented matrix, and solve for the new augmented matrix.
If either of the last two rows of the right-hand-side vector are less than or
equal to zero, we don’t need to do this. These last two entries in the right-
hand-side vector always pertain to the Lagrangians, no matter how many or
how few components there are in total in the matrix. The Lagrangians are
allowed to be negative.
Since the variable returning with the negative answer corresponds to
the weighting of the fourth component, we pull out the fourth column