EE103 HW5 Solution, Winter 2009, Prof SEJ
equate the “slack” variables and
z
(associated with the identity matrix),
4
5
6
,
,
,
x
x
x
z
, with the
right-hand-
side, and to consider all other variables to be zero; such a solution is called a “basic
solution”. That is, the current solution is
( ,
)
(0,0,0,17,9,16,0)
x
z
Because we’re trying to decrease the value of the ob
jective (its current value is
0
z
), we
attempt to increase the value of one of the non-basic variables, in this case
1
2
3
x or x or x
.
However, if we increase the value of
1
x
we’ll actually
increase
the value of the objective (note
that, as a function of the increase of
1
x
, we’ll have
1
1
1
(
)
12
,
0
z x
z
x
z
for x
(why?)
).
However, if we increase the value of either
2
3
x or x
, the value of the objective will, in fact,
decrease (why?).
Note that if we increase the value of
2
x
,
z
will decrease at the rate of 10 per
unit increase of
2
x
, or if we increase the value of
3
x
,
z
will decrease at the rate of 30 per unit
increase of
3
x
.
While it is not necessary to choose the variable that provides the greatest “rate
of decrease”, we’ll do so here.
We choose to in
crease the value of
3
x
as much as possible.
It is
important to note that in this case the value of the current basic variables,
4
5
6
,
,
x
x
x
will each
decrease as
3
x
increases. However, we require that all current basic variables remain non-
negative; therefore, we need to determine the largest value of
3
x
that maintains the non-
negativity of the current basic variables. The largest such value is determined by requiring all of
the below to remain non-negative as
3
x
increases:
4
3
3
5
3
3
6
3
3
(
)
17
8
0
(
)
9
3
0
(
)
16
8
0
x
x
x
x
x
x
x
x
x
That is, the largest allowable increase is given by the so-
called “min ratio test”
min{17 / 8,9 / 3,16 / 8}
16 / 8
.
That is, the current basic variable,
6
x
, “blocks” the increase of
3
x
.
The Simplex Algorithm proceeds by replacing
6
x
with
3
x
in the basis.
That is, we pivot on
the element in the 3
rd