Putting an LP in Standard Form for Karmarkar's Method

604 CHAPTER 10 Advanced Topics in Linear Programming
f (x k ) f (x k1 ) (41)
Inequality (41) states that each iteration of Karmarkar’s method decreases the potential
function by an amount bounded away from 0. Karmarkar shows that if the potential
function evaluated at x k is small enough, then z cx k will be near 0. Because f (x k ) is decreased
by at least per iteration, it follows that by choosing k sufficiently large, we can
ensure that the z-value for x k is less than .
Putting an LP in Standard Form for Karmarkar’s Method
We now show how to convert any LP to the form defined by (31)–(33). To illustrate, we
show how to transform the following LP
max z 3x1 x2 s.t.
s.t.
2x1 x2 2
2x1 2x2 5
(42)
s.t.
into the form defined by (31)–(33).
We begin by finding the dual of (42).
2x1 x1, x2 0
min w 2y1 5y2 s.t.
s.t.
2y1 y2 3
y1 2y2 1
(42)
s.t. 2y1y1, y2 0
From the Dual Theorem (Theorem 1 of Chapter 6), we know that if (x1, x2) is feasible in
(42), (y1, y2) is feasible in (42), and the z-value for (x1, x2) in (42) equals the w-value for
(y1, y2) in (42), then (x1, x2) is optimal for (42). This means that any feasible solution to
the following set of constraints will yield the optimal solution to (42):
3x1 x2 2y1 5y2 0
2x1 x2 2y1 5y2 2
x1 2x2 2y1 5y2 5
3x1 x2 2y1 y2 3
(43)
3x1 x2 y1 2y2 1
3x1 x2 All variables 0
Inserting slack and excess variables into (43) yields
3x1 x2 2y1 5y2 s1 0
2x1 x2 2y1 5y2 s1 2
x1 2x2 2y1 5y2 s2 5
3x1 x2 2y1 y2 e1 3
(44)
3x1 x2 y1 2y2 e2 1
3x1 x2 All variables 0
We now find a number M such that any feasible solution to (44) will satisfy
sum of all variables in (44) M (45)
and add constraint (45) to (44). Being conservative, we can see that any values of the vari-

ables that yield an optimal primal solution to (42) and an optimal dual solution to (42)
will have no variable exceeding 10. This would yield M 10(8) 80. We then add a
slack variable (dummy variable d1) to (45). Our new goal is then to find a feasible solution
to
3x1 x2 2y1 5y2 s1 0
2x1 x2 2y1 5y2 s1 2
x1 2x2 2y1 5y2 s2 5
3x1 x2 2y1 y2 e1 3 (46)
3x 1 x 2 y 1 2y 2
e 2 1
x 1 x 2 y 1 y 2 s 1
s 2 e 1 e 2 d 1 80
All variables 0
We now define a new
dummy variable d2; d2 1.
We can use this new variable
to “homogenize” the constraints
in (46), which have
SUMMARY The Revised Simplex Method and the Product Form
of the Inverse
Step 0 Note the columns from which the current B 1 will be read. Initially B 1 I.
Step 1 For the current tableau, compute c BVB 1 .
Step 2 Price out all nonbasic variables in the current tableau. If (for a max problem) each
nonbasic variable prices out nonnegative, the current basis is optimal. If the current basis
is not optimal, enter into the basis the nonbasic variable with the most negative coefficient
in row 0. Call this variable x k.
Step 3 To determine the row in which x k enters the basis, compute x k’s column in the
current tableau (B 1 a k) and compute the right-hand side of the current tableau (B 1 b).
Then use the ratio test to determine the row in which x k should enter the basis. We now
know the set of basic variables (BV) for the new tableau.
Step 4 Use the column for x k in the current tableau to determine the EROs needed to enter x k
into the basis. Perform these EROs on the current B 1 to yield the new B 1 . Return to step 1.
Alternatively, we may use the product form of the inverse to update B 1 . Suppose we have
found that x k should enter the basis in row r. Let the column for x k in the current tableau be
a 1k
a 2k
a mk
nonzero right-hand sides. To
do this, we add the appropriate
multiple of the constraint
d 2 1 to each constraint in
(46) (except the last constraint)
having a nonzero
right-hand side. For example
we add 2(d 2 1) to the
constraint 2x 1 x 2 s 1 2.
We also replace the last constraint
in (46) by the following
two constraints:
(a) Add d 2 1 to the last
constraint
10.1 Summary 605