1 1 2
 w w w p p p* x RHS
 W+ +



 I A b1



 -1 -A2 -b2



 A3 b3

 T T T
 I -A -A -A -Q 0
 1 2 3


Figure 13.--Internal matrix structure.



3. Using all of the data, create the inverse and solution for the new
 basis. Note, at this point we do not require w2 > 0 as in the
 original Wolfe algorithm.
4. Find the optimal solution to the quadratic problem. Maintain all
 1 
 x, w w variables non-negative.
 a. Find 2 = minimum w2 for w2 < 0. If all w > 0, solution is
 s
 optimal.
 2
 b. Introduce column xs (complement of ws) into the solution. Record
 2, 1 1
 the name of w2. Choose pivot among rows that contain x, w w
 variables and w Pivot according to smallest xi ratio > 0 or
 s aij
 largest aij for all xi = O; all other w2 variables and p* may be
 negative.
 c. If column w2 is dropped from the basis, go back to step 4a and
 s
 repeat.
 2 2
 d. Introduce a column w2 into the solution. If w is dropped from
 r s
 the basis, go to step 4a and repeat. If a primal variable xr is
 dropped, introduce its complement w2 into the basis.
 rl