1. 6(1).1 Chapter 6 The Revised Simplex Method  This method is a modified version of the that we studied in Chapter 5.  This method is a modified version of the Primal Simplex Method that we studied in Chapter 5.  It is designed to exploit the fact that in many practical applications the coefficient matrix {a ij } is very, namely most of its elements are equal to zero.  It is designed to exploit the fact that in many practical applications the coefficient matrix {a ij } is very sparse, namely most of its elements are equal to zero.

2. 6(1).2  Bottom line: – Don’t update all the columns of the simplex tableau: update only those columns that you need!

3. 6(1).3 Standard Form opt=maxopt=max ~ ~  b i ≥ 0,b i ≥ 0, for all i. for all i.

4. 6(1).4 More Convenient Form

5. 6(1).5 Canonical Form As in the standard format, b i ≥0 for all i.

6. 6(1).6 System P (6.4)

7. 6(1).7 Example 6.1.1

8. 6(1).8 System P

9. 6(1).9 System P’ After a number of Pivot Operations

10. 6(1).10 ObservationObservation  After any iteration of the simplex method the columns of the m basic variables comprise the columns of the mxm  After any iteration of the simplex method the columns of the m basic variables comprise the columns of the mxm identity matrix.  The in which these columns are arranged for this purpose is important.  The order in which these columns are arranged for this purpose is important.  This order is specified in the of the simplex tableau.  This order is specified in the BV column of the simplex tableau.

11. 6(1).11 Example 6.1.2

12. 6(1).12 6.2 The Transformation  How can we compute S’ from S ?  From Linear Algebra we know that any finite sequence of pivot operations is equivalent to (left).  From Linear Algebra we know that any finite sequence of pivot operations is equivalent to (left) multiplication by a matrix.  In other words, S’ = TS  The question is then: T = ??????

13. 6(1).13 What is T ???  Observation 1: After any number of iterations of the simplex method, the columns of the coefficient matrix corresponding to the at that iteration, comprise the. After any number of iterations of the simplex method, the columns of the coefficient matrix corresponding to the basic variables at that iteration, comprise the identity matrix.  Observation 2: Initially, the of the coefficient matrix comprise the identity matrix. Initially, the last m columns of the coefficient matrix comprise the identity matrix.

14. 6(1).14 AnalysisAnalysis  If we group the columns of the basic variables into I and the nonbasic variables into D’, then S’ = [I,D’]  If we do the same for the initial matrix S, we have S = [B,D] where B is the matrix constructed from the columns of the initial matrix corresponding to the current basic variables.

15. 6(1).15  Since S’ = TS, it follows that S’ = [I,D’] = TS = [TB,TD] hence I = TB from which we conclude that T = B -1

16. 6(1).16 Example 6.2.1

17. 6(1).17

18. 6(1).18

19. 6(1).19

20. 6(1).20

21. 6(1).21 Notation:Notation:  I B = Indices of the basic elements (in canonical form)  I D = indices of the nonbasic variables in increasing order  c B = Initial cost vector of the basic variables  c D = cost vector of the nonbasic variables  c D = Initial cost vector of the nonbasic variables

22. 6(1).22  r = reduced costs vector  D = columns of the coefficient matrix in the simplex tableau corresponding to the current  D = columns of the coefficient matrix in the initial simplex tableau corresponding to the current nonbasic variables.

23. 6(1).23 Behind the Formula (NILN)  Each column of the coefficient matrix in the new tableau is equal to B -1 times the corresponding initial column, i.e new column = B -1 initial column  This is also true for the right-hand-side vector, i.e new RHS = B -1 initial RHS  Observe that the is in this formulation (why?)  Observe that the z-row is not included in this formulation (why?)

24. 6(1).24

25. 6(1).25 r D = C’ B = (0,0,...,0) correction

26. 6(1).26 (NILN) But how do we compute B -1 ?  :  Bad news:  We have to compute it as we go along  :  Good News:  We do not have to compute it from scratch  Observation: S’ = B -1 S = B -1 [M,I] = [B -1 M,B -1 I] = [B -1 M,B -1 ]  Hence, B -1 is equal to the matrix comprising the of the  Hence, B -1 is equal to the matrix comprising the last m columns of the LHS matrix.