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P. 418

402                Chapter  Ten:  Linear  Programming


           is  called  a  6-ratio  for  Xj.  Hence  the  value  of  Xj  cannot  be
           increased  by  more  than  the  smallest  non-negative  #-ratio  of
           XJ;  for  otherwise  one  of the  constraints  will  be  violated.
                Suppose  that  the  smallest  non-negative  #-ratio  for  Xj  oc-
           curs  in  the  ith  row:  this  is  called  the  pivotal  row.  One  then
           applies  row operations to the tableau,  with the  aim  of  making
           the ith  entry  of column j  equal to  1 and  all other  entries  of the
           column  equal to  0.  (This  is called  pivoting  about  (i,j)  entry).
           The  choice  of  i  and  j  guarantees  that  no  negative  entries  will
           appear  in  the  right  most  column.  Replace  Xi  (the  departing
           variable)  by  Xj  (the  entering  variable).  Now  Xj  becomes  a
           basic  variable  with  value  —*-, replacing  X{. With  this  value  of
           Xj,  the  value  of  z  will  increase  by  biCj/aij,  at  least  if  6;  >  0.
                After  substituting  Xj  for  Xi  in the  list  of  basic  variables,
           we obtain  the  second tableau.  This  is treated  in the same  way
           as the  first  tableau,  and  if  it  is not  optimal,  one  proceeds  to  a
           third  tableau.  If at  some point  in the  procedure  all the  entries
           of the  objective  row become  non-negative,  an optimal  solution
           has  been  reached  and  the  algorithm  stops.

           Summary     of  the  simplex   algorithm
                Assume   that  a  linear  programming  problem  is  given  in
           standard  form
                                maximize:   z  =  C  X
                                 , .  .  .      (AX<B
                              subject  to:  <  x  >  Q


           where  B  >  0.  Then  the  following  procedure  is to  be  applied.
                1.  Convert  the  program  to  canonical  form  by  introducing
                slack  variables.  With  the  slack  variables  as  basic
                variables,  construct  the  initial  tableau.
                2.  If  no  negative  entries  appear  in the  objective  row,  the
                solution  is optimal.  Stop.
                3.  Choose  the  column  with  the  most  negative  entry  in
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