Consider this new problem

Consider this new problem min -1 + 2- = N (1) 21 + 2 +03 4 -201 + 22 + 03 3 IV IV IV IA IA 0 CC 2 {x20, x3 = 0,21 = 4} (2) In standard form min -21 + 22 - 23 Z (3) 4 -1+ 22 + 23 + 0x3 + 24 3 C1 0 IVIV IV IV IA IA II C2 EC 3 CC 4 0 The format for Mathematica is -21 + 22 -23 + 024 + 035 (4) 1+ 2+3+ 34+ 036 = 4 -201 + 202 + 203 + 024 + 20'6 = 0 3 2 0 0 IVIV IV IV 0 0 (5) The optimal solution, found by software, is 2 = 0, 14 = 0, 26 = 0,21= 2:43 =Initialization and starting solution: = -1 1 1 " ) (6) BON = IN = N = C B and CB = C4 05 = 0 0 0 O and ON = ( C1 C2 C3 ) = (-1 1 -1 ) Optimality check of the initial solution based on reduced costs: ON CN - CBB-IN = ( -1 1 -1 ) -0B-IN = ( -1 1 -1 ) >0 = 1" or 3" nonbasic wishes to enter the basis 's enters N = -1 1 = g { min . 1 4 3 - 3= 2nd basic leaves as leaves the basis Update the basic and nonbasic variables in this first pivot: C 3 CB 1 = OC 2 = C 3We now update all information: A 1 1 0 1 B = 0 and B-1 = 1 1 ) N = 1 -1 1 1 BON = -1.0 1.0 = 1 2.0 0 -1.0 We need the cost vectors for current basis: 4 3 3 CB -CB O CN O = (03 04 ) = ( -1 0 ) 1 = (01 02 C4 ) = ( -1 1 0 ) Now we check reduced costs: ON = CN - CBB-IN = ( -1 1 0 ) -(-1 0 ) (1 3 )( 1 19) CN = ON - CBB-IN = ( -1 1 0 ) -(-1 0 )( -1.0 1.0 1 2.0 0 -1.0 = (0 2 0 )20= optimality CB = B-1b = 1 -1) ( 3) 3.0 1.0 Assignment: repeat using , instead of 23 as the entering variable of the first pivot. Comment on and explain what you find