Page 420 - A Course in Linear Algebra with Applications

404 Chapter Ten: Linear Programming

Xi *x 2 X3 £4 £5 x 6
£4 1 1 2 1 0 0 2
* * £5 2 3 4 0 1 0 3
XQ 3 3 1 0 0 1 4
-8 -9 -5 0 0 0 0

Here the z-column has been suppressed. The initial basic
feasible solution £4 = 2, £5 = 3, XQ = 4 is not optimal since
there are negative entries in the objective row; the most neg-
ative entry occurs in column 2, so £2 is the entering variable,
(indicated in the tableau by *).
|
2
The #-values for £2 are ,1, , corresponding to £4, £5, x&.
The smallest (non-negative) 9-value is 1, so £5 is the departing
variable (indicated in the tableau by **). Now pivot about the
(2, 2) entry to obtain the second tableau.

*X\ X2 X3 £4 £5 x 6
£4 1/3 0 2/3 1 -1/3 0 1

X2 2/3 1 4/3 0 1/3 0 1
* * £ 6 1 0 -3 0 -1 1 1
-2 0 7 0 3 0 9

The objective row still has a negative entry, so this is not
optimal: the entering variable is x\. The smallest 9-value for
£1 is 1, occurring for £6, so this is the departing variable. Now
pivot about the (3,1) entry to get the third tableau.

£l x 2 X3 £4 £5 £6
£4 0 0 5/3 1 0 -1/3 2/3
X2 0 1 10/3 0 1 -2/3 1/3
£l 1 0 -3 0 -1 1 1
0 0 1 0 1 2 11