Below are the

blue

Since one INDICATOR

Named below are

results of the numbers

(namely -1/5) remains

Below is the result

4 row operations

row operations are the

negative, we must

of changing our pivot

needed to pivot

named above new ISM

repeat steps 3-9

to "1"

on the number(16/5)

encircled in red

RATIOS

0 0

8

SEE

0

0

0

5

8

1 - 8/2 = R

0 1

#2

1 16

1642 -5 - BELOW

0

5

16

0

18

0 0

5

6

( 8 ) 12 = R2

10 05

6 5

Er2 = Rg

0 3 0 09

0

3 0 0 3

1 12 4+5 2-84

PIVOT COLUMN

(PIVOT ROW )

TO

UINDICATOR, ROW

MATRIX

BELOW

X1 X2

X3

$1

$2

0

0

0

Above there was a tie for least non-negative ratio:

either row 1 or row 2 could have become the pivot row, and either

5

0

Col

16

choice leads to the final tableau after one additional pivoting At the

right is the result of the final 3 row operations

Col-

0

0

16

13

INDICATOR ROW

All indicators (0. 0. 0. . and , } are now zero or bigger ("13" is NOT an indicator)

Thus, as in step 8 of the SIMPLEX METHOD, the last tableau is a FINAL TABLEAU

Row operations of SIMPLEX METHOD are done

Thus, the basic solution for the tableau above is the solution to our