I have the following linear program:
$f(x)=2x_{1}+18x_{2} -> min$
$80x_{1}+100x_{2}>=100$
$20x_{1}+200x_{2}>=300$
$80x_{1}>=1.5$
$x_{1},x_{2}>=0$
In standard form:
$f(x)=2x_{1}+18x_{2} -> min$
$4x_{1}+5x_{2}-x_{3}=5$
$1x_{1}+10x_{2}+x_{4}=15$
$x_{1}+x_{5}=1.5$
$x_{1},x_{2},x_{3},x_{4},x_{5}>=0$
Ok, seems fairly simple, the A matrix contains the base matrix already ($x_{4},x_{5},x_{6}$).
The first row contains $-1$, the other two rows 1 (in the base matrix), so I'm just gonna multiply the row in the simplex table by $(*-1)$
The answer should be $f(x)=2.5$, but I always get $0$ myself. Let me post my simplex table:
(I don't know why does it upload like that, you'll just have to download it and rotate)
Now, as you can see, all $x_{i}$ values ( the 3rd row from the top, $(z_{i}*C_{i}+z_{i+1}*C_{i+1}...) - c_{j}$ ) are negative or equal $0$, which according to my knowledge means that this table is already optimal, and 0 is the outcome.
What's wrong?
Since your second and third constraint is >= you should add negative slack variable. You also missed 80 next to x1 in your 3rd constarint.