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I'm supposed to solve the following LP:

$(P_2) \text{ } \text{max}\left\{ 33x_1+13x_2+18x_3 \mid \begin{matrix} 8x_1+3x_2+4x_3 \leq 32\\ 12x_1+5x_2+7x_3 \leq 51\\ 5x_1+2x_2+3x_3 \leq 21\\ x_1+x_2+x_3 \geq 3 \end{matrix} \right\}$

I have formed the inequations to equations:

$8x_1+3x_2+4x_3+y_1=32$

$12x_1+5x_2+7x_3+y_2=51$

$5x_1+2x_2+3x_3+y_3=21$

$-x_1-x_2-x_3+y_4=-3$

And the solution problem (on my paper "Zielfunktion") is $33x_1+13x_2+18x_3$

I'm supposed to solve it by using simplex algorithm. As you will see below, my solution is very long... I needed about 8 tables and our teacher said that it's solvable with maybe 3 tables. But how? :(

The yellow marks are the pivot elements. enter image description here

enter image description here

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  • $\begingroup$ Are you familiar with the "künstliche Variable". It is commonly use if the constraint is a $\geq$ or a $=$ constraint. $\endgroup$
    – callculus42
    May 7, 2017 at 3:33
  • $\begingroup$ There is an easier method to calculate the values for the next table. See here math.stackexchange.com/questions/2182662/… Your method is too time-consuming. $\endgroup$
    – callculus42
    May 7, 2017 at 4:25

1 Answer 1

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See picture below:

enter image description here

Hope this helps !

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  • $\begingroup$ Thank you very much for your help! Maybe you can shortly explain your calculation steps? Because I don't understand why you calculated it like that. $\endgroup$
    – cnmesr
    May 6, 2017 at 20:56
  • $\begingroup$ @cnmesr I've tried to find a way to get only one coefficient for each row. This was obvious for the second row, but I must agree that I had to search a little bit for the other rows. $\endgroup$
    – Curious
    May 7, 2017 at 5:17
  • $\begingroup$ In class we have never talked about your way, but it seems very fast :) Is it still simplex algorithm though? Tyvm. $\endgroup$
    – cnmesr
    May 8, 2017 at 20:44
  • $\begingroup$ @cnmesr Have you read my comments ? $\endgroup$
    – callculus42
    May 9, 2017 at 4:25

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