Primal-Dual basic (feasible) solution?












1














I have been struggling with the following questions, especially the first two:




  1. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a feasible solution?

  2. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a basic solution?

  3. Find the optimal solution $[y_1*, y_2*]^T$.


The following primal is given:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & 2x_1 -4x_2 + x_3 -8x_4 leq 22 \
& x_1 + x_2 + 3x_3 + x_4 leq 8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



If I am not mistaken, this should be equivalent to:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & -2x_1 +4x_2 - x_3 +8x_4 geq -22 \
& -x_1 - x_2 - 3x_3 - x_4 geq -8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



Which should corresponds to the following dual:
$$begin{array}{cc}
text{max } & w = -22y_1 -8y_2\
text{s.t. } & -2y_1 - y_2 leq -4 \
& 4y_1 -y_2 leq -7\
&-y_1 -3y_2 leq 5\
&8y_1 -y_2 leq -14\
& y_1, y_2 geq 0 \
end{array}$$



My thoughts:




  1. It is not a feasible solution, since for $y_1 = 2, y_2 = 1$ we have
    $$begin{array}{cc}
    & -2y_1 - y_2 = -5 neq -4 \
    & 4y_1 -y_2 = 7 neq -7\
    &-y_1 -3y_2 = -5neq 5\
    &8y_1 -y_2 = 15neq -14\
    end{array}$$

  2. By that same argument, it is not a basic solution, since there is no basis $B$ of row vectors in $A^T$ such that $By_B = c$.

  3. I assume this is simply done by the Dual simplex algorithm, starting with a basic solution. But it feel like I did something incorrect at (1) and (2) and I actually need to use $[2,1]$ for this.


I am familiar with complementary slackness, but I don't think this is useful here?










share|cite|improve this question




















  • 1




    Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
    – callculus
    Nov 20 '16 at 4:44












  • So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
    – user260710
    Nov 20 '16 at 4:51












  • A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
    – callculus
    Nov 20 '16 at 4:58












  • If there is no typo number 2 can be answered immediately.
    – callculus
    Nov 20 '16 at 5:03










  • @callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
    – GNUSupporter 8964民主女神 地下教會
    Feb 20 '17 at 10:24


















1














I have been struggling with the following questions, especially the first two:




  1. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a feasible solution?

  2. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a basic solution?

  3. Find the optimal solution $[y_1*, y_2*]^T$.


The following primal is given:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & 2x_1 -4x_2 + x_3 -8x_4 leq 22 \
& x_1 + x_2 + 3x_3 + x_4 leq 8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



If I am not mistaken, this should be equivalent to:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & -2x_1 +4x_2 - x_3 +8x_4 geq -22 \
& -x_1 - x_2 - 3x_3 - x_4 geq -8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



Which should corresponds to the following dual:
$$begin{array}{cc}
text{max } & w = -22y_1 -8y_2\
text{s.t. } & -2y_1 - y_2 leq -4 \
& 4y_1 -y_2 leq -7\
&-y_1 -3y_2 leq 5\
&8y_1 -y_2 leq -14\
& y_1, y_2 geq 0 \
end{array}$$



My thoughts:




  1. It is not a feasible solution, since for $y_1 = 2, y_2 = 1$ we have
    $$begin{array}{cc}
    & -2y_1 - y_2 = -5 neq -4 \
    & 4y_1 -y_2 = 7 neq -7\
    &-y_1 -3y_2 = -5neq 5\
    &8y_1 -y_2 = 15neq -14\
    end{array}$$

  2. By that same argument, it is not a basic solution, since there is no basis $B$ of row vectors in $A^T$ such that $By_B = c$.

  3. I assume this is simply done by the Dual simplex algorithm, starting with a basic solution. But it feel like I did something incorrect at (1) and (2) and I actually need to use $[2,1]$ for this.


I am familiar with complementary slackness, but I don't think this is useful here?










share|cite|improve this question




















  • 1




    Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
    – callculus
    Nov 20 '16 at 4:44












  • So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
    – user260710
    Nov 20 '16 at 4:51












  • A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
    – callculus
    Nov 20 '16 at 4:58












  • If there is no typo number 2 can be answered immediately.
    – callculus
    Nov 20 '16 at 5:03










  • @callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
    – GNUSupporter 8964民主女神 地下教會
    Feb 20 '17 at 10:24
















1












1








1







I have been struggling with the following questions, especially the first two:




  1. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a feasible solution?

  2. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a basic solution?

  3. Find the optimal solution $[y_1*, y_2*]^T$.


The following primal is given:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & 2x_1 -4x_2 + x_3 -8x_4 leq 22 \
& x_1 + x_2 + 3x_3 + x_4 leq 8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



If I am not mistaken, this should be equivalent to:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & -2x_1 +4x_2 - x_3 +8x_4 geq -22 \
& -x_1 - x_2 - 3x_3 - x_4 geq -8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



Which should corresponds to the following dual:
$$begin{array}{cc}
text{max } & w = -22y_1 -8y_2\
text{s.t. } & -2y_1 - y_2 leq -4 \
& 4y_1 -y_2 leq -7\
&-y_1 -3y_2 leq 5\
&8y_1 -y_2 leq -14\
& y_1, y_2 geq 0 \
end{array}$$



My thoughts:




  1. It is not a feasible solution, since for $y_1 = 2, y_2 = 1$ we have
    $$begin{array}{cc}
    & -2y_1 - y_2 = -5 neq -4 \
    & 4y_1 -y_2 = 7 neq -7\
    &-y_1 -3y_2 = -5neq 5\
    &8y_1 -y_2 = 15neq -14\
    end{array}$$

  2. By that same argument, it is not a basic solution, since there is no basis $B$ of row vectors in $A^T$ such that $By_B = c$.

  3. I assume this is simply done by the Dual simplex algorithm, starting with a basic solution. But it feel like I did something incorrect at (1) and (2) and I actually need to use $[2,1]$ for this.


I am familiar with complementary slackness, but I don't think this is useful here?










share|cite|improve this question















I have been struggling with the following questions, especially the first two:




  1. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a feasible solution?

  2. Is the solution $[y_1, y_2]^T = [2,1]$ of the dual a basic solution?

  3. Find the optimal solution $[y_1*, y_2*]^T$.


The following primal is given:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & 2x_1 -4x_2 + x_3 -8x_4 leq 22 \
& x_1 + x_2 + 3x_3 + x_4 leq 8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



If I am not mistaken, this should be equivalent to:
$$begin{array}{cc}
text{min } & z = -4x_1 - 7x_2 + 5x_3 -14x_4\
text{s.t. } & -2x_1 +4x_2 - x_3 +8x_4 geq -22 \
& -x_1 - x_2 - 3x_3 - x_4 geq -8 \
& x_1, x_2, x_3, x_4 geq 0 \
end{array}$$



Which should corresponds to the following dual:
$$begin{array}{cc}
text{max } & w = -22y_1 -8y_2\
text{s.t. } & -2y_1 - y_2 leq -4 \
& 4y_1 -y_2 leq -7\
&-y_1 -3y_2 leq 5\
&8y_1 -y_2 leq -14\
& y_1, y_2 geq 0 \
end{array}$$



My thoughts:




  1. It is not a feasible solution, since for $y_1 = 2, y_2 = 1$ we have
    $$begin{array}{cc}
    & -2y_1 - y_2 = -5 neq -4 \
    & 4y_1 -y_2 = 7 neq -7\
    &-y_1 -3y_2 = -5neq 5\
    &8y_1 -y_2 = 15neq -14\
    end{array}$$

  2. By that same argument, it is not a basic solution, since there is no basis $B$ of row vectors in $A^T$ such that $By_B = c$.

  3. I assume this is simply done by the Dual simplex algorithm, starting with a basic solution. But it feel like I did something incorrect at (1) and (2) and I actually need to use $[2,1]$ for this.


I am familiar with complementary slackness, but I don't think this is useful here?







optimization convex-optimization linear-programming






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edited Feb 20 '17 at 10:10









GNUSupporter 8964民主女神 地下教會

12.8k72445




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asked Nov 20 '16 at 4:04









user260710user260710

1008




1008








  • 1




    Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
    – callculus
    Nov 20 '16 at 4:44












  • So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
    – user260710
    Nov 20 '16 at 4:51












  • A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
    – callculus
    Nov 20 '16 at 4:58












  • If there is no typo number 2 can be answered immediately.
    – callculus
    Nov 20 '16 at 5:03










  • @callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
    – GNUSupporter 8964民主女神 地下教會
    Feb 20 '17 at 10:24
















  • 1




    Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
    – callculus
    Nov 20 '16 at 4:44












  • So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
    – user260710
    Nov 20 '16 at 4:51












  • A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
    – callculus
    Nov 20 '16 at 4:58












  • If there is no typo number 2 can be answered immediately.
    – callculus
    Nov 20 '16 at 5:03










  • @callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
    – GNUSupporter 8964民主女神 地下教會
    Feb 20 '17 at 10:24










1




1




Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
– callculus
Nov 20 '16 at 4:44






Note that you have inequalities in your dual problem. Thus only the second and the fourth constraint are not fullfilled. But this alone is sufficient that (2,1) is not a feasible solution.
– callculus
Nov 20 '16 at 4:44














So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
– user260710
Nov 20 '16 at 4:51






So a basic solution, that is not necessarily feasible, has to satisfy all constraints of a LP? Does it also have to satisfy the non-negativity constraint? So if for example $[2,-1]^T$ satisfies the first four inequalities, then it's still not a basic solution because $y_2 < 0$?
– user260710
Nov 20 '16 at 4:51














A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
– callculus
Nov 20 '16 at 4:58






A basic solution is a feasible solution. Because it is a feasibille solution it has to fullfill all constraints. $(2,-1)^T$ is not a basic (feasible) solution, because it does not fullfill the non-negativity condition.
– callculus
Nov 20 '16 at 4:58














If there is no typo number 2 can be answered immediately.
– callculus
Nov 20 '16 at 5:03




If there is no typo number 2 can be answered immediately.
– callculus
Nov 20 '16 at 5:03












@callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
– GNUSupporter 8964民主女神 地下教會
Feb 20 '17 at 10:24






@callculus I searced "basic infeasible solution", and found "basic infeasible solution" on P.9 of this OR notes $unicode{x1F517}$
– GNUSupporter 8964民主女神 地下教會
Feb 20 '17 at 10:24












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  1. $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.

  2. $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
    begin{align}
    4y_1 -y_2 = 7 > -7 &implies [y_1, y_2]^T text{ infeasible} \
    4y_1 -y_2 = 7 ne -7 &implies [y_1, y_2]^T text{ non-basic}
    end{align}
    The second implication follows since we actually transform the LP into
    begin{array}{lrr}
    text{max} & w = -22y_1 -8y_2 &\
    text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \
    & 4y_1 -y_2 +s_2 =& -7\
    &-y_1 -3y_2 +s_3 =& 5\
    &8y_1 -y_2 +s_4 =& -14\
    & y_1, y_2, s_1, s_2, s_3, s_4 geq 0
    end{array}
    The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.

  3. You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.


Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.



Current basis: $s_1, s_2, s_3, s_4$



begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \
s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \
s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \
s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \ hline
& 22 & 8 & 0 & 0 & 0 & 0 & 0 \
text{ratio} & 11/4 & -8 & & & & 0 &
end{array}



Leaving variable: $s_4$, entering variable: $y_2$



Current basis: $s_1, s_2, s_3, y_2$



begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \
s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \
s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \
y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \ hline
& color{blue}{86} & color{blue}{0} & color{red}{0} & color{red}{0} & color{red}{0} & color{red}{bbox[2px, border: solid 1px]{8}} & -112 \
end{array}



Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.



By adding slack variabes $t_1,t_2$ to the primal



begin{array}{rlr}
min z = & color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\
text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +color{blue}{t_1} &= 22 \
& x_1 + x_2 + 3x_3 + x_4 +color{blue}{t_2} &= 8 \
& color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2} geq 0,
end{array}



we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2}]^T = [color{red}{0,0,0, bbox[2px, border: solid 1px]{8}}, color{blue}{86,0}]$.






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    1. $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.

    2. $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
      begin{align}
      4y_1 -y_2 = 7 > -7 &implies [y_1, y_2]^T text{ infeasible} \
      4y_1 -y_2 = 7 ne -7 &implies [y_1, y_2]^T text{ non-basic}
      end{align}
      The second implication follows since we actually transform the LP into
      begin{array}{lrr}
      text{max} & w = -22y_1 -8y_2 &\
      text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \
      & 4y_1 -y_2 +s_2 =& -7\
      &-y_1 -3y_2 +s_3 =& 5\
      &8y_1 -y_2 +s_4 =& -14\
      & y_1, y_2, s_1, s_2, s_3, s_4 geq 0
      end{array}
      The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.

    3. You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.


    Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.



    Current basis: $s_1, s_2, s_3, s_4$



    begin{array}{rrrrrrr|r}
    & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
    s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \
    s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \
    s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \
    s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \ hline
    & 22 & 8 & 0 & 0 & 0 & 0 & 0 \
    text{ratio} & 11/4 & -8 & & & & 0 &
    end{array}



    Leaving variable: $s_4$, entering variable: $y_2$



    Current basis: $s_1, s_2, s_3, y_2$



    begin{array}{rrrrrrr|r}
    & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
    s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \
    s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \
    s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \
    y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \ hline
    & color{blue}{86} & color{blue}{0} & color{red}{0} & color{red}{0} & color{red}{0} & color{red}{bbox[2px, border: solid 1px]{8}} & -112 \
    end{array}



    Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.



    By adding slack variabes $t_1,t_2$ to the primal



    begin{array}{rlr}
    min z = & color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\
    text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +color{blue}{t_1} &= 22 \
    & x_1 + x_2 + 3x_3 + x_4 +color{blue}{t_2} &= 8 \
    & color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2} geq 0,
    end{array}



    we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2}]^T = [color{red}{0,0,0, bbox[2px, border: solid 1px]{8}}, color{blue}{86,0}]$.






    share|cite|improve this answer




























      0















      1. $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.

      2. $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
        begin{align}
        4y_1 -y_2 = 7 > -7 &implies [y_1, y_2]^T text{ infeasible} \
        4y_1 -y_2 = 7 ne -7 &implies [y_1, y_2]^T text{ non-basic}
        end{align}
        The second implication follows since we actually transform the LP into
        begin{array}{lrr}
        text{max} & w = -22y_1 -8y_2 &\
        text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \
        & 4y_1 -y_2 +s_2 =& -7\
        &-y_1 -3y_2 +s_3 =& 5\
        &8y_1 -y_2 +s_4 =& -14\
        & y_1, y_2, s_1, s_2, s_3, s_4 geq 0
        end{array}
        The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.

      3. You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.


      Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.



      Current basis: $s_1, s_2, s_3, s_4$



      begin{array}{rrrrrrr|r}
      & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
      s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \
      s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \
      s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \
      s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \ hline
      & 22 & 8 & 0 & 0 & 0 & 0 & 0 \
      text{ratio} & 11/4 & -8 & & & & 0 &
      end{array}



      Leaving variable: $s_4$, entering variable: $y_2$



      Current basis: $s_1, s_2, s_3, y_2$



      begin{array}{rrrrrrr|r}
      & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
      s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \
      s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \
      s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \
      y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \ hline
      & color{blue}{86} & color{blue}{0} & color{red}{0} & color{red}{0} & color{red}{0} & color{red}{bbox[2px, border: solid 1px]{8}} & -112 \
      end{array}



      Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.



      By adding slack variabes $t_1,t_2$ to the primal



      begin{array}{rlr}
      min z = & color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\
      text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +color{blue}{t_1} &= 22 \
      & x_1 + x_2 + 3x_3 + x_4 +color{blue}{t_2} &= 8 \
      & color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2} geq 0,
      end{array}



      we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2}]^T = [color{red}{0,0,0, bbox[2px, border: solid 1px]{8}}, color{blue}{86,0}]$.






      share|cite|improve this answer


























        0












        0








        0







        1. $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.

        2. $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
          begin{align}
          4y_1 -y_2 = 7 > -7 &implies [y_1, y_2]^T text{ infeasible} \
          4y_1 -y_2 = 7 ne -7 &implies [y_1, y_2]^T text{ non-basic}
          end{align}
          The second implication follows since we actually transform the LP into
          begin{array}{lrr}
          text{max} & w = -22y_1 -8y_2 &\
          text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \
          & 4y_1 -y_2 +s_2 =& -7\
          &-y_1 -3y_2 +s_3 =& 5\
          &8y_1 -y_2 +s_4 =& -14\
          & y_1, y_2, s_1, s_2, s_3, s_4 geq 0
          end{array}
          The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.

        3. You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.


        Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.



        Current basis: $s_1, s_2, s_3, s_4$



        begin{array}{rrrrrrr|r}
        & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
        s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \
        s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \
        s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \
        s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \ hline
        & 22 & 8 & 0 & 0 & 0 & 0 & 0 \
        text{ratio} & 11/4 & -8 & & & & 0 &
        end{array}



        Leaving variable: $s_4$, entering variable: $y_2$



        Current basis: $s_1, s_2, s_3, y_2$



        begin{array}{rrrrrrr|r}
        & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
        s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \
        s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \
        s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \
        y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \ hline
        & color{blue}{86} & color{blue}{0} & color{red}{0} & color{red}{0} & color{red}{0} & color{red}{bbox[2px, border: solid 1px]{8}} & -112 \
        end{array}



        Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.



        By adding slack variabes $t_1,t_2$ to the primal



        begin{array}{rlr}
        min z = & color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\
        text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +color{blue}{t_1} &= 22 \
        & x_1 + x_2 + 3x_3 + x_4 +color{blue}{t_2} &= 8 \
        & color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2} geq 0,
        end{array}



        we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2}]^T = [color{red}{0,0,0, bbox[2px, border: solid 1px]{8}}, color{blue}{86,0}]$.






        share|cite|improve this answer















        1. $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.

        2. $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
          begin{align}
          4y_1 -y_2 = 7 > -7 &implies [y_1, y_2]^T text{ infeasible} \
          4y_1 -y_2 = 7 ne -7 &implies [y_1, y_2]^T text{ non-basic}
          end{align}
          The second implication follows since we actually transform the LP into
          begin{array}{lrr}
          text{max} & w = -22y_1 -8y_2 &\
          text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \
          & 4y_1 -y_2 +s_2 =& -7\
          &-y_1 -3y_2 +s_3 =& 5\
          &8y_1 -y_2 +s_4 =& -14\
          & y_1, y_2, s_1, s_2, s_3, s_4 geq 0
          end{array}
          The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.

        3. You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.


        Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.



        Current basis: $s_1, s_2, s_3, s_4$



        begin{array}{rrrrrrr|r}
        & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
        s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \
        s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \
        s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \
        s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \ hline
        & 22 & 8 & 0 & 0 & 0 & 0 & 0 \
        text{ratio} & 11/4 & -8 & & & & 0 &
        end{array}



        Leaving variable: $s_4$, entering variable: $y_2$



        Current basis: $s_1, s_2, s_3, y_2$



        begin{array}{rrrrrrr|r}
        & y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \ hline
        s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \
        s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \
        s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \
        y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \ hline
        & color{blue}{86} & color{blue}{0} & color{red}{0} & color{red}{0} & color{red}{0} & color{red}{bbox[2px, border: solid 1px]{8}} & -112 \
        end{array}



        Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.



        By adding slack variabes $t_1,t_2$ to the primal



        begin{array}{rlr}
        min z = & color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\
        text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +color{blue}{t_1} &= 22 \
        & x_1 + x_2 + 3x_3 + x_4 +color{blue}{t_2} &= 8 \
        & color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2} geq 0,
        end{array}



        we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[color{red}{x_1, x_2, x_3, bbox[2px, border: solid 1px]{x_4}}, color{blue}{t_1,t_2}]^T = [color{red}{0,0,0, bbox[2px, border: solid 1px]{8}}, color{blue}{86,0}]$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 21 '17 at 9:30

























        answered Feb 20 '17 at 17:12









        GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

        12.8k72445




        12.8k72445






























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