Given a problem:

Find the dual:

$$ \text{Primal} =\begin{Bmatrix} \max \ \ \ \ 5x_1 - 6x_2 \\ \text{s.t.} \ \ \ \ 2x_1 -x_2 = 1\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_1 +3x_2 \leq9\\ \ \ \ \ x_1 \geq 0\\ \end{Bmatrix} $$

i) Introduce a slack variable ($+x_3$) into the 2nd constraint

$$ \text{Primal} =\begin{Bmatrix} \max \ \ \ \ 5x_1 - 6x_2 \\ \text{s.t.} \quad \quad2x_1 -x_2 \quad \quad \quad \quad \ = 1\\ \quad \quad \quad \quad x_1 +3x_2 + x_3 \quad \quad =9\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x_1 \geq 0\\ \end{Bmatrix} $$

iii) Problem becomes a minimization problem with $\geq$ constraints

iv) Multiply the constraints by $y_1,y_2$

$$ \text{Dual} =\begin{Bmatrix} \min \ \ \ \ y_1 + 9y_2 \\ \text{s.t.} \ \ \ \ 2y_1 + y_2 \geq5\\ \quad \quad \quad \quad \quad-y_2+3y_2 \geq -6\\ \quad \quad \quad y_2 \geq 0\\ \end{Bmatrix} $$

v) Change sign from $(-)$ to $(+)$ in second constraint

$$ \text{Dual} =\begin{Bmatrix} \min \ \ \ \ y_1 + 9y_2 \\ \text{s.t.} \ \ \ \ 2y_1 + y_2 \geq5\\ \quad \quad \quad \quad \quad y_2-3y_2 \leq 6\\ \quad \quad \quad y_2 \geq 0\\ \end{Bmatrix} $$

Is this correct or am I missing something?


  • 1
    $\begingroup$ You can't add slack variables to $x_i \geq 0$ type constraints. If you introduce $x_4$ into the system as you did, how do you keep track of $x_4\geq 0$? If you make one more slack variable for this, you will do so infinitely. If you don't, in your current primal, $x_1$ is as good as free. Read your textbook on how to handle non-negativity constraints. $\endgroup$
    – Inquest
    Commented Nov 11, 2012 at 20:28

2 Answers 2


Maybe it's worthwhile to talk through where the dual comes from. This will take a while, but hopefully the dual won't seem so mysterious when we're done.

Suppose we want to use the primal's constraints as a way to find an upper bound on the optimal value of the primal. If we multiply the first constraint by $9$, the second constraint by $1$, and add them together, we get $9(2x_1 - x_2) + 1(x_1 +3 x_2)$ for the left-hand side and $9(1) + 1(9)$ for the right-hand side. Since the first constraint is an equality and the second is an inequality, this implies $$19x_1 - 6x_2 \leq 18.$$ But since $x_1 \geq 0$, it's also true that $5x_1 \leq 19x_1$, and so $$5x_1 - 6x_2 \leq 19x_1 - 6x_2 \leq 18.$$ Therefore, $18$ is an upper-bound on the optimal value of the primal problem.

Surely we can do better than that, though. Instead of just guessing $9$ and $1$ as the multipliers, let's let them be variables. Thus we're looking for multipliers $y_1$ and $y_2$ to force $$5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9).$$

Now, in order for this pair of inequalities to hold, what has to be true about $y_1$ and $y_2$? Let's take the two inequalities one at a time.

The first inequality: $5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2)$

We have to track the coefficients of the $x_1$ and $x_2$ variables separately. First, we need the total $x_1$ coefficient on the right-hand side to be at least $5$. Getting exactly $5$ would be great, but since $x_1 \geq 0$, anything larger than $5$ would also satisfy the inequality for $x_1$. Mathematically speaking, this means that we need $2y_1 + y_2 \geq 5$.

On the other hand, to ensure the inequality for the $x_2$ variable we need the total $x_2$ coefficient on the right-hand side to be exactly $-6$. Since $x_2$ could be positive, we can't go lower than $-6$, and since $x_2$ could be negative, we can't go higher than $-6$ (as the negative value for $x_2$ would flip the direction of the inequality). So for the first inequality to work for the $x_2$ variable, we've got to have $-y_1 + 3y_2 = -6$.

The second inequality: $y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9)$

Here we have to track the $y_1$ and $y_2$ variables separately. The $y_1$ variables come from the first constraint, which is an equality constraint. It doesn't matter if $y_1$ is positive or negative, the equality constraint still holds. Thus $y_1$ is unrestricted in sign. However, the $y_2$ variable comes from the second constraint, which is a less-than-or-equal to constraint. If we were to multiply the second constraint by a negative number that would flip its direction and change it to a greater-than-or-equal constraint. To keep with our goal of upper-bounding the primal objective, we can't let that happen. So the $y_2$ variable can't be negative. Thus we must have $y_2 \geq 0$.

Finally, we want to make the right-hand side of the second inequality as small as possible, as we want the tightest upper-bound possible on the primal objective. So we want to minimize $y_1 + 9y_2$.

Putting all of these restrictions on $y_1$ and $y_2$ together we find that the problem of using the primal's constraints to find the best upper-bound on the optimal primal objective entails solving the following linear program:

$$\begin{align*} \text{Minimize }\:\:\:\:\: y_1 + 9y_2& \\ \text{subject to }\:\:\:\:\: 2y_1 + y_2& \geq 5 \\ -y_1 + 3y_2& = -6\\ y_2 & \geq 0. \end{align*}$$

And that's the dual.

It's probably worth summarizing the implications of this argument for all possible forms of the primal and dual. The following table is taken from p. 214 of Introduction to Operations Research, 8th edition, by Hillier and Lieberman. They refer to this as the SOB method, where SOB stands for Sensible, Odd, or Bizarre, depending on how likely one would find that particular constraint or variable restriction in a maximization or minimization problem.

             Primal Problem                           Dual Problem
             (or Dual Problem)                        (or Primal Problem)

             Maximization                             Minimization

Sensible     <= constraint            paired with     nonnegative variable
Odd          =  constraint            paired with     unconstrained variable
Bizarre      >= constraint            paired with     nonpositive variable

Sensible     nonnegative variable     paired with     >= constraint
Odd          unconstrained variable   paired with     = constraint
Bizarre      nonpositive variable     paired with     <= constraint
  • $\begingroup$ wonderful explanation, thank you so much $\endgroup$
    – lhk
    Commented Mar 14, 2018 at 17:20

The primal formulation is:

$ \max < \mathbf c, \mathbf x > $

s.t. $\left\{ \begin{array}{l} < \mathbf a_1, \mathbf x > = b_1\\ < \mathbf a_2, \mathbf x > \leq b_2\\ x_1 \geq 0\\ \end{array} \right. $

The dual formulation is expressed as:

$ \min (b_1y_1 + b_2y_2) $

s.t. $\left\{ \begin{array}{l} a_{11}y_1 + a_{21}y_2 \geq c_1\\ a_{12}y_1 + a_{22}y_2 \leq c_2\\ y_1, y_2 \geq 0\\ \end{array} \right. $

Let $\mathbf c = (5,-6) $, $\mathbf a_1 = (2,-1) $, $\mathbf a_2 = (1,3) $, $\mathbf b = (1,9) $, $\mathbf x = (x_1,x_2) $

Solution of the primal is $\mathbf x^* = (x_1^*,x_2^*) = (0,-1) $ with objective function equal to 6.

Solution of the dual is $\mathbf y^* = (y_1^*,y_2^*) = (6,0) $ with objective function equal to 6.


The dual formulation of the primal problem can be derived by writing the Lagrangian function $L$ of the primal problem and connecting heuristically such function to minimax theorem of John von Neumann. The Lagrangian function to be considered is: $ L(\mathbf x,y_1,y_2) = < \mathbf c, \mathbf x > + < y_1, b_1 – \mathbf a^t_1 \cdot \mathbf x > + < y_2, b_2 – \mathbf a^t_2 \cdot \mathbf x > $

Minimax theorem says that:

$ \min_{y_1,y_2} \max_\mathbf x L(\mathbf x,y_1,y_2) = \max_\mathbf x \min_{y_1,y_2} L(\mathbf x,y_1,y_2)$

Taking advantage of the linearity of the dot product and putting in evidence variable $ \mathbf x $, we get $ L = < \mathbf c, \mathbf x > + < y_1, b_1> - < y_1, \mathbf a^t_1 \cdot \mathbf x > + < y_2, b_2> - < y_2, \mathbf a^t_2 \cdot \mathbf x > = y_1 b_1 + \sum_{i=1}^2 c_i x_i – y_1 (\sum_{i=1}^2 a_{1i} x_i) + y_2 b_2 - y_2 (\sum_{i=1}^2 a_{2i} x_i) $

Maximizing $ L(\mathbf x,y_1,y_2) $ with respect to variable $ \mathbf x $ where $x1 \ge 0$ and $x_2 \in R$: $ \max_\mathbf x L(\mathbf x,y_1,y_2) = < y_1, b_1> + < y_2, b_2> + \max_\mathbf x [\sum_{i=1}^2 (c_i-y_1 a_{1i} - y_2 a_{2i} )x_i] $

The maximum of $[\sum_{i=1}^2 (c_i-y_1 a_{1i} - y_2 a_{2i} )x_i] $ is finite and equal to 0 if and only if

$\left\{ \begin{array}{l} c_1-y_1 a_{11} - y_2 a_{21} \leq0\\ c_2-y_1 a_{12} - y_2 a_{22} \geq0\\ y_1, y_2 \geq 0\\ \end{array} \right. $

As a consequnce the dual formulation is expressed as:

$ \min (b_1y_1 + b_2y_2) $

s.t. $\left\{ \begin{array}{l} a_{11}y_1 + a_{21}y_2 \geq c_1\\ a_{12}y_1 + a_{22}y_2 \leq c_2\\ y_1, y_2 \geq 0\\ \end{array} \right. $


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