# weak duality theorem

considering the primal a minimize problem, if $x$ and $p$ are feasible solution to the primal and to the dual then $p^tb \leq c^tx$

for any vector $x$ and $p$ we define

$u_i=p_i(a^t_ix-b_i)$

$v_j=x_j(c_j-p^tA_j)$

for the definition of the dual problem the sign of $p_i$ must be the same of the sign of $(a^t_ix-b_i)$ and the sign of $x_j$ must be the same of the sign of $(c_j-p^tA_j)$ so $u_i$ and $v_j$ are both $\ge 0$ ... etc

I don't understand why the sign must be the same and what these amounts mean.

In order for you to understand, I'll first type out the primal and dual problem first in vector form, then prove the weak duality theorem, and finally explain what $u_i$ and $v_j$ mean, so as to clear your doubts.

\begin{array}{ll} \text{min} & c^t x \\ \text{s.t.} & Ax \ge b \\ & x \ge 0 \tag{P} \end{array}

\begin{array}{ll} \text{max} & b^t p \\ \text{s.t.} & A^t p \le c \\ & p \ge 0 \tag{D} \end{array}

If you think of $u_i$ and $v_j$ as components of vectors, from the RHS of these two equations, it's not hard to see that the scalar $x^t A^t p$ is the "man in the middle".

$$b^t p \le x^t A^t p \le c^t x \tag{1}\label{pf}$$

The first half of the above inequality is from the primal constraint $Ax \ge b$ and the second half is from $A^t p \le c$. We've finished the proof the weak duality theorem. From the above inequality, we have

\begin{array}{r@{}c@{}l} 0 &{}\le (x^t A^t - b^t) p &{}= p^t (Ax - b) \\ 0 &{}\le c^t x - x^t A^t p &{}= x^t (c - A^t p) \tag{2} \label{eq2} \end{array}

Therefore, comparing these two inequality with $u_i=p_i(\color{red}{A^t_i} x-b_i)$ and $v_j=x_j(c_j-p^tA_j)$, we know that

1. To calculate $Ax - b$, we need to multiply the $i$-th row of $A$ by $x$. (i.e. The $i$-th column of $A^t$, which is denoted by $A_i^t$. $p_i$ is simply the $i$-th component of $p$, and $u_i$ is the $i$-th term of the sum $p^t (Ax - b)$.
2. To calculate $c - A^t p$, we need to multiply the $j$-th row of $A^t$ by $p$. (i.e. The $j$-th column of $A$, which is denoted by $A_j$. $x_j$ is simply the $j$-th component of $x$, and $v_j$ is the $j$-th term of the sum $x^t (c - A^t p)$.

• $p_i,x_j$ are due to $x,p \ge 0$
• $\color{red}{A^t_i} x-b_i \ge 0$ and $c_j-p^tA_j \ge 0$ are due to $Ax \ge b$ and $A^t p \le c$.
• i.e. We have $u_i=p_i(\color{red}{A^t_i} x-b_i) \ge 0$ and $v_j=x_j(c_j-p^tA_j) \ge 0$

The importance of $u_i$ and $v_j$ is that they allow us to get \eqref{eq2}, which implies \eqref{pf}.