# Congested network笔记(3)

\left\{\begin{align}\frac{\partial\mathcal{Z(F)}}{\partial\underline{\mathcal{F}}}\cdot(\mathcal{F}-\underline{\mathcal{F}})&\ge 0&\forall \mathcal{F}\in S&\text{ 最优解}\\ \mathcal{C}(\mathcal{F}^*)\cdot(\mathcal{F}-\mathcal{F}^*)&\ge 0&\forall \mathcal{F}\in S&\text{ 均衡解}\end{align}\right.


$$A=\left(\frac{\partial\mathcal{C}_i}{\partial\mathcal{F}_j}\right)$$

$$A=\left(\frac{\partial^2\mathcal{Z}}{\partial\mathcal{F}_i\partial\mathcal{F}_j}\right)$$

\left\{\begin{align}x^tAx&\gt 0\\ \forall x\in R^n, x&\not=0\end{align}\right.

$$x\cdot 2\cdot x=2x^2\gt 0 \quad \text{if } x\not=0$$

$$A=\left(\begin{array}{}2 & 0 & 1\\ 0 & 3 & 0\\ 1 & 0 & 2\end{array}\right)$$

$$x=\left(\begin{array}{x}x_1\\ x_2\\ x_3\end{array}\right)$$

\begin{align}x^tAx&=x^t\left(\begin{array}{A}2 & 0 & 1\\ 0 & 3 & 0\\ 1 & 0 & 2\end{array}\right)\left(\begin{array}{x}x_1\\ x_2\\ x_3\end{array}\right)\\ &=\left(x_1 x_2 x_3\right)\left(\begin{array}{}2x_1 + x_3\\ 3x_2\\ x_1 + 2x_3\end{array}\right)\\ &=2x^2_1+x_1x_3+3x^2_2+x_1x_3+2x^2_3\\ &=(x^2_1+2x_1x_3+2x^2_3)+x^2_1+x^2_3+3x^2_2\\ &=(x_1+x_3)^2+x^2_1+x^2_3+3x^2_2\\ &\gt 0\end{align}


$$B=\left(\begin{array}{}1 & 2 & 0\\ 2 & 3 & 0\\ 0 & 0 & 1\end{array}\right)$$

\begin{align}x^tBx&=x^t\left(\begin{array}{}1 & 2 & 0\\ 2 & 3 & 0\\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{x}x_1\\ x_2\\ x_3\end{array}\right)\\ &=\left(x_1 x_2 x_3\right)\left(\begin{array}{}x_1 + 2x_2\\ 2x_1+3x_2\\ x_3\end{array}\right)\\ &=x^2_1+2x_1x_2+2x_1x_2+3x^2_2+x^2_3\\ &=(x^2_1+2x_1x_2+x^2_2)+2x_1x_2+3x^2_2+x^2_3\\ &=(x_1+x_2)^2+2x^2_2+x^2_3+2x_1x_2\end{align}


\left\{\begin{align}x_1&=-1\\ x_2&=1\\ x_3&=0\end{align}\right.

\begin{align}x^tBx&=(x_1+x_2)^2+3x^2_2+x^2_3+2x_1x_2\\ &=0+2+0+(-2)\\ &=0\end{align}


$$a_{ii}\gt \sum_{i\not=k}\left| a_{ik} \right|\quad \forall i$$

\left\{\begin{align}\mathcal{c}_1&=f_1+5\\ \mathcal{c}_2&=2f_2+10\\ \mathcal{c}_3&=f_3+15\end{align}\right.

$$\mathcal{C}=\left(\begin{array}{}f_1+5\\ 2f_2+10\\ f_3+15\end{array}\right)$$

\begin{align}A&=\left(\frac{\partial\mathcal{C}_i}{\partial\mathcal{F}_j}\right)\\ &=\left(\begin{array}{}\frac{\mathcal{C}_1}{\mathcal{F}_1} & \frac{\mathcal{C}_1}{\mathcal{F}_2} & \frac{\mathcal{C}_1}{\mathcal{F}_2}\\ \frac{\mathcal{C}_2}{\mathcal{F}_1} & \frac{\mathcal{C}_2}{\mathcal{F}_2} & \frac{\mathcal{C}_2}{\mathcal{F}_2}\\ \frac{\mathcal{C}_3}{\mathcal{F}_1} & \frac{\mathcal{C}_3}{\mathcal{F}_2} & \frac{\mathcal{C}_3}{\mathcal{F}_2}\end{array}\right)\\ &=\left(\begin{array}{}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1\end{array}\right) \end{align}

$$\mathcal{C}_i=\mathcal{F}^2_i+i\mathcal{F}_i$$

\begin{align}\mathcal{Z}&=\mathcal{C}_1+\mathcal{C}_2+\mathcal{C}_3+\mathcal{C}_4+\mathcal{C}_5\\ &=(f^2_1+f_1)+(f^2_2+2f_2)+(f^2_3+3f_3)+(f^2_4+4f_4)+(f^2_5+5f_5)\end{align}


\left\{\begin{align}\frac{\partial\mathcal{Z}}{\partial\mathcal{F}_1}&=2f_1+1\\ \frac{\partial\mathcal{Z}}{\partial\mathcal{F}_2}&=2f_2+2\\ \frac{\partial\mathcal{Z}}{\partial\mathcal{F}_3}&=2f_3+3\\ \frac{\partial\mathcal{Z}}{\partial\mathcal{F}_4}&=2f_4+4\\ \frac{\partial\mathcal{Z}}{\partial\mathcal{F}_5}&=2f_5+5\end{align}\right.

$$A=\left(\begin{array}{A}2 & 0 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 0 & 2\end{array}\right)$$

$$A=\left(\begin{array}{}1 & 1 & 0\\ 2 & 1 & 2\\ 0 & 0 & 1\end{array}\right)$$

\begin{align}x^tAx&=\left(\begin{array}{}x_1 & x_2 & x_3\end{array}\right)\left(\begin{array}{}1 & 1 & 0\\ 2 & 1 & 2\\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{}x_1\\ x_2\\ x_3\end{array}\right)\\ &=\left(\begin{array}{}x_1 & x_2 & x_3\end{array}\right)\left(\begin{array}{}x_1+x_2\\ 2x_1+x_2+2x_3\\ x_3\end{array}\right)\\ &=x^2_1+x_1x_2+2x_1x_2+x^2_2+2x_2x_3+3x^2_3\\ &=(x^2_1+2x_1x_2+x^2_2)+x_1x_2+2x_2x_3+x^2_2\\ &=(x_1+x_2)^2+x^2_2+x_1x_2+2x_2x_3\end{align}


\left\{\begin{align}\mathcal{C}_1&=f_1+f_2\\ \mathcal{C}_2&=2f_1+f_2+2f_3\\ \mathcal{C}_3&=f_3\end{align}\right.

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