2020年“深圳杯”数学建模挑战赛C题-无线可充电传感器网络充电路线规划
深圳杯2020——数学建模模拟赛——C题
文章目录
- 深圳杯2020——数学建模模拟赛——C题
- Github链接:[www.github.com/lifuguan/szcup2020_simulation]
- 依赖库
- 计算公式
- 第一题
- 使用first solution strategy获得计算近似解
- 使用guided local search获得最优解
- 第二问：数值解法（numerical）
- 参数定义
- 等式
- 约束条件
- 根据约束条件得出线性方程组
- 参数设置
- 数值解结果
- 第二问：符号解法（symbolic）
Github链接:[www.github.com/lifuguan/szcup2020_simulation]
依赖库
- Google or-tools
- xlrd
- matplotlib
- sys
- numpy
- math
- sympy (符号计算)
计算公式
-
经度（东西方向）1M实际度：31544206M*cos(纬度)/360°=
31544206 ⋅ cos ( l a t i t u d e = 36 ) / 360 = 708883.29 m / l o n g t i t u d e 31544206\cdot\cos(latitude=36)/360 = 708883.29m/longtitude 31544206⋅cos(latitude=36)/360=708883.29m/longtitude
-
纬度（南北方向）1M实际度：40030173M360°=
40030173 / 360 = 111194.92 m / l a t i t u d e 40030173/360 = 111194.92m/latitude 40030173/360=111194.92m/latitude
第一题
使用first solution strategy获得计算近似解
求得结果
Route:
Route:
0 -> 10 -> 16 -> 27 -> 12 -> 8 -> 9 -> 15 -> 7 -> 11 -> 6 -> 14 -> 25 -> 18 -> 26 -> 19 -> 20 -> 1 -> 2 -> 17 -> 29 -> 21 -> 23 -> 24 -> 28 -> 22 -> 3 -> 4 -> 5 -> 13 -> 0
Distance: 39410m
使用guided local search获得最优解
求得结果
Route:
0 -> 29 -> 17 -> 2 -> 1 -> 20 -> 19 -> 26 -> 18 -> 25 -> 14 -> 6 -> 11 -> 7 -> 15 -> 9 -> 8 -> 12 -> 27 -> 16 -> 10 -> 13 -> 5 -> 4 -> 3 -> 22 -> 28 -> 24 -> 23 -> 21 -> 0
Distance: 39305m
第二问：数值解法（numerical）
要使传感器一直工作的最低要求：在移动电源走完一整个回路后，电池容量刚好达到最低要求为最优
假设初始条件：当到达该节点时，剩余电池容量刚好为最低要求
参数定义
- x 1 , x 2 , ⋯ , x 30 x_1,x_2,\cdots,x_{30} x1,x2,⋯,x30：假设每个节点的电池容量
- c 1 , c 2 , ⋯ , c 30 c_1,c_2,\cdots,c_{30} c1,c2,⋯,c30：每个节点的电池消耗速度
- r ( m A / s ) r(mA/s) r(mA/s)：电池充电速度
- f f f：最低工作电量
- v ( m / s ) v(m/s) v(m/s)：移动充电器移动速度
- d s t dst dst：总路程
等式
- 时间总花费： t t o t = d s t / v + ∑ i = 1 30 ( x i − f ) / r i t_{tot}=dst/v+\sum_{i=1}^{30}{(x_i - f)/r_i} ttot=dst/v+∑i=130(xi−f)/ri
- 电池容量推导： x i = t t o t ⋅ c i + f x_i=t_{tot}\cdot c_i+f xi=ttot⋅ci+f
约束条件
去掉数据中心节点的充电计算
x i = [ d s t / v + ∑ i = 2 30 ( x i − f ) / r i ] ⋅ c i + f x_{i} = [dst/v+\sum_{i=2}^{30}{(x_{i} - f)/r_i}]\cdot c_i+f xi=[dst/v+i=2∑30(xi−f)/ri]⋅ci+f
根据约束条件得出线性方程组
组合结果为一个29*29的矩阵：
[ x 2 ⋮ x 30 ] = [ d s t ⋅ c 2 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ⋮ d s t ⋅ c 30 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ] + [ x 2 ⋅ c 2 r + ⋯ + x 30 ⋅ c 30 r ⋮ x 2 ⋅ c 2 r + ⋯ + x 30 ⋅ c 30 r ] \left[ \begin{array}{l} \boldsymbol{x}_2\\ \vdots\\ \boldsymbol{x}_{30}\\ \end{array} \right] =\left[ \begin{array}{c} \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_2}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots\\ \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_{30}}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \end{array} \right] +\left[ \begin{array}{c} \frac{\boldsymbol{x}_2\cdot \boldsymbol{c}_2}{\boldsymbol{r}}+\cdots +\frac{\boldsymbol{x}_{30}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots\\ \frac{\boldsymbol{x}_2\cdot \boldsymbol{c}_2}{\boldsymbol{r}}+\cdots +\frac{\boldsymbol{x}_{30}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \end{array} \right] ⎣⎢⎡x2⋮x30⎦⎥⎤=⎣⎢⎡vdst⋅c2+f−rf⋅c2−⋯−rf⋅c30⋮vdst⋅c30+f−rf⋅c2−⋯−rf⋅c30⎦⎥⎤+⎣⎢⎡rx2⋅c2+⋯+rx30⋅c30⋮rx2⋅c2+⋯+rx30⋅c30⎦⎥⎤
化简：
[ x 2 ⋮ x 30 ] = [ d s t ⋅ c 2 v + f − f ⋅ c 2 r 2 − ⋯ − f ⋅ c 30 r 30 ⋮ d s t ⋅ c 30 v + f − f ⋅ c 2 r 2 − ⋯ − f ⋅ c 30 r 30 ] + [ x 2 ⋅ c 2 r 2 + ⋯ + x 30 ⋅ c 30 r 30 ⋮ x 2 ⋅ c 2 r 2 + ⋯ + x 30 ⋅ c 30 r 30 ] \left[ \begin{array}{l} \boldsymbol{x}_2\\ \vdots\\ \boldsymbol{x}_{30}\\ \end{array} \right] =\left[ \begin{array}{c} \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_2}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}_2}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}_{30}}\\ \vdots\\ \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_{30}}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}_2}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}_{30}}\\ \end{array} \right] +\left[ \begin{array}{c} \frac{\boldsymbol{x}_2\cdot \boldsymbol{c}_2}{\boldsymbol{r}_2}+\cdots +\frac{\boldsymbol{x}_{30}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}_{30}}\\ \vdots\\ \frac{\boldsymbol{x}_2\cdot \boldsymbol{c}_2}{\boldsymbol{r}_2}+\cdots +\frac{\boldsymbol{x}_{30}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}_{30}}\\ \end{array} \right] ⎣⎢⎡x2⋮x30⎦⎥⎤=⎣⎢⎡vdst⋅c2+f−r2f⋅c2−⋯−r30f⋅c30⋮vdst⋅c30+f−r2f⋅c2−⋯−r30f⋅c30⎦⎥⎤+⎣⎢⎡r2x2⋅c2+⋯+r30x30⋅c30⋮r2x2⋅c2+⋯+r30x30⋅c30⎦⎥⎤
转化成 A ⋅ X = b A\cdot X = b A⋅X=b形式：
[ c 2 r − 1 ⋯ c 30 r ⋮ ⋱ ⋮ c 2 r ⋯ c 30 r − 1 ] ⋅ [ x 2 ⋮ x 30 ] = − [ d s t ⋅ c 2 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ⋮ d s t ⋅ c 30 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ] \left[ \begin{matrix} \frac{\boldsymbol{c}_2}{\boldsymbol{r}}-1& \cdots& \frac{\boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots& \ddots& \vdots\\ \frac{\boldsymbol{c}_2}{\boldsymbol{r}}& \cdots& \frac{\boldsymbol{c}_{30}}{\boldsymbol{r}}-1\\ \end{matrix} \right] \cdot \left[ \begin{array}{l} \boldsymbol{x}_2\\ \vdots\\ \boldsymbol{x}_{30}\\ \end{array} \right] =-\left[ \begin{array}{c} \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_2}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots\\ \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_{30}}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \end{array} \right] ⎣⎢⎡rc2−1⋮rc2⋯⋱⋯rc30⋮rc30−1⎦⎥⎤⋅⎣⎢⎡x2⋮x30⎦⎥⎤=−⎣⎢⎡vdst⋅c2+f−rf⋅c2−⋯−rf⋅c30⋮vdst⋅c30+f−rf⋅c2−⋯−rf⋅c30⎦⎥⎤
其中
A = [ c 2 r − 1 ⋯ c 30 r ⋮ ⋱ ⋮ c 2 r ⋯ c 30 r − 1 ] , b = − [ d s t ⋅ c 2 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ⋮ d s t ⋅ c 30 v + f − f ⋅ c 2 r − ⋯ − f ⋅ c 30 r ] \boldsymbol{A}=\left[ \begin{matrix} \frac{\boldsymbol{c}_2}{\boldsymbol{r}}-1& \cdots& \frac{\boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots& \ddots& \vdots\\ \frac{\boldsymbol{c}_2}{\boldsymbol{r}}& \cdots& \frac{\boldsymbol{c}_{30}}{\boldsymbol{r}}-1\\ \end{matrix} \right] , \boldsymbol{b}=-\left[ \begin{array}{c} \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_2}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \vdots\\ \frac{\boldsymbol{dst}\cdot \boldsymbol{c}_{30}}{\boldsymbol{v}}+\boldsymbol{f}-\frac{\boldsymbol{f}\cdot \boldsymbol{c}_2}{\boldsymbol{r}}-\cdots -\frac{\boldsymbol{f}\cdot \boldsymbol{c}_{30}}{\boldsymbol{r}}\\ \end{array} \right] A=⎣⎢⎡rc2−1⋮rc2⋯⋱⋯rc30⋮rc30−1⎦⎥⎤,b=−⎣⎢⎡vdst⋅c2+f−rf⋅c2−⋯−rf⋅c30⋮vdst⋅c30+f−rf⋅c2−⋯−rf⋅c30⎦⎥⎤
参数设置
- 速度 v = 100 m / s v=100m/s v=100m/s
- 充电速度 r = 2000 m A / s r = 2000mA/s r=2000mA/s
- 最低工作值 f = 4 m A f=4mA f=4mA
数值解结果
| nodes | battery capacity |
|---|---|
| 0 | 2298.415 |
| 1 | 3241.665 |
| 2 | 1944.696 |
| 3 | 2337.717 |
| 4 | 1590.977 |
| 5 | 1944.696 |
| 6 | 2691.436 |
| 7 | 1983.998 |
| 8 | 1944.696 |
| 9 | 2337.717 |
| 10 | 1944.696 |
| 11 | 3084.457 |
| 12 | 2730.738 |
| 13 | 1944.696 |
| 14 | 1669.581 |
| 15 | 1944.696 |
| 16 | 2337.717 |
| 17 | 3123.759 |
| 18 | 2337.717 |
| 19 | 1944.696 |
| 20 | 1551.675 |
| 21 | 2337.717 |
| 22 | 3123.759 |
| 23 | 1551.675 |
| 24 | 2337.717 |
| 25 | 1866.092 |
| 26 | 1590.977 |
| 27 | 2691.436 |
| 28 | 2298.415 |
第二问：符号解法（symbolic）
代码已经实现，但是矩阵渲染太大导致计算机崩溃，故此可以得出利用符号解求解该问题并没有意义。