-
youxianyuanfachenyong
有限元常用子程序模块,WORD版内含各种有限元法C++编程的子模块(finite element commonly used subroutine modules, Word version containing various Finite Element Method C Programming Module)
- 2006-12-15 09:36:52下载
- 积分:1
-
yuanmian
圆面上的点到圆心的距离 圆面上的点到圆心的距离 圆面上的点到圆心的距离 圆面上的点到圆心的距离(Round surface of the distance from point to circle center point of the surface of the distance to the center circle to the center point of the surface of the distance from the surface point to the center circle distance)
- 2011-04-26 00:41:31下载
- 积分:1
-
lxcad-stress-umat12
利用abaqus二次开发平台,利用for语言编写的混凝土弹性本构(Use abaqus secondary development platform, the use of language for concrete elastic constitutive)
- 2020-11-23 20:09:34下载
- 积分:1
-
Ansys-bridge
ANSYS 分析一座预应力连续刚构桥的命令流,很实用哦(ANSYS analysis of a prestressed continuous rigid frame bridge command stream, and very practical!)
- 2012-05-30 12:01:35下载
- 积分:1
-
BFFusion2a
采用贝叶斯推论的数据融合,将CHAN算法和泰勒算法的结结果经过处理,得到更好的定位结果,已通过测试。
(Bayesian inference using the data fusion, will CHAN Taylor algorithm and the results of algorrithm processed and better positioning of the results of the full source code, has been tested.
)
- 2012-05-20 23:52:30下载
- 积分:1
-
A numerical method for analysis of Korres2003
说明: A numerical method for analysis of Korres2003
- 2020-06-18 17:20:02下载
- 积分:1
-
FFP
计算水声传播的快速场(FFP)程序,基于传播矩阵,迭代和快速傅里叶变换方法。(underwater acoustic field computing Fast FIELD Program, based on TMM, FFT etc.)
- 2021-03-19 20:29:19下载
- 积分:1
-
ActivdeSet
有效集法利用数学规划的对偶理论,将所求双层规划转化为一个下层只有一个无约束凸二次子规划的双层规划问题.然后根据两个双层规划的最优解和最优目标值之间的关系,提出一种简单有效的算法来解决非增值型凸二次双层规划问题.(Quadratic Bilevel Programming )
- 2009-10-05 22:08:44下载
- 积分:1
-
GA-DF2
说明: 利用GA遗传算法解决欺骗函数最优问题,具体问题描述如下,如有问题请与我联系(The deceptive functions are a family of functions in which there exists
low-order building blocks that do not combine to form the higher-order
building blocks. Here, a deceptive problem that consists of 25 copies of
the order-4 fully deceptive function DF2 is constructed for this paper.
DF2 can be described as follows:
f(0000)=28 f(0001)=26 f(0010)=24 f(0011)=18
f(0100)=22 f(0101)=6 f(0110)=14 f(0111)=0
f(1000)=20 f(1001)=12 f(1010)=10 f(1011)=2
f(1100)=8 f(1101)=4 f(1110)=6 f(1111)=30
This problem has a maximal function value of 750.)
- 2020-05-10 09:50:49下载
- 积分:1
-
huxiangguan
基于互相关分析的时延估计算法;基于互相关分析的时延估计算法(cross-correlation analysis for time delay calculation)
- 2020-10-25 16:40:00下载
- 积分:1