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fcm
模糊C均值方法，将模糊的方法用到K均值方法里面(Fuzzy C Means method, the fuzzy approach method which used the mean K)
- 2010-06-15 14:23:35下载
- 积分：1
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Mobilebroadband
This is book aboud Mobile Broadband
Including WiMAX and LTE in English version
- 2010-07-11 15:06:10下载
- 积分：1
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MATERIAL-DMSPSO
source codes for bayes classifier
- 2012-04-03 03:58:06下载
- 积分：1
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sPso_Samp1NoNoise
在不加噪声的情况下利用PSO（粒子群算法）辨识NARMAX模型(In the case without noise, using PSO (PSO) model identification NARMAX)
- 2011-05-17 15:38:34下载
- 积分：1
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reading-the-file
digicommunications 2
- 2013-03-30 13:43:51下载
- 积分：1
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3zuoye
某市有一码头，每次仅容一辆船停泊装卸货，由于经常有船等候进港，部分人提出要扩建码头。经过调查历史资料发现，码头平均每月停船24艘，每艘船的停泊时间为24±20小时，相邻两艘船的到达时间间隔为30±15小时，如果一艘船因有船在港而等候1小时，其消耗成本为1000元。经预算，扩建码头大约需要1350万元，故市长决策如下：如果未来五年内停泊船只因等候的成本消耗总和超过扩建码头花费则扩建码头，否则，不予扩建。因此，希望你能够帮助市长做出决策。此问题已知到达的大概时间和大概停泊时间，对于此问题用概率统计的方法来做比较复杂，可用程序随机产生到达时间和停泊时间来模拟未来五年内船的停泊，多次模拟预测停泊情况，以做出决策(。-Problem in a city with a terminal expansion dock, each only allow a boat moored loading and unloading, as often there are waiting boat into port, some people proposed to expand pier. Terminal stopping 24 per month, per vessel berthing time of 24 ± 20 hours, the arrival of two adjacent vessels interval 30 ± 15 hours, if a ship due to boat in the harbor while waiting for one hour, and its consumption cost of 1,000 yuan. The budget of about 13.5 million yuan expansion dock, so the mayor decision as follows: If a ship is moored next five years, the cost of waiting more than the sum of consumption spending is expanding marina pier extension, otherwise, not expanded.)
- 2015-04-17 18:59:46下载
- 积分：1
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multiSoundRecorder
try this for you frieds.....
- 2012-06-14 19:56:05下载
- 积分：1
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OFDM
document for basic learning of OFDM concepts.
- 2014-02-26 05:00:45下载
- 积分：1
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LDA
数据挖掘的线性判别分析,用matlab编写(Data Mining linear discriminant analysis, using matlab write)
- 2009-11-06 21:49:19下载
- 积分：1
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MATLAB
matlab函数参考，包括各种常用的函数开发工具，功能很好很强大啊(matlab function reference, including a variety of functions commonly used development tools, features a very good strong ah)
- 2009-01-05 14:50:44下载
- 积分：1
