Based on the analysis and comparison, the basic skeleton of the ADSL Modem based on USB interface is determined, and than the detailed design of the Modem's hardware and software is given.

1、

Then, continue to work out and analyze the model, and find out the non-transshipment model of X telecom ADSL Modem is a typical newsboy model.

其次，对该模型进行了求解与分析，得出了无转运条件下的X电信的ADSL Modem订货模型实为一经典的报童模型。

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2、

Many communication products such as SOHO router, ADSL modem, T1/ E1 terminal equipment and ATM line card can be developed based on this platform.

基于此平台可以开发诸如SOHO路由器、ADSL Modem、T1／E1终端、ATM线卡等通信产品。

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3、

ADSL Modem is the key equipment in ADSL system, and DSP is the core of the ADSL Modem.

ADSL Modem是ADSL系统的关键配置，而DSP又是ADSL Modem的核心。

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4、

ADSL MODEM Remote Management Solutions

ADSL Modem远程管理方案浅析

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5、

It will be beneficial to research on the implementation of the new DSP in ADSL Modem.

探讨最新DSP技术在ADSL Modem上的应用对它的性能提升有很大的帮助。

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6、

ADSL Modem Realize to Share the Internet Access

ADSL Modem实现共享上网

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7、

Misuse of a patent occurs when a patent holder uses the patent to achieve something illegally.

当一个专利权人利用专利去非法的完成某些事时，专利滥用就发生了。

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8、

Fourthly, establish particular antitrust law or guideline to regulate the misuse of patent in technology standard and set up specialized agency to enforce it.

第四，制定专门的技术标准中专利权滥用方面的反垄断法规章或者指南，并设立专门机构进行执法。

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9、

Secondly, bring RAND principle which has been widely accepted by countries and international organizations into the system to regulate the misuse of patent.

第二，在技术标准中引入世界范围内被各个国家及国际组织广泛接受和认可的RAND原则，来规制专利权滥用问题。

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10、

Misuse of Patent and Regulation in Law

专利权滥用的法律规制研究

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11、

From this day, the Certificate will have a validity of additional three years minus one day.

从今天起，该证书将由有效期三年减去一天。

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12、

And that gives you the ratio of k1 plus k minus one.

那给出k1和k负1的比。

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13、

Minus one equals the total velocity of red dwarf.

等于红矮人号的完全速率。

provided by jukuu

14、

Square the imaginary unit, I, and you get minus one.

虚数单位i的平方等于-1。

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15、

Property minus one or is negative.

属性的值减去一，或者为负值。

provided by jukuu

16、

So this gives me p minus one constraints.

这给了我们p-1个约束条件。

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17、

So the ratio of the logs is just minus one.

所以这两个log的比是负一。

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18、

Then F is one minus one plus two, is equal to two.

那么F等于1减去1再加上2。

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19、

So for a total of C times p minus one total constraints.

所以一共C个约束条件。

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20、

So p minus one's constraint for one component times C components.

那么每个组分p-1个约束条件，乘以C各组分。

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