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ML8204 Datasheet | OKI electronic componets
Part No. : ML8204
Description : Video Signal Noise Reduction and Rate Conversion IC with a Built-in 3.9 Mbit Field Memory
Download : ML8204 Click to Download
Page Number : 14
Manufacturer : OKI electronic componets
File Size : 136 Kb

ML8204 Article About

This year in April, Baidus search lasted two years with great concentration to create the formal launch of the promotion of professional version. Baidu had more than 300 companies have been invited to beta testing, the company participated in the trial that "the new platform better grasp of the details of marketing in place, almost did everything, from a practical effect, the use of new platforms and signing after the volume of telephone inquiries has significantly improved. " Industry experts believe that Baidu search extension platform upgrade, and search marketing from a "precision" to "fine" by leaps and bounds, from the first phase of search marketing emphasis put on accuracy, the effect can be statistical "precision" feature, upgrade emphasis on promoting the second phase can be managed to optimize the "fine" features.
Hester his 30 years of advanced system design and enterprise computing experience to the AMD. Prior to joining AMD, Hester was Newisys Sanmina-SCI companys co-founder and CEO. Prior to working at IBM, 23 years, he held several key technical positions in the leadership and management positions, and is the IBM corporate technical committee of 15 members, presided over multiple systems, including RS/6000 technology, including development projects. Information: Amdahls law coefficient and the measured speed multi-processor system performance, it is usually called an indicator of use to speed up factor, defined as follows: S = single-processor execution time / use p processors with the Amdahl execution time required representation of the traditional law of equation: S = p / * f) where S p, said that the processor speed factor that the serial number of the portion f the proportion of execution time when f = 5%, p = 20 when S = 10.256 so when f = 5%, p = 100 time, S = 16.8 or so

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