网络计算 [英] Network Calculations

问题描述

我需要帮助以下计算我遇到了麻烦，我不想让别人这样做，所以我如何解决它们如下


？在给定系统中，IP分组用于传送由各种源生成的语音样本。对于每个源，每125微秒生成一个样本，并使用8个二进制位对样本进行编码。对于13 Mbps的比特率，可以容纳多少来源？

？在给定系统中，IP分组用于传送由各种源生成的CD质量音乐样本。每个源的采样率为44 KHz（每秒产生44000个样本），每个样本使用16个二进制位进行编码。假设带宽为100 Mbps，可以容纳多少来源？


？计算与从源到目的地的数据包传输（15千字节长）相关的延迟，物理距离

Hi,

I need help with the following calculations i am having trouble with i dont want someone to do them just so me how to work them out they are as follows

? In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?
? In a given system, IP packets are used to convey CD quality music samples generated by various sources. The sampling rate at each source is 44 KHz (44000 samples are made per second) and each sample is encoded using 16 binary bits. Assuming the bandwidth to be 100 Mbps, how many sources can be accommodated?

? Calculate the delay associated with the transmission of a packet (15 Kbytes long) from source to destination, with a physical distance

推荐答案

alrite dude！她怎么了？这是我的第一个回复，失去了我的童贞，首先是一切：P。在我的拙见中... 1微秒是百万分之一秒（10-6）。如果每秒有13mbps或13631488位，或每微秒13.631488位......在125微秒......或一个源中传输1703.936位。


是啊？！大声笑......已经有一段时间了！


这是一个大师，还是博士问题！现在，这是本地化的！

alrite dude! hows she going? this is my first reply, losing my virginity, first for everything :P. in my humble opinion... 1 microsecond is one millionth of a second(10−6). if there is 13mbps or 13631488 bits per second, or 13.631488 bits per microsecond... 1703.936 bits transferred in 125 microseconds... or one source.

yeah?! lol... been a while!

is this a masters, or phd question! now way this is undergrad!

对不起男人一件事我们（微秒）是1MHz的循环时间。

sorry man one more thing 1 us(microsecond) is the cycle time of 1MHz.

1微秒是百万分之一秒（10-6）。如果每秒有13mbps或13631488位，或每微秒13.631488位...... 在125微秒内传输1703.936位...或一个源。

不幸的是，所有这些都是正确的，除了最后一个（突出显示的步骤）是错误的。

Unfortunately all of this is correct except the last (highlighted step) which is wrong.

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