Math 通过网络电缆计算数据包的时间

Math 通过网络电缆计算数据包的时间,math,networking,distance,data-transfer,Math,Networking,Distance,Data Transfer,我正在准备网络方面的考试 在之前的一次考试中,给出了以下问题: Assume you're sending a packet of length 4000 bit through a cable of length 1000 km. The signal in the cable is moving at 200000 km/s. The signal bandwidth is 10 Mbit/s. Calculate how much time it would take for the pa

我正在准备网络方面的考试

在之前的一次考试中,给出了以下问题:

Assume you're sending a packet of length 4000 bit
through a cable of length 1000 km.
The signal in the cable is moving at 200000 km/s.
The signal bandwidth is 10 Mbit/s.
Calculate how much time it would take for the packet to arrive.
考虑到道路长度和车速,如果我用一辆车来做这件事, 可能需要200秒。虽然我不知道如何在计算中应用mbit/s和位

这是正确的做法吗

(10 mbit/s / 4000 bit) * (200000 km/s / 1000 km) = seconds packet needs to arrive

传输时间等于
发送时间+扩展时间

(为了简单起见,我使用Mbit作为10^6位,而不是2^20位,原理保持不变)

总计为
0.0054


奖金:

一个好的做法是查看单元,并确保在最后找到正确的单元,因此上面实际上是:

SEND_TIME = #bits / #bits_per_sec = 4000[bit] / 10*10^6[bit/sec] = 4*10^-4 [bit/bit * sec] = 0.0004 [sec]
EXPANSION_TIME = length / speed = 1000 [km] / 200000 [km/sec] = 5 * 10^-3 [km / km * sec] = 0.005 [sec]

非常感谢你!这比我最初的想法更有意义。
SEND_TIME = #bits / #bits_per_sec = 4000[bit] / 10*10^6[bit/sec] = 4*10^-4 [bit/bit * sec] = 0.0004 [sec]
EXPANSION_TIME = length / speed = 1000 [km] / 200000 [km/sec] = 5 * 10^-3 [km / km * sec] = 0.005 [sec]