TCP 协议头字段大小
这是我网络课的家庭作业问题。我并不是在寻找确切的答案（因为这是我的作业），但如果能提供一些帮助就很好了：
您被雇用来设计一个使用滑动窗口的类似 TCP 的可靠字节流协议。该协议将在 100Mbps 网络上运行。网络中的RTT为100ms，最大分段生存时间为60秒。假设协议头中的 AdvertizedWindow 和 SequenceNum 字段都计算字节数，那么这两个字段所需的最小位数是多少？
我不知道如何计算这个。我知道标头字段是什么，并且我明白它的意思，但是有人可以解释这是如何计算的吗？
This is a homework question for my networking class. I'm not looking for an exact answer (since this is my homework), but some help would be nice:
You are hired to design a TCP-like reliable byte-stream protocol that uses a sliding window. This protocol will run over a 100Mbps network. The RTT on the network is 100ms, and the maximum segment lifetime is 60 seconds. What is the minimum number of bits required for the AdvertisedWindow and SequenceNum fields in your protocol header, assuming that those two fields count bytes?
I don't have a clue how to calculate this. I know what the header fields are, and I understand what it's saying, but could anyone explain how this is calculated?
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您主要需要弄清楚的是可以传输的最大数据量，即带宽延迟乘积。然后是包含该值所需的位数。
What you primarily need to figure out is the maximum amount of data that can be in flight, i.e. the bandwidth-delay product. Then, the number of bits required to contain that value.
带宽延迟乘积 (BDP) 根据最大带宽和往返时间计算得出-时间。
BDP = 每秒带宽 * rtt-秒
为了使吞吐量不受窗口大小的限制，窗口大小必须能够表达 BDP。
序列号必须大于窗口大小。阅读滑动窗口。
您可能还应该查看 TCP Tuning 和 窗口缩放。
The Bandwidth-Delay Product (BDP) is calculated from the maximum bandwidth and the round-trip-time.
BDP = bandwidth-in-bits-per-second * rtt-in-seconds
In order that throughput is not constrained by window-size the window-size must be able to express the BDP.
The sequence number must be bigger than the window-size. Read up on Sliding Windows.
You should probably also look at TCP Tuning and Window Scaling.