The default window size for modern implementations of TCP on modern operating systems is bigger than 4080 bytes anyway. No change will probably be necessary. The window size varies as clients and servers ACK data, but it should start out at 8192 or greater, depending on the OS.
Are you wondering about throughput for user data or throughupt for all bytes including overhead that are being transferred? By user data I mean the actual file data that is being transferred. The maximum throughput for all bytes is about 64 kbps for the file size you mentioned (1 MB) and a window size of 4080 bytes or bigger. With that window size, he can fill the pipe. With such a big file size, the delay of .5 seconds isn't a big deal. It's going to take a couple minutes anyway to transfer the file. If you add on that delay to the entire time and use it in the kbps calculation, it essentially disappears. It doesn't disappear for a small file, however. You want to send 8,000,000 bits at 64 Kbps. This will take 125 seconds. The first bit will take .5 seconds. So the entire time is 125.5 seconds because of your long delay. Guess what 8,000,000 divided by 125.5 is? About 64,000 bits per second. If you want a better calculation, then you have to know the amount of time that elapses for the stuff I mentioned earlier, disk access speed etc. If you consider throughput just for user data, i.e. application-layer or file data, then you have to pay attention to all that other stuff I wrote too. Priscilla Cliff Cliff wrote: > > HI All, > > Thx for your reply and I really appreciate your explaination. > > I know that my customer is very difficult to use the whole BW > in the high delay time environment. > > But I need to tell him what's the max throughput. > > So that's why I need to tell him how they can do in their > computer / network change in order to get the max throughput - > just like the network optimization (I don't think they can get > 64k, it is only theory can occur). > > So according to s vermill, > > I will ask customer to tune the window to the following figure: > > 64,000/8 * 0.51 = 4,080 Bytes > > to get the max throughput, am I right? > > In here, I assume that the satellite part is very stable > (always 510ms), my customer only transfer IP stuff. Not IPX or > other protocol. Also I assume that their MTU size is by default > using cisco router default and their transfer file size is 1M. > > So is they can get the max throughput after setting the window > size to 4080 bytes? and how it can be calculate base on above > assumption (I mean the max throughput)? > > Kindly advice..... Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=57265&t=57158 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
