🚀 go-pugleaf

James Knott (the.lights.are.on.but.no.one.is@home.com) wrote:
>
> Here's the relavent text from that link.
>
>    "If the reverse bottleneck link has deep queues so that ACKs do not 
>    get dropped on the reverse path, then performance is a strong 
>    function of the normalized bandwidth ratio, k, defined in [LMS97]. k 
>    is the ratio of the raw bandwidths divided by the ratio of the 
>    packet sizes used in the two directions. For example, for a 10 Mbps 
>    forward channel and a 50 Kbps reverse channel, the raw bandwidth 
>    ratio is 200. With 1000-byte data packets and 40-byte ACKs, the 
>    ratio of the packet sizes is 25. This implies that k is 200/25 = 8. 
>    Thus, if the receiver acknowledges more frequently than one ACK 
>    every k = 8 data packets, the reverse bottleneck link will get 
>    saturated before the forward bottleneck link does, limiting the 
>    throughput in the forward direction."
>
> Notice that variable k for the packet size (they shouldn't
> have used the same letter for the 2 values).  

They didn't -- they only use k for the normalized bandwidth ratio.  Read
it again.

> It's the ratio
> of data/ack packet size.  As you go to larger packets, k
> gets bigger. i.e. if you had a 2000 byte packet, k becomes
> 16.  

No, you're completely misreading it.  k is the bandwidth ratio divided
by the packet size ratio.  As the downstream packet size increases, the
divisor increases and k gets smaller.  For 2000 byte data packets, the
packet size ratio is 50 instead of 25 and k is 200/50 = 4.

-Larry Jones

See if we can sell Mom and Dad into slavery for a star cruiser. -- Calvin