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CCNP Route - Bandwidth Delay Product
Hi Everybody,
The theory behind bandwidth delay product seems very hard to apply. The basic example is 768kpbs x 100ms equals 76,800 bits.
However applying this is very hard. Once different conversions are needed the theory seems to go out the window. For example: 128 kpbs with a delay of 2.5 seconds should be what? My math would say 320kbps.
Can someone walk me through this?
Thanks!
Node
The theory behind bandwidth delay product seems very hard to apply. The basic example is 768kpbs x 100ms equals 76,800 bits.
However applying this is very hard. Once different conversions are needed the theory seems to go out the window. For example: 128 kpbs with a delay of 2.5 seconds should be what? My math would say 320kbps.
Can someone walk me through this?
Thanks!
Node
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joetest
Member Posts: 99 ■■□□□□□□□□
you're doing it right, it's just not kbps but just kb. You can send 320k bits before the first bit reaches the destination. Meaning you can send up to 40k Bytes at any given time on the link. -
Optionsjames43026
Member Posts: 303 ■■□□□□□□□□
The product of this calculation gives you the max number of bytes or bits that can be on a link at any given time, before an ack is expected back from the destination. TCP calls this the receive window. In your example of 128kbps and 2.5 seconds of RTT, the product would be 320kb not kbps. Meaning you would be able to send a total of 320kb over the period of 2.5 seconds before needing to receive a tcp ack from the destination. Joetest your math may be a bit off, I believe in this example we are sending at line speed, meaning you could have up to 128kb on the line at any given time, otherwise you would be wasting bandwidth, which completely goes against the math that we just calculated.
Some great sources:
How to Calculate TCP throughput for long distance WAN links
https://en.wikipedia.org/wiki/Bandwidth-delay_product
https://networklessons.com/network-management/bandwidth-delay-product/ -
Optionsjoetest
Member Posts: 99 ■■□□□□□□□□
128000bps * 2.5sec = 320kilobits / 8 to get bytes = 40k bytes is just the bandwidth delay product(and TCP rwin).
You can't send anymore unacknowledged data at a time than that in one TCP session - We're writing the same thing I believe
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Optionsjames43026
Member Posts: 303 ■■□□□□□□□□
Ahhh yeah, for some reason I didn't read your last statement correctly. My bad, we are saying the exact same thing.