Stop and wait means that each frame is sent upon acknowledgement of the last frame. You are sending 32KB per frame, and you have a round trip defined by the distance and SoL. So you can work out how many frames you can send per second and so how much data per second, and this will be a fraction of 64Mbps.

so 32kb/64 MB thats 32/64000?

32KB is bytes, 64Mbps is bits, so you need to convert. Also you obviously cannot ignore the distance and SoL.

oh right you are, so 32/8000, but what is the formula of utilization i should use? I have found various on the net, and there is no one such formula in the book

You have to work it out. How long does it take to get the frame to the planet?

there is no delay mentioned in the text book

The question you posted clearly defines a delay introduced by the fact that the other endpoint is far away and the speed of light is the speed limit.

Hi paul sorry i was away from my computer, presidential debate was on

i understand that there is a delay introduced however i'm not sure how we can be sure of the latency that is produced between the two points

Oh hello... With any connection there is bandwidth and latency. The bandwidth defines how much data you can put onto a connection in a second - 64Mbps. Latency defines how long it takes that data to get to the other endpoint

In this case, the other endpoint is a long way away, and data can travel at the speed of light maximum

So the latency is how long it take a frame to get there

If we simplify it, lets say the other end is 100 meters away, and it take 10 minutes to send a packet over 100 meters

Because we are using a stop and wait protocol, each packet will take 10 minutes to get to the other end, and the acknowledgement will take 10 minutes to get back to us

So that means we can only send 32KB every 20 minutes, even though our pipe can send 64Mbps

32KB is 256kbits, and 20 minutes is 1200 seconds. So we can send 256kbits every 1200 seconds, or .213 kbps - this is a tiny fraction of 64Mbps

So the connection would be 0.0003% utilised. You can do the same calculations for the planet

I'm certain of everything you've explained there paul but in the calculations i have seen elsewhere

they involve further claculations of latency

for instance let me show you

i'm' trying to find the link

No problem

there is this notion of a roundtrip propagation delay

that is not only governed by the latency experienced by the signal itself

there is a transmission time of the packet

i mean frame

Sure - but given the distance we are talking about, and that you have been given no information about the time costs with getting a packet into the device for sending, you can assume these values are negligable compared to the latency due to the distance

that is fair enough

you think given the question i should just stick tot he latency of the actual data back and forth?

fair enough

Yeah I would say, figure out the time it takes for a roundtrip of a packet regardless of anything else - if it is a big number, then the rest doesn't matter much

Alright thank you for the time you took

if you like, you can formulate an answer that explains what you have typed and i will accept it