1. Hey,
I have a question about GBN protocol or Go-Back-N protocol, i have not yet received my text book, long story but was wondering if anyone knows how to figure out this algorithm or even has a working of how to do it? So i need to find out over a 1g per second data rate, with a window size of 4 over 500km the efficiency say 80% or 90%? probably make no sense lol, also how could i do this for other window sizes 1 or 10 if i wanted? k any direction to sites or posting of information would be appreciated!

Problemed

2.

3. 1 gram of data? window size of 4 over 500 Kilometres?

I think something weird has fallen in your coffee....

4. 1 gig bit per second not a gram

5. Windows size 400-500 kilometres ??

Er.........what are you talking about ?

I think something weird has also fallen into your coffee, not only that but you enjoyed it so much you are now consuming acres of raw coffee beans !

6. We send data at 1G bit second using per the GBN protocol (window size = 4) from Here to Somwhere (500 km).What must be the packet size in order to have an efficiency of at least 80%?

Hopefully a little clearer now?

7. I suppose that would depend on what TTL you give each of the packets.

8. How is the data sent over 500Km ? radio, wire, light, or semaphore?

HOw wide is the stream? - what is the carrier frequency?

9. For the LTT i was just going to use the speed of light!

How wide is the stream? - what is the carrier frequency?
this however im not sure! I assumed stream was effected by window size!

Is there a standard formula you guys know that you could paste so i could try and play with it abit? Or a few formulas then i could try a few. My book shop said again that the text book would be in by Friday but they said it would be in by Monday over a week ago, so fingers crossed.

Problemed

10. What is LTT ??

I said TTL !!!

And I really dont understand what bandwidth or frequency has to do with window size ??

11. He's not talking about 'windows' on screen, he's talking about communications windows....

12. Hey, yeah as in the transport layer. ive figured this from what i have grasped from the net so far!

Where Tframe = L/R

And L is the packet size in bits.
R is the transmission rate of the channel.
Tframe = L (bits/packet) / 109 (bits/sec)
So Ttotal = 1000000 meters / 3 x 108 meters/sec + Tframe

13. Snag is those in computer communications have deamt up their own terms for what in telecomms are standard - which makes it very confusing. At the end of the day, data is data, grouping it into frames/packets/groups/cells is irrelevant there is only one optimum maximum transfer rate consistent with low errors.

If you can transmit 100Kbits/second divide it by the packet size and you have packets per second. the inverse [as you say] is time per packet.

I am not sure why you are adding tframe to the time it takes for the signal to travel through the medium though, it is usually considered instantaneous. If you really must add this time in the the signals will NOT travel at the speed of light, you need to multiply 'c' by the propagation factor, which is only 1 for free space, in a fibre optic medium it is somewhere between 0.6 and 0.8 depending on cable construction, for a cable it will depend on many things but again can be between 0.6 & 0.9

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