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Consider sending a file of 640,000 bits from Host A to Host B over a circuit-switched network. Suppose that all links in the network use TDM with 24 slots and have a bit rate of 1.536 Mbps. Also suppose that it takes 500 ms to establish an end-to-end circuit before Host A can begin to transmit the file. Ignoring the propagation, processing, and queuing delays, calculate the amount of time that elapses from when the source begins to send the file until the destination has received the entire file.

The answer: Total transmission time = 0.5s + (640000/((1.536x10^6)/24)) = 10.5s

If I am not mistaken, Host A should get full bandwidth when transmitting the file because time division multiplexing is based in time frame and time slots for each hosts.

May I know why the bandwidth of the link have to be divided by 24?

Calvin Low
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  • "use TDM with 24 slots" – doublesharp Dec 17 '16 at 08:10
  • But for each slot in TDM, the host should transmit file in full bandwidth(1.536Mbps) no? If is in FDM I understand why, but I really don't get it in TDM. Can you explain more in details? – Calvin Low Dec 17 '16 at 08:17
  • Sorry for being so short, but you ended up getting a much better answer than I would have typed :) "Time divisional" means it will only be able to access one of the slots at a time, in this case 1/24. – doublesharp Dec 17 '16 at 09:35

1 Answers1

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On a TDM link with 24 slots, it is apparently assumed your connection will occupy a single slot, which means the throughput you'll be getting will not be full 1.536 Mbps, but 1/24th of it.

I'm saying "assumed", because you might be as well using multiple time slots on that link. I think i.e. GPRS does that - if no one else is using other time slots, and you'd need more bandwidth, you might be assigned them. That would proportionally increase the bandwidth.

In other words, you'd get 1.536 Mbps if you'd be transmitting, over that link, all the time. But you won't - it's TDM, so you'll be transmitting 1/24th of the time.

Day-to-day world analogy would be: imagine if your tap can pour 20 litres of water per minute. If you open it, and let it be open for 10 minutes, you'd get 200 litres of water out. But if you open it for just 5 seconds, then close for 55, then open again for 5 and close for 55, and over again for 10 full minutes, you won't get 200 litres, you'd get only 16.(6) litre.

16.(6) = 20 litres/minute (tap throughput)
         * 10 minutes     (pouring time)
         * 5/60           (proportion of time when water was being poured)

Moving back from analogy to your case:

640k bits = 1.536 Mbits/s (link throughput)
            * 10 seconds  (transmission time)
            * 1/24        (proportion of time when data was being transmitted)
kamituel
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