How long does it take to send a file of 640,000 bits from host A to host B over a circuit-switched network?
Just going over an exam question and can't figure out the correct method to solve it. The question is as follows :
How long does it take to send a file of
640,000 bits from host A to host B over a
circuit-switched netw开发者_运维知识库ork?
 - All links are 1.536 Mb/s
 - Each link uses TDM with 24 slots/sec (TDM = Time Division Multiplexing)
 - 500 msec to establish end-to-end circuit  
To my mind, then, the answer would be
 x = (1.536 * 10^6) / 24
 y = 640,000 / x.
 Answer : y + 500 * 10^-3
I have a feeling this is catastrophically wrong though. Can anyone help me out please ? :)
The link is 1,536 Mbps, but is time divisioned multiplexed. One circuit uses one timeslot, which means it gets 1/24th of the link bandwidth. So, one circuit has a bandwidth of 64 Kbps (=1.536/24).
So the actual transfer time to move 640.000 bits (=640 Kb) over that circuit is 10 seconds (=640/64).
Add the connection setup time of 500 msec and you get a total of 10.500 msec or 10,5 seconds.
One circuit uses one time slot in TDM, therefore, in this case its 1/24th of the link bandwidth. 1.536 Mbps = 1536 Kbps. 1536/24=64 Kb can be transmitted in each time slot.The file is 640000 bits = 80 Kb. 80/64= 1.25 sec to transmit the file. Add the connection setup time 1.25sec+500msec.
- Each circuit has a transmission rate of (1.536 Mbps)/24 = 64kbps
- It takes 640,000 bits/64 kbps = 10 seconds to transmit the file.
- To this time, we have to add the time taken to establish the connection. That makes it 10.5 seconds.
- Each circuit has a transmission rate of (1.536 Mbps)/24 = 64 kbps. 
- It takes (640,000 bits)/(64 kbps) = 10 seconds to transmit the file. 
- To this 10 seconds we add the circuit establishment time, giving 10.5 seconds to send the file. 
- Note that the transmission time is independent of the number of links: The transmission time would be 10 seconds if the end-to-end circuit passed through one link or a hundred links 
 
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