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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