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The first answer is this:

"We require to send 1000 frames. We ignore the overhead due to the header and trailer.

Data frame Transmission time = 1000 bits / 1,000,000 bits ( 1 Mbps = 1,000,000 bps)

= 1 ms

The distance between the sender and receiver is 5000 km.

Data frame trip time = 5000 km / 200,000 km

= 25 ms

ACK transmission time = 0

ACK trip time = 5000 km / 200,000 km

= 25 ms

Delay for 1 frame = 1 + data frame trip time + ACK trip time

= 1 + 25 + 25

= 51 ms.

Total delay = 1000 51ms

= 1000bits 51 10-3

= 51s"

The second answer is this:

"The propagation delay is given by tp = d/c

Where d = distance between transmitter and receiver, c = speed of light, d = 5000 and c =3000000

So, tp = 5000/300,000 = 0.02

Transmission delay tt = L/λ

Where L = number of bits in the frame and λ = transmission rate in bits per sec.

Here L = 1000 and λ = 200000000

So, tt = 1000/200000000 = 0.000005

Total time taken T = tt + 2tp

T = 0.000005+0.02

T = 0.020005"

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Q: A system uses the stop-and-wait arq protocolif each packet carries 1000 bits of data determine the time taken to send 1 million bits of data if the distnce between te sender and receiver is 5000 km?
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