Question

Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 200 kbps, R2 = 2 Mbps, and R3 = 1 Mbps. Assuming no other traffic in the network, what is the average throughput for the file transfer? Suppose the file is 4 million bytes, on average, how long will it take to transfer the file to Host B? Repeat (a) and (b), but now with R2 reduced to 100 kbps.

Answer 1

Solution

a)

Consider givend data:
R1 = 200 kbps
R2 = 2 Mbps
R3 = 1 Mbps

The throughput for the file transfer
=min{R1,R2,R3}
=min{200 kbps, 2 Mbps, 1 Mbps}
=200 kbps

So, the throughput for the file transfer=200 kbps

b)

Consider given data:

The file size= 4 million bytes

Convert million bytes to bits

=32000000 bits.

From (a),
Throughput for the file transfer=200 Kbps
=200000 bps

Dividing the file size by the throughput,roughly how long will it take to transfer the file to Host B:

=file size/throughput for the file transfer

=32000000 bits/200000 bps

=160 seconds

c)

Consider the given data:

Repeat (a) and (b), but now with R2 reduced to 100 kbps.

That means
R2 = 100 kbps
R1 = 200 kbps
R3 = 1 Mbps

The throughput for the file transfer
=min{R1,R2,R3}
=min{200 kbps, 100 kbps, 1 Mbps}
=100 kbps

So, the throughput for the file transfer=100 kbps

Dividing the file size by the throughput,roughly how long will it take to transfer the file to Host B:

=file size/throughput for the file transfer

=32000000 bits/100000 bps

=320 seconds

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