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DedPeter [7]
3 years ago
15

How long does it take to send a 15 MiB file from Host A to Host B over a circuit-switched network, assuming: Total link transmis

sion rate
Computers and Technology
1 answer:
alexdok [17]3 years ago
7 0

Answer:

The answer is "102.2 milliseconds".

Explanation:

Given:

Size of file = 15 MiB

The transmission rate of the total link= 49.7 Gbps

User = 7

Time of setup = 84.5 ms

calculation:

1\ MiB = 2^{20} = 1048576\ bytes\\\\1\ MiB = 2^{20 \times 8}= 8388608\  bits\\\\15\ MiB= 8388608 \times 15 = 125829120\ bits\\\\

So,

Total Number of bits = 125829120 \ bits

Now

The transmission rate of the total link= 49.7 Gbps

1\ Gbps = 1000000000\ bps\\\\49.7 \ Gbps = 49.7 \times 1000000000 =49700000000\ bps\\\\FDM \ \ network

\text{Calculating the transmission rate for 1 time slot:}

=\frac{ 49700000000}{7} \ bits / second\\\\= 7100000000 \ bits / second\\\\ = \frac{49700000000}{(10^{3\times 7})} \ in\  milliseconds\\\\ =7100000 \ bits / millisecond

Now,

\text{Total time taken to transmit 15 MiB of file} = \frac{\text{Total number of bits}}{\text{Transmission rate}}

= \frac{125829120}{7100000}\\\\= 17.72\\\\

\text{Total time = Setup time + Transmission Time}\\\\  

                 = 84.5+ 17.72\\\\= 102.2 \ milliseconds

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