Question

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 transmission rate

Answer 1

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,

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

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

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

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