Question

How long does it take to send a 8 MiB file from Host A to Host B over a circuit-switched network, assuming: Total link transmission rate = 44.2 Gbps. Network is FDM, with 10 permitted users, each with an equal bandwidth share. A link connection requires a setup time of 85.1 miss. Your answer should be in milliseconds (ms) with one decimal place, and without the unit

Answer 1

Answer:

Given values are:

$File size= 8 MiB$

$Total Link Transmission Rate = 44.2 Gbps$

$Users = 10$

$Setup Time=85.1 ms$

We know that

1 MiB = $2^{20}$ = 1048576 bytes

1 MiB = $2^{20}$x8= 8388608 bits

8MiB= 8388608x8 = 67108864 bits

So,

Total Number of bits = 67108864 bits

Now

Total link transmission rate = 44.2 Gbps

1 Gbps = 1000000000 bps

44.2 Gbps = 44.2 x 1000000000 =44200000000 bps

For FDM Network

Transmission rate for one time slot = $\frac{44200,000,000}{10}$ bits per second

Transmission rate for one time slot = 4420000000 bit per second

Transmission rate for one time slot = $\frac{44200,000,000}{10^{3} .10 }$

   in milliseconds

Transmission rate for one time slot in =4420000 bits per millisecond

Now,

Total time taken to transmit 8 MiB of file = $\frac{Total Number of Bits}{Transmission Rate}$

Total time taken to transmit 8 MiB of file = $\frac{67108864}{4420000}$

As setup time is also given So,

Total time = Setup time + Transmission Time

Total time = 85.1 + 15.18

Total time = 100.2 milliseconds

Hence it is the time required to send a 8MiB of file from Host A to Host B.

I hope it will hep you!