edited by
340 views
0 votes
0 votes

Based on the following passage, answer the Questions :

A $3000 \mathrm{~km}$ long trunk operates at $1.536 \mathrm{mbps}$ and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is $6 \mu \mathrm{sec} / \mathrm{km}$.
The transmission and propagation delays are respectively

  1. $\mathrm{T}_{\mathrm{t}}=333.33 \mu \mathrm{sec}, \mathrm{T}_{\mathrm{p}}=18000 \mu \mathrm{sec}$
  2. $T_{t}=300 \mu \mathrm{sec}, T_{p}=15360 \mu \mathrm{sec}$
  3. $\mathrm{T}_{\mathrm{t}}=33.33 \mu \mathrm{sec}, \mathrm{T}_{\mathrm{p}}=1800 \mu \mathrm{sec}$
  4. $T_{\mathrm{t}}=1800 \mu \mathrm{sec}, T_{\mathrm{P}}=33.33 \mu \mathrm{sec}$
edited by

1 Answer

0 votes
0 votes
  • distance (d)=$3000$km
  • BW=$1.536\times10^6$bps
  • Length(L)=$64\times8$bit
  • velocity=$6\mu sec/km$

Transmission time $(T_t)=\frac{L}{BW}=\frac{64*8bit*sec}{1.536*10^6 bit}=333.33\mu sec$

Propagation delay$(T_p)=3000km*6\mu sec/km=18000\mu sec$

Option (A) is correct.

Related questions

1 votes
1 votes
0 answers
2
admin asked Oct 23, 2022
896 views
Using $\text{'RSA'}$ algorithm, if $p=13, q=5$ and $e=7$. the value of $d$ and cipher value of $'6'$ with $(e, n)$ key are$7, 4$$7, 1$$7, 46$$55,1$
0 votes
0 votes
1 answer
4
admin asked Oct 23, 2022
316 views
Size and complexity are a part ofPeople MetricsProject MetricsProcess MetricsProduct Metrics